A body of mass $0.04 \mathrm{~kg}$ excutes simple harmonic motion (SHM) about $\mathrm{x}=0$ under the…

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
  1. $2 \pi \mathrm{s}$
  2. $0.2 \pi \mathrm{s}$
  3. $\pi \mathrm{s}$
  4. $\frac{\pi}{2} \mathrm{~s}$

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 $\mathrm{T}=2 \pi \sqrt{\frac{0.04}{4}}=0.2 \pi \mathrm{s}$ .

Asked in: MHT CET 2023 (13 May Shift 1)

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