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Two blocks of masses $m_{1} and m_{2}$ are kept on a smooth horizontal table as shown in 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 a distance d, the amplitude of oscillation of block of mass $m_{1}$ after the system is released is

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

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

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

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

answer is A.

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Detailed Solution

Block of mass

m_{2}

shoots off carrying. So m KE away from the system

PE of spring = maximum KE of block

$\frac{{kd}^{2}}{2} = (m_{1} + m_{2}) \frac{v^{2}}{2}; υ^{2} = \frac{{kd}^{2}}{m_{1} + m_{2}} \to (1)$

With $m_{1}$ alone on the spring

$P {.E}_{\max} = {KE}_{\max} of m$

$\frac{1}{2} {kA}^{2} = \frac{1}{2} m_{1} v^{2}$

From (1) & (2)

$A = d \sqrt{\frac{m_{1}}{m_{1} + m_{2}}}$

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