Two bodies A and B of equal mass are suspended from two massless springs of spring constant $k_1$ and $k_2$,…

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. $\frac{\mathrm{k}_1}{\mathrm{k}_2}$
  2. $\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}$
  3. $\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}$
  4. $\frac{\mathrm{k}_2}{\mathrm{k}_1}$

Solution

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

Asked in: JEE Main 2025 (29 Jan Shift 2)

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