Two masses (m_1) and (m_2) are suspended together by a massless spring of spring constant (k). When the masses are in equilibrium, (m_1) is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of (m_2).
.

Options

A. `(m_1g)/(k)`
B. `(m_2g)/(k)`
C. `((m_1+m_2)g)/(k)`
D. `((m_2-m_1)g)/(k)`

Answer

A

Explanation

With mass `m_2` alone, the extension of the spring l is given as
`m_2g=kl`
With mass `(m_1+m_2)`, the extension `l'` is given by
`(m_1+m_2)g=kl'=k(l+trianglel)`
Hence `trianglel` is the amplitude of vibration.
subtracting Eq. (i) from Eq. (ii) we get `m_1g=ktrianglel`
or `trianglel=(m_1g)/(k)`