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 $5T/3$. then the ratio of $m/M$ is

• A. $3/5$
• B. $25/9$
• C. $16/9$
• D. $5/3$

Answer 1

The correct option is C $16/9$

Time period of oscillation of the block executing SHM $T=2\pi \sqrt{\frac{M}{K}}$

where $M$ and $K$ is the mass of the block and spring constant of the spring.

Thus new time period $T^{\prime }=\frac{5T}{3}=2\pi \sqrt{\frac{M+m}{K}}$

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

Squaring both sides $\frac{25}{9}=1+\frac{m}{M}$

$⟹\frac{m}{M}=\frac{16}{9}$