Question

A particle is executing simple harmonic motion with a period of T seconds and amplitude a meter. 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

Answer 1

Answer: (C)

T/8

The standard equation for SHM is denoted as:

$y=\sin \frac{2\pi }{T}t$

So, $\Rightarrow \frac{a}{\sqrt{2}}=a\sin \frac{2\pi }{T}.t$

So, the time taken to reach the point $\frac{a}{\sqrt{2}}$

$\Rightarrow \frac{2\pi }{T}.t=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}$

$\Rightarrow \frac{2\pi }{T}t=\frac{\pi }{4}$

$\Rightarrow t=\frac{T}{8}$