Two masses M and m are suspended together by massless spring of force constant -k. When the masses are in equilibrium, M is removed without disturbing the system. The amplitude of oscillations.

Options

A. `Mg//k`
B. `mg//k`
C. `((M+m)g)/(k)`
D. `((M-m)g)/(k)`

Answer

A

Explanation

For equilibrium of (M+n), `x_(1)=((M+m)g)/(k)`
For equilibrium of m `x_(2)=(mg)/(k)`, Amplitude=shift in equilibrium positioin =`(Mg)/(K)`