Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant K 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 $m_{2}$ .

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Solution

If only mass $m_{2}$ is present the extension of spring $l$ is given by $m_{2} g = k l \Rightarrow l = \frac{m_{2} g}{k}$ , where $k$ =Spring constant
Now with $m_{1}$ and $m_{2}$ the extension be $l^{'}$ $(m_{1} + m_{2}) g = k l^{'}$
$l^{'} = \frac{m_{1} + m_{2}}{k}$
When $m_{1}$ is removed, $(l^{'} - 1)$ acts as amplitude of oscillation
$A = (l^{'} - l) = (\frac{m_{1} + m_{2}}{k}) g - \frac{m_{2} g}{k}$
$\Rightarrow A = \frac{m_{1} g}{k}$
Now, if $m_{2}$ is removed the angular frequency of oscillation $ω = \sqrt{\frac{k}{m_{2}}}$ (This is same as the case of $m_{2}$ attach to spring)
$*$ N.B: When $m_{1}$ is removed the spring mass system will oscillates with the same angular frequency as the case of spring mass system with only mass $m_{2}$

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Two masses m1 and m2 are suspended together by a massless spring of constant K as shown in figure. When the masses are in equilibrium, m1 is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of m2.

Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of constant K 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 $m_{2}$ .