Question

A particle is executing SHM of period 4s.Then the time taken by it to move from thr extreme position to $\frac{\sqrt{3}}{2}$ of the amplitude is:

• A. $\frac{1}{3}$s
• B. $\frac{2}{3}$s
• C. $\frac{3}{4}$s
• D. $\frac{4}{3}$s

Answer 1

The correct option is A $\frac{1}{3}$s

We know ,

$x=A\cos \omega t$

Now,

$\begin{array}{c}\Rightarrow \frac{\sqrt{3}}{2}A=A\cos \omega t\\ \Rightarrow \omega t=\frac{\pi }{6}\\ \Rightarrow \frac{2\pi }{T}.t=\frac{\pi }{6}\\ \Rightarrow t=\frac{T}{12}=\frac{4}{12}=\frac{1}{3}\sec \\ =\frac{1}{3}\sec \end{array}$

$\therefore$ Option $A$ is correct.