MODULE-1. Chapter 1: SIMPLE HARMONIC MOTION. 1.1 Introduction

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1 MODULE- Chapter : SIMPLE HARMONIC MOTION. Introduction You are familiar with many examples of repeated motion in your daily life. If an object returns to its original position a number of times, we call its motion repetitive. Typical examples of repetitive motion of the human body are heartbeat and breathing. Many SIMPLE objects move in a repetitive way, a swing, a rocking chair, and a clock pendulum, for example. Probably the first understanding the ancients had of repetitive motion grew out of their observations of the motion of the sun and the phases of the moon. Strings undergoing repetitive motion are the physical basis of all stringed musical instruments. What are the common properties of these diverse examples of repetitive motion? In this chapter we will discuss the physical characteristics of repetitive motion, and we will develop techniques that can be used to analyze this motion quantitatively.. Kinematics of Simple Harmonic Motion: One common characteristic of the motions of the heartbeat, clock pendulum, violin string, and the rotating phonograph turntable is that each motion has a well-defined time interval for each complete cycle of its motion. Any motion that repeats itself with equal time intervals is called periodic motion. Its period is the time required for one cycle of the motion. Let us analyze the periodic motion of the turntable of a phonograph. Suppose that we place a marker on a turntable that is rotating about a vertical axis at a uniform rate in a counterclockwise direction when viewed from above. If you observe the motion of the marker in a horizontal plane-that is, viewing the turntable edge-on-the marker will seem to be moving back and forth along a line. The motion you see is the projection of uniform circular motion onto a diameter and is called simple harmonic motion (Figure.). Fig..: Simple Harmonic Motion. Projection of uniform circular motion upon a diameter

2 To derive the equation for simple harmonic motion, project the motion of the marker upon the diameter AB. The displacement is given relative to the center of the path O and is represented by x = OC. From Figure. we see that x = R cos θ, where R is the distance of the marker from the axis of rotation. The maximum displacement of the motion is called the amplitude of the motion and is represented by the symbol A. The displacement of the marker in a direction parallel to the diameter AOB is then given by the following equation, x = A cos θ (.) where θ is the angle through which the marker on the turntable has turned. Since we know that the turntable is rotating with a constant angular velocity ω, we can write an expression for the angular displacement θ as the angular speed times the time plus the starting angle, θ = ω t + φ (.) where t is the time of rotation and φ, the phase angle, is the angular displacement at the beginning, t =. If we choose the starting position along the line DOB, then φ = at t =. In general, the equation for the x-displacement is given by x = A cos (ω t + φ). The velocity of the marker for the position shown in Figure. is tangential to the circle of motion of the marker and in the direction shown in Figure.. The magnitude of the velocity is given by v = πrn = ωr = ωa (.3) where n is the number of revolutions of the turntable in one second, ω is the angular speed in radians per second, and R is the radius of the circle of motion and is equal to the amplitude of displacement A. Fig..: Velocity in Simple Harmonic Motion. Projection of the velocity of uniform circular motion upon a diameter EXAMPLE What is the velocity of a point on the rim of the standard -inch long-playing phonograph record? R = 5. cm ω = 33 (/3) rpm = 5.56 x - rev/sec = 3.49 rad/sec. v = ω R = 53. cm/sec. The velocity of the marker in a direction parallel to the line DOB is shown by

3 the component υ x in Figure., where v x = -v sin θ = ω A sin θ. The negative sign indicates that the direction of motion is in the negative x-direction. When sin θ is positive, the velocity v x is in the negative x-direction. Notice sin θ is always positive for θ 8 ; so v x is negative for those angles and positive for the rest of the motion of one cycle. The acceleration of the marker for the position shown in Figure. is perpendicular to the velocity, toward the center of circular motion as shown in Figure.3. Fig..3: Acceleration in Simple Harmonic Motion. Projection of the acceleration of uniform circular motion upon a diameter. v The magnitude of the acceleration is a =. For the case we are considering, we can write the R acceleration in terms of the angular speed and the amplitude, a = v R = ω R /R = ω R =ω A. (.4) Fig..5: Displacement, Velocity, Acceleration of SHM as a function of time. Red line denotes displacement, Blue line denotes velocity and Magenta line denotes acceleration.

4 The projection of the acceleration vector onto the line DOB is shown by a x in Figure.3 where a x = -ω A cos θ (.5) The negative sign indicates that the acceleration is in the negative x -direction. The cos θ is positive for 9 o θ 9 o ; so a x is in the negative x-direction for these angles and positive for the rest of each cycle of motion. Notice that from Equation we know that A cos θ is the x- displacement. The equation for the acceleration can be rewritten in terms of the x-displacement a x =-ω x (.6) If we substitute the equation for the angular displacement as a function of time, θ = ωt + φ, into the equations for x-displacement, x-velocity, and x-acceleration, then the linear displacement, velocity, and acceleration are given in general by the following equations: x = A cos (ω t + φ) (.7) v x = ω A sin ( ω t + φ ) (.8) a x = -ω A cos (ω t + φ) (.9) Assume x = A at time t =. Then φ =. Then these equations can be represented in graphical form as shown in Figure.5. The curves in Figure.5 show that at the time of zero velocity the acceleration and the displacement are maximum. At a time of maximum velocity the acceleration and the displacement are zero. For simple harmonic motion the acceleration is proportional to the displacement x and is oppositely directed (6). If the displacement is to the right of the equilibrium position, then the acceleration is to the left, and vice versa. The angular speed ω is a constant, a characteristic of the motion. The angular speed can be expressed in terms of the frequency, or the period, of the motion. In uniform circular motion we defined the number of revolutions per second as the frequency f. Then ω = πn = πf (.) where f is the number of cycles per second (hertz). The period T of one vibration is equal to the reciprocal of f. T = /f (.) Then the angular speed is proportional to the reciprocal of the period of the motion, ω = π/t (.) where T is the period of the motion. By substituting the expression for ω from Equation 5.6 into

5 Equation 5.8 and solving for f, we get the relationship between the frequency, the acceleration and the displacement, f a =/T (.3) x T x (.4) a The value of - a/x is always positive because a and x are in opposite directions. The above equation can be used to calculate either the frequency (period) or the acceleration or the displacement, if you know the other two variables..3 Dynamics of Simple Harmonic Motion: We have considered simple harmonic motion without regard to the forces that produce such motion. We now want to discover the common characteristics of the forces that produce simple harmonic motion (SHM). In many cases, the recognition of this SHM force not only allows you to predict harmonic motion, but it allows you to predict the frequency of the motion. What kind of forces produces simple harmonic motion? Look at Equation 6. According to this equation, the acceleration in a SHM system must be proportional to the displacement of the system from equilibrium and in the opposite direction. Let us combine this equation with our knowledge of Newton's second law F x = ma x (.5) where a x = - ω x so F x = -mω x (.6) according to Equation 5, where the mass m and the angular speed ω are constants of the system. The component of the force F x in the direction of motion is a restoring force, acting opposite in direction to the displacement as indicated by the negative sign. The dx Asin( t) dt d x dt Acos( t) (.7) magnitude of F x is proportional to the displacement; F x is a Hooke's law force. Therefore, Hooke's law force systems produce simple harmonic motion. If the Hooke's law force equation is written as follows, F = -kx then the force constant k is equal to mω. We can derive equations for

6 the period and frequency of the simple harmonic motion that arises from a Hooke's law force by making use of the equality between the ratio of the force constant to the mass of the system and the square of the angular speed k/m = ω = (π f) = (π/t) (.8) frequency = k m (.9) m TimePeriod= (.) k In summary, all Hooke's law force systems will produce simple harmonic motion. Any system of simple harmonic motion can be used to deduce a Hooke's law force. There are many examples of Hooke's law systems such as spring balances and simple pendulums. In fact, in the enrichment section we will use Taylor's theorem to show that almost all equilibrium systems exhibit simple harmonic motion near equilibrium. EXAMPLE It is known that a load with a mass of g will stretch a spring. cm. The spring is then stretched an additional 5. cm and released. Find: a. the spring constant b. the period of vibration and frequency c. the maximum acceleration d. the velocity through equilibrium positions e. the equation of motion SOLUTIONS a. If the force acting mg =. x 9.8 N and x =. m, then the spring constant is k = F/x =. x 9.8/. = 9.6 N/m b. To find the period of vibration let T = π SQR RT [m/k]. Then, T/m = πsqr RT [./9.6] = π/7 SQR RT [] = SQR RT []/7 π =.634 s The frequency is f = /T = /.634 =.59 Hz c. Given that the amplitude = 5. cm, ω = k/m = (9.6 N/m)/. kg = 98./s Then the maximum acceleration is amax = ω A x.5 = 4.9 m/s d. To find the velocity through the equilibrium position let vmax =A ω = (.5 m) (SQR RT [98./s] =.495 m/s e. A = 5. cm, ω = SQR RT [98]/sec = 9.9 sec-, and φ =, since x =A at time t =. Therefore, x = 5. cos (9.9t) cm.4 Energy Relationships in Simple Harmonic Motion: You will recall that in previous chapters we have been able to use the concept of conservation of energy to solve mechanical problems. The conservation of energy has been especially useful in problems that involve systems whose total energy is a constant. Perhaps you wonder if energy analyses of SHM systems will yield worthwhile results. Let us consider the energy relationships in simple harmonic motion. We can show that the potential energy associated with a -kx force is equal to (½) kx for a displacement of x. Since potential energy is equal to the work, PE = W = (F ave )(x) = (kx) x/ = (½)kx (.) The total energy of the system is equal to the sum of the potential and kinetic energy,

7 Energy = KE + PE = (½) mv + (½) kx (.) where v is the velocity of the moving object whose mass is m. We can use the equations for displacement and velocity as functions of time (Equations 5. and 5.7) to write an expression for the total energy as a function of time: E = (½)m [-ωa sin (ωt + φ)] + (½) k [ω cos (ωt +φ)] E = (½)mω A sin (ωt + φ) + (½)kA cos (ωt + φ) where ω =k/m for Hooke's law system. Therefore, E = (½)kA [sin (ωt +φ) + cos (ωt +φ)] From our knowledge of trigonometry we know that for any angle θ, sin θ + cos θ =, E = (½) ka (.3) The total energy of a SHM system is a constant, independent of time, and equal to one half of the product of the force constant and the square of the amplitude of oscillation. The relationship between the potential energy and the kinetic energy as a function of time is shown in Figure.6. Fig..6: The energy distribution curve of an SHO. When the potential energy has its maximum value (½)kA, the kinetic energy is zero, and the object is instantaneously at rest at its position of maximum displacement, x =A. When the kinetic energy has its maximum value (½)mω A or (½)kA, the potential energy is zero, and the object is moving with its maximum velocity at zero displacement. At all other positions neither the potential energy nor kinetic energy is zero. Since the total energy is a constant equal to (½) ka, we can use Equation to find an expression for the velocity as a function of displacement, (½) ka = (½) kx + (½) mv which we use to solve for velocity at any displacement x,

8 v = k/m (A - x ) (.4) Remember that k/m is equal to ω ; so the magnitude of the velocity is given by v = ω A x (.5) The maximum velocity occurs as the object passes through its equilibrium position, x =, v max = ±ωa (.6) This is the same result we derived earlier from Equation 5.7. In summary, the universality and the simplicity of simple harmonic motion is very appealing. The total energy of a SHM system is conserved. From the dynamics of the system, that is, the force constant and mass, all of the properties of the system can be calculated if the maximum displacement is known. From the kinematics of the system, the angular speed, and the amplitude, all the properties of the system can be calculated if the mass is known. Then if the position of the object is known at any time, the system is completely determined..5 Superposition of SHM s Let us consider two SHMs of amplitudes a and a having same angular velocity ω which act respectively alone x and y axis. Then their amplitudes can be represented by x= a cos (ωt + φ ) (.7) y= a cos (ωt + φ ) y a = cos {(ωt + φ ) - (φ -φ )} φ ) = cos (ωt + φ ) cos (φ -φ ) + sin (ωt + φ ) sin (φ - x = a cos (φ -φ ) + x a sin (φ -φ ) Or, y x a cos (φ -φ ) = a - x a sin (φ -φ ) Squaring both sides of the equation we get,

9 y x xy x a a a a a cos ( ) cos sin y xy cos x {cos ( ) sin } sin ( ) a a a a cos sin ( ) y xy x (.8) a a a a Where φ= φ φ Case If then sin = and cos = Therefore Equation (8) gets reduced to y xy x a a a a x y a a a y x a This represents a straight line passing through the origin and making an angle of inclination δ= tan - (a /a ). Case If then sin = and cos = then Equation (8) gets reduced to y a x a

10 This is the equation of an ellipse which is symmetrical about the two axes and a and a are the semi- axes. Case 3 If then sin = and cos =- Then Eq. 8 reduces to y xy x a a a a x y a a a y x a This equation represents a straight line which passes through the origin making an angle of inclination δ= tan - (-a /a ) to the x axis. Case 4 If and a =a =a then sin φ = and cos φ=

11 Therefore Eq..8 reduces to x + y = a This equation represents a circle which is symmetrical about the two axes with a as radius.

12 Chapter : Free and Damped Vibration. Introduction : The equation of motion for a mass m vibrating on the end of a spring of force constant k, in the absence of any damping, is mẍ = -kx (.) Here, I am assuming that the displacement x is a function of time, and a dot denotes d/dt. However, in most real situations, there is some damping, or loss of mechanical energy, which is dissipated as heat. In the case of our example of a mass oscillating on a horizontal table, damping may be caused by friction between the mass and the table. For a mass hanging vertically from a spring, we might imagine the mass to be immersed in a viscous fluid. These are obvious examples. Slightly less obvious, it may be that the constant bending and stretching of the spring produces heat, and the motion is damped from this cause. In any case, in this analysis we shall assume that, in addition to the restoring force kx, there is also a damping force that is proportional to the speed at which the particle is moving.. Differential equation of damped vibration : I shall denote the damping force by bẋ. The equation of motion is then mẍ = kx bẋ (.) If I divide by m and write for k/m and γ for b/m, we obtain the equation of motion in its usual ẍ + γẋ + x = (.3) Here γ is the damping constant. We are ready to solve the differential equation (.3). Indeed, we know that the general solution is x = Ae k t + Be k t (.4) where k and k are the solutions of the quadratic equation. k + γk + = (.5) k and k are given by 4 k = (.6) 4 k = The nature of the solution depends on whether is less than, equal to or greater than These cases are referred to, respectively, as u n d e r damped, critically damped and overdamped. We shall start by considering underdamping. Light damping: γ < ω Since γ < ω, we have to write equations 6 as k i 4 k i 4 (.7)

13 Further let ω = 4 Therefore Eq. (.4) becomes (.8) x = A e t it + B e t it = t it it e Ae Be (.9) it it If x is to be real, A and B must be complex. Also, * the asterisk denotes the complex conjugate. a ib a ib Let A and B= t e e B must equal A*, where where a and b are real. Then equation (9) can be written as x e acost bsint (.) And if C a b a b, sin, cos a b a b The equation can be written as t x C e sin t (.) Equations (.9), (.) or (.) are three equivalent ways of writing the solution. Each has two arbitrary integration constants (A, B), (a, b) or (C,α), whose values depend on the initial conditions - i.e. on the values of x and ẋ when t=. Equation (.) shows that the motion is a sinusoidal oscillation of period a little less than ω, with an exponentially decreasing amplitude. To find C and in terms of the initial conditions, differentiate equation (.) with respect to the time in order to obtain an equation showing the speed as a function of the time: t x Ce cos t sin t (.) By putting t = in equations. and. we obtain x Csin (.3) cos x C sin (.4) From these we easily obtain x cot (.5) x

14 And C = x cos ec (.6) Figure.: Displacement vs time curve of a damped harmonic oscillator The quadrant of can be determined from the signs of cot and cosec, C always being positive. Note that the amplitude of the motion falls off with time as e t, but the mechanical energy, which is proportional to the square of the amplitude, falls off as e t. Although the motion of a damped oscillator is not strictly "periodic", in that the motion does not repeat itself exactly, we could define a "period" P = π/ω' as the interval between two consecutive ascending zeroes. Extrema do not occur exactly halfway between consecutive zeroes, and the reader should have no difficulty in showing, by differentiation of equation, that extrema occur at times given by tan ( t ). However, provided that the damping is not very large, consecutive extrema occur approximately at intervals of P/. The ratio of consecutive maximum displacements is, then, t xˆ n e t P n xˆ (.7) e Further we see that the logarithmic decrement is ln x x n n P 4 (.8) from which the damping constant can be determined. The constant Pγ/4 is called the logarithmic decrement. Heavy damping: γ>ω օ

15 The motion is given by equations 4 and 6 where, this time, k and k are each real and negative. For convenience, I am going to write k and positive, with given by and k. and are both real, (.9) The general solution for the displacement as a function of time is t t x Ae Be (.) The velocity is given by x A e Be (.) t t The constants A and B depend on the initial conditions. Thus: x = A + B From these we obtain A ( x )= - (A +B ) ( x ) x, B x x Therefore x = ( x ) x e t x x t e (.) Critically Damped: γ=ω օ In this case and are each equal to. Therefore the general solution is of the form t t x Ce Dte t x Ce t Either way, there are two arbitrary constants, which can be determined by the initial values of the displacement and speed. It is easy to show that

16 x C x and a (.3) x.3 : Electrical Analogue Figure.: Nature of critically damped, overdamped, underdamped motion A charged capacitor of capacitance C is connected in series with a switch and an inductor of inductance L. The switch is closed, and charge flows out of the capacitor and hence a current flows through the inductor. Thus while the electric field in the capacitor diminishes, the magnetic field in the inductor grows, and a back electromotive force (EMF) is induced in the inductor. Let Q be the charge in the capacitor at some time. The current I flowing from the positive plate is equal to -Q. The potential difference across the capacitor is Q/C and the back EMF across the inductor is LI LQ. The potential drop around the whole circuit is zero, so that Q/C= LQ Q The charge on the capacitor is therefore governed by the differential equationq. LC which is simple harmonic motion with ω =/ (LC). If there is a resistor of resistance R in the circuit, while a current flows through the resistor there is potential drop RI=-R Q across it, and the differential equation governing the charge on the capacitor is then LCQ RCQ Q (.4) This is damped oscillatory motion, the condition for critical damping being R = 4L/C. In fact, it is not necessary actually to have a physical resistor in the circuit. Even if the capacitor and inductor were connected by superconducting wires of zero resistance, while the charge in the circuit is slopping around between the capacitor and the inductor, it will be radiating electromagnetic energy into space and hence losing energy. The effect is just as if a resistance were in the circuit.

17 If a battery of EMF E were in the circuit, the differential equation for Q would be LCQ RCQ Q EC This equation is identical to the equation of damped SHM.

18 CHAPTER 3: FORCED VIBRATION 3. Introduction: In order to sustain vibrations, energy must be supplied to a system to make up for the energy transferred out of the vibrating system due to dissipative forces or damping. The resulting vibrations are called forced vibrations. In a clock or watch the pushes that maintain the vibrations are applied at the frequency at which the pendulum or balance wheel normally vibrates, i.e. at its natural frequency. Pushing at the natural frequency is the most efficient way to transfer energy into a vibrating system, as your childhood experiences on a swing should confirm. Musical instruments and other objects are set into vibration at their natural frequency when a person hits, strikes, strums, plucks or somehow disturbs the object. For instance, a guitar string is strummed or plucked; a piano string is hit with a hammer when a pedal is played; and the tines of a tuning fork are hit with a rubber mallet. Whatever the case, a person or thing puts energy into the instrument by direct contact with it. This input of energy disturbs the particles and forces the object into vibrational motion - at its natural frequency. If you were to take a guitar string and stretch it to a given length and a given tightness and have a friend pluck it, you would hear a noise; but the noise would not even be close in comparison to the loudness produced by an acoustic guitar. On the other hand, if the string is attached to the sound box of the guitar, the vibrating string is capable of forcing the sound box into vibrating at that same natural frequency. The sound box in turn forces air particles inside the box into vibrational motion at the same natural frequency as the string. The entire system (string, guitar, and enclosed air) begins vibrating and forces surrounding air particles into vibrational motion. The tendency of one object to force another adjoining or interconnected object into vibrational motion is referred to as a forced vibration. In the case of the guitar string mounted to the sound box, the fact that the surface area of the sound box is greater than the surface area of the string means that more surrounding air particles will be forced into vibration. This causes an increase in the amplitude and thus loudness of the sound. 3. Analysis of Forced Vibration The forces acted upon the particle are: (i) A restoring force proportional to the displacement but oppositely directed given by kx where k is known as force constant. (ii) A frictional force proportional to velocity but oppositely directed given by bdx/dt where b is the frictional force per unit velocity and (iii) The external periodic force, represented by F sin ωt where F is the maximum value of this force and ω/π is its frequency. So the total force acting on the particle is given by -kx bdx/dt + F sin ωt

19 By Newton s second law of motion this must be equal to the product of mass m of the particle ans its instantaneous acceleration i.e. m d x/dt Hence where d x dx ' m b kx F sin t dt dt d x b dx k F ' x sin t dt m dt m m d x dx ' b x f sint dt dt d x dx ' or m kx b F sin t dt dt or (3.) k ' k F, b and f m m m Equation (3.) is the differential equation of the motion of the particle. In this case when the steady state is set up, the particle vibrates with the frequency of applied force and not with its own natural frequency. The solution of differential equation (3.) must be of the type x= A sin (ω ' t -ϴ) (3.) where A is the steady amplitude of vibrations and ϴ is the angle by which the displacement x lags behind the applied force F sin ω t. A and ϴ being arbitrary constants. Differentiating Eq. (3.) we have dx A dt d x dt ' cos( ' t ) A Substituting these values in eq. (3.), we get ' sin ( ' t ) A ' sin ( ' t ) +b ' A cos( ' t ) + A sin (ω ' t -ϴ) = f sin t = f sin {( t-θ) +ϴ}

20 Or A ( ) sin ( t ) + b A ' cos( ' t ) = f sin ( t - ϴ) cos ϴ + f cos ( t- ϴ) sin ϴ If this relation holds true for all values of t the coefficients of sin ( t- ϴ) and cos ( t- ϴ) terms on both sides of this equation must be equal i.e. comparing the coefficients of sin ( t- ϴ) and cos ( t- ϴ) on both sides we have Squaring eqs. (3) and (4) A ( ) fcos (3.3) ba f sin (3.4) A ( ) 4b A f ' A [( ) 4 b ] f f A ( ) 4b (3.5) While on dividing eq (3.4) by eq. (3.3) we get ba tan ϴ = ( ) b tan (3.6) Equation (3.5) gives the amplitude of forced vibration while (3.6) its phase. Depending upon the relative values of and So the complete solution at the starting is given by bt x = cos pe b t + f ( ) 4b cos t tan b (3.7) 3.3 Energy of a forced vibrator In the steady state under the influence of a periodic force, the motion of a particle is obtained in eq. () as given below: x= A sin (ω 't -ϴ)

21 where A f ( ) 4b b tan So, dx A ' cos( ' t ) dt Hence, the kinetic energy of the forced system at any instant is given by dx T= m ma cos ( t ) dt m f T= ( ) 4b cos ( t ) (3.8) Since the motion of the particle is steady SHM, the total energy of the system at any instant is equal to the maximum kinetic energy T max m f E total = T max = ( ) 4b (3.9) Eq. (3.9) is the expression for total energy. 3.4 Energy or velocity resonance We can rewrite Eq. (3.9) as follows: E total = = mf mf 4b 4b

22 mf = 4b (3.) Where Δ= If ω=ω` then Δ= and the energy of the system is maximum for a given value of b. Thus we observe that when the frequency of the forcing system coincides with that of the natural frequency of the forced system, the energy of vibration of the forced system is maximum. This phenomenon is known as energy resonance or velocity resonance or simply resonance. And the energy at resonance E r is given by E r = E max = mf (4 b ) (3.) 3.5 Sharpness of Resonance Since the frequency of the influencing applied periodic force differs from the natural frequency of the forced system, the response of the forced system diminishes. The energy of response of the forced system falls off rapidly for slight deviation from the resonance. The E total versus Δ graph has been plotted in the following figure. The graph shows that for slight variation of from ω the total energy gets reduced rapidly. The sharpness of resonance curve is a measure of the rate of fall of energy of resonance with departure from equality between the frequencies. From eqs. (3.) and (3.) we obtain Emax 4b (3.) E 4b 4b total Now dropping the suffix of E total, we get E E max (3.3) 4b The rate of decrease of total energy E with respect to Δ is greater the smaller the value of b. The sharpness of resonance can be quantitatively defined as the reciprocal of Δ when E reduces to half of its resonance value. Let us assume that Δ=Δ for which E= E max Then from Eq. (3.3) we get

23 = + 4b or 4b = b So the sharpness of resonance (S) is given by= b (3.4) The sharpness of resonance is important in connection with a radio receiver. 3.6 Amplitude resonance In case of forced vibration the amplitude is given by A f ( ) 4b A f ( ) 4b (3.5) A will be maximum when the denominator becomes minimum. For a given system, b and ω are constants. And for the deonominator to be minimum, we must have the first derivative of it with respect to to be equal to zero, i.e.,

24 d d or or or ( ) 4b d d d d ( ) 4b 4 8 b 4 ( ) b or b or or b b or, b or, b Thus for the amplitude to be maximum the frequency must be b. This is called amplitude resonance. This frequency is neither equal to the natural frequency of the system ( ) nor equal to the frequency of the damped vibrator But it is slightly lower than d. The maximum amplitude at resonance is given by A r A max f Ar b b f or Ar b b b 4b b f d b Hence the maximum amplitude is greater lower is the value of the damping factor b. The sharpness is greater if the damping factor b is smaller..

25 3.7 Power supplied by the driving force Since energy is dissipated in each cycle due to frictional force, this loss must be made up by the energy of the driving force to maintain the steady vibration. Suppose at any instant the force F sin t moves through a distance x in time t. Then work done is F sint x. Hence rate of work done T T dx F sint dt dt F sint A cos( t ) dt x Asin t T T T T FA(sint cost cos sin t sin ) dt T FA sin T FAsin Also work done against frictional resistance for the displacement work done against frictional force x is k dx x dt. Hence rate of T T T T ka cos t dt dx dx k dt dt dt T ka ka T F Asin F A ba f k A Thus the rate of supply of energy by the driving force is equal to the rate of work done against the frictional resistance.