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Wave Equation

Two masses ...

Question

Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of spring constant $k$ . When the masses are in equilibrium, $m_{2}$ is removed without disturbing the system. The amplitude of SHM is

\frac{m_{1} g}{k}

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\frac{m_{2} g}{k}

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\frac{{(m}_{1} + m_{2}) g}{k}

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\frac{m_{2} g}{m_{1} k}

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Solution

The correct option is B $\frac{m_{2} g}{k}$
The displacement from equilibrium position (natural length of spring)
$k y_{1} = (m_{1} + m_{2}) \times g$
So,
$y_{1} = \frac{(m_{1} + m_{2}) \times g}{k}$
Now new equilibrium position,
$y_{2} = \frac{m_{1} \times g}{k}$ (when $m_{2}$ is removed)
Therefore,
Amplitude $A = y_{1} - y_{2} = \frac{(m_{1} - m_{2}) \times g}{k} - \frac{m_{1} \times g}{k}$
Amplitude $= \frac{m_{2} \times g}{k}$

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