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_(1)g)/(k)`
B. `(m_(2)g)/(k)`
C. `((m_(1)+m_(2))g)/(k)`
D. `((m_(1)-m_(2))g)/(k)`

Answer

D

Explanation

For equilibrium of `(m_(1)+m_(2)),x_(1)=((m_(1)+m_(2))g)/(k)`
and for equilibrium of `m_(2),x_(2)=(m_(2)g)/(k)`
`therefore` Amplitude of oscillation will be
`A=x_(1)-x_(2)=((m_(1)+m_(2))g)/(k)-(m_(2)g)/(k)=(m_(1)g)/(k)`