MHT CET · Physics · Oscillations
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 \(2 \pi \mathrm{s}\)
- B \(0.2 \pi \mathrm{s}\)
- C \(\pi \mathrm{d}\)
- D \(\frac{\pi}{2} \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(0.2 \pi \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{k}=400 \mathrm{~N} / \mathrm{m} \\
& \mathrm{m}=0.04
\end{aligned}
\)
From the graph,
\(
\mathrm{K}=\frac{\mathrm{F}}{\mathrm{x}}=\frac{8}{2}=4
\)
From \(T=2 \pi \sqrt{\frac{M}{K}}\),
we get
\(
\mathrm{T}=2 \pi \sqrt{\frac{0.04}{4}}=0.2 \pi \mathrm{s}
\)
\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
\(
\mathrm{T}=2 \pi \sqrt{\frac{0.04}{4}}=0.2 \pi \mathrm{s}
\)
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