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JEE Mains · Physics · STD 11 - 13. oscillations

Two bodies A and B of equal mass are suspended from two massless springs of spring constant \(k_1\) and \(k_2\), respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of \(A\) to the maximum velocity of \(B\) is _______.

  1. A \(\frac{\mathrm{k}_1}{\mathrm{k}_2}\)
  2. B \(\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}\)
  3. C \(\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}\)
  4. D \(\frac{\mathrm{k}_2}{\mathrm{k}_1}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \mathrm{V}_1=\mathrm{A}_1 \omega_1 \\ & \mathrm{~V}_2=\mathrm{A}_2 \omega_2 \\ & \mathrm{~A}_1=\mathrm{A}_2 \\ & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\frac{\omega_1}{\omega_2}=\frac{\sqrt{\frac{\mathrm{K}_1}{\mathrm{~m}}}}{\sqrt{\frac{\mathrm{~K}_2}{\mathrm{~m}}}…

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