Two bodies A and B of equal mass are suspended from two separate massless springs of spring constants…
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
$\frac{\mathrm{K}_1}{\mathrm{~K}_2}$
$\frac{\mathrm{K}_2}{\mathrm{~K}_1}$
$\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}$
$\sqrt{\frac{\mathrm{K}_2}{\mathrm{~K}_1}}$
Solution
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
$\begin{aligned}
& \frac{1}{2} \mathrm{~K}_1 \mathrm{~A}_1^2=\frac{1}{2} \mathrm{~K}_2 \mathrm{~A}_2^2 \\
\therefore \quad & \frac{\mathrm{~A}_2}{\mathrm{~A}_1}=\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}
\end{aligned}$