A mass M attached to a horizontal spring executes S.H.M. of amplitude $\mathrm{A}_{1}$. When the mass M…

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
  1. $\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}$
  2. $\left(\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}\right) \frac{1}{2}$
  3. $\left(\frac{M+m}{M}\right) \frac{1}{2}$
  4. $\frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}$

Solution

$M V=(M+m) V^{\prime}$ $\frac{1}{2}(M+m) V^{\prime 2}=\frac{1}{2} k A_{2}^{2}$ $\frac{1}{2} M V^{2}=\frac{1}{2} k A_{1}^{2}$ $\frac{M+m}{M} \cdot \frac{V^{\prime 2}}{V^{2}}=\frac{A_{2}^{2}}{A_{1}^{2}}$ $\frac{M+m}{M} \cdot\left(\frac{M}{M+m}\right)^{2}=\frac{A_{2}^{2}}{A_{1}^{2}}$ $\frac{A_{1}^{2}}{A_{2}^{2}}=\frac{M+m}{M}$ $\therefore \frac{A_{1}}{A_{2}}=\left(\frac{M+m}{M}\right)^{\frac{1}{2}}$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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