MHT CET · Physics · Oscillations
Two bodies A and B of equal mass are suspended from two separate massless springs of spring constants \(\mathrm{K}_1\) and \(\mathrm{K}_2\) respectively. The two bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitude of B to that of A is
- A \(\frac{\mathrm{K}_1}{\mathrm{~K}_2}\)
- B \(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\)
- C \(\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}\)
- D \(\sqrt{\frac{\mathrm{K}_2}{\mathrm{~K}_1}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}\)
Step-by-step Solution
Detailed explanation
If the maximum velocities of the bodies are equal, then their total energy would also be the same. If \(\mathrm{A}_1, \mathrm{~A}_2\) are the amplitudes of A and B respectively, then
\(\begin{aligned}
& \frac{1}{2} \mathrm{~K}_1 \mathrm{~A}_1^2=\frac{1}{2} \mathrm{~K}_2 \mathrm{~A}_2^2 \\
\therefore \quad & \frac{\mathrm{~A}_2}{\mathrm{~A}_1}=\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}
\end{aligned}\)
\(\begin{aligned}
& \frac{1}{2} \mathrm{~K}_1 \mathrm{~A}_1^2=\frac{1}{2} \mathrm{~K}_2 \mathrm{~A}_2^2 \\
\therefore \quad & \frac{\mathrm{~A}_2}{\mathrm{~A}_1}=\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}
\end{aligned}\)
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