Q168 / 180 · COMEDK 2015 · Physics
A mass $M$ is suspended from a light spring. An additional mass $m$ added displaces the spring further by a distance $X$. Now, the combined mass will oscillate on the spring with period
- A$T=2 \pi \sqrt{\frac{m g}{X(M+m)}}$
- Correct answerB$T=2 \pi \sqrt{\frac{(M+m) X}{m g}}$
- C$T=\frac{\pi}{2} \sqrt{\frac{m g}{X(M+m)}}$
- D$T=2 \pi \sqrt{\frac{(M+m)}{m g}}$
Solution
At initial condition of spring,
$$
M g=k X_{0} \quad \text{...(i)}
$$
When mass $m$ is added, then at equilibrium, $(M+m) g=k\left(X+X_{0}\right)$
$\Rightarrow M g+m g=k X+k X_{0}$
$\Rightarrow M g+m g=k X+M g \quad \text{[from Eq. (i)]}$
$\Rightarrow \quad m g=k X$
$\Rightarrow \quad k=\frac{m g}{X} \quad \text{...(ii)}$
$\therefore$ Time period,
$$
\begin{aligned}
T &=2 \pi \sqrt{\frac{\text { Total mass of the block }}{\text { Force constant } k}} \\
&=2 \pi \sqrt{\frac{M+m}{m g / X}} \\
&=2 \pi \sqrt{\frac{(M+m) X}{m g}}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.