Assignment 6 Solutions 4u136f

Vibrations and Waves MP205, Assignment 6 Solutions 1. Two identical pendulums are connected by a light coupling spring. Each pendulum 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 two normal modes? (b) What is the time interval between successive maximum possible amplitudes of one pendulum after one pendulum is drawn aside and released? (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,pFA are −mω02 xA −kxA , where ω0 is the natural pendulum frequency given by ω0 = g/`. The EOM for mass A is d 2 xA d 2 xA d 2 xA 2 2 2 + mω + ω + (ω02 + ωC2 )xA = 0 x + kx = 0 x + ω x = 0 A A A A 0 0 C dt2 dt2 dt2 p where ωc = k/m is the coupling frequency. We note that the above equation is 2 now in the form d2 x/dt = −ω 2 x and p p we can read the new angular frequency of 0 2 2 the system to be ω = ω0 + ωC = g/` + k/m. A solution to the above EOM is xA = D cos(ω 0 t). We are given the period of this system to be 1.25 sec, therefore

m

TA =

2π = 1.25 ω0

2π 1.25 = p 2 ω0 + ωC2

⇒ ω02 + ωC2 = 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 between them at allptimes. So it is like they are uncoupled and we have their natural frequency ω0 = g/`. So the EOM for mass A is simply d 2 xA + ω02 xA = 0 dt2

and a solution is xA = C p cos(ω0 t). So we know that the first normal angular frequency is just ω1 = ω0 = g/` and hence the first normal period is s r 2π ` 0.4 T1 = = 2π = 2π = 1.27 s ω1 g 9.8 (ii) The second normal mode occurs when both masses are free and move in opposite 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, FA are −mω02 xA − 2kxA and the EOM for A is just m

d 2 xA + mω02 xA + 2kxA = 0 dt2

d 2 xA + (ω02 + 2ωC2 )xA = 0 dt2

Again we directly read off the angular frequency of this is the angular p system, it p frequency of the second normal mode given by ω2 = ω02 + 2ωC2 = g/` + 2k/m. Also a solution to this EOM is xA = D cos(ω2 t) and xB = −D cos(ω2 t) for B since it is just the mirror image. We find the period of the second normal mode. It is T2 =

2π 2π =p 2 ω2 ω0 + 2ωC2

We already know that ω02 + ωC2 = 25.2661, so ωC2 = 25.2661 − ω02 = 25.661 − g/` = 0.7661. So we have T2 =

2π = 1.23 s 5.102

(b) This is the case described in the first diagram. A is set into motion, pulling and pushing B. Spring also pulls + pushes A, sometimes helping, sometimes hindering its motion. The motion is like beats of two SHMs of same amplitude and different frequency. The time between successive maximum amplitudes is just the beat period. T2 T1 (1.27)(1.23) 2π 2π = T = = = = 39 s |ω1 − ω2 | 2π|1/T1 − 1/T2 | T2 − T1 1.27 − 1.23 2. Two equal masses on an effectively frictionless horizontal air track are held between rigid s by three identical springs, as shown. The displacements from equilibrium along the line of the springs are described by 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 normal modes of the system? Sketch graphs of xA and xB versus t in each mode. At t = 0, mass A is at its normal resting position and mass B is pulled aside 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 as a function of time. (c) What length of time (in seconds) characterises the periodic transfer of the motion from B to A and back again? (a) When one is clamped the restoring force on the other is F = −2kx since the spring is stretched a distance 2x. The EOM is then just d2 x + 2ω 2 x = 0 dt2 p 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 restoring forces on A, FA are −kxA − k(xA − xB ) and the restoring forces on B, FB are −kxB − k(xB − xA ). So the equations of motion are m

d2 x + 2kx = 0 dt2

d 2 xA m 2 + kxA + k(xA − xB ) = 0 dt

d 2 xB m 2 + kxB + k(xB − xA ) = 0 dt

Adding both together gives the first normal mode angular frequency. (occurs when xA = xB ) d2 d2 q1 (x + x ) + k(x + x ) = 0 + ω12 q1 = 0 A B A B dt2 dt2 p where ω1 = k/m and q1 = xA + xB . Subtracting one from the other gives the second normal mode frequency. (occurs when xA = −xB ) m

d2 q2 d2 (x − x ) + 3k(x − x ) = 0 + ω22 q2 = 0 A B A B dt2 dt2 p where ω2 = 3k/m and q2 = xA − xB . The normal mode periods are then simply r √ √ 2π m = 2π = T1 = 2T = 3 2 s ω1 k r r r p r √ 2π m/3k m T2 2 2 2 2π T2 = = 2π = p = T2 = T = ·3= 6 s ω2 3k T 3 3 3 2π m/2k m

(b) Had xA = 12 (q1 + q2 ) = 12 A1 cos(ω1 t + δ1 ) + 21 A2 cos(ω2 t + δ2 ). Also xb = 21 (q1 − q2 ) = 12 A1 cos(ω1 t + δ1 ) − 21 A2 cos(ω2 t + δ2 ). At t = 0, A1 = 0 cm, A2 = 5 cm (=B0 ). Told released from rest so xA = 0, xB = 0, dxA = 0, dxdtB = 0. Filling in these conditions we get that: dt ω1 ω2 dxAB = ± A1 sin(ω1 t + δ1 ) ± A2 sin(ω2 t + δ2 ) dt 2 2 which must equal 0 due to boundary conditions. Set δ1 = δ2 = 0. 1 xA = 0 at t = 0 ⇒ A + 21 A2 = 0. 2 1 1 A − 12 A2 = B0 . xB = 0 at t = 0 ⇒ 2 1 Adding both equations gives A1 = B0 and subtracting one from the other gives A2 = −B0 . Therefore: 1 B0 [cos(ω1 t) − cos(ω2 t)] 2 1 = B0 [cos(ω1 t) + cos(ω2 t)] 2

xA = xB

Using trigonometric identities we can rewrite them as: ω1 − ω2 ω1 + ω2 xA = −B0 sin t sin t 2 2 ω1 − ω2 ω1 + ω2 xB = B0 cos t cos t 2 2 (c) The time taken is just the time between successive maximum amplitudes which is the beat period. √ √ √ T2 T1 (3 2)( 6) 3 2π 3 2π = √ √ = √ = = T = 6−3 2 1− 3 |ω1 − ω2 | 2π|1/T1 − 1/T2 | T2 − T1 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 springs have spring constant k0 . If B is clamped, A vibrates at a frequency νA of 1.81 sec−1 . The frequency ν1 of the lower normal mode is 1.14 sec−1 .

(a) Satisfy yourself that the equations of motion of A and B are d 2 xA = −k0 xA − kc (xA − xB ) dt2 d 2 xB m 2 = −k0 xB − kc (xB − xA ) dt m

p (b) Putting ω0 = k0 /m, show that the angular frequencies ω1 and ω2 of the normal modes are given by ω1 = ω0

,

ω2 = [ω02 + (2kc /m)]1/2 ,

and that the angular frequency of A when B is clamped (xB = 0 always) is given by ωA = [ω02 + (kc /m)]1/2 . (c) Using the numerical data above, calculate the expected frequency (ν2 ) 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 spring constants. (a) Look at the restoring forces on both masses. Restoring forces on A, FA are −k0 xA −kc (xA −xB ) and on B are −k0 xB −kc (xB −xA ), therefore we get their equations of motion to be: d2 xA m 2 = −k0 xA − kc (xA − xB ) dt d2 xB m 2 = −k0 xB − kc (xB − xA ) dt (b) Use simultaneous equations of EOMs to find the normal mode angular frequencies. Adding both together gives the first normal mode angular frequency. (occurs when xA = x B ) d2 q1 d2 (x + x ) + k (x + x ) = 0 + ω12 q1 = 0 A B 0 A B dt2 dt2 p where ω1 = ω0 = k0 /m and q1 = xA + xB . Subtracting one from the other gives the second normal mode frequency. (occurs when xA = −xB ) d2 d2 q2 k0 2kc m 2 (xA − xB ) + k0 (xA − xB ) + 2kc (xA − xB ) = 0 + + q2 = 0 dt dt2 m m d2 q 2 ⇒ + ω22 q2 = 0 dt2 p where ω2 = (ω02 + 2kc /m) and q2 = xA − xB . When B is clamped, xB = 0 and therefore the EOM for A becomes: d2 xA d2 xA k0 kc m 2 + k0 xA + kc xA = 0 ⇒ + + xA = 0 dt dt2 m m r r k0 kc kc So: ωA = + = ω02 + . m m m m

(c) We know that νA = 1.81 sec−1 and ν1 = 1.14 sec−1 .

p ω02 + kc /m 1 ωA kc νA = = = ⇒ ω02 + = 129.335 TA 2π 2π m 1 ω0 kc ν1 = = ⇒ ω0 = 2π(1.14) = 7.1628 s−1 ⇒ = 78.0292 T1 2π m p ω02 + 2kc /m ω2 1 = = = 2.29 s−1 ν2 = T2 2π 2π p (c) We know that ω0 = k0 /m ⇒ k0 = mω02 . Also kc /m = 78.0292 so: kc 78.0292m = = 1.52 k0 mω02