A horizontal spring-block system of mass 2 kg executes S.H.M. When the block is passing through its equilibrium position, an object of mass I kg is put on it and the two move together. The new amplitude of vibration is (A being its initial amplitude)

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$\sqrt{2} A$

$\frac{A}{\sqrt{2}}$

$\sqrt{\frac{2}{3}} A$

$\sqrt{\frac{3}{2}} A$

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Detailed Solution

Conserving momentum: $2 V = 3 V^{'}$

$\Rightarrow V^{'} = \frac{2}{3} V$

$E_{i} = \frac{1}{2} m_{1} {V_{1}}^{2} = \frac{1}{2} 2 . V^{2} \Rightarrow \frac{1}{2} {KA}^{2} = V^{2}$

$E_{f} = \frac{1}{2} m_{2} {V_{2}}^{2} = \frac{1}{2} . 3 . \frac{2}{3} . \frac{2}{3} V^{2} = \frac{2}{3} V^{2}$

$\Rightarrow \frac{1}{2} {KA}^{' 2} = \frac{2}{3} V^{2} = \frac{2}{3} E_{i}$

( $E_{i} = V^{2} from above)$

$\Rightarrow \frac{1}{2} {KA}^{' 2} = \frac{2}{3} (\frac{1}{2} {KA}^{2}) \Rightarrow A^{'} = \sqrt{\frac{2}{3}} A$

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