Question

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 $\text{kg}$ is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{16}{9}$
• D. $\frac{9}{16}$

Answer 1

The correct option is D $\frac{9}{16}$

We know that,
Time period of oscillation $T=2\pi \sqrt{\frac{m}{k}}$
Case I :
$T_1=2\pi \sqrt{\frac{m}{k}}...\left(1\right)$
Case II When the mass $m$ is increased by $1\text{kg}$, then $m_2=m+1$, from $T_2=2\pi \sqrt{\frac{m_2}{K}}$
We get, $T_2=2\pi \sqrt{\frac{m+1}{k}}....\left(2\right)$
From Eqs. $\left(2\right)$ and $\left(1\right)$, we get
$\frac{T_2}{T_1}=\sqrt{\frac{m+1}{m}}\Rightarrow \frac{5}{3}=\sqrt{\frac{m+1}{m}}$
$\Rightarrow \frac{25}{9}=\frac{m+1}{m}\Rightarrow \frac{25}{9}=1+\frac{1}{m}$
$\Rightarrow \frac{1}{m}=\frac{16}{9}$
$\therefore m=\frac{9}{16}\text{kg}$