Q. Two masses $m_{1\ }and\ m_{2}$ are suspended together by a massless spring of constant k. When the masses are in equilibrium, $m_{1}\ $is removed without disturbing the system. Then the angular frequency of oscillation of $m_{2}\ is\ $
A) $\sqrt{k/m_{1}}$
B) $\sqrt{k/m_{2}}$
C) $\sqrt{k/(m_{1} + m_{2})}$
D) $\sqrt{k/(m_{1} - m_{2})}$
Explanation
(b) We both the masses are there, then angular frequency $\omega = \sqrt{k/(m_{1} + m_{1})};$ is there, then $\omega^{'} = \sqrt{k/m_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.