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

A

Explanation

`m_(2)g=kl` . . . (i)
l is the extension in the string when only `m_(2)` is connected.
Now, `(m_(1)+m_(2))g=k(l+Deltal)` . . . (ii)
`Deltal to ` increase in extension when `(m_(1)+m_(2))` is connected
from equation (i) and (ii)
`m_(1)g=k Deltal implies Deltal=(m_(1)g)/(k)`.