A mass $\mathrm{M}$ 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 $\mathrm{m}$, the time period becomes $\frac{5 \mathrm{~T}}{3}$. Then the ratio of $\frac{\mathrm{m}}{\mathrm{M}}$ is

• A

  $\frac{3}{5}$
• B

  $\frac{25}{9}$
• Correct answer
  C

  $\frac{16}{9}$
• D

  $\frac{5}{3}$

Solution

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$
$\frac{5 \mathrm{~T}}{3}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{K}}}$
Dividing equation (ii) by equation (i), $\frac{5}{3}=\sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}}$. Squaring both the sides $\frac{25}{9}=\frac{M+m}{M}=1+\frac{m}{M} \Rightarrow \frac{m}{M}=\frac{25}{9}-1=\frac{16}{9}$

Hence, the correct answer is option: 'C'.