Q. A mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period $T$. If the mass is increased by $m$, the time period become $\frac{5T}{3}$. Then the ratio of $\frac{m}{M}$ is
A) $\frac{3}{5}$
B) $\frac{25}{9}$
C) $\frac{16}{9}$
D) $\frac{5}{3}$
Explanation
(c) $T = 2\pi\sqrt{\frac{M}{k}}$ when mass is increased by $m$ then …(i)
$T = 2\pi\sqrt{\frac{M + m}{k}}$
$\Rightarrow$ $\frac{5T}{3} = 2\pi\sqrt{\frac{M + m}{k}}$ …(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{3}{5} = \sqrt{\frac{M}{M + m}}$
$\frac{9}{25} = \frac{M}{M + m}$
$\Rightarrow$ $9M + 9m = 25M$
$\Rightarrow$ $16M = 9m$
$\frac{m}{M} = \frac{16}{9}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.