Q. A mass $M$, attached to a horizontal spring, executes SHM with amplitude $A_{1}$. When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude $A_{2}$. The ratio of $\left( \frac{A_{1}}{A_{2}} \right)$ is
A) $\frac{M + m}{M}$
B) $\left( \frac{M}{M + m} \right)^{\frac{1}{2}}$
C) $\left( \frac{M + m}{M} \right)^{\frac{1}{2}}$
D) $\frac{M}{M + m}$
Explanation
(c) At mean position, $F_{net} = 0$
$\therefore$ By conservation of linear momentum
$Mv_{1} = (M + m)v_{L}$
$M\omega_{1}A_{1} = (M + m)\omega_{2}A_{2}$
$\omega_{1} = \sqrt{\frac{k}{m}}$
$\omega_{2} = \sqrt{\frac{k}{M + m}}$
On solving
$\frac{A_{1}}{A_{2}} = \sqrt{\frac{M + m}{M}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.