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. 3/5
• B. 25/9
• C. 16/9
• D. 5/3

Answer 1

Answer: (C)

16/9

$T=2\pi \sqrt{\frac{M}{k}}$ ...............(1)

$T^{\prime }=2\pi \sqrt{\frac{M+m}{k}}$

$\Rightarrow \frac{5T}{3}=2\pi \sqrt{\frac{M+m}{k}}$ ......(2)

Dividing Equation (1) by (2), we have

$\therefore \frac{3}{5}=\sqrt{\frac{M}{M+m}}$

$\frac{9}{25}=\frac{M}{M+m}$

$\Rightarrow 9M+9m=25M\Rightarrow 16M=9m$

$\Rightarrow \frac{m}{M}=\frac{16}{9}$