Question

Two particles A and B of masses ${}^{\prime }{m}^{\prime }$ and ${}^{\prime }2m^{\prime }$ are suspended from two massless springs of force constants $K_1$ and $K_2$. During their oscillation, if their maximum velocities are equal, then ratio of amplitudes of A and B is :

• A. $\sqrt{\frac{K_1}{K_2}}$
• B. $\sqrt{\frac{K_2}{2K_1}}$
• C. $\sqrt{\frac{K_2}{K_1}}$
• D. $\sqrt{\frac{2K_1}{K_2}}$

Answer 1

Answer: (B)

$\sqrt{\frac{K_2}{2K_1}}$

Since by energy conservation,

$\frac{1}{2}K{A}^2=\frac{1}{2}M{v}_{max}^2$.

Hence, $\frac{K_1{A}_1^2}{K_2{A}_2^2}=\frac{m}{2m}=\frac{1}{2}$, since maximum velocities are same in both cases.

Hence, $\frac{A_1}{A_2}=\sqrt{\frac{K_2}{2K_1}}$.