A mass ' $m_{1}$ ' connected to a horizontal spring performs S.H.M. with amplitude 'A'. While mass ' $m_{1}$ ' is passing through mean position another mass ' $m_{2}$ ' is placed on it so that both the masses move together with amplitude ' $A_{1}$ '. The ratio of $\frac{A_{1}}{A}$ is $(m_{2} < m_{1})$

{[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

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{[\frac{m_{1} + m_{2}}{m_{1}}]}^{\frac{1}{2}}

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{[\frac{m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}

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{[\frac{m_{1} + m_{2}}{m_{2}}]}^{\frac{1}{2}}

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Solution

The correct option is C ${[\frac{m_{1}}{m_{1} + m_{2}}]}^{\frac{1}{2}}$
Since amplitude of S.H.M. = A
Applying energy conservation
$\frac{1}{2} \times k \times A^{2} = \frac{1}{2} m_{1} v^{2}$ .................(1)

where $v =$ velocity of block at mean position.
When $m_{2}$ will be placed over $m_{1}$ , then their common velocity can be calculated using conservation of linear momentum as
$m_{1} \times v = (m_{1} + m_{2}) \times V_{c o m m o n}$
$V_{c o m m o n} = \frac{m_{1} v}{m_{1} + m_{2}}$

Now when both masses will execute S.H.M. Combinedly, we can find their amplitude using energy conservation
$\frac{1}{2} \times k \times A^{' 2} = \frac{1}{2} \times (m_{1} + m_{2}) (V_{c o m m o n})^{2}$

On solving we will get $\frac{A^{'}}{A} = [\frac{m_{1}}{m_{1} + m_{2}}]^{\frac{1}{2}}$

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