Q. Two bodies $M$ and $N$ of equal masses are suspended from two separate massless springs of force constants $k_{1}$ and $k_{2}$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude $M$ to that of $N$ is
A) $k_{1}/k_{2}$
B) $\sqrt{k_{1}/k_{2}}$
C) $k_{2}/k_{1}$
D) $\sqrt{k_{2}/k_{1}}$
Explanation
(d) Maximum velocity $= a\omega = a\sqrt{\frac{K}{m}}$
Given that $a_{1}\sqrt{\frac{K_{1}}{m}} = a_{2}\sqrt{\frac{K_{2}}{m}} \Rightarrow \frac{a_{1}}{a_{2}} = \sqrt{\frac{K_{2}}{K_{1}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.