Two bodies A and B of equal mass are suspended from two massless springs of spring constant $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$, 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

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

(B) $\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}$

(C) $\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}$

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

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Elasticity & SHM

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

(A) $\frac{k _{1}}{k _{2}}$ ✓ Correct
(B) $\frac{k _{1}}{k _{2}}$
(C) $\frac{k _{2}}{k _{1}}$
(D) $\frac{k _{2}}{k _{1}}$

Answer: (A)

Solution

For simple harmonic motion, the maximum speed is

v_{ma x} = A ω, ω = \frac{k}{m}

Since both bodies have the same mass

m

and the same amplitude

A

v_{A} = A \frac{k _{1}}{m}, v_{B} = A \frac{k _{2}}{m}

\frac{v _{A}}{v _{B}} = \frac{k _{1}}{k _{2}}

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option

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