Two bodies ' $A$ ' and 'B' of equal mass are suspended from two separate massless springs of force constant…

Two bodies ' $A$ ' and 'B' of equal mass are suspended from two separate massless springs of force constant ' $\mathrm{k}_{1}$ ' and ${ }^{\circ} \mathrm{k}_{2}$ ' respectively. The bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitudes of body A to that of body B is
  1. $\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}$
  2. $\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}$
  3. $\sqrt{\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}}$
  4. $\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}$

Solution

If their maximum velocities are equal then their total energy is same. If $A_{1}, A_{2}$ are their amplitudes, then $\begin{aligned} & \frac{1}{2} \mathrm{~K}_{1} \mathrm{~A}_{1}^{2}=\frac{1}{2} \mathrm{~K}_{2} \mathrm{~A}_{2}^{2} \\ \therefore & \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}} \end{aligned}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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