Engi Mechanical Vibrations 1. Consists of a mass, spring and possibly a damper.

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1 Engi Mechanical Vibrations 1 1 Introduction 1.1 Definitions Vibration A motion that repeats itself after a time interval or oscillation (e.g. pendulum, plucked guitar string). Vibrating system Consists of a mass, spring and possibly a damper. Mass (or inertia element for rotating systems) is a rigid body that stores kinetic energy Spring stores potential energy Damper dissipates energy 1.2 Spring Elements A spring element is generally assumed to have no mass and no damping. It will store potential energy due to stretch (or compression) or twist (for torsional springs). For a linear spring the spring force is proportional to the change in length of the spring element: F kx (1) where, x is the measured deformation of the spring from its undeformed length, and k is the spring constant.

2 Engi Mechanical Vibrations 2 The work done deforming a spring from x 1 to x 2 is: U 12 x2 x 1 F dx and this work is stored as potential energy. The spring constant of a helical spring is: x2 x 1 kx dx k 2 (x2 2 x 2 1) (2) k Gd4 8nD 3 (3) wehere G is the shear modulus of the spring material, d is the diameter of the wire, D is the mean coil diameter, and n is the number of active turns of the coil. A rod can act as a spring (or consider an elevator cable when the winding drum suddenly stops). Consider a rod being stretched by a force F : For linear elastic behaviour Hooke s Law applies: σ ɛ E Since σ F/A and ɛ x/l, then: or F L A x E F AE x k x L i.e. the linear spring constant for a rod (or cable in tension) is: k AE L (4)

3 Engi Mechanical Vibrations Damping Elements Damping is the mechanism by which vibrational energy is converted into heat or sound. A damping element is assumed to have zero mass and elasticity. There are three standard models of damping: (1) viscous damping; (2) Coulomb or dry friction damping; and (3) hysteretic damping. Viscous Damping: Fluid dynamic drag is used to dissipate energy (e.g. air drag on a pendulum, drag as a liquid is forced through an orifice (shock absorber)). Viscous damping is the most common form of damping used in vibrating systems. The damping force is proportional to the rate of change of length of the damping element or the relative velocity between the two ends of the damping element. F cẋ (5) Note: Kinematics of a rigid body is used to give the position, velocity and acceleration of various components of a suspension system. The distance between the mounting points of the spring is used to give the spring force. The rate of change of the distance between the mounting points of the damper is used to give the damping force. If the spring and damper are coaxial, only one length is of interest. Coulomb Damping: Coulomb damping is constant in magnitude but opposite in direction to the motion. Due to the friction between rubbing surfaces (dry or with insufficient lubrication). Hysteretic, Solid or Material Damping: F µ k N (6) As a material is deformed, the work done on the material is stored as strain energy or heat. Since energy is absorbed, a body subjected to material damping shows a hysteresis loop on a stress-strain diagram.

4 Engi Mechanical Vibrations A Simple Viscous Damper Consider two parallel plates separated by a distance h. The bottom plate is fixed and the top plate moves to the right at a constant speed v. The space between the plates is occupied by a fluid with dynamic viscosity µ. Solution of the Navier-Stokes equations shows that the velocity profile is linear, i.e. Couette flow. The shear stress in the fluid is: τ µ u y µ v h The drag (i.e. frictional) force exerted by the fluid on the bottom plate is: F Aτ y0 µa h v where c µa/h is the damping constant and the damping force behaves linearly wrt v.

5 Engi Mechanical Vibrations 5 2 Free Vibration of a One Degree of Freedom System 2.1 Undamped Motion of a Single DOF System Consider a mass, m, resting on a flat frictionless surface and connected to a rigid support by a linear spring, k: Equation of Motion We will need an equation governing the motion of the mass as it is moved from its equilibrium position. Such an equation of motion can be derived from Newton s 2nd Law using the following steps: 1. Sketch the system and define a suitable positive co-ordinate direction to define the motion of the mass. 2. Determine the static equilibrium position of the system. It is common (and recommended) to define the positive co-ordinate from the equilibrium position. 3. Draw a free body diagram (FBD) of the mass, assuming positive displacement, velocity and acceleration are given to it. Draw the forces acting on the mass. 4. Use Newton s 2nd Law to derive an equation of motion. + F x ma x or mẍ kx mẍ + kx 0 (7)

6 Engi Mechanical Vibrations 6 Equation (7) is the equation governing the free vibration of the mass. What about a mass, m, suspended from a rigid support by a linear spring, k? + F x ma x But, from statics: so the governing equation is: mẍ kx + (W kδ st ) W kδ st mẍ + kx 0 Note: The same governing equation arises when x is defined from the equilibrium position (this is the advantage of defining x measured from equilibrium). 2.2 Solution of the Governing Equation The governing equation for undamped free vibration of a 1 DOF mass-spring system is an homogeneous ordinary differential equation (ODE): mẍ + kx 0 (8) Assume a solution of the form: then: x(t) Ce st ẋ(t) Cse st ẍ(t) Cs 2 e st Substitution of the assumed solution into Eq. (8) gives: mcs 2 e st + kce st 0 or ms 2 + k 0 (9) which is called the characteristic equation.

7 Engi Mechanical Vibrations 7 The solution of Eq. (9) is: where ( s ± m) k 1/2 ±iω n (10) ( ) k 1/2 ω n (11) m is the natural frequency of the undamped system. It is the frequency of oscillation of the undamped system. The two values of s are the roots or eigenvalues of the characteristic equation. Since both values of s will satisfy the governing equation, the solution for x(t) is written as: x(t) C 1 e iωnt + C 2 e iωnt (12) where C 1 and C 2 are constants to be determined from initial conditions. It is difficult to interpret the motion given by Eq. (12), so let s make it look pretty. Euler s formula states: e ±iωnt cos ω n t ± i sin ω n t Substitution of Euler s formula into Eq. (12) gives: x(t) C 1 cos ω n t + i sin ω n t + C 2 cos ω n t i sin ω n t or x(t) (C 1 + C 2 ) cos ω n t + i(c 1 C 2 ) sin ω n t Since x(t) is a displacement, it must be real, therefore, (C 1 +C 2 ) must be real and (C 1 C 2 ) must be imaginary. For this to be true C 1 and C 2 must be complex conjugates: C 1 a + ib ; C 2 a ib so: x(t) 2a cos ω n t 2b sin ω n t Then the displacement x(t) can be written as: x(t) A 1 cos ω n t + A 2 sin ω n t (13) where A 1 and A 2 are new constants evaluated from the initial conditions of the system. Two conditions are required (i.e. the same number as the order of the governing equation). Specifying the initial displacement and velocity at t 0: x(t 0) x o and solving for A 1 and A 2 : ẋ(t 0) ẋ o A 1 x o ẋ(t) A 1 ω n sin ω n t + A 2 ω n cos ω n t ẋ o ω n A 2 A 2 ẋo ω n

8 Engi Mechanical Vibrations 8 So the solution to Eq. (8) is: x(t) x o cos ω n t + ẋo ω n sin ω n t (14) i.e. harmonic motion. Equation (14) is the motion of any system whose governing equation can be written in the form of Eq. (8). Note: 1. If x o 0: 2. If ẋ o 0: x(t) ẋo ω n sin ω n t x(t) x o cos ω n t Equation (13) can be written in a more convenient form by defining: Square and add Eqs. (15) and (16) to find A: A Divide Eq. (16) by Eq. (15) to find φ: A 1 A cos φ (15) A 2 A sin φ (16) ( ( ) 1/2 A A 2 2 x 2 o + ( ) ) 2 1/2 ẋo ω n ( ) ( ) φ tan 1 A2 tan 1 ẋo A 1 x o ω n Substitution of Eqs. (15) and (16) into Eq. (13) gives: Using the trigonometric identity: x(t) A cos φ cos ω n t + A sin φ sin ω n t cos(x y) cos x cos y + sin x sin y Then the solution for x(t) can be written in the following convenient form: where the amplitude A and phase angle φ are: x(t) A cos(ω n t φ) (17) A ( x 2 o + ( ẋo ω n ( ) φ tan 1 ẋo x o ω n ) 2 ) 1/2 (18) (19)

9 Engi Mechanical Vibrations 9 The motion can be plotted against time, t(s), or frequency, ω n t(rad). Note: The vertical displacement at t ω n t 0 is x o, the initial displacement of the system from equilibrium. The slope of the x(t) curve at t ω n t 0 is ẋ o, the initial velocity of the system. The amplitude, A of the motion is the maximum displacement from equilibrium. The phase angle, φ, can be measured as the horizontal from t ω n t 0 to the first peak. The ω n t plot gives φ directly in radians, while the t plot gives φ in seconds, which will have to be muliplied by ω n to get the angle in radians.

10 Engi Mechanical Vibrations 10 The velocity and acceleration of mass m are: ẋ(t) ω n A sin(ω n t φ) ω n A cos(ω n t φ + π 2 ) (20) ẍ(t) ω 2 na cos(ω n t φ) ω 2 na cos(ω n t φ + π) (21) i.e. ẋ(t) leads x(t) by π/2 (or 90 o ) and ẍ(t) leads x(t) by π (or 180 o ). For a system with ω n π radians, i.e. τ 2 s, with initial conditions x o 0.02 m and ẋ o 0 we get the following motion. Note: Maximum velocity occurs as the system reaches the equilibrium position. Maximum acceleration occurs when the sytem reaches maximum displacement from equilibrium. Here the system is stopping an changing direction. The motion x(t) can also be expressed as a sine wave by defining A 1 and A 2 in Eq. (13) as: A 1 A sin φ o A 2 A cos φ o

11 Engi Mechanical Vibrations 11 Then: A ( ( ) 1/2 A A 2 2 x 2 o + ( ) φ o tan 1 A1 A 2 ( ) ) 2 1/2 ẋo ω n tan 1 ( xo ω n ẋ o ) Substitution of the definitions of A 1 and A 2 into Eq. (13) gives: Using the trigonometric identity: Then x(t) A sin φ o cos ω n t + A cos φ o sin ω n t sin(x + y) sinx cos y + cos x sin y x(t) A sin(ω n t + φ o ) (22) with the same magnitude as Eq. (17), but a new phase angle, φ o : ( ) φ o tan 1 xo ω n ẋ o 2.3 Free Vibration with Viscous Damping Consider a mass, m, resting on a flat frictionless surface and connected to a rigid support by a linear spring, k, and a viscous damper, c: (23) We will need an equation governing the motion of the mass as it is moved from its equilibrium position. Such an equation of motion can be derived using Newton s 2nd Law.

12 Engi Mechanical Vibrations Equation of Motion The FBD for a 1 DOF mass-spring-damper system is shown below: To derive the equation of motion assume a positive displacement, velocity and acceleration of the mass m. Newton s 2nd Law gives: + F x mẍ F s F d kx cẋ where, for a linear spring F s kx, and a viscous damper F d cẋ. The governing equation can be rearranged to give: mẍ + cẋ + kx 0 (24) This equation governs the motion of all 1 DOF systems in translation with a linear spring and viscous damper Solution of the Governing Equation The governing equation for viscously damped vibration of a 1 DOF system is an homogeneous ODE: mẍ + cẋ + kx 0 As for undamped motion, assume a solution of the form: x(t) Ce st Substitution of the assumed solution into the governing equation, Eq. (24), gives the following characteristic equation: or mcs 2 e st + ccse st + kce st 0 ms 2 + cs + k 0 (25) i.e. a quadratic equation in s. The roots of the characteristic equation are: s 1,2 c ± c 2 4mk 2m c 2m ± ( c 2m) 2 k m (26)

13 Engi Mechanical Vibrations 13 So the solution to Eq. (24) is: or x(t) C 1 e s 1t + C 2 e s 2t ( ) ( ) c 2m + ( 2m) c 2 k t c m 2m ( 2m) c 2 k t m x(t) C 1 e + C 2 e where, C 1 and C 2 are constants obtained from initial conditions. This form of the solution is hard to interpret so other forms are used. Define: 1. Critical Damping, c c, as the damping constant c which makes the radical in Eq. (26) zero: ( ) 2 cc k 2m m 0 or k c c 2m m 2 km 2mω n (28) 2. Damping ratio, ζ, as the ratio of the damping constant c to the critical damping constant c c : ζ c/c c (29) (27) Using Eqs. (28) and (29): c 2m c c c c c 2m ζω n (30) Using Eqs. (30) and the definition of ω n, Eq. (11), the roots s 1 and s 2, Eq. (26), can be written in terms of ζ and ω n : ( ) s 1,2 ζ ± ζ 2 1 ω n (31) and solution of Eq. (24) can be rewritten in the following form: ( ) ( ) ζ+ ζ x(t) C 1 e 2 1 ω nt ζ ζ + C2 e 2 1 ω nt (32) Note: if ζ 0, the solution reduces to that for undamped free vibration, Eq. (12). It should be obvious that due to the ζ 2 1 term in the radical different types of solution will arise if ζ < 1, ζ 1 and ζ > 1. These three cases correspond to underdamped, critically damped, and overdamped cases, respectively. The solutions for these different cases will now be considered separately.

14 Engi Mechanical Vibrations Case 1: Underdamped Motion (ζ < 1, c < c c ) If ζ < 1, then ζ 2 1 < 0, and ζ 2 1 i 1 ζ 2, therefore, the roots s 1 and s 2 can be written as: ( ) s 1,2 ζ ± i 1 ζ 2 ω n where ω d is the damped frequency of oscillation: ζω n ± iω d (33) ω d The solution to Eq. (24) can be written in the following form: 1 ζ 2 ω n (34) x(t) C 1 e ( ζωn+iω d)t + C 2 e ( ζωn iω d)t or x(t) e ζωnt C 1 e iω dt + C 2 e iω dt (35) Compare Eq. (35) with Eq. (12) for undamped motion. The solution will be harmonic with an exponentially decaying amplitude. The frequency of oscillation will be ω d 1 ζ 2 ω n, i.e. less than ω n for undamped motion. Due to the similarities of Eqs. (12) and (35) the same techniques may be used to rewrite Eq. (35) in more easily interpreted forms. x(t) e ζωnt (C 1 + C 2 ) cos ω d t + i(c 1 C 2 ) sin ω d t or Then: or x(t) e ζωnt C 1 cos ω d t + C 2 sin ω d t (36) x(t) Xe ζωnt sin (ω d t + φ o ) (37) x(t) Xe ζωnt cos (ω d t φ) (38) Where the final two forms of the solution clearly show that the motion is harmonic with frequency ω d and exponential decay in the amplitude. The constants C 1 and C 2 can be derived from the initial conditions: x(t 0) x o ẋ(t 0) ẋ o Solving for C 1 and C 2 : C 1 x o (39) ẋ(t) ζω n e ζωnt C 1 cos ω d t + C 2 sin ω d t + e ζωnt ω d C 1 sin ω d t + ω d C 2 cos ω d t ẋ o ζω n C 1 + ω d C 2 C 2 ẋo + ζω n x o ω d (40)

15 Engi Mechanical Vibrations 15 The amplitude X and phase angles φ and φ o are defined as follows: X (C 1 )2 + (C 2 )2 (41) φ o tan 1 (C 1/C 2) (42) φ tan 1 (C 2/C 1) (43) The amplitude X is the maximum amplitude that would exist at t 0, for zero phase angle. A typical underdamped system response plotted versus ω d t is shown below. The motion is harmonic, with exponential decay of amplitude and damped frequency of oscillation, ω d, which is always less than ω n since ζ < 1. Vehicle suspension systems are examples of underdamped systems.

16 Engi Mechanical Vibrations 16 Consider two cases of a system with ω n 4π rad/s (τ 0.5 s), with initial conditions x o 0.05 m and ẋ o 1 m/s: 1. Case 1: ζ 0.2 ω d rad/s, τ d s, X m and φ 1.07 rad. 2. Case 2: ζ 0.4 ω d rad/s, τ d s, X m and φ 1.14 rad. Note: Neither case reaches X displacement due to the phase angle. Case 2 has a larger X, but it reaches a lower peak displacement due to the quicker delay cause by ζ in the exponential term. Both overshoot the equilibrium position and oscillate about it. Case 2 more quickly stabilizes at the equilibrium. Case 2 is an example of the behaviour of modern sports car, and Case 1 is the behaviour of a land yacht from the 1960 s. A vehicle with the behaviour of Case 1 would respond more easily to bumps (have a smoother ride) but would not feel like it was stable when driven fast. The motions here are exaggerated but Case 1 is a Toyota Corolla and Case 2 is a VW Golf R...different suspension tuning for different markets.

17 Engi Mechanical Vibrations Case 2: Critically Damped Systems (ζ 1, c c c ) If ζ 1 the roots s 1 and s 2 are not distinct: s 1 s 2 c c 2m ζω n ω n (44) To determine the motion take the limit of Eq. (36) as ζ 0. lim x(t) lim ζ 1 ζ 1 e ζωnt C 1 cos ω d t + C 2 sin ω d t e ωnt (C 1 + C 2ω d t Remember: So the solution can be written as: ω d 1 ζ 2 ω n lim (cos ψ) 1 ψ 0 lim (sin ψ) ψ ψ 0 x(t) (C 1 + C 2ω d t)e ωnt Using Eqs. (39) and (40), the motion of a critically damped system is given by: x(t) (x o + (ẋ o + ω n x o ) t) e ωnt (45) The motion is not harmonic, it is aperiodic and decreases to zero as t increases due to the e ωnt term. Critical damping, c c, is the lowest value of c that will not produce oscillations. Consider two cases of a system with ω n 4π rad/s (τ 0.5 s), with initial conditions x o 0.05 m and ẋ o 1 m/s.

18 Engi Mechanical Vibrations Case 3: Overdamped System (ζ > 1, c > c c ) For this case, the roots s 1 and s 2 are both real, distinct and less than 1. ( ) s 1 ζ + ζ 2 1 ω n (46) ( ) s 2 ζ ζ 2 1 ω n (47) and s 2 < s 1. The solution is: ( ) ( ) ζ+ ζ x(t) C 1 e 2 1 ω nt ζ ζ + C2 e 2 1 ω nt (48) For initial conditions: the constants are: x(t 0) x o ẋ(t 0) ẋ o ( C 1 x oω n ζ + ) ζ ẋ o 2ω n ζ 2 1 ( C 2 x oω n ζ ) ζ 2 1 ẋ o 2ω n ζ 2 1 (49) (50) Consider two cases of a system with ω n 4π rad/s (τ 0.5 s), with initial conditions x o 0.05 m and ẋ o 1 m/s.

19 Engi Mechanical Vibrations 19 The motion is aperiodic. It has a similar form to the critically damped solution, however, it undershoots the maximum amplitude of the critically damped system and takes longer to return to equilibrium. A good example of an overdamped system is the closer on a screen door which takes forever and an age to close Summary Consider three cases of a system with ω n 4π rad/s (τ 0.5 s), with initial conditions x o 0.05 m and ẋ o 1 m/s. Underdamped motion: is harmonic with exponential decay of amplitude. The frequency of oscillation of underdamped motion (ω d 1 ζ 2 ω n ) is always less than the frequency of oscillation of undamped motion (ω n ). Underdamped motion will not reach the same peak amplitude as undamped motion, unless the phase angle is zero. The underdamped case is the only solution that results in oscillatory motion. Underdamped systems return to equilibrium quickly, but continue to oscillate. In the context of a suspension system the wheels would be in position to respond to another bump more quickly than a critically damped system.

20 Engi Mechanical Vibrations 20 Critically damped motion: is aperiodic with exponential decay to equilibrium position. Critically damped systems are quickest to return to the equilibrium postion. A critically damped system would transmit too much force to car passengers. Vehicle suspensions are underdamped. Overdamped systems: take a long time to return to the equilibrium position. This is not practical for suspension systems. These systems are useful for controlled, slow return to equilibrium without oscillation. Screen door closers are overdamped Logarithmic Decrement Consider the motion of an underdamped system. The ratio of displacements at two times separated by one period of damped oscillation is: x 1 x 2 Xe ζωnt1 cos(ω d t 1 φ) Xe ζωnt 2 cos(ωd t 2 φ) Since t 2 t 1 + τ d where τ d 2π/ω d, the angle in the cosine terms would be different by 360 o (2π radians or one full oscillation), therfore, the cosines would be equivalent. x 1 x 2 e ζωnt1 e ζωn(t 1+τ d ) Define the logarithmic decrement as the natural log of the displacement ratios: δ ln x 1 2π ζω n τ d ζω n ζω n x 2 ω d 2π ω n 1 ζ 2 2πζ 1 ζ 2 (51) Note: in the log domain the amplitude decay is constant. In the figure above the dashed line is Xe ζωnt.

21 Engi Mechanical Vibrations 21 The damping ratio, ζ, can be defined as a function of δ: ζ δ (2π) 2 + δ 2 (52) How can logarithmic decrement be used? 1. Given a system with a known ω n and both amplitude ratio and τ d are specified ζ, i.e. the damping ratio and thus the damping constant. Suspension design. 2. Given a system with known ω n and ζ δ, i.e. the amplitude ratio. Suspension analysis. Experimental measurement of time and displacement can be used to determine the damping constant. It is not necessary to use only one period of oscillation, however, as any number of periods may be used: x 1 x m+1 x 1 x 2 x2 x 3 x3 x 4 x m x m+1 where m is the number of cycles. Note: m may be a partial cycle. Since then or and x 1 x 2 e ζωnτ d x 1 x m+1 e mζωnτ d ( ) x1 ln mδ x m+1 δ 1 m ln ( x1 x m+1 ) (53)

22 Engi Mechanical Vibrations 22 3 Forced Vibration - Harmonic Excitation Forced vibration is caused by an external applied force or displacement. Harmonic excitation implies the applied force is of the form F (t) F o e i(ωt+φ) of F (t) F o cos(ωt + φ) or F (t) F o sin(ωt + φ). The phase angle, φ, is often chosen to be zero as only the steady state response of the system is of interest. 3.1 Equation of Motion Consider a 1 DOF spring-mass-damper system subjected to an external appled force. The governing equation is: mẍ + cẋ + kx F (t) (54) Equation (54) is a nonhomogeneous ODE that has solution x(t) that is a sum of an homogeneous solution x h (t) and a particular solution x p (t). The homogeneous solution satisfies: mẍ + cẋ + kx 0 i.e. the equation for free vibration. The solution x h (t) will, therefore, decay to zero for any under, over or critically damped system. The solution to Eq. (54) will, after some transient during which the homogenous solution decays to zero, reduce to the particular solution, x p (t).

23 Engi Mechanical Vibrations Undamped Motion Consider Eq. (54) for the undamped case when F (t) F o cos(ωt): mẍ + kx F o cos ωt (55) The homogeneous solution is: x h (t) C 1 cos ω n t + C 2 sin ω n t (56) where ω n k/m. Since F (t) is harmonic, the particular solution is also harmonic. Propose: x p (t) X cos ωt (57) Substitue Eq. (57) into Eq. (55): mω 2 X cos ωt + kx cos ωt F o cos ωt then X F o k mω 2 δ st ( ) 2 (58) 1 ω ωn where is the static deflection due to force F o. The total solution is: x(t) x h (t) + x p (t) C 1 cos ω n t + C 2 sin ω n t + Using the initial conditions: x(t 0) x o, ẋ(t 0) ẋ o : So: then or δ st F o /k (59) x o C 1 + C 1 x o F o k mω 2 ẋ(t) C 1 ω n sin ω n t + C 2 ω n cos ω n t And the total solution for the motion is: ( x(t) x o F o k mω 2 ẋ o C 2 ω n F o cos ωt k mω2 F o k mω 2 (60) F o ω sin ωt k mω2 C 2 ẋo ω n (61) ) cos ω n t + ẋo ω n sin ω n t + ( ) F o k mω 2 cos ωt (62)

24 Engi Mechanical Vibrations 24 Interesting stuff can be seen by writing Eq.(58) in a non-dimensional form: X δ st 1 ( ) 2 (63) 1 ω ωn which is callled the amplitude ratio (or amplitude factor or magnification factor). It is the ratio of the dynamic amplitude to the static amplitude of the motion. A plot of X/δ st versus ω/ω n yields: As shown by the figure there are three distinct behaviours: 1. Case 1: 0 < ω/ω n < 1 The frequency of the input force is less than the natural frequency of the system and the denominator in Eq. (63) is positive. The steady state motion is given by Eq. (62) and the harmonic response of the system, x p (t), is in phase with the forcing function.

25 Engi Mechanical Vibrations Case 2: 1 < ω/ω n < The frequency of the input force is greater than the natural frequency of the system and the denominator in Eq. (63) is negative, therefore, the amplitiude ratio is negative. The motion of the system, x p (t), Eq. (57), will be out of phase with the external force. As ω/ω n, X 0, i.e. zero response to a high frequency input. In other words there is insufficient time for the system to respond to the input. Later this concept will be used to show why you should always drive faster. 3. Case 3: ω/ω n 1 Resonance The frequency of the input force equals the natural frequency of the system. X, i.e. resonance and if energy is continuously fed into the system it will fail. Let s prove it. Equation(62) can be rewritten as: or x(t) x o cos ω n t + ẋo F o sin ω n t + ω n k mω 2 (cos ωt cos ω nt) x(t) x o cos ω n t + ẋo sin ω n t + δ st cos ωt cos ω nt ( ) ω 2 (64) n 1 ω ωn

26 Engi Mechanical Vibrations 26 Consider the response at ω ω n. The last term in Eq. (64) is undefined when ω ω n, so use l Hôpital s Rule: lim cos ωt cos ω nt d ( ) ω ω n 2 lim dω (cos ωt cos ω nt) 1 ω ω ω n ( ( ) ) 2 d ωn dω 1 ω ωn t sin ωt lim ω ω n 2ω ωn 2 ω nt 2 sin ω nt So, the system response at resonance is: x(t) x o cos ω n t + ẋo sin ω n t + δ stω n t sin ω n t (65) ω n 2 which is harmonic with linear increase in amplitude.

27 Engi Mechanical Vibrations Total Response Similar to free vibration, x(t), can be written as follows: δ st x(t) A cos(ω n t φ) + ( ) 2 cos ωt (66) 1 ω ωn Note: Eq. (3.17) in the text is incorrect (and redundant). I don t know what the author was trying to do there, as Eq. (3.17) is Eq. (3.16)! Beat Phenomena Beating occurs when ω is close to, but not equal to, ω n. The amplitude will build up and diminish in a regular pattern.

28 Engi Mechanical Vibrations Damped Motion Consider a 1 DOF system with a linear spring and viscous damper subjected to a harmonic applied force. The governing equation is: mẍ + cẋ + kx F o cos ωt (67) Assume a particular solution of form: x p (t) X cos(ωt φ) (68) with derivatives: ẋ p (t) ωx sin(ωt φ) ẍ p (t) ω 2 X cos(ωt φ) Substitution of the assumed solution into Eq. (67) gives: X The trigonometric relations: can be used in Eq. (69) to give: X or (k mω 2 ) cos(ωt φ) cω sin(ωt φ) F o cos(ωt) (69) cos(ωt φ) cos(ωt) cos(φ) + sin(ωt) sin(φ) sin(ωt φ) sin(ωt) cos(φ) cos(ωt) sin(φ) (k mω 2 )(cos(ωt) cos(φ) + sin(ωt) sin(φ)) cω(sin(ωt) cos(φ) cos(ωt) sin(φ)) F o cos ωt X cos ωt((k mω 2 ) cos φ + cω sin φ) + sin ωt((k mω 2 ) sin φ cω cos φ) F o cos ωt (70) Equating LHS and RHS coefficients of cos ωt and sin ωt in Eq. (70): Equations (71) and (72) can be solved for X and φ. X X (k mω 2 ) cos φ + cω sin φ F o (71) (k mω 2 ) sin φ cω cos φ 0 (72) To solve for X, square each equation and then add the resultants to eliminate φ: X 2 (k mω 2 ) 2 cos 2 φ + 2cω(k mω 2 ) cos φ sin φ + (cω) 2 sin 2 φ (73) +(k mω 2 ) 2 sin 2 φ 2cω(k mω 2 ) cos φ sin φ + (cω) 2 cos 2 φ Fo 2 or X 2 (k mω 2 ) 2 + (cω) 2 F 2 o

29 Engi Mechanical Vibrations 29 So: and from Eq. (72): X F o (k mω 2 ) 2 + (cω) 2 1/2 (74) ( ) φ tan 1 cω k mω 2 (75) A typical plot of a forcing function and steady-state system response, x p (t), is shown below. Here, the system motion lags the forcing function. Using the definitions of undamped natural frequency, ω n, damping ratio, ζ, static deflection, δ st, and frequency ratio, r: ω n k/m ζ c c c c 2mω n c 2ζmω n δ st F o k r ω ω n (76) the amplitude ratio (or magnification factor, M), X/δ st, can be obtained from Eq. (74): X F o /k ( ) 2 ( 1 mω2 k + cω ) 1/2 2 k δ st ( ( ) ) 2 2 ) 1/2 2 1 ω + (2ζ ω ωn ωn So: M X ( ( ) ) ω (2ζ ω ) 2 δ st ωn ωn 1/2 1 (1 r 2 ) 2 + (2ζr) 2 (77) Similarly, φ, Eq. (75) can be written in terms of r and ζ: φ tan 1 2ζ ω ( ) ω n 2ζr ( ) 2 tan 1 1 ω 1 r 2 ωn (78)

30 Engi Mechanical Vibrations 30 Plots of M X/δ st and phase angle, φ, versus frequency ratio are shown below for various values of ζ. Magnification factor comments: 1. Any amount of damping reduces M at all frequency ratios. Damping slows system response and leads to the phase angle, see Eq. (78). 2. As ζ, M at any r, i.e. more damping less motion. 3. As r, M, i.e. the system does not respond to high frequency inputs. 4. Damping produces a significant reduction in M near r 1, i.e. damping is very important near resonance as it controls the motion. 5. For what value of r would M be a maximum, and what is M max? dm (1 dr 0 r 2 ) 2 + (2ζr) 2 3/2 ( 1 ) 2(1 r 2 )( 2r) + (2ζ) (2r) (1 r 2 ) + (2ζ) 2 So: then M max r 1 2ζ 2 (79) ( ( ) 2 2 1/2 1 (1 2ζ )) 2 + 2ζ 1 2ζ 2 (4ζ 2 (1 ζ 2 )) 1/2 1 2ζ 1 ζ 2 (80) If ζ is known, Eq. (79) gives r and Eq. (80) gives M max and X max M max δ st gives an estimate of the maximum displacement. Measurement of X max will give an estimate of ζ (if F o and k are known).

31 Engi Mechanical Vibrations 31 Phase angle comments: 1. For ζ > 0, 0 < r < 1, then 0 o < φ < 90 o, i.e. the response lags the input. 2. For ζ > 0 r > 1 then 90 o < φ < 180 o, i.e. response is out of phase with input (similar to undamped system) Total Response (Damped System) The complete expression for the response of the damped 1 DOF system to a harmonic input is: x(t) X o e ζωnt cos(ω d t φ o ) + X cos(ωt φ) (81) Note: the first term in Eq. (81) is the homogeneous solution and it will decay to zero, and the solution will reduce to the steady-state solution, i.e. the second term which is the particular solution. X o and φ o are obtained from initial conditions: x(t 0) x o and ẋ(t 0) ẋ o x o X o cos( φ o ) + X cos( φ) X o cos(φ o ) + X cos(φ) (82) ẋ X o ( ζω n e ζωnt cos(ω d t φ o ) ω d e ζωnt sin(ω d t φ o )) Xω sin(ωt φ) ẋ o X o ( ζω n cos( φ o ) ω d sin( φ o )) Xω sin( φ) X o ( ζω n cos(φ o ) + ω d sin(φ o )) + Xω sin(φ) (83) It is easier to solve for X o and φ o with numbers from a particular problem, rather than derive a messy general solution.

32 Engi Mechanical Vibrations 32 4 Harmonic Motion of the Base Consider a 1 DOF viscously damped system subjected to a harmonic motion of the base. The motion of the base is defined by: y(t) Y sin ωt (84) The spring force and damping force will be functions of both x and y: F s k(x y) F d c(ẋ ẏ) Note: I assumed x and y are in the same direction with x > y. The governing equation for the motion is: mẍ + c(ẋ ẏ) + k(x y) 0 (85) Note: there is no forcing function. Since y(t) Y sin ωt, the governing equation can be written as: mẍ + cẋ + kx cẏ + ky cωy cos ωt + ky sin ωt A sin(ωt α) (86) where A Y k 2 + (cω) 2 (87) ( α tan 1 cω ) (88) k Comparison of Eq. (86) with Eq. (67) shows that they are of the same form (i.e. a damped 1 DOF system with a harmonic forcing function). The forcing motion is not transmitted

33 Engi Mechanical Vibrations 33 directly to the mass, however, as it is transmitted through the spring and damper (e.g. a suspension system). The particular solution is then: x p (t) Y k 2 + (cω) 2 ((k mω 2 ) 2 + (cω) 2) 1/2 sin(ωt φ 1 α) (89) where ( ) φ 1 tan 1 cω k mω 2 i.e. similar to Eqs. (74) and (75). (90) Using the trigonometric identities for angle addition, Eq. (89) can be written as: x p (t) X sin(ωt φ) (91) where ( X Y k 2 + (cω) 2 ) 1/2 ( 1 + (2ζr) 2 ) 1/2 (k mω 2 ) 2 + (cω) 2 (1 r 2 ) 2 + (2ζr) 2 (92) ( φ tan 1 mcω 3 ) ( k (k mω 2 ) + (cω) 2 tan 1 2ζr 3 ) 1 + (4ζ 2 1) r 2 (93) The ratio of the amplitudes of the response to base input, X/Y, is called the displacement transmissibility, T d. The variation of T d and φ with frequency ratio are shown below for various values of ζ. T d has a maximum value at: r m 1 1/ ζ 2ζ 2 1

34 Engi Mechanical Vibrations 34 The force, F, acting on the base is F c(ẋ ẏ) + k(x y) mẍ (94) Using Eq. (91) for x: F mω 2 X sin(ωt φ) F T sin(ωt φ) where F T mω 2 X is the amplitude of the force transmitted to the base. Using Eq. (92): F T ky r2 X Y (95) where F T /ky is the force transmissibility. The behaviour of force transmissibilty with frequency ratio is plotted below for various values of damping ratio.

35 Engi Mechanical Vibrations 35 Define z x y as the motion of the mass relative to the base, then Eq. (86) can be written as: m z + cż + kz mÿ mω 2 Y sin ωt (96) Comparison of Eq. (??) with Eq. (86) gives the solution: z(t) Z sin(ωt φ 1 ) (97) where Z mω 2 Y (k mω 2 ) 2 + (cω) Y r 2 2 (98) (1 r 2 ) 2 + (2ζr) 2 ( ) φ 1 tan 1 cω k mω 2 ( ) 2ζr tan 1 1 r 2 (99) The variation of Z/Y with frequency ratio for several values of ζ is shown below.

36 Engi Mechanical Vibrations 36 5 System Modelling 5.1 Combinations of Springs In practical systems several linear springs may be used in combination (e.g. four springs used in a car). An effective spring constant can be developed for spring combinations in parallel and series Springs in Parallel Consider the case where two linear springs support a platform. If a force W is applied to the platform such that it does not rotate, the springs will obtain the same static deflection, δ st. Using the FBD of the platform when it is in static equilibrium: But + F y 0 F 1 + F 2 W F 1 k 1 δ st F 2 k 2 δ st therefore W k 1 δ st + k 2 δ st We want an equivalent spring k eq that will support force W and attain the same static deflection, δ st. W k eq δ st So k eq k 1 + k 2 For any combination of parallel springs: k eq n k i (100) i1

37 Engi Mechanical Vibrations Springs in Series Consider two linear springs k 1 and k 2 attached in series: When load W is applied to spring 2 the springs will deflect a different amount, unless k 1 k 2. Consider an FBD for each spring: Each spring will carry the same load, W : W k 1 δ 1 k 2 δ 2 k eq δ st k eq (δ 1 + δ 2 ) Define an equivalent linear spring k eq that would carry load W, but deflect δ st δ 1 + δ 2. Note: Then: δ 1 W k 1 k eq k 1 δ st δ 2 W k 2 k eq k 2 δ st or δ st δ 1 + δ 2 k eq k 1 δ st + k eq k 2 δ st 1 k eq 1 k k 2

38 Engi Mechanical Vibrations 38 For any combination of springs in series: 1 k eq General Method for Combinations of Springs n i1 1 k i (101) Since springs are used to store potential energy it is possible to equate the potential energy stored due to the deformation of a new equivalent spring to that stored in the original spring system. e.g. springs in parallel e.g. spings in series: 1 2 k 1δ 2 st k 2δ 2 st 1 2 k eqδ 2 st k eq k 1 + k k 1δ k 2δ k eqδ 2 st

39 Engi Mechanical Vibrations 39 But δ 1 W k 1 k eq k 1 δ st δ 2 W k 2 k eq k 2 δ st Then 1 2 k eqδst 2 1 ( ) 2 2 k keq 1 δ st + 1 ( ) 2 k 1 2 k keq 2 δ st k 2 1 k eq 1 k k Combinations of Damping Elements To obtain equivalent damping elements use the method of equivalence of force between the original combination and the equivalent damper Viscous Dampers in Parallel Assume the velocity at the end of each damper is equivalent. Since the ends of each damper will have the same velocities: F 1 c 1 (ẋ 2 ẋ 1 ) F 2 c 2 (ẋ 2 ẋ 1 ) F eq c eq (ẋ 2 ẋ 1 ) But the equivalent damper will exert the force of both original dampers: F eq F 1 + F 2 So: c eq c 1 + c 2 For any combination of dampers in parallel: c eq n c i (102) i1

40 Engi Mechanical Vibrations Viscous Dampers in Series Consider three viscous dampers in series to be replaced by and equivalent viscous damper c eq that will have the same velocites at its end points. Each damper will carry the same force: F 1 F 2 F 3 F eq where F 1 c 1 (ẋ 2 ẋ 1 ) F 2 c 2 (ẋ 3 ẋ 2 ) F 3 c 3 (ẋ 4 ẋ 3 ) F eq c eq (ẋ 4 ẋ 1 ) From kinematics: (ẋ 4 ẋ 1 ) (ẋ 2 ẋ 1 ) + (ẋ 3 ẋ 2 ) + (ẋ 4 ẋ 3 ) Therefore Then F eq c eq F 1 c 1 + F 2 c 2 + F 3 c 3 1 c eq 1 c c c 3 For any combination of viscous dampers in series: 1 c eq n i1 1 c i (103)

41 Engi Mechanical Vibrations Pendulum Consider a pendulum of mass m, on a length of string l, pivoting off a fixed surface at O: Since the pendulum pivots about O, sum the moments about O: MO I O θ mgl sin θ or I O θ + mgl sin θ 0 When θ 0, sin θ θ, therefore: I O θ + mglθ 0 (104) Compare Eq. (104) with Eq. (7): mẍ + kx 0 Both equations have the same form where m I O and k mgl. frequency of the translating system is k/m, then for the pendulum: ω n Since the natural mgl I O (105) From the parallel axis theorem: I O I G + ml 2 where I G 0 for a point mass. So for the pendulum: ω n mgl g ml 2 l (106) i.e. the natural frequency is not dependent on the mass of the pendulum!

42 Engi Mechanical Vibrations Torsional Systems Consider a uniform slender rod of mass m attached to a fixed frame through a linear spring and viscous damper: Assuming a postive direction for θ, θ and θ sum the moments about O: ( ) ( ( ) ( ) l l 3l 3l MO I O θ k θ c θ 4 4) 4 4 ( ) l 2 ( ) 3l 2 k θ c θ 4 4 or Which can be written as: ( ) 3l 2 ( ) l 2 I O θ + c θ + k θ I O θ + ct θ + kt θ 0 (107) where c T and k T are the torsional damping and torsional spring constants. units of c T are (N m s/rad) and for k T (N m/rad). Note: the Compare Eq. (107) with Eq. (24): mẍ + cẋ + kx 0 Both equations have the same form, where m I O, c c T, and k k T. So, the natural frequency is: k T ω n (108) I O Critical damping is: The damping ratio is: c c,t 2I O ω n (109) ζ c T c c,t (110)

43 Engi Mechanical Vibrations 43 And a frequency of damped oscillation is: Using the parallel axis theorem: ω d 1 ζ 2 ω n I O 1 ( ) l 2 12 ml2 + m (111) 4 Due to the similarity of Eqs. (107) and (24) the solutions of the free vibration of the rod are already available. For example, if the system is underdamped: θ(t) Θe ζωnt cos (ω d t φ) (112) The amplitude Θ and phase angle φ are defined from initial conditions: θ(t 0) θ o θ(t 0) θ o Solving for C 1 and C 2 : C 1 θ o (113) θ(t) ζω n e ζωnt C 1 cos ω d t + C 2 sin ω d t + e ζωnt ω d C 1 sin ω d t + ω d C 2 cos ω d t The amplitude X and phase angles φ θ o ζω n C 1 + ω d C 2 Θ C 2 θ o + ζω n θ o ω d (114) (C 1 )2 + (C 2 )2 (115) φ tan 1 (C 2/C 1) (116) The solutions for free vibration of critically damped and overdamped translational systems can also be translated to give the motion of critically and overdamped torsional systems! 5.5 Mass Moment of Inertia The mass moment of inertia, I, is a property of a body that measures the resistance of the body to angular motion. It is analogous to mass, which measures the resistance of a body to acceleration. For general plane motion we will use two equations: F m a M I α where the second equation is a form of Newton s second law for rotational motion. The mass moment of inertia is defined as the integral of the second moment about an axis of all elements of mass dm that compose a body. For example the mass moment of inertia of the body below about the z axis is:

44 Engi Mechanical Vibrations 44 I m r 2 dm (117) The moment arm r is the perpendicular distance from the axis to the element dm. Usually the axis chosen passes through the center of gravity and we obtain I G. The moment of inertia is always positive and has units of mass times length squared (kg m 2 or slug ft 2 ). The integral may be expressed as a volume integral by replacing dm with ρdv. I r 2 ρdv (118) and if density is uniform: V I ρ r 2 dv (119) V The mass moments of inertia for some common shapes are tabulated inside the back cover of your Mechanics text Parallel Axis Theorem If the mass moment of inertia is known for an axis passing through the center of gravity of a body, the parallel axis theorem is used to determine the mass moment of inertia about another parallel axis.

45 Engi Mechanical Vibrations 45 Consider an arbitrary body with center of gravity G. The z axis passes through G. The z axis is parallel to z and located at a distance d from z. The mass moment of inertia of the body about the z axis is: I r 2 dm m (d + x ) 2 + y 2 dm m ( x 2 + y 2) dm + 2d x dm + d 2 dm m m m But x 2 + y 2 r 2 So the first term represents I G. The second term is zero since the z axis passes through G. x dm x dm but x 0, since we are on an axis through G. m So the moment of inertia about the z axis is defined by the parallel axis theorem: m I I G + md 2 (120) where I G is the mass moment of inertia about an axis through G, m is the mass of the body, and d is the distance between the parallel axes Radius of Gyration The mass moment of inertia is sometimes tabulated using the radius of gyration, k: I k (121) m where k has units of length.

46 Engi Mechanical Vibrations 46 The mass moment of inertia is found from k as follows: Composite Bodies I mk 2 (122) The mass moment of inertia of a composite body about an axis may be obtained by algebraically adding the mass moments of inertia of simpler bodies about the same axis. Note: this would probably require use of the parallel axis theorem. I O ( ) ( ) I G1 + m 1 d I G2 + m 2 d 2 2 (123) Confusion/Rant Another term that is defined in mechanics is the moment of inertia: I y 2 da A which is actually the second moment of area about the y axis. Inertia is a misnomer, as inertia requires mass, but there is no mass in this term. Also, it is given the same symbol, I. This is too stoopid! And it leads to much confusion in students. And it is stoopid! So, the real moment of inertia is sometimes called the mass moment of inertia (as I have done) to try to reduce the misunderstanding. This would never happen in thermodynamics! The second moment of area arises in Solid Mechanics when determining normal stresses resulting from moments, and deflection of beams.

47 Engi Mechanical Vibrations Degrees of Freedom Minimum number of independent co-ordinates required to completely specify the motion of a system. Simple pendulum 1 DOF: θ or x or y (Note: x 2 + y 2 l 2 so only one is required) Simple vehicle model 1 DOF: x(t) vertical motion Mass on a shaft 1 DOF: θ

48 Engi Mechanical Vibrations 48 Two mass system (elevator and a passenger) 2 DOF: x 1 and x 2 Two belt-driven pulleys on a shaft 2 DOF: θ 1 and θ 2 Three mass pendulum 3 DOF: (θ 1,θ 2,θ 3 ) or (x 1,x 2,x 3 ) or (y 1,y 2,y 3 )

49 Engi Mechanical Vibrations 49 Three mass system (vehicle or building model) 3 DOF: x 1, x 2 and x Problem Solution The goal of a vibration analysis is to determine the time response of a system to a given forcing function. For example, to determine the comfort of car passengers you would need the displacement, velocity and acceleration of the passengers. Also of importance would be natural frequencies, resonance and beat frequencies. Most systems are very complex, and it is generally impossible to consider all the details of a system for a mathematical analysis. For example, how would you model a passenger in a vehicle, or how would you model the vehicle? The first step in any vibration analysis is to develop a suitable model that will capture the details required for the analysis.

50 Engi Mechanical Vibrations Step 1: Mathematical Modelling Represent the important features of the system to permit the derivation of a tractable mathematical model. The model must incorporate sufficient detail to allow analysis of the appropriate dynamic response of the system. The model must be solvable. Linear models are easy to solve, however, nonlinear models often reveal realistic details that are impossible to reproduce with a linear model. A simple 1 DOF model can be used to give qualitative information on the motion of a system, and it may be sufficient for the desired outputs. A simple model may indicate possible problems (e.g. resonance) which require more detailed study. More complicated models may be developed from an initial simple model. E.g. Develop plausible mathematical models for a motorcycle and rider.

51 Engi Mechanical Vibrations 51

52 Engi Mechanical Vibrations Step 2: Derivation of the Governing Equations Draw an FBD for each mass and inertia element. Use Newton s Second Law, conservation of energy or d Alembert s Principle to derive the governing equations. The result will be a set of ODE s for linear system and a set of PDE s for nonlinear systems Step 3: Solve the Governing Equations Solve the governing equations using standard methods for ODE s, Laplace transforms, matrix methods or numerical techniques. Nonlinear systems with PDE s for governing equations require numerical solution for realistic problems. If a closed form solution (i.e. a functional relationship) is possible many conclusions can be drawn from the results. For example, a 1 DOF system subjected to free or forced (harmonic) motion has a general solution that applies to many systems. The final numerical answers will be functions of the system inputs, initial conditions and forcing function. The actual form of the solution, however, will be the same for different systems Step 4: Interpretation of Results Solution of the governing equations gives the displacement, velocity and acceleration of each mass and inertia element in the mathematical model. This information can be used to determine: - Passenger comfort - Natural frequencies - Resonance (damage) - Beat frequencies (comfort and damage) - Operational parameters - Required spring and damping constants

53 Engi Mechanical Vibrations 53 6 Two Degree of Freedom Systems 6.1 Governing Equations Consider the two degree of freedom (2 DOF) system shown below: Each mass can move in one dimension, therefore the system has two degrees of freedom. This system is the most general case where a forcing function is applied to each mass. Use the FBD for each mass to derive the governing equations: To define directions of forces assume relative magnitudes for the motion of each mass, e.g. x 2 > x 1 and ẋ 2 > ẋ 1. m 1 ẍ 1 c 2 (ẋ 2 ẋ 1 ) + k 2 (x 2 x 1 ) k 1 x 1 c 1 ẋ 1 + F 1 (t) m 2 ẍ 2 c 2 (ẋ 2 ẋ 1 ) k 2 (x 2 x 1 ) k 3 x 2 c 3 ẋ 2 + F 2 (t) or m 1 ẍ 1 + (c 1 + c 2 )ẋ 1 c 2 ẋ 2 + (k 1 + k 2 )x 1 k 2 x 2 F 1 (t) m 2 ẍ 2 c 2 ẋ 1 + (c 2 + c 3 )ẋ 2 k 2 x 1 + (k 2 + k 3 )x 2 F 2 (t) (124) or m1 0 ẍ1 (c1 + c 2 ) c 2 ẋ1 0 m 2 ẍ 2 + c 2 (c 2 + c 3 ) ẋ 2 (k1 + k + 2 ) k 2 x1 k 2 (k 2 + k 3 ) x 2 F1 (t) F 2 (t) (125)

54 Engi Mechanical Vibrations 54 Note: The governing equation for any 2 DOF system can be written in the following general form: m11 m 12 ẍ1 c11 c + 12 ẋ1 k11 k + 12 x1 F1 (t) (126) m 21 m 22 ẍ 2 c 21 c 22 ẋ 2 k 21 k 22 x 2 F 2 (t) Here: m11 m 12 m1 0 m 21 m 22 c11 c 12 0 m 2 (c1 + c 2 ) c 2 c 21 c 22 k11 k 12 k 21 k 22 c 2 (c 2 + c 3 ) (k1 + k 2 ) k 2 k 2 (k 2 + k 3 ) For different systems, the entries in the matrices on the RHS of the equations would change. 6.2 Free Vibration of an Undamped 2 DOF System Simplify the system above for the case of undamped free vibration: From the general form of the governing equation for a 2 DOF system Eq. equations of motion for this system are: m11 m 12 ẍ1 k11 k + 12 x1 0 m 21 m 22 ẍ 2 k 21 k 22 x 2 0 (126) the (127) For this system, m 12 m This will have significance later. So, the governing equations are: m 11 ẍ 1 + k 11 x 1 + k 12 x 2 0 m 22 ẍ 2 + k 21 x 1 + k 22 x 2 0 (128)

55 Engi Mechanical Vibrations 55 Can m 1 and m 2 oscillate at the same frequency and phase angle, but with different amplitude? x 1 (t) X 1 cos(ωt + φ) x 2 (t) X 2 cos(ωt + φ) (129) The same phase angle would imply each mass passes through its equilibrium position at the same time. From Eq. (129): ẍ 1 ω 2 X 1 cos(ωt + φ) ẍ 2 ω 2 X 2 cos(ωt + φ) (130) Substitution of Eqs. (129) and (130) into Eq. (127): ) ) (( m 11 ω 2 + k 11 X 1 + k 12 X 2 cos(ωt + φ) 0 ) ) (( m 22 ω 2 + k 22 X 2 + k 21 X 1 cos(ωt + φ) 0 (131) or ( m11 ω 2 + k 11 ) k 12 k 21 ( m 22 ω 2 + k 22 ) X1 X (132) For a nontrivial solution: ( m 11 ω 2 + k 11 ) k 12 k 21 ( m 22 ω 2 + k 22 ) 0 (133) So: (m 11 m 22 )ω 4 (m 11 k 22 + m 22 k 11 )ω 2 + (k 11 k 22 k 12 k 21 ) 0 (134) This equation is quadratic in ω 2. For convenience, rewrite the equation in the following form: aω 4 + bω 2 + d 0 (135) where: a m 11 m 22 b (m 11 k 22 + m 22 k 11 ) d k 11 k 22 k 12 k 21 (136) Solving for the two roots gives the two natural frequencies of the system: ω 1 ω 2 ( ) b b 2 1/2 4ad 2a (137) ( ) b + b 2 1/2 4ad 2a (138)

56 Engi Mechanical Vibrations 56 Substitution of Eqs. (137) and (138) into Eq. (153) allows determination of amplitude ratios at the two natural frequencies. ( m 11 ω k 11 )X 1 + k 12 X 2 0 ( m 22 ω k 22 )X 2 + k 21 X 1 0 and ( m 11 ω k 11 )X 1 + k 12 X 2 0 ( m 22 ω k 22 )X 2 + k 21 X 1 0 Then the amplitude ratios: r 1 r 2 ( X2 X 1 ( X2 X 1 ) (1) ( m 11ω1 2 + k 11) k 12 ) (2) ( m 11ω2 2 + k 11) k 12 ω 1, ω 2, r 1 and r 2 define the natural modes of the system. indicate the first and second natural modes of the system. k 21 ( m22 ω k 22) (139) k 21 ( m22 ω k 22) (140) The bracketed superscripts Use of the amplitude ratios allows for a general solution for the motion of the system to be written as: (1) x (1) x (t) 1 (t) (1) X x (1) 2 (t) 1 cos(ω 1 t + φ 1 ) r 1 X (1) first mode (141) 1 cos(ω 1 t + φ 1 ) x (2) (t) x (2) 1 (t) x (2) 2 (t) X (2) 1 cos(ω 2 t + φ 2 ) r 2 X (2) 1 cos(ω 2 t + φ 2 ) Using superposition, the motion of each mass can be written as: x 1 (t) X (1) 1 cos(ω 1 t + φ 1 ) + X (2) 1 cos(ω 2 t + φ 2 ) second mode (142) x 2 (t) r 1 X (1) 1 cos(ω 1 t + φ 1 ) + r 2 X (2) 1 cos(ω 2 t + φ 2 ) (143) where the general solution for X (1) 1, X(2) 1, φ 1 and φ 2 is given by: X (1) 1 1 {r 2 x 1 (0) x 2 (0)} 2 + { r 2ẋ 1 (0) + ẋ 2 (0)} 2 r 2 r 1 ω1 2 X (2) 1 1 { r 1 x 1 (0) x 2 (0)} 2 + {r 1ẋ 1 (0) + ẋ 2 (0)} 2 r 2 r 1 ω2 2 ( ) φ 1 tan 1 r2 ẋ 1 (0) + ẋ 2 (0) ω 1 r 2 x 1 (0) x 2 (0) ( ) φ 2 tan 1 r 1 ẋ 1 (0) + ẋ 2 (0) ω 2 r 1 x 1 (0) x 2 (0) See the derivation for Eq. (5.18) in the text. 1/2 1/2 (144) The i-th mode can be excited by setting initial conditions as follows: x 1 (t 0) X (i) 1 ẋ 1 (t 0) 0 x 2 (t 0) r i X (i) 1 ẋ 2 (t 0) 0 (145)

57 Engi Mechanical Vibrations 57 So what s that about? Consider the following simple system: From Eq. (125) the governing equation is: m 0 0 m ẍ1 ẍ 2 2k k + k 2k x1 x (146) So, for reference: m11 m 12 m 0 m 21 m 22 k11 k 12 0 m 2k k k 21 k 22 k 2k From Eq. (136): From Eqs. (137) and (138): ω 1 ω 2 a m 2 b (2km + 2km) 4km d 4k 2 k 2 3k 2 ( ) 4km (4km) 2 4(m 2 3k 2 1/2 ) k 2m 2 m ( ) 4km + (4km) 2 4(m 2 3k 2 1/2 ) 3k 2m 2 m From Eqs. (139) and (140) r 1 r 2 ( ) (1) X2 X 1 ( ) (2) X2 X 1 ( ( ) ) m k m + 2k k ( ( m 3k m k ) ) + 2k k ( ( ) ) 1 m k m + 2k k ( ( ) ) 1 m 3k m + 2k

58 Engi Mechanical Vibrations 58 The bracketed superscripts indicate the 1st and 2nd natural modes. - 1st mode - amplitudes of both masses are equivalent. - 2nd mode - amplitudes of both masses are the same, but the motion is out of phase (due to the negative sign). The modes are often shown on a mode diagram. As shown in the mode diagram for the second mode, the point in the middle of the second spring remains fixed for all time. This point is called a node. The general solution for the motion of the system is (using Eq. (143)). x 1 (t) X (1) k 1 cos m t + φ 1 + X (2) 3k 1 cos m t + φ 2 x 2 (t) X (1) k 1 cos m t + φ 1 X (2) 3k 1 cos m t + φ 2 (147) where φ 1, φ 2, X (1) 1 and X (2) 1 are dependent on the intial conditions: x 1 (0), x 2 (0), ẋ 1 (0) and ẋ 2 (0). See Eq. (144) or Eq. (5.18) in the text. Let s use some numbers. Consider the case where m 100 kg, k 10 kn/m. Then: ω 1 k 10 rad/s m ω 2 3k rad/s m Consider the following three initial states: 1. First mode: x 1 (0) 0.02 m x 2 (0) r 1 x 1 (0) 0.02 m ẋ 1 (0) ẋ 2 (0) 0 From Eq. (144) or Eq. (5.18) in the text: φ 1 φ 2 0 X (1) 1 1 ( 1 (0.02) 0.02) 0.02 m 1 1 X (2) 1 1 ( 1 (0.02) ) 0 m 1 1