Question

A particle executes $\text{SHM}$ with an angular frequency $4\pi \text{rad/s}$. If it is at its positive extreme position initially, find the first instant when it is at a distance of $\frac{1}{\sqrt{2}}$ times its amplitude from the mean position in positive direction

• A. $\frac{1}{24}\text{s}$
• B. $\frac{1}{16}\text{s}$
• C. $-\frac{1}{16}\text{s}$
• D. $\frac{3}{16}\text{s}$

Answer 1

The correct option is B $\frac{1}{16}\text{s}$

Let $A$ be the amplitude and $\varphi$ be the phase constant for the oscillation.
From the data given in the question, the displacment equation can be written as
$x=\text{A sin}(4\pi t+\varphi )...\left(1\right)$
$(\because \omega =4\pi \text{rad/s})$
Since at $t=0;x=+A$
$\therefore 4\pi \left(0\right)+\varphi =\frac{\pi }{2}$
$\Rightarrow \varphi =\frac{\pi }{2}$
$\therefore$ equation $\left(1\right)$ can be written as $x=\text{A sin}(4\pi t+\frac{\pi }{2})=\text{A cos}\left(4\pi t\right)$
Evidently, $\frac{A}{\sqrt{2}}=\text{A cos}\left(4\pi t\right)$
$\Rightarrow 4\pi t=\frac{\pi }{4}$
$\Rightarrow t=\frac{1}{16}\text{s}$