Two identical balls A and B , each of mass 0.1 kg , are attached to two identical massless springs. The spring - mass system is constrained to move inside a rigid smooth pipe of bent in the form of circle , as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.03 m . Each spring has a natural length of `0.06pi` m and force constant `0.1 Nm^(-1)` . Initially , both the balls are displaced by an angle `theta=(pi)/6` rad , with respect to the diameter PQ of the circle , and released from rest . The frequency of oscillation of the ball B is

Answer

A, B, D

Explanation

(i) This system can be reduced to

Where `mu = (m_(1)m_(2))/(m_(1) + m_(2)) = ((0.1)(0.1))/(0.1+0.1) = 0.05 kg`
and `k_(eq) = k_(1) + k_(2) = 0.1 + 0.1 = 0.2 Nm^(-1)`
`rArr f = (1)/(2pi) sqrt((k_(eq.))/(mu)) = 1/2pi sqrt((0.2)/(0.05)) = 1/(pi) Hz`
(ii) Compression in one spring is equal to extension in other spring
`= 2Rtheta = 2(0.6)(pi/6) = (pi)/(50)m`
`= 2Rtheta = 2(0.06)(pi)/(6) = (pi)/(50)m`
Total energy of the system
`e = 1/2k_(1)(2Rtheta)^(2) +1/2k_(2)(2Rtheta)^(2)`
`= k(2Rtheta)^(2) = (0.1)(pi/5)^(2) = 4pi xx 10^(-5)J`
(iii) From mechanical energy conservation
`1/2m_(1)v_(1^(2)) + (1)/(2)m_(2)v^(2^(2)) = E`
`rArr 0.1v^(2) = 4pi^(2) xx 10^(-5) rArr v = 2pi xx 10^(-2)ms^(-1)`