Solution:
$T \propto \sqrt{m} \Rightarrow \frac{3}{5} = \sqrt{\frac{m}{m+1}} \Rightarrow \frac{9}{25} = \frac{m}{m+1} \Rightarrow 25m = 9m + 9 \Rightarrow m = \frac{9}{16} \text{ kg}$.
A body of mass $m$ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released, it oscillates with a time period of $3 \text{ s}$. When the mass $m$ is increased by $1 \text{ kg}$, the time period of oscillations becomes $5 \text{ s}$. The value of $m$ in kg is: (2016 - II)
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