Assignment 6 Solutions

7/16/2019 Assignment 6 Solutions

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Vibrations and Waves MP205, Assignment 6 Solutions

1. Two identical pendulums are connected by a light coupling spring. Eachpendulum has a length of 0.4 m, and they are at a place where g =9.8 ms−2. With the coupling spring connected, one pendulum is clamped

and the period of the other is found to be 1.25 sec exactly.

(a) With neither pendulum clamped, what are the periods of the twonormal modes?

(b) What is the time interval between successive maximum possible am-plitudes of one pendulum after one pendulum is drawn aside andreleased?

(a) We find the EOM for mass A when B is clamped and A is free as in the figure.

The restoring forces on A, F A are −mω2

0xA−kxA, where ω0 is the natural pendulum

frequency given by ω0 =

g/. The EOM for mass A is

md2xA

dt2+ mω2

0xA + kxA = 0

d2xA

dt2+ ω2

0xA + ω2

C xA = 0d2xA

dt2+ (ω2

0+ ω2

C )xA = 0

where ωc =

k/m is the coupling frequency. We note that the above equation isnow in the form d2x/dt2 = −ω2x and we can read the new angular frequency of the system to be ω =

ω20

+ ω2

C =

g/ + k/m. A solution to the above EOM isxA = D cos(ωt).We are given the period of this system to be 1.25 sec, therefore

T A =2π

ω= 1.25 1.25 =

2π ω2

0+ ω2

C

⇒ ω2

0+ ω2

C = 25.2661

(i) We now can study the normal modes. The first one occurs when both masses

are free and move in the same direction as in the figure below.

In this case it is clear that the spring exerts no force on either mass, we have thesame distance between them at all times. So it is like they are uncoupled and we

have their natural frequency ω0 =

g/. So the EOM for mass A is simply

d2xA

dt2+ ω2

0xA = 0



and a solution is xA = C cos(ω0t). So we know that the first normal angularfrequency is just ω1 = ω0 =

g/ and hence the first normal period is

T 1 =2π

ω1

= 2π

g= 2π

0.4

9.8= 1.27 s

(ii) The second normal mode occurs when both masses are free and move in oppo-site directions as in the figure below.

In this case we see that the spring is stretched a distance 2x (x on each side) and

so the restoring forces on A, F A are −mω20xA − 2kxA and the EOM for A is just

md2xA

dt2+ mω2

0xA + 2kxA = 0

d2xA

dt2+ (ω2

0+ 2ω2

C )xA = 0

Again we directly read off the angular frequency of this system, it is the angularfrequency of the second normal mode given by ω2 =

ω20

+ 2ω2

C =

g/ + 2k/m.Also a solution to this EOM is xA = D cos(ω2t) and xB = −D cos(ω2t) for B sinceit is just the mirror image.We find the period of the second normal mode. It is

T 2 = 2πω2

= 2π ω20

+ 2ω2

C

We already know that ω2

0+ ω2

C = 25.2661, so ω2

C = 25.2661 − ω2

0= 25.661 − g/ =

0.7661. So we have

T 2 =2π

5.102= 1.23 s

(b) This is the case described in the first diagram.

A is set into motion, pulling and pushing B. Spring also pulls + pushes A, sometimeshelping, sometimes hindering its motion. The motion is like beats of two SHMs of same amplitude and different frequency. The time between successive maximumamplitudes is just the beat period.

T =2π

|ω1 − ω2| =2π

2π|1/T 1 − 1/T 2| =

T 2T 1T 2 − T 1

=(1.27)(1.23)

1.27 − 1.23= 39 s

2. Two equal masses on an effectively frictionless horizontal air track areheld between rigid supports by three identical springs, as shown. Thedisplacements from equilibrium along the line of the springs are describedby coordinates xA and xB, as shown. If either of the masses is clamped,the period T for one complete vibration of the other is 3 sec.



(a) If both masses are free, what are the periods of the two normalmodes of the system? Sketch graphs of xA and xB versus t in eachmode.

At t = 0, mass A is at its normal resting position and mass B is pulledaside a distance of 5 cm. The masses are released from rest at this instant.

(b) Write an equation for the subsequent displacement of each mass asa function of time.

(c) What length of time (in seconds) characterises the periodic transferof the motion from B to A and back again?

(a) When one is clamped the restoring force on the other is F = −2kx since thespring is stretched a distance 2x.

The EOM is then just

md2x

dt2+ 2kx = 0

d2x

dt2+ 2ω2x = 0

We are told that the period is 3 sec so T = 2π/ω = 2π

m/2k = 3 s.We now find the periods of the two normal modes. When coupled the restoringforces on A, F A are

−kxA

−k(xA

−xB) and the restoring forces on B, F B are

−kxB − k(xB − xA). So the equations of motion are

md2xA

dt2+ kxA + k(xA − xB) = 0 m

d2xB

dt2+ kxB + k(xB − xA) = 0

Adding both together gives the first normal mode angular frequency. (occurs whenxA = xB)

md2

dt2(xA + xB) + k(xA + xB) = 0

d2q 1dt2

+ ω2

1q 1 = 0

where ω1 =

k/m and q 1 = xA + xB. Subtracting one from the other gives thesecond normal mode frequency. (occurs when xA = −xB)

md2

dt2(xA − xB) + 3k(xA − xB) = 0

d2q 2dt2

+ ω2

2q 2 = 0

where ω2 =

3k/m and q 2 = xA − xB.The normal mode periods are then simply

T 1 =2π

ω1

= 2π

m

k=

√ 2T = 3

√ 2 s

T 2 = 2πω2

= 2π

m3k

T 2T

= 2π

m/3k

2π

m/2k=

23

T 2 =

23

T =

23· 3 = √ 6 s



(b) Had xA = 1

2(q 1 + q 2) = 1

2A1 cos(ω1t + δ 1) + 1

2A2 cos(ω2t + δ 2).

Also xb = 1

2(q 1 − q 2) = 1

2A1 cos(ω1t + δ 1) − 1

2A2 cos(ω2t + δ 2).

At t = 0, A1 = 0 cm, A2 = 5 cm (=B0). Told released from rest so xA = 0, xB = 0,dxAdt

= 0, dxBdt

= 0. Filling in these conditions we get that:

dxAB

dt= ±ω1

2A1 sin(ω1t + δ 1) ± ω2

2A2 sin(ω2t + δ 2)

which must equal 0 due to boundary conditions. Set δ 1 = δ 2 = 0.xA = 0 at t = 0 ⇒ 1

2A1 + 1

2A2 = 0.

xB = 0 at t = 0 ⇒ 1

2A1 − 1

2A2 = B0.

Adding both equations gives A1 = B0 and subtracting one from the other givesA2 = −B0. Therefore:

xA =1

2B0[cos(ω1t) − cos(ω2t)]

xB = 12

B0[cos(ω1t) + cos(ω2t)]

Using trigonometric identities we can rewrite them as:

xA = −B0

sin

ω1 − ω2

2

t sin

ω1 + ω2

2

t

xB = B0

cos

ω1 + ω2

2

t cos

ω1 − ω2

2

t

(c) The time taken is just the time between successive maximum amplitudes whichis the beat period.

T =2π

|ω1 − ω2| =2π

2π|1/T 1 − 1/T 2| =

T 2T 1T 2 − T 1

=

(3√

2)(√

6)√ 6 − 3

√ 2

=3√

3

1 −√ 3

3. Two objects, A and B, each of mass m, are connected by springs as shown.The coupling spring has a spring constant kc, and the other two springshave spring constant k0. If B is clamped, A vibrates at a frequency ν A of

1.81 sec

−1

. The frequency ν 1

of the lower normal mode is 1.14 sec

−1

.

(a) Satisfy yourself that the equations of motion of A and B are

md2xA

dt2= −k0xA − kc(xA − xB)

md2xB

dt2

=

−k0xB

−kc(xB

−xA)



(b) Putting ω0 =

k0/m, show that the angular frequencies ω1 and ω2 of the normal modes are given by

ω1 = ω0 , ω2 = [ω2

0+ (2kc/m)]1/2,

and that the angular frequency of A when B is clamped (xB = 0

always) is given by

ωA = [ω2

0+ (kc/m)]1/2.

(c) Using the numerical data above, calculate the expected frequency(ν 2) of the higher normal mode. (The observed value was 2.27 sec−1).

(d) From these same data calculate the ratio kc/k0 of the two springconstants.

(a) Look at the restoring forces on both masses.

Restoring forces on A, F A are −k0xA−kc(xA−xB) and on B are −k0xB−kc(xB−xA),therefore we get their equations of motion to be:

md2xA

dt2= −k0xA − kc(xA − xB)

md2xB

dt2= −k0xB − kc(xB − xA)

(b) Use simultaneous equations of EOMs to find the normal mode angular frequen-cies.

Adding both together gives the first normal mode angular frequency. (occurs whenxA = xB)

md2

dt2(xA + xB) + k0(xA + xB) = 0

d2q 1dt2

+ ω2

1q 1 = 0

where ω1 = ω0 =

k0/m and q 1 = xA + xB.Subtracting one from the other gives the second normal mode frequency. (occurswhen xA = −xB)

m d2

dt2(xA − xB) + k0(xA − xB) + 2kc(xA − xB) = 0 d

2

q 2dt2

+

k0

m+ 2kc

m

q 2 = 0

⇒ d2q 2dt2

+ ω2

2q 2 = 0

where ω2 =

(ω2

0+ 2kc/m) and q 2 = xA − xB.

When B is clamped, xB = 0 and therefore the EOM for A becomes:

md2xA

dt2+ k0xA + kcxA = 0 ⇒ d2xA

dt2+

k0

m+

kcm

xA = 0

So: ωA =

k0

m + kcm =

ω20 + kcm .

(c) We know that ν A = 1.81 sec−1 and ν 1 = 1.14 sec−1.



ν A =1

T A=

ωA

2π=

ω20

+ kc/m

2π⇒ ω2

0+

kcm

= 129.335

ν 1 =1

T 1=

ω0

2π⇒ ω0 = 2π(1.14) = 7.1628 s−1 ⇒ kc

m= 78.0292

ν 2 =1

T 2=

ω2

2π

= ω2

0+ 2kc/m

2π

= 2.29 s−1

(c) We know that ω0 =

k0/m ⇒ k0 = mω2

0. Also kc/m = 78.0292 so:

kck0

=78.0292m

mω2

0

= 1.52

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