Rankers Physics

Spring Block System: Practice Problem & Solution

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)
$\frac{16}{9}$
$\frac{9}{16}$
$\frac{3}{4}$
$\frac{4}{3}$

Solution Explained:

To solve this problem, we apply the core principles of Spring Block System. Understanding the underlying formula is key to arriving at the correct answer below:

$T \propto \sqrt{m} \Rightarrow \frac{3}{5} = \sqrt{\frac{m}{m+1}} \Rightarrow \frac{9}{25} = \frac{m}{m+1} \Rightarrow 25m = 9m + 9 \Rightarrow m = \frac{9}{16} \text{ kg}$.

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