Vibrations and Waves MP205, Assignment 6 Solutions 1. Two iden tical pendulums are connected by a light coupling spring. Each pendulum ha s a leng th of 0.4 m, an d th ey ar e at a pl ace wh er e g = 9.8 ms −2 . With the couplin g 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 two normal modes? (b) What is the time interv al between successive maximum possible am- plitudes of one pendulum after one pendulum is drawn aside and released? (a) We find the EOM for mas s A when B is clamped and A is free as in the figure. The restoring forces on A,FA ar e −mω 2 0 x A − kx A , where ω 0 is the natural pendulum frequency given by ω 0 = g/. The EOM for mass A is m d 2 x A dt 2 +mω 2 0 x A +kx A = 0 d 2 x A dt 2 + ω 2 0 x A + ω 2 Cx A = 0 d 2 x A dt 2 + (ω 2 0 + ω 2 C)x A = 0 where ω c = k/m is the coupling frequenc y . We note that the above equat ion is now in the form d 2 x/dt 2 = −ω 2 x and we can read the new angular frequency ofthe system to beω = ω 2 0 + ω 2 C= g/+k/m. A solution to the above EOM is x A =D cos(ω t). We are given the period of this system to be 1.25 sec, therefore TA = 2π ω = 1.25 1.25 = 2π ω 2 0 + ω 2 C⇒ ω 2 0 + ω 2 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 the same distance betw een them at all times. So it is like they are uncoup led and we have their natural frequency ω 0 = g/. So the EOM for mass A is simply d 2 x A dt 2 + ω 2 0 x A = 0
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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
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.
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