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A particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is $t_1$, and the minimum time taken by the particle to move from mean position to half of the amplitude is $t_2$, then
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Verified Answer
The correct answer is:
$t_1=2 t_2$
For the particle to move from extreme position to half of the amplitude
$\begin{aligned}
& \mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}_1 \\
& \frac{\mathrm{A}}{2}=\mathrm{A} \cos \omega \mathrm{t}_1 \\
& \mathrm{t}_1=\frac{\pi}{3 \omega}
\end{aligned}$
For the particle to move from mean position to half of the amplitude
$\begin{aligned}
& x=A \sin \omega t_2 ; \frac{A}{2}=A \sin \omega t_2 \\
& t_2=\frac{\pi}{6 \omega} \\
& \frac{t_1}{t_2}=\frac{\frac{\pi}{3 \omega}}{\frac{\pi}{6 \omega}} \Rightarrow t_1=2 t_2
\end{aligned}$
$\begin{aligned}
& \mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}_1 \\
& \frac{\mathrm{A}}{2}=\mathrm{A} \cos \omega \mathrm{t}_1 \\
& \mathrm{t}_1=\frac{\pi}{3 \omega}
\end{aligned}$
For the particle to move from mean position to half of the amplitude
$\begin{aligned}
& x=A \sin \omega t_2 ; \frac{A}{2}=A \sin \omega t_2 \\
& t_2=\frac{\pi}{6 \omega} \\
& \frac{t_1}{t_2}=\frac{\frac{\pi}{3 \omega}}{\frac{\pi}{6 \omega}} \Rightarrow t_1=2 t_2
\end{aligned}$
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