DYNAMICS OF MACHINERY

Transcription

1 DYNAMICS OF MACHINERY

2 B.TECH. DEGREE COURSE SCHEME AND SYLLABUS (00-03 ADMISSION ONWARDS) MAHATMA GANDHI UNIVERSITY KOTTAYAM, KERALA DYNAMICS OF MACHINERY Module 1 Balancing: - Balancing of rotating masses, static balancing and dynamic balancing, Balancing of several masses rotating in same plane, Balancing of several masses rotating in several planes, Balancing machines. Balancing of reciprocating masses: - The effect of inertia force of the reciprocating mass on the engine. Partial primary balance. Partial balancing of locomotive, Hammer blow, Variation of tractive effort, Swaying couple. Coupled locomotives, Balancing of multi cylinder inline engines, v-engines, Radial engines, Direct and Reverse cranks Module Vibrations: - Definitions, simple harmonic motion. Single degree freedom systems: Undamped free vibrations: - Equations of motion Natural frequency, Energy method, Equilibrium methods, Rayleigh s methods, Equivalent stiffness of spring combinations. Damped free vibrations: - Viscous damping, Free vibrations with viscous damping, over-damped system, critically damped system, under-damped system, Logarithmic decrement, viscous dampers, coulomb damping. Forced Vibrations: - Forced harmonic excitation Rotating unbalance, Reciprocating unbalance. Energy dissipated by damping, vibration isolation and Transmissibility. Vibration measuring instruments. Module 3 Two degree freedom systems: - Principal modes of vibration, Rectilinear and angular modes, systems with damping, vibration absorbers, centrifugal pendulum damper, dry friction damper, untuned viscous damper. Multi-degree of freedom system: - Free vibrations, equations of motion, Influence coefficients method, lumped mass and distributed mass systems, Stodola method, Dunkerly s method, Holzer s method, Matrix iteration method.

3 Torsional Vibrations: - Torsionally equivalent shaft, torsional vibration of tworotor, three-rotor, and geared systems. Module 4 Critical speeds of shafts: - Critical speed of a light shaft having a single disc without damping. Critical speeds of a light cantilever shaft with a large heavy disc at its end. Transient vibration: - Laplace transformation, response to an impulsive input, response to a step input, response to a pulse input, phase plane method, shock spectrum. Non-linear vibrations: - Phase plane, undamped free vibration with non-linear spring forces, hard spring, soft spring, Perturbation method, Forced vibration with nonlinear forces, Duffings equation, self excited vibrations. Module 5 Noise control: - Sound propagation, decibels, acceptance noise levels, Air columns, Doppler effect, acoustic measurements, microphones and loud speakers, Recording and reproduction of sound, fourier s theorem and musical scale, Acoustics of buildings, Acoustic impedence filters and human ear.

4 MODULE 1 Introduction The high speed of engines and other machines is a common phenomenon now-a-days. It is, therefore, very essential that all the rotating and reciprocating parts should be completely balanced as far as possible. If these parts are not properly balanced, the dynamic forces are set up. These forces not only increase the loads on bearings and stresses in the various members, but also produce unpleasant and even dangerous vibrations. In this chapter we shall discuss the balancing of unbalanced forces caused by rotating masses, in order to minimize pressure on the main bearings when an engine is running. BALANCING OF ROTATING MASSES Static and dynamic unbalance: A rotor can in general have two types of unbalance viz., static and dynamic. It is of course to be appreciated that practical systems will all have dynamic unbalance only and considering it as static unbalance is a good-enough approximation for some cases. Fig 1.1 A thin Rotor Disc - Illustration of Static Unbalance If the rotor is thin enough (longitudinally) as shown in Fig. 1.1 the unbalance force can be assumed to be confined to one plane (the plane of the disc). Such a case is known as static unbalance. Such a system when mounted on a knife-edge as shown in Fig. 1 will always come to rest in one position only where the centre of gravity comes vertically below the knife-edge point. Thus in order to balance out, all we need to do is to attach an appropriate balancing mass exactly opposite to this position.

5 Fig 1. Thin Rotor on a knife edge - Illustration of Static Unbalance Thus we first mount the disc on a knife edge and allow it to freely oscillate. Mark the position when it comes to rest. Choose a radial location (180 0 opposite to this position) where we can conveniently attach a balancing mass. By trial and error the balancing mass can be found out. When perfectly balanced, the disc will exhibit no particular preferred position of rest. Also when the disc is driven to rotate by a motor etc., there will be no centrifugal forces felt on the system (for example, at the bearings). Thus the condition for static balance is simply that the effective centre of gravity lie on the axis. Fig 1.3 A Case of Dynamic Unbalance Consider the rotor shown in Fig It is easily observed that mass distribution cannot be approximately confined to just one plane. So unbalance masses and hence unbalance forces are in general present all along the length of the rotor. Such a case is known as dynamic unbalance. The fundamental difference between static and dynamic unbalance needs to be clearly appreciated. When a rotor as shown in Fig.1. 3 is mounted on a knife edge and allowed to oscillate freely, it too may come to rest in one particular position all the time the position corresponding to the resultant unbalance mass (centre of gravity) vertically below the knife edge. We could, like earlier, mount an appropriate balance mass exactly opposite to this position. It would then have no preferred position of rest when mounted on a knife-edge. Thus effective center of gravity lies on the axis.

6 Fig1.4 Example of unbalance masses leading to unbalance force that for a resultant couple because of axial. However, when mounted in bearings and driven by a motor etc., it could still wobble due to the unbalanced moments of these forces as shown in Fig1.4. This becomes apparent only when the rotor is driven to rotate and hence the name dynamic unbalance. Thus it is not, in general, sufficient to do just static balance but achieving good dynamic balance is more difficult. We will discuss one important method of achieving dynamic balance in the next lecture. Two-plane balancing technique Consider the turbo-machine rotor that was discussed earlier wherein each stage contains several blades around the circumference of a disk. Eventhough typically each stage is balanced in itself to the extent possible, it has a likely net unbalance. When the rotor is set to spin, it will cause dynamic forces and moments on the bearings that support the shaft. Therefore it is of interest to achieve good balance of this shaft so that the fluctuating forces on the bearings are reduced. Conceptually our strategy can be simply stated as follows: Step 1: Consider the shaft supported on its bearings. For each unbalance mass, there will be a centrifugal force set-up when the rotor spins at some speed. This would cause some reactions at the supports. Estimate these support reactions that would come onto the bearings. Step : Estimate the balancing mass that needs to be placed in the plane of bearings, to counter this reaction force due to unbalance mass. Repeat steps 1 and for each unbalance mass in the system and each time add the balancing masses obtained in step vectorially to determine the resultant balancing mass required. Let us now understand the details of the technique mentioned earlier. Firstly we choose to place balancing or correcting masses on the shaft (rotating along with the shaft) to counter-act the unbalance forces. We understand that this is to be done on

7 the rotor on site, perhaps during a maintenance period. From the point of view of accessibility, we therefore choose the balancing masses to be kept near the bearings. Figure 1.5 Two plane balancing technique The calculations proceed as shown in Fig1.5. For an unbalance mass m i situated at an angular location in a plane at an axial distance from the left end bearing and rotating at a radius as shown in the figure, the unbalance force is. It is resolved into X and Y components as shown in the figure. These forces are represented by EQUIVALENT FORCES in the balancing planes ( shown in blue ). These forces can be readily calculated (based on calculations similar to those involved in finding support reactions for a simply supported beam). In order to counterbalance this force, we need to place a balancing mass at a radius in the balancing plane such that it creates an equal and opposite force. Now we need to repeat the calculations for ALL the unbalance masses m i (i = 1,,3,..) and find the resultant equivalent force in the balancing plane as shown in blue in Fig This resultant force is balanced out by placing a suitable balancing mass creating an equal and opposite force (shown in red). Since all the masses are rotating at the same speed along with the shaft, we can drop in our calculations i.e., a rotor balanced at one speed will remain balanced at all speeds or in other words, our technique of balancing is independent of speed. We will review this towards the end of the lecture. While these calculations can be done in any manner perceived to be convenient, a tabular form is commonly employed to organize the computations. While doing this, it is also common practice to include the two balancing masses in the balancing planes as indicated in the table.

8 Table 1.1 Tabular form of organizing the computations for two-plane balancing technique Sr. No Cos( Sin( Cos( ) ) ) Balancing 0 Plane 1 Balancing L Plane TOTAL FORCES ) Sin( It is observed in Table 1 that the balancing masses and their locations (radial as well as angular) are unknowns while the location of the balancing plane itself is treated as a known (any accessible location near the bearings etc). The resultant total forces and moments must sum up to ZERO and therefore we have four equations but six unknowns. Thus any two of the six unknowns can be freely chosen and the other four determined from the computations given in the table. This method of balancing is known as the two-plane balancing technique since balancing masses are kept in two planes. Balancing of reciprocating masses: Figure 1.6 Slider-crank Mechanism of IC Engine

9 A typical crank-slider mechanism as used in an IC Engine is shown in Fig 1.6 It essentially consists of four different parts viz., frame(i.e., cylinder),crank,connecting rod and reciprocating piston. The frame is supposedly stationary; crank is undergoing purely rotary motion while the piston undergoes to-and-fro rectilinear motion. The connecting rod undergoes complex motion its one end is connected to the crank (undergoing pure rotation) and the other end is connected to the piston (undergoing pure translation). We know that the inertia forces are given by mass times acceleration and we shall now estimate the inertia forces (shaking forces and moments) due to the moving parts on the frame (cylinder block). CONNECTING ROD One end of the connecting rod is circling while the other end is reciprocating and any point in between moves in an ellipse. It is conceivable that we derive a general expression for the acceleration of any point on the connecting rod and hence estimate the inertia forces due to an elemental mass associated with that point. Integration over the whole length of the connecting rod yields the total inertia force due to the entire connecting rod. Instead we try to arrive at a simplified model of the connecting rod by replacing it with a dynamically equivalent link as shown in Fig 1.7 Figure 1.7 Dynamically Equivalent link for a connecting rod. In order that the two links are dynamically equivalent, it is necessary that: Total mass be the same for both the links Distribution of the mass be also same i.e., location of CG must be same and the mass moment of inertia also must be same. Thus we can write three conditions:

10 For convenience we would like the equivalent link lumped masses to be located at the big and small end of the original connecting rod and if its center of mass (G) location is to remain same as that of original rod, distances AG and GB are fixed. Given the mass m and mass moment of inertia of the original connecting rod, the problem of finding dynamically equivalent link is to determine, and. An approximate equivalent link can be found by simply ignoring and treating just the two lumped masses and connected by a mass-less link as the equivalent of original connecting rod. In such a case we take: = (GB)/L = (AG)/L Thus the connecting rod is replaced by two masses at either end (pin joints A and B) of the original rod. rotates along with the crank while purely translates along with the piston. It is for this reason that we proposed use of crank's effective rotating mass located at pin A, which can now be simply added up to part of connecting rod's mass. On the shop floor, can be immediately determined by mounting the existing connecting rod on two weighing balances located at A and B respectively. The readings of the two balance give and directly Dynamic Model of a single cylinder IC Engine Mechanism Figure 1.8 Dynamic Model of slider-crank Mechanism Based on our discussion thus far, we can arrive at a simplified model of the crankslider mechanism for the purpose of our dynamic analysis as shown in Fig. 1.8 Thus we have either purely rotating masses or purely translating masses and these are given by:

11 where the first term in the rotating masses is due to the effective crank mass at pin A and the second term is due to the part of equivalent connecting rod mass located at pin A. Similarly the first term in reciprocating masses is due to the mass of the piston and the second is due to the part of equivalent connecting rod mass located at pin B. There are inertia forces due to. The inertia forces due to can be nullified by placing appropriate balancing masses. Thus the effective force transmitted to the frame due to rotating masses can ideally be made zero. Figure 1.9Counter balancing of rotating masses Figure 1.10 Opposed position configuration However it is not so straight forward to make the unbalanced forces due to reciprocating masses vanish completely. As given in Equation and depicted in Fig. 1.9 there are components of the force which are at the rotational speed and those at twice this speed. It is conceivable to use a configuration as shown in Fig.1.10 to completely balance out these forces but the mechanism becomes too bulky. Thus a single cylinder engine is inherently unbalanced.. Partial Balancing of Locomotives The locomotives, usually, have two cylinders with cranks placed at right angles to each other in order to have uniformity in turning moment diagram. The two cylinder locomotives may be classified as :

12 1. Inside cylinder locomotives ; and. Outside cylinder locomotives. In the inside cylinder locomotives, the two cylinders are placed in between the planes of two driving wheels whereas in the outside cylinder locomotives, the two cylinders are placed outside the driving wheels, one on each side of the driving wheel. The locomotives may be (a) Single or uncoupled locomotives ; and (b) Coupled locomotives. A single or uncoupled locomotive is one, in which the effort is transmitted to one pair of the wheels only ; whereas in coupled locomotives, the driving wheels are connected to the leading and trailing wheel by an outside coupling rod. Effect of Partial Balancing of Reciprocating Parts of Two Cylinder Locomotives We have discussed in the previous article that the reciprocating parts are only partially lanced. Due to this partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke and also an unbalanced primary force perpendicular to the line of stroke. The effect of an unbalanced primary force along the line of stroke is to produce; couple. 1. Variation in tractive force along the line of stroke ; and. Swaying The effect of an unbalanced primary force perpendicular to the line of stroke is to produce variation in pressure on the rails, which results in hammering action on the rails. The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as a jammer blow. We shall now discuss the effects of an unbalanced primary force in the following articles. Variation of Tractive Force The resultant unbalanced force due to the two cylinders, along the line of stroke, is known as tractive force. Let the crank for the first cylinder be inclined at an angle with the line of stroke. Since the crank for the second cylinder is at right angle to the first crank, therefore the angle of inclination for the second crank will be (90 + ).

13 Let m = Mass of the reciprocating parts per cylinder, and c = Fraction of the reciprocating parts to be balanced. We know that unbalanced force along the line of stroke for cylinder 1 1 cm..rcos Similarly, unbalanced force along the line of stroke for cylinder, o 1 cm..rcos 90 As per definition, the tractive force, F T = Resultant unbalanced force along the line of stroke 1 cm..rcos o 1 cm..rcos 90 1 cm..r cos sin The tractive force is maximum or minimum when (cos -sin ) is maximum or minimum. For (cos -sin ) to be maximum or minimum, d cos sin 0 d or sin cos 0 or sin cos tan 1 or = 135 o or 315 o Thus, the tractive force is maximum or minnimum when = 135 o or 315 o. tractive force Maximum and minimum value of the tractive force or the variation in o o 1 c m..r cos135 sin135 1 c m..r Swaying Couple The unbalanced forces along the line of stroke for the two cylinders constitute a couple.this couple has swaying effect about a vertical axis, and tends to

14 sway the engine alternate in clockwise and anticlockwise directions. Hence the couple is known as swaying couple. Let a = Distance between the centre lines of the two cylinders. Swaying couple a 1 cm..rcos 1 cm..rcos 90 o a 1 cm..r cos sin a The swaying couple is maximum or minimum when (cos + sin ) to be maximum or minimum. d cos sin 0 d or sin cos 0 or sin cos tan 1 or = 45 o or 5 o Thus, the swaying couple is maximum or minimum when = 45 o or 5 o. Maximum and minimum value of the swaying couple a o o a 1 cm..r cos45 sin 45 1 cm..r Note: In order to reduce the magnitude of the swaying couple, revolving balancing masses are introduced. But, as discussed in the previous article, the revolving balancing masses cause unbalanced forces to act at right angles to the line of stroke. These forces vary the downward pressure of the wheels on the rails and cause oscillation of the locomotive in a vertical plane about a horizontal axis. Since a swaying couple is more. harmful than an oscillating couple, therefore a value of c from /3 to 3/4, in two-cylinder locomotives with two pairs of coupled wheels, is usually used. But in large four cylinder locomotives with three or more pairs of coupled wheels, the value of c is taken as /5.

15 Hammer Blow We have already discussed that the maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as hammer blow. We know that the unbalanced force along the perpendicular to the line of stroke due to the balancing mass B, at a radius b, in order to balance reciprocating parts only is B. when = 90 or 70..b sin. This force will be maximum when sin is unity, i.e. Hammer blow = B..b (Substituting sin = 1) The effect of hammer blow is to cause the variation in pressure between the wheel and the rail. Let P be the downward pressure on the rails (or static wheel load). Net pressure between the wheel and the rail P B..b If P B..b is negative, then the wheel will be lifted from the rails. Therefore the limiting condition in order that the wheel does not lift from the rails is given by P B..b and the permissible value of the angular speed, P Bb Balancing of Coupled Locomotives The uncoupled locomotives as discussed in the previous article, are obsolete now-a-days. In a coupled locomotive, the driving wheels are connected to the leading and trailing wheels by an outside coupling rod. By such an arrangement, a greater portion of die engine mass is utilised by tractive purposes. In coupled locomotives,

16 the coupling rod cranks are placed diametrically opposite to the adjacent main cranks (i.e. driving cranks). The coupling rods together with cranks and pins may be.treated as rotating masses and completely balanced by masses in the respective wheels. Thus in a coupled engine, the rotating and reciprocating masses must be treated separately and the balanced masses for the two systems are suitably combined in the wheel. It may be noted that the variation of pressure between the wheel and the rail (i.e., hammer blow) may be reduced by equal distribution of balanced mass (B) between the driving, leading and trailing wheels respectively. Balancing of Primary Forces of Multi-cylinder In-line Engines The multi-cylinder engines with the cylinder centre lines in the same plane and on the same side of the centre line of the crankshaft, are known as In-line engines. The following two conditions must be satisfied in order to give the primary balance of the reciprocating parts of a multi-cylinder engine: Figure 1.11 Typical Inline engine 1. The algebraic sum of the primary forces must be equal to zero. In other words, the primary force polygon must *close; and. The algebraic sum of the couples about any point in the plane of the primary forces must be equal to zero. In other words, the primary couple polygon must close. We have already discussed, that the primary unbalanced force due to the

17 reciprocating masses is equal to the component, parallel to the line of stroke, of the centrifugal force produced by the equal mass placed at the crankpin and revolving with it. Therefore, in order to give the primary balance of the reciprocating parts of a multi-cylinder engine, it is convenient to imagine the reciprocating masses to be transferred to their respective crankpins and to treat the problem as one of revolving masses. Notes: 1. For a two cylinder engine with cranks 180 o, condition (1) may be satisfied, but this will result in an unbalanced couple. Thus the above method of primary balancing cannot be applied in this case.. For a three cylinder engine with cranks at 10 o and if the reciprocating masses per cylinder are same, then condition (1) will be satisfied because the forces may be represented by the sides of an equilateral triangle. However, by taking a reference plan through one of the cylinder centre lines, two couples with non-parallel axes will remain and these cannot vanish vectorially. Hence the above method of balancing fails in this case also. 3. For a four cylinder engine, similar reasoning will show that complete primary balance is possible and it follows that For a multi-cylinder engine, the primary forces may be completely balanced by suitably arranging the crank angles, provided that the number of cranks are not less than four'. Balancing of Secondary Forces of Multi-cylinder In-line Engines When the connecting rod is not too long (i.e. when the obliquity of the connecting rod is considered), then the secondary disturbing force due to the reciprocating mass arises. We have the secondary force, cos FS m..r n This expression may be written as r FS m. cos 4n

18 As in case of primary forces, the secondary forces may be considered to be equivalent to be component, parallel to the line of stroke, of the centrifugal force produced by an equal mass placed at the imaginary crank of length r/4n and revolving at twice the speed of the actual crank.thus, in multi-cylinder in-line engines, each imaginary secondary crank with a mass attached to the crankpin s inclined to the line of stroke at twice the angle of the actual crank. The values of the secondary forces and couples nay be obtained by considering the revolving mass. This is done in the similar way as discussed for primary forces, the following two conditions must be satisfied in order to give a complete secondary balance of an engine : 1. The algebraic sum of the secondary forces must be equal to zero. In other words, die secondary force polygon must close, and. The algebraic sum of the couples about any point in the plane of the secondary forces must be equal to zero. In other words, the secondary couple polygon must close. Balancing of Radial Engines (Direct and Reverse Cranks Method) The method of direct and reverse cranks is used in balancing of radial or V- engines, in which the connecting rods are connected to a common crank. Since the plane of rotation of the various cranks (in radial or V-engines) is same, therefore there is no unbalanced primary or secondary couple. Fig. 1.1 Typical Radial Engine (Not to scale) Consider a reciprocating engine mechanism. Let the crank known as the direct crank) rotates uniformly at radians per second in a clockwise direction. Let

19 at any instant the crank makes an angle with the line of stroke. The indirect or reverse crank is the image of the direct crank when seen through the mirror placed at the line of stroke. A little consideration will show that when the direct crank revolves in a clockwise direction, the reverse crank will revolve in the anticlockwise direction. We shall now discuss the primary and secondary forces due to the mass of the reciprocating parts. Considering the primary forces We have already discussed that primary force is m..rcos. This force is equal to the component of the centrifugal force along the line of stroke, produced by a mass placed at the crank pin. Now let us suppose that the mass of the reciprocating parts is divided into two parts, each equal to m/. It is assumed that m/ is fixed at the direct crank (termed as primary direct crank) pin and m/ at the reverse crank (termed as primary reverse crank) pin. We know that the centrifugal force acting on the primary direct and reverse crank m.r Component of the centrifugal force acting on the primary direct crank m.rcos and, the component of the centrifugal force acting on the primary reverse crank m.rcos Total component of the centrifugal force along the line of stroke m.r cos m..r cos = Primary force, F P Hence, for primary effects, the mass m of the reciprocating parts may be replaced by two masses each of magnitude m/.

20 MODULE Vibration (Oscillation) Any motion which repeats itself after an interval of time is called vibration. Eg: Swinging of simple pendulum. Causes of vibration Unbalanced forces in the machine. External excitations applied on the system. Elastic nature of the system Winds, Earthquakes etc. Effect of Vibration Produces unwanted noise, high stresses, wear, poor reliability and premature failure of one or more of the parts. Inspite of these harmful effects, it is used in musical instruments, vibrating conveyors etc. Elimination of Vibrations Using shock absorbers Using vibration absorbers Resting the machinery on proper type of isolation. Definitions Frequency Number of cycles per unit time. Natural Frequency (fn) Frequency of free vibration of the system. Expressed in Hz or rad/sec. Amplitude The maximum displacement of a vibrating body from its equilibrium position.

21 Resonance When the frequency of external excitation is equal to the natural frequency of a vibrating body, the amplitude of vibration becomes excessively large. This concept is known as resonance. Periodic Motion Time Period A motion which repeats itself after equal intervals of time. Time taken to complete one cycle. Fundamental mode of vibration The fundamental mode of vibration of a system is the mode having the lowest natural frequency. Degree of Freedom The minimum number of independent co-ordinates required to specify the motion of system at any instant is known as degrees of freedom. It is equal to the number of independent displacements that are possible. This number varies from zero to infinity. Zero degree of freedom The body at rest is said to have zero degree of freedom. Single Degree of freedom Here there is only one independent co-ordinate to specify the configuration. Eg: A mass supported by a spring. Two degree of freedom There are two independent co-ordinates to specify the configuration. Eg: Springs supported Rigid mass. (It can move in the direction of springs and also have angular motion in one plane) Multi degrees of freedom A cantilever beam has inifinite degrees of freedom.

22 Types of Vibration 1. Free (Natural) Vibration eg: Simple pendulum. After disturbing the system the external excitation is removed, then the system vibrates on its own. This type of vibration is known as free vibrations. 3 types:- a) Longitudinal vibrations. When the particles of the shaft or disc move parallel to the axis of the shaft, then the vibrations are known as longitudinal vibrations. In this case the shaft is elongated and shortened alternately and thus the tensile and compressive stresses are induced alternately in the shaft. b) Transverse vibrations When the particles of the shaft move approximately perpendicular to the axis of the shaft, then the vibrations are known as transverse vibrations. In this case the shaft is straight and bent alternately and bending stresses are induced in the shaft. c) Torsional vibrations When the particles of the shaft move in a circle about the axis of the shaft, then the vibrations are known as torsional vibrations. In this case the shaft is twisted and untwisted alternately and torsional shear stresses are induced in the shaft to.. Forced Vibration Eg: Machine tools, Electric bells. The vibration which is under the influences of external force is called forced vibration. The external force applied to the body is periodic disturbing force created by unbalance. The vibrations have the same frequency as the applied force. Due to the application of external forces the amplitude of these vibrations is maintained almost constant.

23 3. Damped vibration When there is a reduction in amplitude over every cycle of vibration, the motion is said to be damped vibration. That is if the vibrators system has a damper. The motion of the system will be opposed by it and the energy of the system will be dissipated in friction. 4. Undamped vibration. There is no damper. There is no loss of energy due to friction. 5. Deterministic vibration If in the vibratory system the amount of external excitation is known in magnitude it is deterministic vibration. 6. Random vibration Non deterministic vibrations 7. Steady state vibrations In ideal systems, the free vibrations continue indefinitely as there is no damping. Such vibration is termed as steady state vibration. 8. Transient vibrations In real systems, the amplitude of vibration decays continuously because of damping and vanishes finally. Such vibration is real system is called transient vibration. 9. Linear vibration A vibratory system basically consists of there elements: Mass Spring Pamper Fig. - If in a vibratory system mass, spring and damper behave in a linear manner, the vibrations caused are known as linear vibrations. - Linear vibrations are governed by linear differential equations. - They follow the law for superposition.

24 10. Non linear vibrations - if any of the basic components of a vibratory system behaves non linearly, the vibration is called non-linear vibration. - it does not follow the law of superposition. Linear vibration becomes non linear for very large amplitude of vibration. Forced vibrations are also known as excitations. The excitation may be: a) Periodic b) Impulsive c) Random Vibrations because of impulsive forces are called transient. Earth quake is because of random forces. - External force keeps the system vibrating. This force is called external excitation. Harmonic motion : Simplest form of periodic motion is harmonic motion and it is called simple harmonic motion (SHM). It can be expressed as where A is the amplitude of motion, t is the time instant and T is the period of motion. Harmonic motion is often represented by projection on line of a point that is moving on a circle at constant speed.

25 From Figure.1, we have Figure.1: The Simple Harmonic Motion where x is the displacement and is the circular frequency in rad/sec. where T is the period (sec) and f is the frequency (cycle/sec) of the harmonic motion. The SHM repeats itself in radians. Displacement can be expressed as So that the velocity and acceleration can be written as FREE VIBRATIONS In absence of damping, the system can be considered as conservative and principle of conservation of energy offers another approach to the calculation of the natural frequency. The effect of damping is mainly evident in diminishing of the vibration amplitude at or near the resonance

26 Undamped Free Vibration A spring mass system as shown in Figure. is considered. For simplicity at present the damping is not considered. Figure. The direction of x in the downward direction is positive. Also velocity,, acceleration,, and force, F, are positive in the downward direction as shown in Figure.. From Figure.(d) on application of Newton's second law, we have or From Figure.(b), we have (i.e. spring force due to static deflection is equal to weight of the suspended mass), so the above equation becomes The choice of the static equilibrium position as reference for x axis datum has eliminated the force due to the gravity. Equation can be written as or

27 where is the natural frequency (in rads/sec).this Equation satisfies the simple harmonic motion condition. The undamped free vibration executes the simple harmonic motion as shown in Figure.3. Figure.3: Simple harmonic motion Since sine & cosine functions repeat after radians (i.e. Frequency Time period = ), we have The time period (in second) can be written as The natural frequency (in rads/sec or Hertz) can be written as From Figure.(b), we have

28 On substitution we get Here T, f, are dependent upon mass & stiffness of the system, which are properties of the system. Above analysis is valid for all kind of SDOF system including beam or torsional members. For torsional vibrations the mass may be replaced by the mass moment of inertia and stiffness by stiffness of torsional spring. For stepped shaft an equivalent stiffness can be taken or for distributed mass an equivalent lumped mass can be taken. The undamped free response can also be written as where A & B are constants to be determined from initial conditions. Equivalent Stiffness of Series and Parallel Springs : For this system having springs connected in series or parallel, this equation is still valid with the equivalent stiffness as shown in Figures.4 and.5. Figure.4 Figure.5

29 Energy method : In a conservative system (i.e. with no damping) the total energy is constant, and differential equation of motion can also be established by the principle of conservation of energy. For the free vibration of undamped system: Energy=(partly kinetic energy + partly potential energy). Kinetic energy T is stored in mass by virtue of its velocity. Potential energy U is stored in the form of strain energy in elastic deformation or work done in a force field such as gravity, magnetic field etc. Our interest is to find natural frequency of the system, writing this equation for two positions where, 1 & represents two instants of time. Let 1 represents a static equilibrium position (choosing this as the reference point of potential energy, here U 1 =0 ) and represents the position corresponding to maximum displacement of mass and at this position velocity of mass will be zero and hence T = 0. Damped Free Vibration Vibration systems may encounter damping of following types: 1. Internal molecular friction.. Sliding friction 3. Fluid resistance Generally mathematical model of such damping is quite complicated and not suitable for vibration analysis. Simplified mathematical model (such as viscous damping or dash-pot) have been developed which leads to simplified formulation. A mathematical model of damping in which force is proportional to displacement i.e., F d = cx is not possible because with cyclic motion this model will encounter an area of magnitude equal to zero as shown infigure.1(a). So dissipation of energy is not possible with this model.

30 The damping force (non-linearly related with displacement) versus displacement curve will enclose an area, it is referred as the hysteresis loop (Figure.1(b)), that is proportional to the energy lost per cycle. (a): Linear relation (b): Non-linear relation Figure.6: Variation of damping force vs displacement Viscously damped free vibration : Viscous damping force is expressed as, c is the constant of proportionality and it is called damping co-efficient. Figure.7 shows spring-damper-mass system with free body diagram.

31 From free body diagram, we have (1) Let us assume a solution of equation(1) of the following form () where s is a constant (can be a complex number) and t is time. So that and, on substituting in equation (1), we get, From the condition that equation () is a solution for all values of t, above equation gives a characteristic equation (Frequency equation) as Equation (3) has the following form (3) solution of which is given as Solution of equation (3) can be written as Hence the general solution of equation (1) from equations () and (4) is given by the equation (4) (5)

32 where A and B are integration constants to be determined from initial conditions. Substituting equation (4) into equation (5). The term outside the bracket in RHS is an exponentially decaying function. The term (6) can have three cases. (i) : exponents in equation (6) will be real numbers. No oscillation is possible as shown in Figure.8. This is an overdamped system (Figure.8). Figure.8: Overdamped system ii) : exponents in equation (6) are imaginary numbers : we can write Hence the equation (6) takes the following form

33 Let and, equation (6) can be written as (7) where (iii) Critical case between oscillatory and non-oscillatory motion : Damping corresponding to this case is called critical damping, c c (8) Any damping can be expressed in terms of the critical damping by a non-dimensional number called the damping ratio Response corresponding to the critical damping case is shown in Figure.9 for various initial conditions. (9) Figure.9: Critical damping

34 Equation of motion for damped system can be expressed in terms of and as This form of equation is useful in identification of natural frequency and damping of system. It is useful in modal summation of MDOF system also. The roots of characteristic equation (10) can be written as (10) (11) with Depending upon value of damping ratio we can have the following cases, overdamped condition, underdamped condition, critical damping, undamped system 1) Oscillatory motion : [, underdamped case] General solution equation (1) becomes: (1) (3.30)

35 (13) where and and, where C & D and X, are arbitrary constants to be determined from initial conditions, x (0) and (0). From equation (13), we have On application of initial conditions, we get x(0)=c and Hence, equation (13), becomes ; which gives (14) Equation (14) indicates that the frequency of damped system is equal to, (15)

36 It should be noted that for small ( which is the case of most engineering systems) ) Non-oscillatory motion : ( over damped case) Two roots remain real with one increases and another decreases. The general solution becomes so that (16) On application of initial conditions, we have and x(0)=a + B or which gives and

37 3) Critically damped systems : We obtained two roots Two terms in solution combines to give one constant From equation (14) for critically damped case (when ), we have (17) Hence the general solution will be (18) so that, On application of initial conditions, we get x(0)=a and

38 The necessary and sufficient conditions for crossing once can be obtained as (19) or is a necessary condition for crossing time axis once but sufficient conditions is given by equation (19) as shown in Figure.9. Logarithmic Decrement : Rate of decay of free vibration is a measure of damping present in a system. Greater is the decay, larger will be the damping. Damped (free) vibration, general equation of the response is given as Defining a term logarithmic decrement which is defined as the natural logarithm of the ratio of any two successive amplitudes as shown Figure.11. since T d = damped period, where = damped natural frequency We have damped period decrement as, we get logarithmic Since, the above equation reduces to

39 Experimental determination of natural frequency and damping ratio : rad/sec, T d can be obtained from displacement-time free vibration oscillations., where x 1 and x are two consecute amplitudes in the free vibration displacement-time curve. Figure.10 The above illustration shows for two successive amplitude. But in case, the amplitude are recorded after "n" cycles, the formula is modified as Taking log, Forced Harmonic Vibration: Steady State Response due to Harmonic Oscillation : Consider a spring-mass-damper system as shown in figure.11. The equation of motion of this system subjected to a harmonic force can be given by

40 where, m, k and c are the mass, spring stiffness and damping coefficient of the system, F is the amplitude of the force, w is the excitation frequency or driving frequency. Figure.11 Harmonically excited system Figure.1: Force polygon The steady state response of the system can be determined by solving the above equation in many different ways. Here a simpler graphical method is used which will give physical understanding to this dynamic problem. From solution of differential equations it is known that the steady state solution (particular integral) will be of the form As each term of equation (4.1) represents a forcing term viz., first, second and third terms, represent the inertia force, spring force, and the damping forces. The term in the right hand side of equation is the applied force. One may draw a close polygon as shown in figure.1 considering the equilibrium of the system under the action of these forces. Considering a reference line these forces can be presented as follows. Spring force = (This force will make an angle with the reference line, represented by line OA). Damping force = (This force will be perpendicular to the spring force, represented by line AB).

41 Inertia force = (this force is perpendicular to the damping force and is in opposite direction with the spring force and is represented by line BC). Applied force = which can be drawn at an angle with respect to the reference line and is represented by line OC. From equation the resultant of the spring force, damping force and the inertia force will be the applied force, which is clearly shown in figure.1 It may be noted that till now, we don't know about the magnitude of X and which can be easily computed from Figure. Drawing a line CD parallel to AB, from the triangle OCD of Figure, or As the ratio is the static deflection of the spring, is known as the magnification factor or amplitude ratio of the system Figure.13 shows the magnification factor frequency ratio plot. frequency ratio and phase angle

42 Following observation can be made from these plots. For undamped system ( i.e. ) the magnification factor tends to infinity when the frequency of external excitation equals natural frequency of the system. But for underdamped systems the maximum amplitude of excitation has a definite value and it occurs at a frequency For frequency of external excitation very less than the natural frequency of the system, with increase in frequency ratio, the dynamic deflection ( X ) dominates the static deflection reaches a maximum value at resonant frequency., the magnification factor increases till it For, the magnification factor decreases and for very high value of frequency ratio ( say ) One may observe that with increase in damping ratio, the resonant response amplitude decreases. Irrespective of value of, at, the phase angle. For, phase angle. For,, phase angle approaches for very low value of. Figure.13 : (a) Magnification factor ~ frequency ratio for different values of damping ratio.

43 Figure.13 : (b) Phase angle ~frequency ratio for different values of damping ratio. For a underdamped system the total response of the system which is the combination of transient response and steady state response can be given by The parameter will depend on the initial conditions. It may be noted that as remains., the first part of equation tends to zero and second part Rotating Unbalance: One may find many rotating systems in industrial applications. The unbalanced force in such a system can be represented by a mass m with eccentricity e, which is rotating with angular velocity as shown in Figure 4.1. Figure.14: Vibrating system with rotating unbalance

44 Figure.15 Freebody diagram of the system Let x be the displacement of the nonrotating mass (M-m) from the static equilibrium position, then the displacement of the rotating mass m is From the freebody diagram of the system shown in figure.15, the equation of motion is or This equation is same as equation (1) where F is replaced by force polygon as shown in figure.16. So from the or or

45 Figure.16: Force polygon or and So the complete solution becomes Figure.17 : plot for system with rotating unbalance

46 Figure.18 : Phase angle ~ frequency ratio plot for system with rotating unbalance From figure following observations may be made for a rotating unbalanced system. For very low value of frequency ratio (say ) ), the response of the system is very small. For frequency ratio between 0.5 and1, there is a sharp increase in system response with increase in frequency of excitation of the system. At frequency ratio equal to 1, the phase angle is Maximum response amplitude occurs at a frequency slightly greater than. With increase in damping, the response of the system decreases. For higher value of (say >), the response amplitude approaches and phase angle approaches Vibration Isolation & Transmissibility: In many industrial applications, one may find the vibrating machine transmit forces to ground which in turn vibrate the neighbouring machines. So in that contest it is necessary to calculate how much force is transmitted to ground from the machine or from the ground to the machine. Figure.19 : A vibrating system

47 Figure..19 shows a system subjected to a force. and vibrating with This force will be transmitted to the ground only by the spring and damper. Force transmitted to the ground It is known that for a disturbing force oscillation, the amplitude of resulting Substituting these equations and defining the transmissibility TR as the ratio of the force transmitted Force to the disturbing force one obtains Comparing equations for support motion, it can be noted that When damping is negligible

48 to be used always greater than Replacing To reduce the amplitude X of the isolated mass m without changing TR, m is often mounted on a large mass M. The stiffness K must then be increased to keep ratio K/(m+M) constant. The amplitude X is, however reduced, because K appears in the denominator of the expression Figure.0: Transmissibility ~frequency ratio plot Figure.0 shows the variation transmissibility with frequency ratio and it can be noted that vibration will be isolated when the system operates at a frequency ratio higher than

49 Equivalent Viscous Damping : In the previous sections, it is assumed that the energy dissipation takes place due to viscous type of damping where the damping force is proportional to velocity. But there are systems where the damping takes place in many other ways. For example, one may take surface to surface contact in vibrating systems and take Coulomb friction into account. Also in many cases energy is dissipated in joints also, which is a form of structural damping. In these cases one may still use the derived equations by considering an equivalent viscous damping. This can be achieved by equating the energy dissipated in the original and the equivalent system. The primary influence of damping on the oscillatory systems is that of limiting the amplitude at resonance. Damping has little influence on the response in the frequency regions away from resonance. In case of viscous damping, the amplitude at resonance is For other type of damping, no such simple expression exists. It is possible to however, to approximate the resonant amplitude by substituting an equivalent damping Ceq in the foregoing equation. The equivalent damping Ceq is found by equating the energy dissipated by the viscous damping to that of the nonviscous damping with assumed harmonic motion. Where must be evaluated from the particular type of damping. Structural Damping : When materials are cyclically stressed, energy is dissipated internally within the material itself. Experiments by several investigators indicate that for most structural metals such as steel and aluminum, the energy dissipated per cycle is independent of the frequency over a wide frequency range and proportional to the square of the amplitude of vibration. Internal damping fitting this classification is called solid damping or structural damping. With the energy dissipation per cycle proportional to the square of the vibration amplitude, the loss coefficient is a constant and the shape of the hysteresis curve remains unchanged with amplitude and independent of the strain rate. Energy dissipated by structural damping can be written as

50 Where is a constant with units of force displacement. By the concept of equivalent viscous damping Coulomb Damping : or Coulomb damping is mechanical damping that absorbs energy by sliding friction, as opposed to viscous damping, which absorbs energy in fluid, or viscous, friction. Sliding friction is a constant value regardless of displacement or velocity. Damping of large complex structures with non-welded joints, such as airplane wings, exhibit coulomb damping. Work done per cycle by the Coulomb force For calculating equivalent viscous damping From the above equation equivalent viscous damping is found Summary Some important features of steady state response for harmonically excited systems are as follows- The steady state response is always of the form. Where it is having same frequency as of forcing. X is amplitude of the response, which is strongly dependent on the frequency of excitation, and on the properties of the spring mass system.

51 There is a phase lag between the forcing and the system response, which depends on the frequency of excitation and the properties of the spring-mass system. The steady state response of a forced, damped, spring mass system is independent of initial conditions. In this chapter response due to rotating unbalance, support motion, whirling of shaft and equivalent damping are also discussed. Magnification Factor (Dynamic magnifier) or Amplitude Ratio The ratio of the maximum displacement of the forced vibration (x max ) to the static deflection under the static force F 0 (x o ) is known as Magnification factor. Denoted by M.F. i.e, M.F = x x max 0 F We have 0 /s xmax 1 n n x max x 0 1 n n M.F 1 1 n n From this equation, it is clear that the magnification factor depends upon -The ratio of circular frequencies n -The damping factor ( )

52 Energy Dissipated by damping The energy lost per cycle due to a damping force F d is calculated by wd F d.d x where damping force, Fd = dx C. dt Energy dissipated/cycle, W c x d Vibration Measuring Instruments The instruments which are used to measure the displacement, velocity or acceleration of a vibrating body are called vibration measuring instruments. Widely used vibrometers (low frequency transducer) Accelerometers (high frequency transducer) Example: Accelerometers a) Bonded strain gauge accelerometer i) Cantilever beam type accelerometer ii) Solid cylinder accelerometer b) Piezoelectric accelerometer c) Servo accelerometer (force balance accelerometer) Piezo electric accelerometers Certain crystals exhibit the property that they generate a charge across their faces when a stress is applied to them. This property is made use of in piezoelectric accelerometer. The change generated to the device is given by q = f.d. When f applied force d Piezoelectric constant When the device is subjected to acceleration the mass exerts a variable force on the Piezoelectric disc, which is proportional to the acceleration. The charge developed across the disc is in turn proportional to the acceleration of the mass.

53 MODULE 3 TWO-DEGREE-OF-FREEDOM-SYSTEMS A single degree of freedom system has only one natural frequency and requires only one independent co-ordinate to define the system completely. But a Two Degree freedom system has two natural frequencies and the free vibration of any point in the system, in general, is a combination of two harmonies of these two natural frequencies respectively. Under certain conditions, any point in the system may execute harmonic vibrations, at any of the two natural frequencies, and these are known as the principal moles of vibrations. In Two Degree freedom systems, there are two independent co-ordinates to specify the configuration. The vibrating systems, which require two coordinates to describe its motion, are called two-degrees-of -freedom systems. These coordinates are called generalized coordinates when they are independent of each other and equal in number to the degrees of freedom of the system. Unlike single degree of freedom system, where only one co-ordinate and hence one equation of motion is required to express the vibration of the system, in two-dof systems minimum two co-ordinates and hence two equations of motion are required to represent the motion of the system. For a conservative natural system, these equations can be written by using mass and stiffness matrices. One may find a number of generalized co-ordinate systems to represent the motion of the same system. While using these co-ordinates the mass and stiffness matrices may be coupled or uncoupled. When the mass matrix is coupled, the system is said to be dynamically coupled and when the stiffness matrix is coupled, the system is known to be The set of co-ordinates for which both the mass and stiffness matrix are uncoupled, are known as principal co-ordinates. In this case both the system equations are independent and individually they can be solved as that of a single-dof system. A two-dof system differs from the single dof system in that it has two natural frequencies, and for each of the natural frequencies there corresponds a natural state of vibration with a displacement configuration known as the normal mode. Mathematical terms associated with these quantities are eigenvalues and eigenvectors. Normal mode vibrations are free vibrations that depend only on the mass and stiffness of the system and how they are distributed. A normal mode

54 oscillation is defined as one in which each mass of the system undergoes harmonic motion of same frequency and passes the equilibrium position simultaneously. The study of two-dof- systems is important because one may extend the same concepts used in these cases to more than -dof- systems. Also in these cases one can easily obtain an analytical or closed-form solutions. But for more degrees of freedom systems numerical analysis using computer is required to find natural frequencies (eigenvalues) and mode shapes (eigenvectors). Derivation of Equation of Motion Few examples of two-degree-of-freedom systems :: Figure 3.1(a) shows two masses m 1 and m with three springs having spring stiffness k 1, k and k 3 free to move on the horizontal surface. Let x 1 and x be the displacement of mass respectively. Figure 3.1(a) As described in the previous lectures one may easily derive the equation of motion by using d'alembert principle or the energy principle (Lagrange principle or Hamilton 's principle) Figure 3.1(b): Free body diagrams Using d'alembert principle for mass m 1 from the free body diagram shown in figure 3.1(b)

55 and similarly for mass m Derivation of Equation of Motion and Coordinate Coupling Noting, the above two equations in matrix form can be written as Now depending on the position of point C, few cases can are studied below. Case 1 : Considering be written as, i.e., point C and G coincides, the equation of motion can Figure 3. So in this case the system is statically coupled and if disappears, and we obtained uncoupled x and vibrations., this coupling Case : If,, the equation of motion becomes Hence in this case the system is dynamically coupled but statically uncoupled. Case 3: If we choose motion will become, i.e. point C coincide with the left end, the equation of

56 Here the system is both statically and dynamically coupled. Normal Mode Vibration Again considering the problem of the spring-mass system in figure with,,, the equation of motion can be written as We define a normal mode oscillation as one in which each mass undergoes harmonic motion of the same frequency, passing simultaneously through the equilibrium position. For such motion, we let Hence, or, in matrix form Hence for nonzero values of and (i.e., for non-trivial response) Now substituting, equation yields Hence, and

57 So, the natural frequecies of the system are and Now it may be observed that for these frequencies, as both the equations are not independent, one can not get unique value of and. So one should find a normalized value. One may normalize the response by finding the ratio of to. From the first equation. the normalized value can be given by and from the second equation the normalized value can be given by Now, substituting in equation the same values, as both these equations are linearly dependent. Here, and similarly for It may be noted If one of the amplitudes is chosen to be 1 or any number, we say that amplitudes ratio is normalized to that number. The normalized amplitude ratios are called the normal modes and designated by. The two normal modes of this problem are:

58 In the 1 st normal mode, the two masses move in the same direction and are said to be in phase and in the nd mode the two masses move in the opposite direction and are said to be out of phase. Also in the first mode when the second mass moves unit distance, the first mass moves units in the same direction and in the second mode, when the second mass moves unit distance; the first mass moves.73 units in opposite direction. VIBRATION ABSORBER Tuned Vibration Absorber Consider a vibrating system of mass, stiffness, subjected to a force. As studied in case of forced vibration of single-degree of freedom system, the system will have a steady state response given by (1) which will be maximum when Now to absorb this vibration, one may add a secondary spring and mass system as shown in figure.

59 The equation of motion for this system can be given by As we know for steady state vibration, the system will vibrate with a frequency of the external excitation; we can assume the solution to be () Substituting Equation (3) in equation () one may write (3) (4) Or, (5) Using Cramer's rule one may write (6) (7) where

60 Now Here are the roots of the characteristic equation. One may note that these roots are the normal mode frequency for this two-degrees of freedom system. These free-vibration frequencies can be given by From equation (6), it is clear that, Hence, if a system called the primary system with a stiffness mass is subjected to an exciting force or base motion to vibrate, it is possible to completely eliminate the vibration of the primary system by suitably designing an attached spring-mass system (secondary system) with stiffness and mass such that the natural frequency of the secondary system coincide with the exciting frequency.. This is the principle of dynamic vibration absorber. From equation (1) it may be noted that the primary system will have resonance when the natural frequency of the primary system coincide with that of the excitation frequency. Hence to reduce the vibration at resonance of the primary system one should design the secondary system such that the natural frequency of both the components coincides.

61 For this condition Substituting and, the above equation reduces to or, For,, and

62 To keep the displacement of secondary mass small, the stiffness of the secondary spring should be very large. To have this the secondary mass should also be large which is not desirable from practical point of view. Hence a compromise is usually made between the amplitude and the mass ratio. The mass ratio is usually kept between 0.05 and 0.5. Resonant frequency of the vibration absorber Centrifugal Pendulum Vibration Absorber The centrifugal pendulum vibration absorber was devised and patented in France about 1935 and at the same time it was independently conceived and put into practice by E. S. Taylor. Its purpose was to overcome serious torsional vibration problem inherent in geared radial aircraft-engine propeller system. Later it was modified and incorporated into automobile IC engines in order to reduce the torsional vibrations of the crankshaft. This was done by integrating the absorber mass with crankshaft counter balance mass. The tuned vibration absorber is only effective when the frequency of external excitation equals to the natural frequency of the secondary spring and mass system. But in many cases, for example in case of an automobile engine, the exciting torques are proportional to the rotational speed n' which may vary over a wide range. For the absorber to be effective, its natural frequency must also be proportional to the speed. The characteristics of the centrifugal pendulum are ideally suited for this purpose. Placing the coordinates through point O', parallel and normal to r, the line r rotates with angular velocity ( ).

63 The acceleration of mass Since the moment about is zero, Assuming to be small,, so If we assume the motion of the wheel to be a steady rotation sinusoidal oscillation of frequency, one may write plus a small Hence the natural frequency of the pendulum is and its steady-state solution is t may be noted that the same pendulum in a gravity field would have a natural frequency of. So it may be noted that for the centrifugal pendulum the gravity field is replaced by the centrifugal field.

64 Torque exerted by the pendulum on the wheel With the component of equal to zero, the pendulum force is a tension along, given by times the component of. Now assuming small angle of rotation Now substituting Hence the effective inertia can be written as which can be at its natural frequency. This possesses some difficulties in the design of the pendulum. For example to suppress a disturbing torque of frequency equal to four times the natural speed n, the pendulum must meet the requirement. Hence, as the length of the pendulum becomes very small it will be difficult to design it. To avoid this one may go for Chilton bifilar design. MULTIDEGREE OF FREEDOM SYSTEMS It must be appreciated that any real life system is actually a continuous or distributed parameter system (i.e. infinitely many d.o.f). Hence to derive its equation of motion we need to consider a small (i.e., differential) element and draw the free body diagram and apply Newton 's second Law. The resulting equations of motion are partial differential equations and more complex than the simple ordinary differential equations we have been dealing with so far. Thus we are interested in modeling the real life system using lumped parameter models and ordinary differential equations.

65 The accuracy of such models (i.e. how well they can model the behavior of the infinitely many d.o.f. real life system) improves as we increase the number of d.o.f. Thus we would like to develop mult-d.o.f lumped parameter models which still yield ordinary differential equations of motion as many equations as the d.o.f. We would discuss these aspects in this lecture. Derivation of Equations of Motion Fig 3.3 Typical multi-d.o.f. system Consider a typical multi-d.o.f system as shown in Fig As mentioned earlier our procedure to determine the equations of motion remains the same irrespective of the number of d.o.f of the system and is recalled to be: Step 1 : Consider the system in a displaced Configuration Step : Draw Free Body diagrams and Step 3 : Use Newton 's second Law to write the equation of motion From the free body diagrams shown in Fig. 3.3, we get the equations of motion as follows: Rewriting the equations of motion in matrix notation, we get: Or in compact form,

66 There are n equations of motion for an n d.o.f system. Correspondingly the mass and stiffness matrices ([M] and [K] respectively) are square matrices of size (n x n). If we consider free vibrations and search for harmonic oscillations,. Substituting these in equation, we get, ie For a non-trivial solution to exist, we have the condition that the determinant of the coefficient matrix must vanish. Thus, we can write, In principle this (n x n) determinant can be expanded by row or column method and we can write the characteristic equation (or frequency equation) in terms of, solution of which yields the n natural frequencies of the n d.o.f. system just as we did for the two d.o.f system case. We can substitute the values of in eqn and derive a relation between the amplitudes of various masses yielding us the corresponding normal mode shape. Typical mode shapes are schematically depicted in Figure for a d.o.f system. Dunkerly's method of finding natural frequency of multi- degree of freedom system We observed in the previous lecture that determination of all the natural frequencies of a typical multi d.o.f. system is quite complex. Several approximate methods such as Dunkerly's method enable us to get a reasonably good estimate of the fundamental frequency of a multi d.o.f. system. Basic idea of Dunkerly's method Fig 3.4 A typical multi d.o.f. system Consider a typical multi d.o.f. system as shown in fig 3.4 Dunkerly's approximation to the fundamental frequency of this system can be obatined in two steps:

67 Step1: Calculate natural frequency of all the modified systems shown in Fig 3.5 These modified systems are obatined by considering one mass/inertia at a time. Let these frequencies be Step: Dunkerly's estimate of fundamental frequency is now given as: Fig 3.5 Modified system considered in Dunkerly's Method Fig 3.6 A Typical two d.o.f. example Consider a typical two d.o.f. system as shown in Fig 3.6 and the equations of motion are given as: For harmonic vibration, we can write: Thus, Inverting the stiffness matrix and re-writing the equations

68 The equation characteristic can be readily obtained by expanding the determinant as follows: As this is a two d.o.f. system, it is expected to have two natural frequencies viz and., Thus we can write Equation as: Comparing coefficients of like terms on both sides, we have: It would appear that these two equations can be solved exactly for and. While this is true for this simpleexample, we can't practically implement such a scheme for an n- d.o.f system, as it would mean similar computational effort as solving the original problem itself. However, we could get an approximate estimate for the fundamental frequency. If >>, then we can approximately write Let us now study the meaning of and. It is easily verified that These can be readily verified to be the reciprocal of the equivalent stiffness values for the modified systems.

69 Thus, we can write: Holzer's method of finding natural frequency of a multi-degree of freedom system Holzer's Method. This method is an iterative method and can be used to determine any number of frequencies for a multi-d.o.f system. Consider a typical multi-rotor system as shown in Fig. 3.7 Fig 3.7 Typical multi-rotor system The equations of motion for free vibration can be readily written as follows: For harmonic vibration, we assume Thus:

70 Summing up all the equations of motion, we get: This is a condition to be satisfied by the natural frequency of the freely vibrating system. Holzer's method consists of the following iterative steps: Step 1: Assume a trial frequency Step : Assume the first generalized coordinate say Step 3: Compute the other d.o.f. using the equations of motion as follows: Step 4: Sum up and verify if this equation is satisfied to the prescribed degree of accuracy. If Yes, the trial frequency is a natural frequency of the system. If not, redo the steps with a different trial frequency. In order to reduce the computations, therefore one needs to start with a good trial frequency and have a good method of choosing the next trial frequency to converge fast. Two trial frequencies are found by trial and error such that is a small positive and negative number respectively than the mean of these two trial frequencies(i.e. bisection method) will give a good estimate of for which. Holzer's method can be readily programmed for computer based calculations. TORSIONAL VIBRATIONS When the particles of a shaft or disc move in a circle about the axis of a shaft, then the vibrations are known as torsional vibrations. In this case the shaft is twisted and untwisted alternately and torsional shear stresses are introduced in the shaft.

71 Torsional vibrations may result in shafts from following forcings: Inertia forces of reciprocating mechanisms (such as pistons in Internal Combustion engines) Impulsive loads occurring during a normal machine cycle (e.g. during operations of a punch press) Shock loads applied to electrical machineries (such as a generator line fault followed by fault removal and automatic closure) Torques related to gear tooth meshing frequencies, turbine blade passing frequencies, etc. For machines having massive rotors and flexible shafts (where system natural frequencies of torsional vibrations may be close to, or within, the source frequency range during normal operation) torsional vibrations constitute a potential design problem area. In such cases designers should ensure the accurate prediction of machine torsional frequencies and frequencies of any of the torsional load fluctuations should not coincide with torsional natural frequencies. Hence, determination of torsional natural frequencies of a dynamic system is very important. Simple systems with a single disc mass: Consider a rotor system as shown in Figure 4.1. The shaft is considered as massless and it provides the stiffness only. The disc is considered as rigid and it has no flexibility. If we give a small initial disturbance to the disc in the torsional mode and allow it to oscillate its own, it will execute free vibrations. The oscillation will be simple harmonic motion (SHM) with a unique frequency, which is called natural frequency of the rotor system. From the theory of torsion of shaft, we have

72 where, k t is the torsional stiffness of shaft, I p is the rotor polar mass moment of inertia, J is the shaft polar second moment of area, l is the length of the shaft and q is the angular displacement of the rotor. From free body diagram of the disc External torque on the disc or The free (or natural) vibration has the simple harmonic motion (SHM). For the simple harmonic motion of the disc, we have so that where is the amplitude of the torsional vibration and is natural frequency of the torsional vibration. On substitution we get or Hence, the torsional natural frequency is given by square root of the ratio of torsional stiffness to the polar mass moment of inertia. A two-disc torsional system In a two-discs torsional system as shown in Figure 4.3, whole of the rotor is free to rotate i.e. the shaft being mounted on frictionless bearings.

73 From free body diagrams of discs in Figure 4.4, we can write External torque on disc 1 and External torque on disc and or and For SHM, and where is the torsional natural frequency. On substitution we get and which can be written in the matrix form, as

74 with and This equation is a homogeneous equation and the non-trial solution is obtained by taking determinant of the matrix [k] as k = 0 which gives frequency equation of the following form which can be simplified as Roots of equation are given as and From equation corresponding to first natural frequency for = 0, we get From equation it can be concluded that, the first root of equation represents the case when both discs simply rolls together in phase with each other as shown in Figure 4.5 i.e. the rigid body mode, which is of a little practical significance. From equation (4.9), for, we get On substituting for in the above equation we get

75 which can be simplified as which gives relative amplitudes of two discs as or The second mode represents the case when both masses vibrate in anti-phase with one another. Figure 4.6 shows the second mode shape of two-rotor system, showing two discs vibrating in opposite directions. From the second mode shape, i.e. from Figure 4.6 and noting equation we have

76 Since both masses are always moving in the opposite direction, so there must be a point on the shaft where the torsional vibration is not taking place i.e. a torsional node. The location of the node may be established by treating each end of the real rotor system as a separate single-disc cantilever system as shown in Figure 4.5. The torsional node being treated as the point where the shaft is rigidly fixed. Since the natural frequency of the system is known and the frequency of oscillation of each of the single-disc system must be same, hence we write and Lengths l 1 and l then can be obtained by, noting equation as and which must satisfy Torsional Systems with a Stepped Shaft

77 Figure 4.7(a) shows a two-disc stepped shaft. The polar mass moment of inertia of the shaft is negligible as compared to discs. In such cases the actual shaft should be replaced by an unstepped equivalent shaft for the purpose of the analysis as shown in Figure 4.7(b). The equivalent shaft diameter may be same as the smallest diameter of the real shaft. The equivalent shaft must have the same torsional stiffness as the real shaft, since for the present case torsional springs are connected in series. The equivalent torsional spring can be written as we have which gives where where, and are equivalent lengths of shaft segments having equivalent shaft diameter d 3 and l e is the total equivalent length of unstepped shaft having diameter d 3 as shown in Figure 4.7(b). From Figure 4.7(b) and noting equations in the equivalent shaft the node location can be obtained as with and From equations the node position a & b can be obtained in the equivalent shaft length. Now the node location in real shaft system can be obtained as follows:

78 we have Since above equation is for shaft segment in which node is assumed to be present, we can write and It can be combined as So once a & b are obtained the location of the node in the actual shaft can be obtained i.e. the final location of the node on the shaft in real system is given in the same proportion along the length of shaft in the equivalent system in which the node occurs.

79 MODULE 4 WHIRLING OF SHAFTS CRITICAL SPEED Consider a typical shaft, carrying a rotor (disk) mounted between two bearings as shown in Figure.Let us assume that the overall mass of the shaft is negligible compared to that of the rotor (disk) and hence we can consider it as a simple torsional spring. The rotor (disk) section has a geometric centre i.e., the centre of the circular cross-section and the mass centre due to the material distribution. These two may or may not coincide in general, leading to eccentricity. The eccentricity could be due to internal material defects, manufacturing errors etc. As the shaft rotates, the eccentricity implies that the mass of the rotor rotating with some eccentricity will cause in-plane centrifugal force. Due to the flexibility of the shaft, the shaft will be pulled away from its central line as indicated in the figure. Let us assume that the airfriction damping force is negligible. The centrifugal force for a given speed is thus balanced by the internal resistance force in the shaft-spring and the system comes to an equilibrium position with the shaft in a bent configuration as indicated in the figure. Thus the shaft is rotating about its own axis and the plane containing the bent shaft and the line of bearings rotates about an axis coinciding with the line of bearings. We consider here only the case, wherein these two rotational speeds are identical, called the synchronous whirl.

80 Centrifugal force shaft resistance force wherein, the shaft stiffness k is the lateral stiffness of a shaft in its bearings i.e., considering the rotor a span, this is the force required to cause a unit lateral displacement at mid-span of the simply supported Thus Where E is the Young's modulus, I is the second moment of area, and L is the length between the suppo Thus, for equilibrium, ie where we have used to represent the natural frequency of the lateral vibration of the sp shaft-rotor system. Thus when the rotational speed of the system coincides with the natural frequen lateral vibrations, the shaft tends to bow out with a large amplitude. This speed is known as the critical and it is necessary that such a resonance situation is avoided in actual practice. As discussed earlier case of resonance, it takes some time for the amplitude to build up to a large value. Some of the tu rotors whose operating speeds go beyond the critical speed are able to use this fact and rush-throug critical speed. It is necessary to observe that, in synchronous whirl. the heavier side remains all the tim the outer side. Thus when the shaft bends, an inner fibre is under compressive stress and outer fibre is tensible stress but there is NO REVERSAL of stress. Rayleighs Method Fig 4.1 Multiple disks on a shaft

81 Rayleigh's method is based on the principle of conservation of energy. The energy in an undamped system consists of the kinetic energy and the potential energy. The kinetic energy T is stored in the mass and is proportional to the square of the velocity. The potential energy U includes strain energy that is proportional to elastic deformations and the potential of the applied forces. For a conservative system, the total energy must remain constant. That is Differentiating this expression, we get the equation of motion as follows. Note that the amounts of kinetic and potential energy in the system may change with time but their sum must remain constant. Thus if and are energies at time and and are energies at time, then For a shaft as shown in Fig.4. the potential energy is zero at the specific instant of time when the mass is passing through its static equilibrium position and kinetic energy is at its maximum. Similarly at the instant when the mass is at its extreme position the kinetic energy is zero and the potential energy is at its maximum. Thus we have the following relationship. Therefore we have, considering all the disks on the shaft, Where i=1, n represents summation over all the "n" disks. So we get the frequency of natural vibration as,

82 Dunkerley's Empirical Method When a shaft carries multiple disks it is always efficient to use this method. Fig 4. Dunkerly's approximation for shaft We consider only one force (wt of disk) acting on the shaft at a time. For each disk, we find the corresponding natural frequency as. The Nutural frequency of the shaft when all the loads(disks) act on the shaft simultaneously can be found out by the using the formulae:

83 For each of the dub-systems ie shaft with only one disk, natural frequency is obtained as where k is the lateral stiffness of the shaft in its bearings and m is the mass of the disk. To understand the basis of this method, we need to appreciate multi-d.o.f system vibrations. CRITICAL SPEEDS OF A LIGHT CANTILEVER SHAFT WITH A LARGE HEAVY DISC AT ITS END If a light shaft having two end supports has a central disc then the system has been shown to have one critical speed. Even if the disc is not central, the system will have one critical speed as long as we assume the mass of the disc to be concentrated. If, however, the disc has mass as well as moment of inertia, and is not central, then the system will have two critical speeds. The treatment given below is for a light cantilever shaft having a disc which has mass as well as moment of inertia. Since the critical speed is numerically equal to the natural frequency of lateral vibrations, we will find the later for this system. Consider the beam so as to be displaced from the equilibrium position as shown in Fig. In this figure, M Mr = mass of the disc, = moment of inertia of the disc about an axis passing through the CG of the disc and perpendicular to the plane of the paper.

84 y, y = displacement and acceleration of the CG of the disc,, = angular displacement and angular acceleration of the axis of the disc due to bending Further let a 11 a a 1 where l I E = deflection of the CG of the disc per unit force acting on it in the lateral direction = l 3 /3EL = slope at the free end of the beam per unit moment acting on the CG of the disc, in the plane of the paper = l/ei = deflection of the CG of the disc per unit moment acting on it = slope at the free and of the beam per unit force acting on it in a lateral direction = l /EI = length of the beam, = moment of inertia of the section of the beam about the neutral axis. = modulus of elasticity of the material of the beam. The inertia force and the inertia torque on the disc in the displaced position are shown in Fig.8.6.1, along with their directions. These are as follows: Inertia force My M y Inertia torque Mr Mr (1 ) system. where is natural frequency for the principal mode of vibration of the Then the deflection at the CG of the disc and the rotation of the disc in the plane of the paper are given by 11 1 y a M y a Mr () 1 a M y a Mr (3) Eliminating y and from the above two equations, and putting

85 g h (4) 3r l 3 Ml 3EI we have hg 4 4(h + 1)g + 4 = 0 (5) giving the two natural frequencies as g 1, (h 1) (h 1) h h (6) Figure () is a plot of the above equation and shows the variation of the two 3r natural frequencies of the system with the change in h l ; it may be recalled that r is radius of gyration of the disc about an axis passing through its CG and perpendicular to the axis of the disc. h = 0, corresponds to the concentrated mass h, corresponds to the disc having large radius of gyration with the change in 3r h l. Transient Vibrations INTRODUCTION A system subjected to periodic excitation has two components of motion, the transient and the steady state. In most of such cases the transient part is not important as it dies out soon, and the steady state part is the one that persists. However, where the excitation is of the aperiodic nature like a shock pulse or a transient excitation, the response of the system is purely transient. After the duration of the excitation, the system undergoes vibrations with its natural frequency with an amplitude depending upon the type and duration of the excitation. It is in such cases that the transient vibrations have importance. The practical examples of shock excited transient vibrations are rock explosions, gunfires, loading or unloading of packages by dropping them on hard floors, punching operations, automobiles at high

86 speeds passing over pits or curbs on the road, etc. The use of Laplace transform is introduced in this chapter for the analysis of systems subjected to shock pulses. The usual differential equations method or the socalled classical method becomes very lengthy and cumbersome with transient excitations of different shapes. LAPLACE TRANSFORMATION Laplace transform is a powerful mathematical tool that is extremely useful in the solution of differential equations, and especially so, where transients are involved. It is that branch of operational calculus wherein a function is transformed from t (time) domain to a new s domain. The original differential equation in t domain, by use of Laplace tranform, changes itself into an algebraic equation in s domain. The solution of an algebraic equation is very easy as compared to that of a differential equation. Once the solution is s domain is obtained, the process of inverse transformation gives the solution back in t domain. Manipulation with transformation and inverse transformation is facilitated by the use of table of transform pairs which is given later in this section. Laplace transform F(s) of a function f(t) is defines as st F(s) f (t)d dt 0 In shorthand it is generally written as L [f(t)] = F(s) The use of the basic definition of Laplace transform is illustrated below by actually transforming a few common functions. RESPONSE TO AN IMPULSIVE INPUT

87 Consider a damped spring mass system subjected to an impulse ˆF (t), the strength of the impulse being ˆF. Since the impulse acts for an extremely small duration, its effect is to give an initial velocity to the mass given by ˆF mdv where dv is the change in velocity of the mass due to the impulse ˆF. If the system is initially at rest, impulse gives it a starting velocity of ˆF dv m The initial displacement of the mass form the equilibrium position is zero because of the extremely small duration of the impulse. Thus the initial conditions for the mass are x(0) 0 ˆF (1) x(0) m The differential equation for the system can now be written as mx cx kx 0 () The forcing function on the right has been taken to be zero since the impulse effectively gives only the initial conditions obtained in Eq. (1). Dividing Eq. () by m through out, it can be written as n n x x x 0 (3) Taking the Laplace transform of the above equation, we have

88 n n [s X(s) sx(0) x(0)] [sx(s) x(0)] X(s) 0 Substituting the initial conditions of Eq. (1), and re-arranging, gives ˆF 1 X(s) m s ns n (4) In order to obtain the inverse transformation fro the above equation, the expression on the right has to be re-arranged in one of the forms corresponding to the transform pairs, for direct inversion. If 1, the above equation is re-written in the following form F 1 n X(s) m 1 n (s n) 1 n (5) The inverse transform is, ˆF x(t) e n t sin 1 nt m 1 n (6) which is the response of the system to an impulsive input. RESPONSE TO A STEP INPUT Figure shows a spring-mass-dashpot system subjected to a step force F 0 u(t). The magnitude of the force is constant at a value F 0 for all time greater than or equal to zero. The force is zero for 1 < 0. The differential equation of motion can be written as mx cx kx F0 u(t)

89 or F0 x nx n x u(t) (1) m Taking the Laplace transform of the above equation, we have F0 1 [s X(s) s x(0) x(0)] n[s X(s) x(0)] n X(s). m s A second order system subjected to a finite step cannot have any initial velocity or displacement. So, putting all initial conditions zero in the above equation, and re-arranging, we have F 0 1 X(s) m s(s ns n ) () The inverse transform of the above equation cannot be obtained straightway from the tables. Hence splitting the right hand side into partial fractions, we have F 1 1 s X(s) m s s s 0 n n n n The right had expression in the bracket above is still not invertable directly. Assuming an underdamped system, i.e. 1, the above equation is written as follows:. 1 n F (s 1 n ) X(s) m n s (s n ) 1 n (s n ) 1 n (3) Inverse the transform of Eq. (3) can now be obtained directly from the table and is given below F 0 nt nt x(t) 1 e cos 1 nt e sin 1 nt m n 1 Putting m n = k, we finally have

90 F 0 n t x(t) 1 e cos 1 nt sin 1 nt k 1 (4) For an undamped case, response equation can be written from the above equation by putting 0, or F x(t) [1 cos t] k 0 n (5) RESPONSE TO A PULSE INPUT Pulse applications in engineering practice are very common. An explosion occurring on a system with a comparatively larger natural period will be an impulse while the same explosion occurring on a system with a smaller natural period will be pulse. In this section two important types of pulses, rectangular and half sinusoidal, are treated. The method lends itself for the analysis of any type of pulse for which a mathematical equation can be written. The vibratory systems considered in this section have been taken as undamped systems to make the response equations simpler. Further, since most physical systems are lightly damped and in most cases we are interested in maximum displacements and accelerations, we will be slightly erring on the safer side in neglecting small amount of damping. Rectangular pulse Consider a spring-mass system subjected to a rectangular pulse to height F 0 and duration as shown. The response equation can be written directly by

91 comparing the response of the system to a multi-step input by considering in this case two equal and opposite steps, one at t = 0 and the other at t. Therefore, F0 F0 x(t) [1 cos nt]u(t) [1 cos n(t )]u(t ) k k (1) The above equation can be written as the following two equations F x(t) [1 cos t] for 0 t k 0 n F 0 [cos n (t ) cos n t] for t k () PHASE PLANE METHOD A spring mass system with initial conditions X 0 and V 0, has its differential equation written as n x x 0 (1) Its solution may be written as x Asin( nt ) where A X V0 0 n tan V0 1 and nx0 Differentiating equation (1) for velocity, we have

92 x Ancos( nt ) or x n Acos( t ) n () Squaring and adding Eq. (1) and (), we have x x n A (3) The above equation is a circle in a plane with coordinate axes x and (x / n ) Its radius is A and centre at the origin. This is shown in Figure.. The starting point on this displacement velocity plot is marked P 1. At t 1 seconds later the displacement and velocity of the system are represented by point P where P1 OP nt1 radians. From this diagram, the displacement and velocity phase of the motion are available from the single point which corresponds to a particular time. This is the phase plane plot. The horizontal projection of the phase trajectory on a time base gives the displacement-time plot of the motion and is shown in Figure Similarly the vertical projection on the time base will give the velocity-time plot of the motion. It may be noted that the centre of the phase trajectory always lies on the x- axis at a distance equal to the static equilibrium displacement of the system. In the case discussed the static equilibrium displacement was zero and therefore the centre of the circle was located at the origin. In case of a step force input F 0, the static equilibrium position suddenly changes through a distance F 0 /k. Thus the phase plane plot for such a motion will be a circle whose centre lies F 0 /k above the centre. The radius of this circle will be F 0 /k so that the trajectory starts from the origin corresponding to zero initial conditions. The use of the phase plane method is illustrated by the following examples for systems subjected to multiple steps. SHOCK SPECTRUM The response of a spring-mass system to a particular pulse depends upon the natural frequency of the system. The plot of the maximum response of the system against the natural frequency of the system is called the shock spectrum of the

93 particular disturbance. The shock spectrum shows at a glance the natural frequencies which cause large response amplitudes for the particular disturbance. Non-Linear Vibrations INTRODUCTION Most physical systems can be represented by linear differential equations, the types of which have been dealt with in the previous chapters. A general equation of this type is mx cx kx F(t) (1) In this equation which is for a linear system, the inertia force, the damping force and the spring force are linear functions of x,x and x respectively. This is not so in the case of non-linear systems. A general equation for a non-linear system is mx ( )x f(x) F(t) () in which the damping force and the spring force are not linear functions of x and x. There are quite some physical systems which have non-linear spring and damping characteristics. Rubber springs and other similar isolators have spring stiffness which increases with amplitude. Cast iron and concrete have spring stiffness which decreases with amplitude. Examples of non-linear damping are dry friction damping and material damping. Even so called linear systems tend to become non-linear with larger amplitudes of vibration. The analysis of non-linear systems is comparatively difficult. In certain cases there is no exact solution. One major difference between the linear and non-linear systems is that the law of superposition does not hold good for non-linear systems. Mathematically speaking, if x 1 is a solution of and x is a solution of mx cx kx F 1(t) mx cx kx F (t) then (x 1 + x ) is a solution of

94 mx cx kx F 1(t) F (t) This is not so in the case of non-linear systems. Even for the case of free vibration any two known solutions of the non-linear system cannot be superimposed to obtain a general solution. PHASE PLANE Phase plane was introduced in Sec. 9.6 for the case of linear systems. Here we extend it for the case of non-linear systems. Consider the differential equation mx f (x) 0 (1) The acceleration x can also be written as dv x v dx where v is the velocity of the particle. Substituting it in Eq. (1), we have dv mv f (x) 0 dx or mv dv = f(x) dx () The above equation is intergrable directly. If v = V 0 when x = X 0, then the integration of Eq. () gives v x mvdv f (x)dx (a) V0 X0 or 0 mv mv [F(x) F(X 0)] The above equation is in accordance with the Law of Conservation of

95 Energy. The left hand side is the increase in kinetic energy of the system and the right hand side is the decrease in potential energy. The equation can also be written as 0 mv mv F(x) F(X 0) (3) which states that the total energy of the system at any instant is equal to the total initial energy of the system. Curve in x v plane can be drawn from Eq. (3) and this will be a curve of constant energy. A set of such curves can be drawn, each for different total energy. These curves are known as Energy Curves or Integral Curves in phase plane. We had taken the phase with x along the ordinate and v along the abscissa, the trajectory was always counter-clockwise. Here, for convenience we will take x along the abscissa and v along the ordinate. The trajectories here will be clockwise. Systems which have periodic motion would be represented on the phase plane by means of a set of closed curves, each curve for different energy of the system. Consider the linear case when f(x) = kx. Equation (a) then integrates to mv mv0 kx kx0 0 kx0 mv kx mv E(say) (4) The phase plane trajectories are clearly a set of ellipses with the origin as the centre. The right hand side of Eq. (4) is the total initial energy of the system. As the value of this initial energy (depending upon initial conditions) increases, the size of the ellipse also increases. In general a point P(x, v) in the phase plane, called the representative point of the system, represents the state of the system at any instant t ad the trajectory traced gives the history of the system. Through any and every point of the phase plane passes one and only one trajectory and thus the trajectories in the plane do not

96 intersect one another. With the passage of time the representative point moves along the trajectory in a clockwise direction with what is known as the phase velocity given by u x v (5) The velocity is always non-zero at points of equilibrium. Origin is a point of stable equilibrium.most non-linear equations cannot be solved explicitly. However, their phase-plane plots can be drawn graphically and these diagrams give several important conclusions regarding motion of the systems. Consider the general case of a system with non-linear damping and nonlinear spring. Let the differential equation of motion be Letting x following two equations. mx (x) f (x) 0 (6) v, the above equation can be written down in the form of the dx v dt dv (v) f (x) dt m (7) Differential dt can be eliminated from Eq. (7) to give dv (v) f (x) (8) dx mv Equation (8) gives the slope of the trajectory at any point (x, v) in the phase plane and is useful for constructing the plots. The slope of the trajectory is directly obtainable at all points except the ones where v = 0 and (v) f (x) 0. At these points the slope becomes indeterminate and these points are called singular points. From Eq. (7), these points correspond to the conditions v = 0 and dv/dt = 0, i.e. the points of equilibrium. Singular points exist on x-axis (v = 0) wheredv/dt = 0. Origin is always a singular point. Other singular points may or may not be there for the system. At singular points the phase velocity is zero.

97 UNDAMPED FREE VIBRATION WITH NON-LINEAR SPRING FORCES Let the system be represented by mx f (x) 0 (1) The above equation can be written as dv mv f (x) 0 dx () Integrating Eq. (10.4.) we have mv F(x) E (3) where F(x) is the integral of f(x) and so represents the potential energy of the system, and E is the total energy of the system and depends upon the initial conditions.from Eq. (3), we have v [E F(x)] (4) m Let the system have cubic non-linearity represented by 3 f (x) x x ( 0) (5) then, 4 x x F(x) (6) 4 From Eq. (10.4.3) and (10.4.6), we get 4 mv x x E (7) 4 The above equation gives the plot in the phase plane for different values of E. These are closed curves when 0 for any amplitude. For the case when 1, the phase plane plots are closed curves upto a certain amplitude and beyond that they are unstable. For the case of closed curves, the system has periodic motion. Let xmax a be the amplitude of vibration. When x xmax a, v = 0; Eq. (7) becomes 4 a a E 0 (8) 4 giving 4E a (9)

98 In the above equation only positive sign before the radical is applicable whether or 0. Writing v as dx dt Eq. (4) is written as dx m dx dt [E F(x)] [E F(x)] m (10) For the case of periodic motion of amplitude a, the time period per cycle of vibration is given by integrating the above equation over a quarter of a cycle and multiplying it by 4. Thus, a m dx 4 (11) E F(x) 0 Substituting for F(x) from Eq. (6) in the above equation, we have a m dx (1) E ( x / ) ( x / 4) The quadratic expression in the radical sign of the above equation can easily be factorized since comparing it with Eq. (8) it shows that (a x ) is one of the factors of this quadratic function. Therefore, let 4 4 x x x E (a x ) b 4 4 (13) have Comparing the coefficient of x and constant terms in the above equation we E a b a (14) b 4 The second of the above equations gives b x (15) 4

99 Equation (1) can now be written with the help of Eq. (13) and (15) as a m dx 4 0 a x (a x ) 4 4 or a m dx (16) 8 0 (a x )( a x ) It is possible to convert the above complex integral into complete elliptic integral of the firs kind, the value of which can be obtained from tables of Elliptic integrals. Discussed below are the two cases falling under this type of non-linearity. PERTURBATION METHOD This is a very useful method for obtaining solutions of non-linear systems to any degree of accuracy by successive approximations. Consider the system to be represented by the differential equation 3 0 x x x 0 (1) where 0 is the natural frequency of the linear system. Assuming that the solution can be written in the form of a Taylor series in terms of the parameter (known as perturbation parameter), we can write 0 1 x x x x... () where all x s are functions of time t. The only restriction in the above way of writing is that be a small quantity. Since the frequency of vibration which is dependent upon amplitude of vibration is also unknown, we can write write But as only 0 appears in the differential equation, it is more convenient to (3)

100 Substituting equations () and (3) in Eq. (1) we have (x x x...) (...)(x x x...) 3 (x0 x1 x...) 0 (4) The above equation after expanding can be written as (x x ) (x x x x ) (x x x x 3x x )... 0 (5) Since the above equation must hold good for any small value of, it means that each of the terms in the parenthesis must individually be zero, therefore. x0 x0 0 3 x1 x1 1x 0 x0 0 (6) x x x x 3x x Now let the initial conditions be, x a x 0 at t = 0 Substituting these in Eq. () and its derivative, we have 0 1 a x (0) x (0) x (0) x (0) x (0) x (0)... Again, since these equations must be satisfied for any small value of, we have x 0(0) a x 0(0) 0 x 1(0) 0 x 1(0) 0 (7) x (0) 0 x (0) With the first set of initial conditions in Eq. (7), the solution of the first differential equation in Eq. (6), is x0 acos t (8)

101 Substituting the above in the right side of the second differential equation in Eq. (6), we get x x a cos t a cos t (9) Applying the relation cos t cos t cos3 t 4 4 to Eq. (10.5.9), it becomes x1 x 1 ( 1a a )cos t a cos3 t (10) 4 4 In the above equation the forcing function 3 ( a a )cos t will cause resonance to the system since the left hand side shows the natural frequency of the system as, the same as that of the first part of excitation. In order to avoid this absurdity, we must have a a 0 (11) Therefore, 1 3 x1 x1 a cos3 t 4 the solution of which is a x 1 (A1cos t Asint) cos3t 3 3 (1) Applying the zero initial conditions from Eq. (7), we get the constants as A 3 a 3 1 A = 0 Eq. (1) becomes 3 a x 1 [cos t cos3t] 3 (13) Substituting Eq. (8) and (13) in the first two terms of Eq. (), the solution

102 upto first order correction is obtained as a x x0 x1 a cos t [cos t cos3 t] 3 3 (14) with given by Eq. (3) upto first order correction as a 4 (15) [from Eq. (11)] The above process of successive approximations can be continued to include the higher order corrections. FORCED VIBRATIONS WITH NON-LINEAR SPRING FORCES (DUFFING S EQN.) Consider a system represented by the differential equation 3 n 0 x x x F cos t (1) This equation is known as Duffings Equation after the name of the mathematician who made an exhaustive study of this equation. Rewrite Eq. (1) as 3 n 0 x x x F cos t () Considering only small values of and F 0, it is known from experience that the frequency of steady state vibration will be the same as that of excitation plus some higher harmonics. So, the first approximate solution can be written as (), we have x1 acos t (3) Substituting the above approximate solution in the right hand side of Eq.

103 3 n 0 x a cos t a cos3 t F cos t (4) The double integration of the above equation will give x which will be a better approximation than x 1 given in Eq. (3). Using the relationship cos t cos t cos3 t 4 4 Eq. (4), is rewritten as x (F0 n a a )cos t a cos3 t (5) 4 4 Integrating the above equation twice and dropping out the constants of integration to ensure periodic motion, we have a 0 n x {F a } a cost cos3t 4 36 (6) If x 1 were a good first approximation of the system then the coefficient of cos t in Eq. (3) should be approximately the same as the coefficient of cos t in Eq. (6). Equating the two a ( n a a F 0) 4 (7) or 3 F0 n a (7a) 4 a Then the solution x(t) can be written as a x(t) x a cos t cos3t 36 3 (8) Equation (7) is a cubic in a and therefore for any value of there are in general three values of a; one root is always real, the other may be real or complex conjugate. In the non-linear systems there is no resonance like we have in linear systems. The amplitude never becomes infinite. The dotted lines in these figures show the relationship between the amplitude of vibration and the natural frequency. This is obtained from Eq. (7a) by setting F 0 is zero. These lines show that the natural frequency of a hard spring system increases with the amplitude and for a soft spring

104 system the natural frequency decreases with increase in amplitude. Since the natural frequency is different at different amplitudes, the resonance can not build up. When damping is present in the non-linear systems, the skewed peaks wind up at a certain stage.when the frequency of excitation is gradually increasing from zero, the response varies along the points there being a sudden change in amplitude from to 3 at the corresponding frequency. The portion of the response curve is never traced.this phenomenon of sudden change in amplitude from to 3 while the frequency is gradually increasing and the sudden change from 5 to 6 when the frequency is gradually decreasing is known as Jump Phenomenon. SELF EXCITED VIBRATIONS The self excited vibrations differ from forced vibrations in that the fluctuating force that sustains the motion is controlled by some part of the motion itself. The exciting force may be a function of a displacement, velocity or acceleration of the motion. When motion is stopped by some means, the fluctuating force also disappears. The forcing function is thus dependent motion itself unlike forced vibrations. Tool chatter and aeroplane wing flutter are some of the common examples of self excited vibration. It is better here to define what is known as Stability of Oscillations. If the system is such that when disturbed from its equilibrium position it comes back there after the transients die out, the system is known as dynamically stable. In case any disturbance cause the amplitude to build up with time, the system is said to be dynamically unsteady. Effectively, the system becomes unstable when negative damping appears in its differential equation of motion. A more general definition of

105 stability is that the roots of the characteristic equation of the system should either be negative real number or complex with negative real parts. Consider a third order system with its roots of the characteristic equation as s 1 = a 1 s = a + jb s 3 = a jb The solution of the equation would then be given by a1t at 1 x C e C e cos(b t ) If either a 1 or a is positive, the response would build up with time, giving instability to the system. If on the other hand both a 1 and a are negative, the response would die out with time giving stability to the system. Consider the system where a spring-mass system is supported on a horizontal belt moving with a constant velocity V. The coefficient of friction between the mass and the belt material is such that it decreases slightly with the increase in relative velocity. When mass is stationary, the friction coefficient between the mass and the belt is a. When mass is moving towards the right, the relative velocity decreases and the coefficient of friction increases. On the other hand when the mass is moving towards the left, the relative increases and therefore the coefficient of friction decreases. Since the friction force on the mass is always towards right, the helping friction force when mass moves towards right is always greater than the opposing friction force when mass moves towards left. That means a certain net energy is put into the system in each cycle. The amplitude continues to increase. If however, the mass is brought to rest in the equilibrium position, it stays in that position. The least disturbance will set it vibrating with increasing amplitude. This is a case of self excited vibration. The frequency of vibration in the cases of self excited system is approximately equal to the natural frequency of the system provided damping is not large. This problem can also be tackled analytically. At any instant when the

106 displacement of mass is x and its velocity x, the relative velocity between the mass and the belt is (V x). At this instant coefficient of friction is ( a x). The normal reaction on the mass is mg. The differential equation of motion is then written as mx kx mg( a x) mx mgx kx mg (1) a Equation (1) gives an effective negative damping to the system which sends it into large increasing amplitudes. The static displacement is mg / k. In case there is viscous damping between the mass and the belt, the slope of the equivalent friction line is no longer negative and there will not be any self excited vibrations. There are numerous other examples of self excited vibrations caused by dry friction, a few of which are:- (i) Excitation of violin string by a bow. (ii) Screeching of door joints when dry. (iii) Shaft whirl due to dry friction. a MODULE 5

107 PROPOGATION OF SOUND Sound is produced by a vibrating body. A material medium is necessary for the propagation of sound waves. In wave motion momentum and energy are transferred. Characteristics of wave motion. 1. Wave motion is a disturbance produced in the medium by the repeated periodic motion of the particles of the medium... Only the wave travels forward whereas the particles of the medium vibrate about their mean positions. 3. There is a regular phase change between the various particles of the medium. The particle ahead starts vibrating a little later than the particles preceding it. 4. The velocity of the wave is different from the velocity with which the particles of the medium are vibrating about their mean positions. The wave travels with a uniform velocity whereas the velocity of the particles is different at different positions. It is maximum at the mean position and zero at the extreme positions of the particles. There are two types of wave motions. a. Transverse wave. b. Longitudinal wave Sound wave are longitudinal waves and light waves are transverse waves. Figure 1 shows the formation and propagation of transverse and longitudinal wave. A sound wave is propagated and conveyed to the ear by means of the intervening layers of air. Consider a vibrating tuning fork. Let us confine our attention to the right hand prong only. When it moves towards the right, it compresses the layer of air in front of it and as a consequence the pressure of this layer will be greater than the adjacent layers. It tends to relieve the strain thus created, by compressing them. These in turn hand on the compression. Thus a pulse of compression will travel onwards to the right. Again, when the prong moves towards the left rarefaction is started. These follow one another and as the fork vibrates, compressions and rarefaction are sent out in regular succession. These

108 waves at last reach the car and set the tympanic membrane, which is ultimately transmitted via a system of bones and cords to the brain and causes the mental sensation called sound. MEASUREMENT OF INTENSITY Of SOUND The intensity of sound is defined as the quantity or energy propagating through a unit area per unit time, the direction of propagation being perpendicular to the area. The amount of power transmitted is measured in Watts/m. A convenient unit is microwatts /m. According to Weber Fechner law in psychology, the loudness of a sound as judged by the ear is proportional to the logarithm of the intensity. If l 1 and l 0 represents the intensities of two sounds of a particular frequency and L 1 and L 0 the corresponding measure of loudness then L 1 = K log l 1 and L 0 = K log 1 0. The difference in loudness technically known as Intensity Level, L between them is given by L = L 1 - L = K[log 1 - log 1 0 ] l1 L Klog( ) l Where K is a constant that depends on the units and I 0 is some standard reference intensity arbitrarily taken as 10-1 watt/m, which corresponds to the intensity of the sound which can be just heard at a frequency of about 1000 cycles per second. This is threshold of audibility of a normal ear. Where K in the equation above is taken as I the difference in loudness is expressed in bels, a unit named in honour of Alexander Graham Bell, the inventor of telephone. This unit of loudness is rather too large, one tenth of it, the decibel (db) has become the standard. So in order to express the difference in loudness of a sound of intensity I in decibel, the above relation should be written as l1 L 10log( ) l To build a scale of loudness, we have to fix its zero. The loudness corresponding to the threshold of hearing is the zero of this scale. This occurs when

109 the intensity of sound wave equals 1 0 or 10-1 watt/m. The maximum intensity which the ear can tolerate without sensation of pain is about 10 - watt/m and it corresponds to the intensity level decibels. Source 10 L 10log( ) 10db 1 10 The following table gives the approximate value of some sound measured in Intensity level in Decibel Threshold of hearing 0 Rustle of leaves 10 Whisper 15-0 Ordinary conversation Motor trucks and heavy street traffic Roaring lion at 0 ft. 90 Thunder Painful sounds 130 or above In the above table we have expressed the loudness in decibels on the assumption that the threshold of audibility is the same for all frequencies of the ear and the limits of audibility vary over wide ranges of intensity and frequency, hence the sound of same intensity but different frequencies seem to differ in loudness. Therefore a different unit for measuring loudness is used. It is called the phon. The measure of loudness in phons of any sound is equal to the loudness in decibels of an equally loud pure tone of frequency 1000 cycles/second. Acceptable noise levels: Type of residential Area Rural 5-35 Suburban Residential Urban City Industrial area Acceptable noise level in db Outdoor noise levels in residential areas Type of place/building Acceptable noise level in db

110 Radio and TV studios 5-30 Music room Hospital, classroom, Auditorium Apartments, hotel, home Conference Room, small office, concert room Private offices Libraries, Large public office, banks, stores Restaurants Indoor noise levels in public/private places The above table is as per IS standards. AIR COLUMNS Stationary longitudinal waves can be produced in a column of air by any periodical movement whose frequency coincides with one of the natural frequencies of the column. All wind instruments are provided with a column of air called a resonator, which may be in the form of a rectangular air chamber. The periodic movement is caused by an important part of the musical instrument called the mouthpiece, which is different in construction in different instruments: It is the mouth - piece that acts as a generator and supplies the energy necessary to maintain the vibrations in the column of air. In the theoretical treatment the following assumptions are made: 1. The diameter of the pipe is small compared with the length of the pipe and with the wave length of sound.. The diameter is sufficiently great so that the viscosity effects can be neglected. 3. The walls of the pipe are rigid. pipes. The organ pipes are classified into two groups: Flute or Flue pipes and Reed DOPPLER EFFECT

111 It is commonly observed that the pitch of a note apparently changes when either the source or the observer are in motion relative to each other. When the source approaches the observer or when the observer approaches the source or when both approach each other the apparent pitch is higher than the actual pitch of the sound produced by the source. Similarly when the source moves away from the observer or when the observer moves away from the source or when both move away from each other, the apparent pitch is lower than the actual pitch of the sound produced by the source. This apparent change in pitch due to relative motion between the source and the observer is called Doppler effect. Doppler effect in sound is asymmetric, when the source move towards the observer with a certain velocity, the apparent pitch is different to the case when the observer is moving towards the source with the same velocity. But it is not so in the case of light. Doppler effect in light is symmetric. The apparent pitch in different cases is calculated below Let n - pith of sound - wavelength v - velocity of sound n - apparent pitch Case 1: When the source moves towards the stationary observer with a velocity a v n n v a Case : When the source moves away from the stationary observer with a velocity a v n n v a Case 3: When the observer moves towards a stationary source with a velocity b

112 v b n n v Case 4: When the observer moves away from a stationary source. v b n n v Case 5: When the source moves towards the observer and the observer moves away from the source. v b n n v a Case 6: When the source and the observer move towards each other v b n n v a Case 7: When the source and the observer move away from each other. v b n n v a Case 8: Source moving away from the observer and the observer moving towards the source. v b n n v a MUSICAL SCALES: CHORD: When two notes of different frequencies are sounded simultaneously, their combination is called a chord. In the case of a concord or consonance the combination produce a pleasant or agreeable effect. While other combinations produce a disagreeable or unpleasant effect and it is called dissonance. HARMONY: When the two notes sounded together produce concord their combination is called Harmony. MELODY: If the two notes are sounded one after the other, combination is called melody. DIATONIC MUSICAL SCALE: A series of notes separated by definite

113 and simple intervals so as to produce a musical effect when played in succession is said to constitute a musical scale. The most common musical scale is the Gamut or the Diatonic scale. It consists of a series of eight notes, the interval between last and the first note being /1. It is therefore called an octave. The series of note is denoted as: C D E F G A B C sa re ga ma pa dha ni (sa)' All these notes are arranged in increasing order of frequencies so that they present a regular graduation in pitch and their vibration frequencies are represented by l l i.e., if the frequencies of first note C called the tonic or the key note be taken as 4, the relative frequencies (ratio of successive frequencies) of the various notes of the diatonic scale will be (4) (7) (30) (3) (36) (40) (45) (48) (Major (Minor (Semi (Major (Minor (Major (Semi tone) tone) tone) tone) tone) tone) tone) If the frequencies of the note C be taken as 56 and 64 respectively, the various notes of the scale will be denoted by The above scale consists of three main intervals Major tone, Minor tone and semi major tone respectively, so that the sequence of interval in Diatonic scale is major tone, minor tone, semi-tone, or neglecting the difference between major and minor tones, tone, tone, semi-tone, tone, tone, semi-tone. Since the major tones occur more frequently, this scale is called Major Diatonic scale. It could be extended both

114 above C and below C level. Intervals within an octave with small intervals viz., 4 3, 3, are most consonant and are named as fourth, fifth and octave their names being derived from the positions of these notes on the scale. MICROPHONE The microphone is essentially an arrangement for the conversion of sound vibrations into vibrations of electrical current. In the telephone communication system it is very successfully used as a transmitter and the vibrations of electrical current thus produced are converted into sound by the receiver at the distant end. It is also the first element of a loud speaking equipment or a broadcasting arrangement in which electrical oscillations after proper amplification are reconverted into sound by loud speaker. The general principle of modern carbon microphones is shown in fig. It is based on the variation of the resistance of fine carbon granules when subjected to pressure changes. Carbon granules are enclosed in between two plates one of which is fixed and the other serves as a diaphragm which responds to rapid change in pressure. The plates are placed in series with a key, a battery, and the primary of a transformer, the secondary of the transformer is connected to the telephone receiver. When the key is closed, a steady small current flows through the circuit. If R be resistance of the circuit when there is no displacement of the diaphragm and Ka sin t the varying resistance due to the displacement a sin t at any instant. K being constant of the microphone then the total resistance of the circuit at any instant is (R+Kasin t). Let V be the direct emf of the circuit, then the current at any instant is given by V V Ka i l sint R Ka sin t R R 1 V Ka K a l sin t sin t... R R R The first term indicates a steady current when the diaphragm is at rest, the second an alternating current of the same frequency as is impressed on the diaphragm and the rest of the terms denote its other harmonics. The current is thus modulated.

115 This modulated electrical current passes through P, the primary of a transformer and produces by induction a corresponding varying current in S, its secondary. This amplified current passes through the telephone T and excites its diaphragm. The movements of the latter set the air in corresponding vibrations reproducing the original sound.other commonly used types of microphones are The Electrodynamic Microphone: This microphone is base don the principle of electromagnetic induction consisting of a small coil of wire attached to the back of a freely moving light plate. The coil is situated in the magnetic field between the central pole piece and the peripheral pole piece of a permanent "pot" magnet. The sound wave cause the plate end of the coil to vibrate and varying current thus induced in the coil are amplified and conveyed to the distant end. Ribbon or Velocity Microphone: It is also based on the principle of electromagnetic induction. The Condenser Microphone: It is based on the principal that a charged condenser connected to an electrical circuit be subjected to sound vibrations, they will produce variations in the distance between the plates thus changing its capacity,

116 then an alternating current will be set up in the circuit. The Crystal Microphone: This microphone is based on the principle of piezoelectric effect, according to which certain crystals like quartz, Rochelle Salt produce minute differences of potential between its opposite faces when subjected to pressure. The Hot Wire Microphone: It is based on the principle that resistance of a metallic wire changes with change in temperature. THE LOUDSPEAKER: It is a device for converting electrical energy into sound energy and therefore essentially a microphone worked backwards. It is provided with a horn or a circular board called 'baffle' for effectively transferring the vibrations of the moving part to the external air. The commonest and the most efficient type of speakers nowadays is the moving coil type. It is based on the principle that when a variable current is passed through a conductor in a magnetic field, the conductor is acted on by a variable force in accordance with Fleming's Left Hand Rule and is f the current is oscillatory the conductor is set in vibration. The moving part of the apparatus consists of a small coil called the 'voice coil' wound on a cylindrical strip to which the variable current output of the microphone is fed. The voice coil is free to move in the annular gap between the central S and the peripheral pole piece N, of a 'pot' magnet designed to produce a strong radial magnetic field in it. It is usually magnetized by a steady (DC) current flowing in the coil wound round it. The coil is attached to a conical diaphragm made of parchment with circular corrugation and supported round the periphery by a flexible annular strip of leather or rubber. When the variable current passes through the coil in the magnetic field, it causes varying movement of the coil along the axis with the frequency of the current variations. The diaphragm is thus set into vibrations which are communicated to the external air and the sound is reproduced. The greater the energy supplied to the voice coil, the louder will be the sound emitted by the diaphragm. Completely surrounding the cone and attached to it by silk threads is the 'baffle'. It prevents the air vibrating behind the cone from flowing round to its front. The relation between the current and driving force is linear and force is

117 independent of the position of the coil in the gap for considerable movements. When suitably designed a fairly uniform response of 80cps to 1000cps is secured. It is capable of radiating large power without appreciable asymmetric distortion. ACOUSTICAL MEASUREMENTS: The measurement of airborne and waterborne sound is of increasing interest to engineers. Airborne sound measurements are important in the development of less noisy machinery and equipment, in diagnosis of vibration problems, and in the design and test of sound recording and reproducing equipment. In large rocket and jet engines, the sound pressures produced by the exhaust may be large enough to cause fatigue failure of metal panels because of vibration. Waterborne sound has been applied in underwater direction and range finding equipment like "sonar". Most sound transducers are basically pressure measuring devices. The basic definitions of sound are in terms of the magnitude of fluctuating component of pressure in a fluid medium. The sound pressure level (SPL) is defined by SPL = Sound pressure level = P Log decibels (db) [1] p - root mean square (rms) sound pressure, bar. The rms value of fluctuating component of pressure is used in equation I,