Two masses m1 and m2 are suspended together by a massless spring of constant k. When the masses are in equilibrium, m1 is removed without disturbing the system. The amplitude of oscillations is

Options

A. `(m_(2)g)/(K)`
B. `(m_(1)g)/(K)`
C. `(K)/(m_(1)g)`
D. `(K)/(m_(2)g)`

Answer

B

Explanation

When `m_(1)` is removed, only the mass `m_(2)` is left.
Left `x_(1)` be the extension in equilibrium, when only `m_(2)` is suspended.
Then, `Kx_(1)=m_(2)g`
`therefore x_(1)=(m_(2)g)/(K)`
Let `x_(2)` be the extension when both `m_(1) " and " m_(2)` are suspended, then
`therefore Kx_(2)=((m_(1)+m_(2))g)/(K)`
`therefore` Amplitude of oscillation,
`A=x_(2)-x_(1)=(g)/(K) [m_(1)+m_(2)-m_(2)]`
`A= (m_(1)g)/(K)`