Q. A particle is executing simple harmonic motion with a period of $T\ seconds$ and amplitude a $metre.$ The shortest time it takes to reach a point $\frac{a}{\sqrt{2}}m$ from its mean position in $seconds$ is
A) $T$
B) $T/4$
C) $T/8$
D) $T/16$
Explanation
(c) $y = a\sin{\frac{2\pi}{T}t \Rightarrow \frac{a}{\sqrt{2}} = a\sin{\frac{2\pi}{T}.t}}$
$\Rightarrow \sin{\frac{2\pi}{T}t = \frac{1}{\sqrt{2}} = \sin\frac{\pi}{4} \Rightarrow \frac{2\pi}{T}t = \frac{\pi}{4} \Rightarrow t = \frac{T}{8}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.