A disc is free to rotate about a smooth horizontal axis passing through its centre of mass. A particle is fixed at the top of the disc. A slight push is given to the disc and it starts rotating. During the process.

Options

A. The amplitude of oscillation in the first case changes by a factor of `sqrt((m)/(m+M))`, whereas in the second case it remains unchanged
B. The final time period of oscillation in both the cases is same
C. The total energy decreases in both the cases
D. The instantaneous speed at `x_(0)` of the combined masses decreases in both the cases

Answer

ABD

Explanation

In case-1,
`Mv_(1)=(M+m)v_(2)`, `v_(2)=((M)/(M+m))v_(1)`
`sqrt((K)/(M+m))A_(2)=((M)/(M+m))sqrt((K)/(M))A_(1)`, `A_(2)=sqrt((M)/(M+m))A_(1)`
In case -2
`A_(2)=A_(1)`, `T=2pisqrt((M+m)/(K))` in both cases.
Total energy decreases in first case whereas remain same in `2^(nd)` case. Instantaneous speed at `x_(0)` decreases in both cases.