A mass `m_(1)` connected to a horizontal spring performs a S.H.M. with an amplitude A. While the mass `m_(1)` is pas"sin"g through the mean position, another mass `m_(2)` is placed on it so that both the masses move together with amplitude `A_(1)`. The ratio of `(A_(1))/(A) " is " (m_(2) lt m_(1))`.

Options

A. `[(m_(1))/(m_(1)+m_(2))]^(1//2)`
B. `[(m_(1)+m_(2))/(m_(1))]^(1//2)`
C. `[(m_(2))/(m_(1)+m_(2))]^(1//2)`
D. `[(m_(1)+m_(2))/(m_(2))]^(1//2)`

Answer

A

Explanation

When the S.H.M. particle of mass `m_(1)` passes through the mean position, it has maximum K.E.
`K_("max")=(1)/(2) m_(1)omega^(2)A^(2)` where A is the amplitude.
In the second case, `K_("max")=(1)/(2)(m_(1)+m_(2))omega^(2)A_(1)^(2)`
then `(1)/(2) m_(1)omega^(2)A^(2)=(1)/(2)(m_(1)+m_(2))omega^(2)A_(1)^(2)`
`therefore m_(1)A^(2)=(m_(1)+m_(2))A_(1)^(2)`
`therefore (A_(1))/(A)=((m_(1))/(m_(1)+m_(2)))^(1//2)`