Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of $A$ and $B$ is

(B) $\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$

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

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

(D) $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Elasticity & SHM

Two particles

A

and

B

of equal masses are suspended from two massless springs of spring constants

K_{1}

and

K_{2}

respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of

A

and

B

Answer: (B)

For a mass-spring system, the maximum velocity is

v_{ma x} = A ω = A \frac{K}{m}

. For particles A and B with equal masses and equal maximum velocities:

A_{A} \frac{K _{1}}{m} = A_{B} \frac{K _{2}}{m}

. Therefore

\frac{A _{A}}{A _{B}} = \frac{K _{2}}{K _{1}}

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option