Two masses `m_(1)` and `m_(2)` are suspeded togther by a massless spring of spring constnat `k` (Fig). When the masses are in equilibrium, `m_(1)` is removed. Frequency and amplitude of oscillation of `m_(2)` are

Options

A. `omega =sqrt((K)/(m_(1))), A (m_(2)g)/(K)`
B. `omega =sqrt((K)/(m_(2))), A (m_(1)g)/(K)`
C. `omega =sqrt((K)/(m_(1))), A=(m_(1)g)/(K)`
D. `omega = sqrt((K)/(m_(1))), A =(m_(2)g)/(K)`

Answer

B

Explanation

(i) When `m_(1)` is removed only `m_(2)` is left. Therefore, angular frequency: `omega = sqrt((k)/(m_(2)))`
(ii) Let `x_(1)` be the extension when only `m_(2)` is left.
Then, `kx_(1) = m_(2)g or x_(1) = (m_(2)g)/(k) …(1)`
Similarly, let `x_(2)` be extension in equilibrium when both `m_(1)` and `m_(2)` are suspended. Then,
`(m_(1)+m_(2)) g = kx_(2), :. x_(2) = ((m_(1)m_(2)g))/(k) ...(2)`
From Eqs. (1) and (2) amplitude of oscillation,
`A =x_(2) -x_(1) = (m_(1)g)/(k)`