JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 2)
A particle executes simple harmonic motion between $$x=-A$$ and $$x=+A$$. If time taken by particle to go from $$x=0$$ to $$\frac{A}{2}$$ is 2 s; then time taken by particle in going from $$x=\frac{A}{2}$$ to A is
4 s
1.5 s
3 s
2 s
Explanation
$x=A \sin (\omega t)$
$$ \begin{aligned} & x=\frac{A}{2}=A \sin (\omega t) \\\\ & \frac{1}{2}=\sin (\omega t) \\\\ & t=\left(\frac{\pi}{6 \omega}\right)=2 \\\\ & \frac{\pi}{\omega}=12 \sec \end{aligned} $$
$x=A=A \sin (\omega t)$
$\omega t=\left(\frac{\pi}{2}\right)$
$t=\left(\frac{\pi}{2 \omega}\right)=6$ second
time $=6-2=4$ seconds
$$ \begin{aligned} & x=\frac{A}{2}=A \sin (\omega t) \\\\ & \frac{1}{2}=\sin (\omega t) \\\\ & t=\left(\frac{\pi}{6 \omega}\right)=2 \\\\ & \frac{\pi}{\omega}=12 \sec \end{aligned} $$
$x=A=A \sin (\omega t)$
$\omega t=\left(\frac{\pi}{2}\right)$
$t=\left(\frac{\pi}{2 \omega}\right)=6$ second
time $=6-2=4$ seconds
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