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COE in SHM

The masses in...

Question

The masses in figure slide on a frictionless table. $m_{1}$ but not $m_{2}$ , is fastened to the spring. If now $m_{1}$ and $m_{2}$ are pushed to the left, so that the spring is compressed a distance d, what will be the amplitude of the oscillation of $m_{1}$ after the spring system is released?

A = (\sqrt{\frac{m_{1}}{m_{1} + m_{2}}})

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

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

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Solution

The correct option is B $A = (\sqrt{\frac{m_{1}}{m_{1} + m_{2}}}) d$
While returning to equilibrium position,
$\frac{1}{2} k d^{2} = \frac{1}{2} (m_{1} + m_{2}) v^{2}$
$∴$ $v = (\sqrt{\frac{k}{m_{1} + m_{2}}}) d$
Now after mean position $m_{2}$ is detached from $m_{1}$ and keeps on moving with this constant velocity v towards right. Block $m_{1}$ starts SHM with spring and this v becomes its maximum velocity at mean position.
$∴$ $v = ω A$
$∴$ $(\sqrt{\frac{k}{m_{1} + m_{2}}}) d = (\sqrt{\frac{k}{m_{1}}}) A$
$∴$ $A = (\sqrt{\frac{m_{1}}{m_{1} + m_{2}}}) d$

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The masses in figure slide on a frictionless table. m1 but not m2, is fastened to the spring. If now m1 and m2 are pushed to the left, so that the spring is compressed a distance d, what will be the amplitude of the oscillation of m1 after the spring system is released?

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