Two masses `m_(1)`and `m_(2)` are suspended from a spring of spring constant 'k'. When the masses are in equilibrium, `m_(1)` is gently removed. The angular frequency and amplitude of oscillation of `m_(2)` will be respectively

Options

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

Answer

A

Explanation

Removed mass gives amplitude `m_(1)g = kA rArr A = (m_(1)g)/(k)`
remaining mass gives frequency `omega = sqrt((k)/(m_(2)))`