Question

The displacement of a particle executing SHM at any instant t is $x=0.01\sin 100(t+0.05)$ then its time period will be

• A. 0.06 s
• B. 0.2 s
• C. 0.1 s
• D. 0.02 s

Answer 1

Answer: (A)

0.06 s

$x=0.01\sin (100t+5)$
We have $x=A\sin (\omega t+\theta )$

On comparing we get
Thus, here $\omega =100$
Thus, $T=\frac{2\pi }{\omega }=\frac{2\pi }{100}=0.06\text{seconds}.$