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Question

A mass $$\mathrm{M}$$ attached to a horizontal spring, executes SHM with amplitude $$\mathrm{A}_{1}$$. When the mass $$\mathrm{M}$$ passes through its mean position then a smaller mass $$\mathrm{m}$$ is placed over it and both of them move together with amplitude $$\mathrm{A}_{2}$$. The ratio of $$\left(\displaystyle \frac{\mathrm{A}_{1}}{\mathrm{A}_{2}}\right)$$ is: 


A
MM+m
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B
M+mM
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C
MM+m
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D
M+mM
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Solution

The correct option is D $$\sqrt{\displaystyle \frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}}$$
Energy of simple harmonic oscillator is constant.
$$\displaystyle \frac{1}{2}\times M \omega^2A_1^2= \frac{1}{2}(m+M) \omega^2A_2^2$$
$$\displaystyle \frac{A_1^2}{A_2^2}= \frac{M+m}{M}$$
$$\displaystyle \frac{A_1}{A_2}= \sqrt{\frac{M+m}{M}}$$

Physics

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