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A mass M attached to a horizontal spring...

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:

A

`((M)/(M+m))^(1//2)`

B

`((M+m)/(M))^(1//2)`

C

`(M)/(M+m)`

D

`(M+m)/(M)`

Text Solution

Verified by Experts

The correct Answer is:
B

`n_(1) = (1)/(2pi)sqrt((k)/(m))"……"(i)`
`n_(2) = (1)/(2pi)sqrt((k)/(M + m))"……"(ii)`
according to conservation of linear momentum
`Mv_(1) = (M + m)v_(2) rArr M_(1)A_(1)omega_(1) = (M + m) A_(2)omega_(2)`
From equation (i) & (ii)
`(A_(1))/(A_(2)) = ((M + m)/(M)).(omega_(2))/(omega_(2)) = ((M + m)/(M)) sqrt((M)/(M + m)) = sqrt((M + m)/(M))`
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