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MHT CET · Physics · Oscillations

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. A \(\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}\)
  2. B \(\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}\)
  3. C \(\sqrt{\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}}\)
  4. D \(\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}\)

Step-by-step Solution

Detailed explanation

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