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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
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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
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Verified Answer
The correct answer is:
$\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}$
(A)
If their maximum velocities are equal then their total energy is same. If $A_{1}, A_{2}$ are their amplitudes, then
$\begin{aligned}
& \frac{1}{2} \mathrm{~K}_{1} \mathrm{~A}_{1}^{2}=\frac{1}{2} \mathrm{~K}_{2} \mathrm{~A}_{2}^{2} \\
\therefore & \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}
\end{aligned}$
If their maximum velocities are equal then their total energy is same. If $A_{1}, A_{2}$ are their amplitudes, then
$\begin{aligned}
& \frac{1}{2} \mathrm{~K}_{1} \mathrm{~A}_{1}^{2}=\frac{1}{2} \mathrm{~K}_{2} \mathrm{~A}_{2}^{2} \\
\therefore & \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}
\end{aligned}$
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