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A particle executes SHM with amplitude $0.2 \mathrm{~m}$ and time period $24 \mathrm{~s}$. The time required it to move from the mean position to a point $0.1 \mathrm{~m}$ from the mean position is
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Verified Answer
The correct answer is:
$2 \mathrm{~s}$
As the particle starts from the mean position,
$$
\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}=\mathrm{A} \sin \frac{2 \pi}{\mathrm{T}} \mathrm{t}
$$
or
$$
0.1=0.2 \sin \frac{2 \pi}{24} t
$$
or $\quad \sin \frac{\pi}{12} \mathrm{t}=\frac{1}{2} \Rightarrow \sin \frac{\pi}{12} \mathrm{t}=\sin \frac{\pi}{6}$ or
$$
\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}=\mathrm{A} \sin \frac{2 \pi}{\mathrm{T}} \mathrm{t}
$$
or
$$
0.1=0.2 \sin \frac{2 \pi}{24} t
$$
or $\quad \sin \frac{\pi}{12} \mathrm{t}=\frac{1}{2} \Rightarrow \sin \frac{\pi}{12} \mathrm{t}=\sin \frac{\pi}{6}$ or
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