Question 144 of 194

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$\frac{25}{9}$C$\frac{16}{9}$D$\frac{5}{3}$

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Answer: (C) $\frac{16}{9}$

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$

$\frac{5 \mathrm{~T}}{3}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{K}}}$

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