BSc/MSci EXAMINATION PHY-217 Vibrations and Waves Time Allowed: 2 hours 30 minutes Date: 4 th May, 2011 Time: 14:30-17:00 Instructions: Answer ALL questions in section A. Answer ONLY TWO questions from section B. Section A carries 50 marks, each question in section B carries 25 marks. An indicative marking-scheme is shown in square brackets [ ] after each part of a question. Course work comprises 20% of the final mark. Numeric calculators are permitted in this examination. Please state on your answer book the name and type of machine used. Complete all rough workings in the answer book and cross through any work which is not to be assessed. Important Note: The academic regulations state that possession of unauthorised material at any time when a student is under examination conditions is an assessment offence and can lead to expulsion from the college. Please check now to ensure that you do not have any notes in your possession. If you have any then please raise your hand and give them to an invigilator immediately. Exam papers cannot be removed from the exam room You are not permitted to read the contents of this question paper until instructed to do so by an invigilator. Examiners: Dr L.Cerrito, Dr. A.Brandhuber c Queen Mary, University of London 2011
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SECTION A. Attempt answers to all questions. A1. For a particle of mass m executing simple harmonic motion ( SHM ), its displacement from equilibrium can be written as: x(t) Asin(ω 0 t + φ). (i) Write down the amplitude, angular frequency, period and phase at time t. [4] (ii) Sketch a graph of x(t). [4] (i) amplitude: A, angular frequency: ω 0,period:T 2π/ω 0,phase:ω 0 t + φ (ii) sinusoidal graph A2. Amassm is attached to a massless spring with constant k, andcanmovehorizontallyon africtionlessplanewithoutairresistance. (i) By explicit consideration of the forces involved, derive the differential equation of motion and write down the general solution. [3] (ii) What is the angular frequency, expressed in terms of k and m? [2] (iii) Write down the potential, kinetic, and total energy at time t. [2] (iv) Write down an example of initial conditions. [1] (i) The equation is: mẍ kx ẍ + ω 2 0 x 0withω2 0 k/m general solution: x(t) Acos(ω 0 t + φ) witha, φ constants. (ii) ω 2 0 k/m (iii) E p (t) 1 2 kx2 (t) 1 2 ka2 cos 2 (ω 0 t + φ) ande k (t) 1 2 mẋ2 (t) 1 2 mω2 0 A2 sin 2 (ω 0 t + φ) Total energy: E k + E p 1 2 ka2 (iv) An example is: x(t 0)x 0 and v(t 0)v 0. A3. Show that the function y Acos(4x)+Bsin(4x), where A and B are arbitrary constants, is a general solution of the differential equation: [3] d 2 y +16y 0. dx2 dy dx 4Asin(4x)+4Bcos(4x) and d 2 y 16Acos(4x) 16Bsin(4x) dx2 d2 y +16y 0 dx2 A4. For an object starting from maximum diplacement and released from rest, sketch plots of displacement versus time for: (i) underdamped SHM, (ii) critically damped SHM and (iii) overdamped SHM. (iv) How is the oscillator s Quality Factor, Q, defined? [5] Q is a dimensionless number defined as: Q 1/γ 1/ω 0 ω 0 /γ, whereγ is the damping constant in damped harmonic motion and ω 0 is the angular frequency of the undamped oscillator.
A5. Consider a damped driven simple harmonic oscillator in steady state. Sketch plots of (i) amplitude versus driving frequency, and (ii) average power versus driving frequency. Indicate on each plot the resonance frequency and what effect high or low oscillator s quality factor Q has on the graph. [4] A6. Consider N equal masses coupled by springs: how many normal modes does such a system have? For a normal mode, what characterises the motion of each of the masses? [4] A system of N masses coupled by springs has N normal modes. In normal mode vibration all masses oscillate with the same frequency and are either in phase or 180 out-of-phase. A7. What are non-dispersive, normally dispersive, and anomalously dispersive media? [4] In non-dispersive media the speed of wave propagation is independent on the frequency (or the wavelength). For normally dispersive media the group velocity is lower than the phase velocity. For anomalously dispersive media, the group velocity is higher than the phase velocity. A8. Consider a string fixed at one end. A Gaussian pulse travels towards the fixed end, where it is reflected back. Assuming there are no energy losses and that there is no dispersion of the pulse, describe the appearance of the pulse after the reflection, and explain why it looks as you described. [4]
The pulse is reflected back and is flipped upside down (a mountain comes back as a valley and vice-versa). This is due to the boundary condition imposed by the fixed end. A9. The human ear perceives sounds over a frequency range 30 Hz 15 khz. The speed of sound in air is about 340 ms 1. Show that the corresponding range in wavelengths is 11.3 m to 2.27 cm. [5] v ω k f 2π 2π λ λ L v f 340 30 m 11.3 m λ H v f 340 15 10 0.022 m 3 A10. For a taut string attached at both ends, draw a sketch of the first, second and third harmonic, indicating the nodes and the anti-nodes. [5] c Queen Mary, University of London 2011 Page 3 Please turn to next page
SECTION B. Answer two of the four questions in this section. B1. Two equal masses are connected as shown with two identical massless springs of spring constant k. (i) By explicit consideration of the forces and consideringonlymotioninthevertical direction, derive the equations of motion of the two masses. Do you need to consider gravitational forces? [15] (ii) Show that the angular frequencies of the two normal modes are given by ω 2 (3± 5)k/2m and hence that the ratio of the normal mode frequencies is ( 5+1)/( 5 1). [7] (iii) Find the ratio of amplitudes of the two masses in each separate mode. [3] SOLUTION (i) Let the displacement from the equilibrium positions for masses m 1 and m 2 be x 1 and x 2 respectively. Then the tensions in the two strings are T 1 kx 1 and T 2 k(x 2 x 1 ) repectively. We need not to consider the gravitational forces acting on the masses, because they are independent of the displacements and hence do not contribute to the restoring forces that cause the oscillations. The gravitational forces merely cause a shift in the equilibrium positions of the masses, and you do not have to find what those shifts are. Now m 1 ẍ 1 +k(x 2 x 1 ) kx 1 m 2 ẍ 2 k(x 2 x 1 ) substituting m 1 m 2 m and ω 2 s k/m we get: ẍ 1 ω 2 s(x 2 2x 1 ) ẍ 2 ω 2 s (x 1 x 2 ) (ii) Let: x 1 C 1 cos(ωt) andx 2 C 2 cos(ωt) andsubstitutingabove:
ω 2 C 1 +2ω 2 s C 1 ω 2 s C 2 ω 2 C 2 + ω 2 s C 2 ω 2 s C 1 These can be solved directly or using Cramer s Rule. In the first case: (iii) For ω + ω + ω 3+ 5 3 5 (3 + 5)k/2m: C 1 ωs 2 C 2 2ωs 2 ω ω2 s ω2 2 ωs 2 ωs 4 2ωs 4 3ωsω 2 2 + ω 4 ω 4 3ω 2 sω 2 + ω 4 s 0 ω 2 3ω2 s ± 9ωs 4 4ω4 s ( 3 ± 5 ) ωs 2 2 2 ω 2 ( 3 ± 5 ) k 2m 3+ 5 3 5 3+ 5 3+ 5 3+ 5 5 1 5+1 2 5 1 5 1 C 1 C 2 ω 2 s 2ω 2 s ω 2 + 2ω 2 s 4ω 2 s (3 + 5)ω 2 s 2 1 5 For ω (3 5)k/2m: C 1 C 2 ω 2 s 2ω 2 s ω2 2ω 2 s 4ω 2 s (3 5)ω 2 s 2 1+ 5 c Queen Mary, University of London 2011 Page 5 Please turn to next page
B2. The sketch shows a mass M 1 on a frictionless plane connected to support O by a spring of stiffness k. ThemassM 2 is supported by a string of length l from M 1. OA is the length of the relaxed spring. x 1 and x 2 are the positions of M 1 and M 2,respectively,relativeto point A. (i) Assuming small angles θ, write down the differential equation of motion for M 1. [8] (ii) Assuming small angles θ, write down the differential equation of motion for M 2. [7] (iii) Now let O oscillate harmonically in the horizontal direction, driven by an external force, and its position X(t) isgivenbyx 0 cos(ωt). Let both masses be equal to M. Write down the differential equation of motion for each mass. [4] (iv) What are the amplitudes (steady state) of the two masses as a function of k, M, X 0, l and ω?(hint: use Cramer s rule) [3] (v) There is one frequency for which one mass standsstillandtheotheroscillates. What is that frequency? What is peculiar about this state? [3] SOLUTION (i) The equation of motion for mass M 1 is: M 1 ẍ 1 kx 1 + M 2 g l (x 2 x 1 )
(ii) and for mass M 2 is: M 2 ẍ 2 M 2 gsin(θ) M 2 ẍ 2 M 2 g l (x 2 x 1 ) (iii) The equation of motion for mass M 2 in unchanged, whereas for mass M 1 is: M 1 ẍ 1 k[x 1 X(t)] + M 2 g l (x 2 x 1 ) M 1 ẍ 1 + kx 1 + M 2 g l (x 1 x 2 )kx 0 cos(ωt) (iv) Substituting ω 2 s k/m 2, ω 2 p g/l and M 1 M 2 M we get: ẍ 2 + ω 2 px 2 ω 2 px 1 0 ẍ 1 +(ω 2 s + ω2 p )x 1 ω 2 p x 2 ω 2 s X 0cos(ωt) Let: x 1 C 1 cos(ωt) andx 2 C 2 cos(ωt), and substituting above: ω 2 p C 1 +(ω 2 ω 2 p )C 2 0 ( ω 2 + ω 2 s + ω2 p )C 1 ω 2 p C 2 ω 2 s X 0 0 ω 2 ω 2 p ωs 2 C 1 X 0 ωp 2 ωp 2 ω 2 ω 2 p ω 2 + ωp 2 + ω2 s ωp 2 kx 0 (g lω 2 ) Mlω 4 (2Mg + kl)ω 2 + kg
ωp 2 0 ω 2 + ωp 2 C 2 + ω2 s ωs 2X 0 ωp 2 ω 2 ω 2 p ω 2 + ωp 2 + ωs 2 ωp 2 kgx 0 Mlω 4 (2Mg + kl)ω 2 + kg These are the steady state solutions. The general solution is a linear combination between the transient solution and the steady state solutions. (v) We can note from the functional form of C 2 that it cannot have the value zero (except for ω ). However, C 1 will be zero when: g lω 2 0: g ω ω p l C 1 0 This is the resonance frequency of the pendulum. Thus, at this frequency, the two horizontal forces on the upper mass, kx 0 cos(ωt) andt sin(θ) cancel. Sohowcanthependulumswingif the mass attached to the spring does not move at all: what is driving the pendulum? This is only possible in the approximation of zero damping. In the presence of damping, no matter how little, the peculiar state is unstable. This can easily be seen as follows: assume that the system is in that state. That means that at any moment in time the net horizontal force on the upper mass is zero. Thus the vectorial sum of the spring force and T sin(θ) mustbezero. However, if the mass on the spring is not moving, the pendulum is no longer driven, and thus its amplitude will decay, and the net force on the mass on the spring is no longer zero, and thus that mass will start to move. Thus the peculiar state is unstable. c Queen Mary, University of London 2011 Page 8 Please turn to next page
B3. Apulsetravellingalongastretchedstringisdescribedbythefollowingequation: y(x, t) b 3 b 2 +(2x ut) 2 (i) Sketch the graph of y against x for t 0. [7] (ii) What are the speed of the pulse and its direction of travel? [10] (iii) Calculate the transverse velocity (v y )ofagivenpointofthestringasafunctionof x for the instant t 0,andsketchagraphofv y (t 0)againstx. [8] (i) Indicate with the sketch that the function y(x, 0) is symmetric around x 0,tendstozero for x ± and has a maximum at x 0. Thesketchofy(x, 0) is shown below: y(x,0) (in units of b) 1 0.8 0.6 y(x,0)vs. x 0.4 0.2 0-3 -2-1 0 1 2 3 x [b] (ii) Remember that any pulse or wave traveling in the positive x-direction can be expressed as y(ωt kx), for k 0andthatitsspeedofpropagationisv ω/k. Then,lettingz ωt kx and expressing y(x, t) asafunctionofz: b3 y(z) b 2 + z 2 Hence, z 2x ut. Therefore, for positive values of u, thepulsetravelsinthepositivex direction with a speed v u/2. (iii) The transverse velocity is calculated as y/ t: v y (t 0) y t t0 2b3 u(2x ut) (b 2 +(2x ut) 2 ) 2 t0 4b3 xu (b 2 +4x 2 ) 2 The graph can be easily sketched considering that for functions y f(x vt) thetransverse velocity is proportional, and of opposite sign, to the space (x) derivativeofy, y/ x, whichin turn can be drawn from the previous plot: y t v y x
b1, u1 v y (t0) 0.6 0.4 0.2 0-0.2-0.4-0.6-3 -2-1 0 1 2 3 x Indicate that the function tends to zero for x ±,hasazero-crossingatx 0anda minimum and maximum at particular negative and positive values of x, without necessarily specifying the x position (they correspond to the maximum and minimum slope of the tangent to the y(x, 0) function). c Queen Mary, University of London 2011 Page 9 Please turn to next page
B4. Averylongstringofmassdensityµ and tension T is attached to a small hoop with negligible mass. The hoop slides on a greased vertical rod and experiences a small vertical frictional force F fric b y when it moves. t (i) Apply Newton s law to the hoop to find the boundary condition at the end of the string. Express your result in terms of the partial derivatives of y(x, t) at the location of the rod, where x is an axis along the string and y is perpendicular to it (Hint: include the tension and the frictional force) [8] (ii) Show that for an incident pulse f(x vt) and a reflected pulse g(x+ vt), the boundary condition for the hoop at x 0issatisfiedif: g(vt) T bv T + bv f( vt) [10] (iii) Discuss the behaviour of the hoop for b 0, b and b T/v. [7] (i) The following sketch shows the forces acting on the hoop. Applying Newton s second law gives:
F ma mÿ T sinθ + F friction Since the mass of the hoop is negligible and the frictional force is small, the string will form a small angle θ 0(θ would be equal to zero for b 0andm 0otherwisethehoopwould have an infinite acceleration, which is the known boundary condition for an open string), so using small-angle approximation: Since the mass of the hoop is negligible, y x b T mÿ T y x b y t T y x b y t 0 y (at the hoop for all t) t (ii) Let s take the superposition of an incident wave and a reflected wave: y(x, t) f(x vt)+g(x + vt), where the first is the incident, known, and the second is the reflected, unknown. We now use the boundary condition at the hoop to solve for g(x + vt). The respective derivatives are: If the hoop is at x 0,then: y x f (x vt)+g (x + vt) y t v( f (x vt)+g (x + vt)) f ( vt)+g (vt) bv T (f ( vt) g (+vt)) Letting η vt and integrating with respect to η, g (η)dη g (vt) bv/t 1 bv/t +1 f ( vt). bv/t 1 bv/t +1 f ( η)dη g(η) bv T bv + T ( 1)f( η) g(η) T bv T + bv f( η) The integration constant must equal zero for the limiting cases discussed in the next part to hold. (iii) For b 0,thehoopbehavesasafreeend. Ourresultgivesg(η) f( η), which is correct since the wave is reflected without flipping.
For b,thehoopbehavesasaclampedend. Ourresultgivesg(η) f( η), which is correct since the wave is reflected flipped over. Aspecialcaseiswhenb T/v, g(η) 0. Hence,thereisnoreflectedwave. Thisisknownas amatchedload. c Queen Mary, University of London 2011: Dr L.Cerrito Page 12 End of paper