Two identical balls `A and B` each of mass `0.1 kg` are attached to two identical mass less is springs. The spring-mass system is constrained to move inside a right smooth pipe bent in the form of a circle as shown in figure . The pipe is fixed in a horizontal plane. The center of the balls can move in a circle of radius `0.06 m`. Each spring has a natural length of `0.06 pi m` and force constant `0.1 N//m`. Initially, both the balls are displaced by an angle `theta = pi//6` radians with respect to diameter PQ of the circle and released from rest.
a. Calculate the frequency of oscillation of the ball B.
b. What is the total energy of the system ?
c. Find the speed of the ball A when A and B are at the two ends of the diameter PQ.

Explanation

At any moment when angular displacement of the ball from equilibrium is `theta` total energy of the system is given by

`E =2 ((1)/(2) mv^(2)) + 2(1)/(2) k (2R theta)^(2) = mv^(2) + 4KR^(2) theta^(2)`
`:.` E = constant, hence `(dE)/(dt) = 0`
`rArr 0 = 2mv (dv)/(dt) + BKR^(2) theta (d theta)/(dt) = mv (dv)/(dt) + 4KR theta ((d theta)/(dt) R)`
Putting `R theta = x, and (d theta)/(dt) R = v`
`0 = mv (dv)/(dt) + 4kxv`
`rArr (d^(2)x)/(dt^(2)) = - (4K)/(m)x`
Hence `T = 2pi sqrt((m)/(4k)) = pi sqrt((m)/(k))`, Hence `f = (1)/(T) = (1)/(pi) Hz`