Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of force $k$ constant as shown in figure. When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of mass $m_{2}$

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$ω = \sqrt{\frac{k}{m_{2}}}, A = \frac{m_{1} g}{k}$

$ω = \frac{m_{2} g}{k}, A = \sqrt{\frac{k}{m_{2}}},$

$ω = \sqrt{\frac{k}{m_{2}}}, A = \frac{m_{1}}{k}$

$ω = \sqrt{\frac{k}{m_{1}}}, A = \frac{m_{1} g}{k}$

answer is B.

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Detailed Solution

If $F$ be the force is the spring, then for equilibrium

$F - (m_{1} - m_{2}) g = 0 ∴ F - m_{2} g = m_{1} g$

when $m_{1}$ is removed the net force (restoring force) on the block, m2 ,

$F_{r e s t} = (F - m_{2} g) = m_{1} g$

$∴$ Amplitude $= \frac{m_{1} g}{k}$

Angular frequency, $ω = \sqrt{\frac{k}{m_{2}}}$

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