BSc/MSci EXAMINATION

PHY-217 Vibrations and Waves

Time Allowed: 2 hours 30 minutes

Date: 4th May, 2011

Time: 14:30 - 17:00

Instructions: Answer ALL questions in section A. Answer ONLY TWO ques-tions from section B. Section A carries 50 marks, each questionin section B carries 25 marks. An indicative marking-scheme isshown in square brackets [ ] after each part of a question. Coursework comprises 20% of the final mark.

Numeric calculators are permitted in this examination. Please state on your answer book thename and type of machine used. Complete all rough workings in the answer book and crossthrough any work which is not to be assessed.

Important Note: The academic regulations state that possession of unauthorised materialat any time when a student is under examination conditions is an assessment o!ence and canlead to expulsion from the college. Please check now to ensure that you do not have any notesin your possession. If you have any then please raise your hand and give them to an invigilatorimmediately. Exam papers cannot be removed from the exam room

You are not permitted to read the contents of this question paper until instructedto do so by an invigilator.

Examiners: Dr L.Cerrito, Dr. A.Brandhuber

c! Queen Mary, University of London 2011

This (reverse cover) page is left blank.

SECTION A. Attempt answers to all questions.

A1. For a particle of mass m executing simple harmonic motion (“SHM”), its displacementfrom equilibrium can be written as:

x(t) = Asin(!0t + ").

(i) Write down the amplitude, angular frequency, period and phase at time t. [4]

(ii) Sketch a graph of x(t). [4]

SOLUTION:(i) amplitude: A, angular frequency: !0, period: T = 2#/!0, phase: !0t + "(ii) sinusoidal graph

A2. A mass m is attached to a massless spring with constant k, and can move horizontally ona frictionless plane without air resistance.

(i) By explicit consideration of the forces involved, derive the di!erential equation ofmotion 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]

SOLUTION:(i) The equation is: mx = "kx # x + !2

0x = 0 with !20 = k/m

general solution: x(t) = Acos(!0t + ") with A, " constants.(ii) !2

0 = k/m(iii) Ep(t) = 1

2kx2(t) = 12kA2cos2(!0t + ") and Ek(t) = 1

2mx2(t) = 12m!2

0A2sin2(!0t + ")

Total energy: Ek + Ep = 12kA2

(iv) An example is: x(t = 0) = x0 and v(t = 0) = v0.

A3. Show that the function y = Acos(4x)+Bsin(4x), where A and B are arbitrary constants,is a general solution of the di!erential equation: [3]

d2y

dx2+ 16y = 0.

SOLUTION:

dy

dx= "4Asin(4x) + 4Bcos(4x) and

d2y

dx2= "16Acos(4x) " 16Bsin(4x)

# d2y

dx2+ 16y = 0

A4. For an object starting from maximum diplacement and released from rest, sketch plotsof 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]

SOLUTION:Q is a dimensionless number defined as: Q = 1/!

1/"0= !0/$, where $ is the damping

constant in damped harmonic motion and !0 is the angular frequency of the undampedoscillator.

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 e!ect high or low oscillator’squality factor Q has on the graph. [4]

SOLUTION:

A6. Consider N equal masses coupled by springs: how many normal modes does such a systemhave? For a normal mode, what characterises the motion of each of the masses? [4]

SOLUTION:A system of N masses coupled by springs has N normal modes. In normal mode vibrationall 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]

SOLUTION: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 thephase velocity. For anomalously dispersive media, the group velocity is higher than thephase velocity.

A8. Consider a string fixed at one end. A Gaussian pulse travels towards the fixed end, whereit is reflected back. Assuming there are no energy losses and that there is no dispersionof the pulse, describe the appearance of the pulse after the reflection, and explain why itlooks as you described. [4]

SOLUTION: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 ofsound in air is about 340 ms"1. Show that the corresponding range in wavelengths is11.3 m to 2.27 cm. [5]

SOLUTION:

v =!

k=

f

2#

2#

%# %L =

v

f=

340

30m = 11.3 m

%H =v

f=

340

15 · 103= 0.022 m

A10. For a taut string attached at both ends, draw a sketch of the first, second and thirdharmonic, indicating the nodes and the anti-nodes. [5]

SOLUTION:

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 springconstant k.(i) By explicit consideration of the forces and considering only motion in the verticaldirection, derive the equations of motion of the two masses. Do you need to considergravitational forces? [15](ii) Show that the angular frequencies of the two normal modes are given by !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 m1 and m2 be x1 andx2 respectively. Then the tensions in the two strings are T1 = kx1 and T2 = k(x2 " x1)repectively. We need not to consider the gravitational forces acting on the masses, becausethey are independent of the displacements and hence do not contribute to the restoring forcesthat cause the oscillations. The gravitational forces merely cause a shift in the equilibriumpositions of the masses, and you do not have to find what those shifts are. Now

m1x1 = +k(x2 " x1) " kx1

m2x2 = "k(x2 " x1)

substituting m1 = m2 = m and !2s = k/m we get:

x1 = !2s(x2 " 2x1)

x2 = !2s(x1 " x2)

(ii) Let: x1 = C1cos(!t) and x2 = C2cos(!t) and substituting above:

"!2C1 + 2!2sC1 = !2

sC2

"!2C2 + !2sC2 = !2

sC1

These can be solved directly or using Cramer’s Rule. In the first case:

C1

C2=

!2s

2!2s " !2

=!2

s " !2

!2s

!4s = 2!4

s " 3!2s!

2 + !4

!4 " 3!2s!

2 + !4s = 0

!2 =3!2

s ±!

9!4s " 4!4

s

2=

"3 ±

$5# !2

s

2

!2 ="3 ±

$5# k

2m

!+

!"=

$%%&3 +$

5

3 "$

5=

$%%&3 +$

5

3 "$

5· 3 +

$5

3 +$

5=

3 +$

5

2·$

5 " 1$5 " 1

=

$5 + 1$5 " 1

(iii) For !+ =!

(3 +$

5)k/2m:

C1

C2=

!2s

2!2s " !2

+=

2!2s

4!2s " (3 +

$5)!2

s

=2

1 "$

5

For !" =!

(3 "$

5)k/2m:

C1

C2=

!2s

2!2s " !2

"=

2!2s

4!2s " (3 "

$5)!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 M1 on a frictionless plane connected to support O by a spring ofsti!ness k. The mass M2 is supported by a string of length l from M1. OA is the lengthof the relaxed spring. x1 and x2 are the positions of M1 and M2, respectively, relative topoint A.

(i) Assuming small angles &, write down the di!erential equation of motion for M1. [8](ii) Assuming small angles &, write down the di!erential equation of motion for M2. [7](iii) Now let O oscillate harmonically in the horizontal direction, driven by an externalforce, and its position X(t) is given by X0cos(!t). Let both masses be equal to M . Writedown the di!erential equation of motion for each mass. [4]

(iv) What are the amplitudes (steady state) of the two masses as a function of k, M , X0,l and ! ? (Hint: use Cramer’s rule) [3]

(v) There is one frequency for which one mass stands still and the other oscillates. Whatis that frequency? What is peculiar about this state? [3]

SOLUTION

(i) The equation of motion for mass M1 is:

M1x1 = "kx1 + M2g

l(x2 " x1)

(ii) and for mass M2 is:

M2x2 = "M2gsin(&)

# M2x2 = "M2g

l(x2 " x1)

(iii) The equation of motion for mass M2 in unchanged, whereas for mass M1 is:

M1x1 = "k[x1 " X(t)] + M2g

l(x2 " x1)

M1x1 + kx1 + M2g

l(x1 " x2) = kX0cos(!t)

(iv) Substituting !2s = k/M2, !2

p = g/l and M1 = M2 = M we get:

x2 + !2px2 " !2

px1 = 0

x1 + (!2s + !2

p)x1 " !2px2 = !2

sX0cos(!t)

Let: x1 = C1cos(!t) and x2 = C2cos(!t), and substituting above:

!2pC1 + (!2 " !2

p)C2 = 0

("!2 + !2s + !2

p)C1 " !2pC2 = !2

sX0

C1 =

'''''0 !2 " !2

p

!2sX0 "!2

p

''''''''''

!2p !2 " !2

p

"!2 + !2p + !2

s "!2p

'''''

=kX0(g " l!2)

Ml!4 " (2Mg + kl)!2 + kg

C2 =

'''''!2

p 0"!2 + !2

p + !2s !2

sX0

''''''''''

!2p !2 " !2

p

"!2 + !2p + !2

s "!2p

'''''

=kgX0

Ml!4 " (2Mg + kl)!2 + kg

These are the steady state solutions. The general solution is a linear combination between thetransient solution and the steady state solutions.

(v) We can note from the functional form of C2 that it cannot have the value zero (except for! % &). However, C1 will be zero when: g " l!2 = 0:

! = !p =(

g

lC1 % 0

This is the resonance frequency of the pendulum. Thus, at this frequency, the two horizontalforces on the upper mass, kX0cos(!t) and T sin(&) cancel. So how can the pendulum swing ifthe mass attached to the spring does not move at all: what is driving the pendulum? This isonly possible in the approximation of zero damping. In the presence of damping, no matterhow little, the peculiar state is unstable. This can easily be seen as follows: assume that thesystem is in that state. That means that at any moment in time the net horizontal force onthe upper mass is zero. Thus the vectorial sum of the spring force and T sin(&) must be zero.However, if the mass on the spring is not moving, the pendulum is no longer driven, and thusits amplitude will decay, and the net force on the mass on the spring is no longer zero, andthus 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. A pulse travelling along a stretched string is described by the following equation:

y(x, t) =b3

b2 + (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 (vy) of a given point of the string as a function ofx for the instant t = 0, and sketch a graph of vy(t = 0) against x. [8]

SOLUTION:(i) Indicate with the sketch that the function y(x, 0) is symmetric around x = 0, tends to zerofor x % ±& and has a maximum at x = 0. The sketch of y(x, 0) is shown below:

x [b]-3 -2 -1 0 1 2 3

y(x,

0) (i

n un

its o

f b)

0

0.2

0.4

0.6

0.8

1

y(x,0)vs. x

(ii) Remember that any pulse or wave traveling in the positive x-direction can be expressed asy(!t" kx), for k ' 0 and that its speed of propagation is v = !/k. Then, letting z = !t " kxand expressing y(x, t) as a function of z:

y(z) =b3

b2 + z2

Hence, z = 2x " ut. Therefore, for positive values of u, the pulse travels in the positive xdirection with a speed v = u/2.

(iii) The transverse velocity is calculated as 'y/'t:

vy(t = 0) ='y

't

'''''t=0

=2b3u(2x" ut)

(b2 + (2x " ut)2)2

'''''t=0

=4b3xu

(b2 + 4x2)2

The graph can be easily sketched considering that for functions y = f(x " vt) the transversevelocity is proportional, and of opposite sign, to the space (x) derivative of y, 'y/'x, which inturn can be drawn from the previous plot:

'y

't= "v

'y

'x

x-3 -2 -1 0 1 2 3

(t=0)

yv

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

b=1, u=1

Indicate that the function tends to zero for x % ±&, has a zero-crossing at x = 0 and aminimum and maximum at particular negative and positive values of x, without necessarilyspecifying the x position (they correspond to the maximum and minimum slope of the tangentto the y(x, 0) function).

c! Queen Mary, University of London 2011 Page 9 Please turn to next page

B4. A very long string of mass density µ and tension T is attached to a small hoop withnegligible mass. The hoop slides on a greased vertical rod and experiences a small verticalfrictional force Ffric = "b#y

#t when it moves.

(i) Apply Newton’s law to the hoop to find the boundary condition at the end of thestring. Express your result in terms of the partial derivatives of y(x, t) at the location ofthe rod, where x is an axis along the string and y is perpendicular to it (Hint: includethe tension and the frictional force) [8]

(ii) Show that for an incident pulse f(x"vt) and a reflected pulse g(x+vt), the boundarycondition for the hoop at x = 0 is satisfied if:

g(vt) =T " bv

T + bvf("vt)

[10]

(iii) Discuss the behaviour of the hoop for b % 0, b % & and b = T/v. [7]

SOLUTION:(i) The following sketch shows the forces acting on the hoop.

Applying Newton’s second law gives:

F = ma

"my = "T sin& + Ffriction

Since the mass of the hoop is negligible and the frictional force is small, the string will form asmall angle & ( 0 (& would be equal to zero for b = 0 and m = 0 otherwise the hoop wouldhave an infinite acceleration, which is the known boundary condition for an open string), sousing small-angle approximation:

"my = "T'y

'x" b

'y

't

Since the mass of the hoop is negligible,

"T'y

'x" b

'y

't= 0

# 'y

'x= " b

T

'y

't(at the hoop for all 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 usethe boundary condition at the hoop to solve for g(x + vt). The respective derivatives are:

'y

'x= f #(x " vt) + g#(x + vt)

'y

't= v("f #(x " vt) + g#(x + vt))

If the hoop is at x = 0, then:

f #("vt) + g#(vt) =bv

T(f #("vt) " g#(+vt))

g#(vt) =bv/T " 1

bv/T + 1f #("vt).

Letting ( = vt and integrating with respect to (,

)g#(()d( =

) bv/T " 1

bv/T + 1f #("()d(

g(() =bv " T

bv + T("1)f("()

g(() =T " bv

T + bvf("()

The integration constant must equal zero for the limiting cases discussed in the next part tohold.

(iii) For b = 0, the hoop behaves as a free end. Our result gives g(() = f("(), which is correctsince the wave is reflected without flipping.

For b % &, the hoop behaves as a clamped end. Our result gives g(() = "f("(), which iscorrect since the wave is reflected flipped over.A special case is when b = T/v, g(() = 0. Hence, there is no reflected wave. This is known asa matched load.

c! Queen Mary, University of London 2011: Dr L.Cerrito Page 12 End of paper