Q115 / 150 · MHT CET 2023 (13 May Shift 1) · Physics
A body of mass $0.04 \mathrm{~kg}$ excutes simple harmonic motion (SHM) about $\mathrm{x}=0$ under the influence of force $\mathrm{F}$ as shown in graph. The period of
- A$2 \pi \mathrm{s}$
- Correct answerB$0.2 \pi \mathrm{s}$
- C$\pi \mathrm{d}$
- D$\frac{\pi}{2} \mathrm{~s}$
Solution
$$
\begin{aligned}
& \mathrm{k}=400 \mathrm{~N} / \mathrm{m} \\
& \mathrm{m}=0.04
\end{aligned}
$$
From the graph,
$$
\mathrm{K}=\frac{\mathrm{F}}{\mathrm{x}}=\frac{8}{2}=4
$$
From $T=2 \pi \sqrt{\frac{M}{K}}$,
we get
$$
\mathrm{T}=2 \pi \sqrt{\frac{0.04}{4}}=0.2 \pi \mathrm{s}
$$
Hence, the correct answer is option: 'B'.