Question

A particle executes SHM with a period of $T$ second and amplitude $A$ metre. The particle is at its mean position initially The shortest time it takes to reach point $\frac{A}{\sqrt{2}}$ metre from its mean position in seconds is

• A. $T$
• B. $\frac{T}{4}$
• C. $\frac{T}{8}$
• D. $\frac{T}{16}$

Answer 1

The correct option is C $\frac{T}{8}$

Let $y=A\text{sin}\omega t$
i.e. $\frac{A}{\sqrt{2}}=A\text{sin}\omega t$
$\Rightarrow \text{sin}\omega t=\frac{1}{\sqrt{2}}=\text{sin}\frac{\pi }{4}$
i.e. $\omega t=\frac{\pi }{4}$
But $\omega =\frac{2\pi }{T}$
Therefore, $t=\frac{T}{8}$