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A mass $M$, attached to a horizontal spring, executes S.H.M. with amplitude $A_1$. When the mass $M$ passes through its mean position then a smaller mass $\mathrm{m}$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\left(\frac{A_1}{A_2}\right)$ is :
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Verified Answer
The correct answer is:
$\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}\right)^{1 / 2}$
$\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}\right)^{1 / 2}$
Energy of simple harmonic oscillator is constant.
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} M \omega^2 A_1^2=\frac{1}{2}(m+M) \omega^2 A_2^2 \\
& \frac{A_1^2}{A_2^2}=\frac{M+m}{M} \\
& \therefore \frac{A_1}{A_2}=\sqrt{\frac{M+m}{M}}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} M \omega^2 A_1^2=\frac{1}{2}(m+M) \omega^2 A_2^2 \\
& \frac{A_1^2}{A_2^2}=\frac{M+m}{M} \\
& \therefore \frac{A_1}{A_2}=\sqrt{\frac{M+m}{M}}
\end{aligned}
$$
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