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The two block...

Question

The two blocks of masses $m_{1}$ and $m_{2}$ are kept on a smooth horizontal table as shown in the figure. Block of mass $m_{1}$ but not $m_{2}$ is fastened to the spring. If now both the blocks are pushed to the left, so that the spring is compressed at a distance $d$ . The amplitude of oscillation of block of mass $m_{1}$ after the system released, is :

d \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}

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d \sqrt{\frac{m_{2}}{m_{1} + m_{2}}}

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d \sqrt{\frac{2 m_{2}}{m_{1} + m_{2}}}

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d \sqrt{\frac{2 m_{1}}{m_{1} + m_{2}}}

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Solution

The correct option is A $d \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}$
Block of mass $m_{2}$ shoots off carrying some kinetic energy away from the system. To find its speed:
potential energy of spring $=$ maximum kinetic energy of blocks.
$\frac{k d^{2}}{2} = (m_{1} + m_{2}) \frac{v^{2}}{2}$ [ $k =$ force constant of spring]
$v^{2} = \frac{k d^{2}}{m_{1} + m_{2}}$ with $m_{1}$ along on the spring
Maximum potential energy $=$ Maximum kinetic energy of $m_{2}$
$\Rightarrow \frac{1}{2} k A^{2} = \frac{1}{2} m_{1} v^{2} \Rightarrow k A^{2} = \frac{k m_{1} d^{2}}{m_{1} + m_{2}}$
$\Rightarrow A = d \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}$

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