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A mass M, attached to a horizontal sprin...

A mass `M`, attached to a horizontal spring, excutes `SHM` with a amplitude `A_(1)`. When the mass `M` passes through its mean position then a smaller mass `m` is placed over it and both of them move together with amplitude `A_(2)`, the ratio of `((A_(1))/(A_(2)))` is :

A

`(M)/( M + m)`

B

`(M + m)/(M)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
The net force becomes zero at the mean point .
Hence , linear momentum, must be conserved
`Mv_(1) = (M + m) v_(2) rArr MA_(1) sqrt(( k)/(M)) = (M + m) A_(2) sqrt(( k)/( m + M))`
`rArr A_(2) = MA_(1) sqrt(( k)/(M)) . (1)/( M + m) sqrt(( m + m)/( k))` [ as ` v = A omega = a sqrt( k // m)` ]
`rArr A_(2) = A_(1) sqrt(( M)/(( m + M))) rArr (A_(1))/( A_(2)) = ((m + M)/( M))^((1)/(2))`
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