Question

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 becomes $\frac{5T}{3}$. Then the ratio of $\frac{m}{M}$ is

• A. $\frac{25}{9}$
• B. $\frac{5}{3}$
• C. $\frac{16}{9}$
• D. $\frac{3}{5}$

Answer 1

Answer: (C)

From formula 1, we can say
$T=2\pi \sqrt{\frac{M}{k}}$ ...(1)
Also,
$\frac{5T}{3}=2\pi \sqrt{\frac{M+m}{k}}$ ...(2)
Dividing eqn 2 and 1, we get
$\frac{5}{3}=\sqrt{\frac{M+m}{M}}$
$\therefore \frac{25}{9}=\frac{M+m}{M}$
$\therefore \frac{25}{9}=1+\frac{m}{M}$
Hence, answer=16/9, i.e. option B