A mass `M` attached to a horizontal spring executes `SHM` with an amplitude `A_(1)`. When mass `M` passes through its mean position a smaller mass `m` is placed over it and both of them move togther with amplitude `A_(2)`. Ratio of `((A_(1))/(A_(2)))` is:

Options

A. `(M+m)/(M)`
B. `((M)/(M+m))^(1//2)`
C. `((M+m)/(M))^(1//2)`
D. `(M)/(M+m)`

Answer

C

Explanation

At mean position, `F_("net")=0`
`therefore` By conservation of linear momentum,
`Mv_(1)=(M+m)v_(L)`,
`M omega_(1)A_(1)=(M+m)omega_(2)A_(2)`
`omega_(1)=sqrt((k)/(m))and omega_(2) = sqrt((k)/(M+m))`
On solving, `(A_(1))/(A_(2))=(sqrt((M+m)/(M)))^(1//2)`