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A mass M attached to a horizontal spring executes S.H.M. of amplitude $\mathrm{A}_{1}$. 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 $\mathrm{A}_{2}$. The ratio of $\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)$ is
Options:
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 $\mathrm{A}_{2}$. The ratio of $\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)$ is
Solution:
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Verified Answer
The correct answer is:
$\left(\frac{M+m}{M}\right) \frac{1}{2}$
$M V=(M+m) V^{\prime}$
$\frac{1}{2}(M+m) V^{\prime 2}=\frac{1}{2} k A_{2}^{2}$
$\frac{1}{2} M V^{2}=\frac{1}{2} k A_{1}^{2}$
$\frac{M+m}{M} \cdot \frac{V^{\prime 2}}{V^{2}}=\frac{A_{2}^{2}}{A_{1}^{2}}$
$\frac{M+m}{M} \cdot\left(\frac{M}{M+m}\right)^{2}=\frac{A_{2}^{2}}{A_{1}^{2}}$
$\frac{A_{1}^{2}}{A_{2}^{2}}=\frac{M+m}{M}$
$\therefore \frac{A_{1}}{A_{2}}=\left(\frac{M+m}{M}\right)^{\frac{1}{2}}$
$\frac{1}{2}(M+m) V^{\prime 2}=\frac{1}{2} k A_{2}^{2}$
$\frac{1}{2} M V^{2}=\frac{1}{2} k A_{1}^{2}$
$\frac{M+m}{M} \cdot \frac{V^{\prime 2}}{V^{2}}=\frac{A_{2}^{2}}{A_{1}^{2}}$
$\frac{M+m}{M} \cdot\left(\frac{M}{M+m}\right)^{2}=\frac{A_{2}^{2}}{A_{1}^{2}}$
$\frac{A_{1}^{2}}{A_{2}^{2}}=\frac{M+m}{M}$
$\therefore \frac{A_{1}}{A_{2}}=\left(\frac{M+m}{M}\right)^{\frac{1}{2}}$
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