Question

A particle is executing SHM along a straight line $8\text{cm}$ long. While passing through mean position its velocity is $16{\text{cms}}^{-1}$. Its time period will be:

• A. $0.0157\text{s}$
• B. $0.157\text{s}$
• C. $1.57\text{s}$
• D. $15.7\text{s}$

Answer 1

The correct option is C $1.57\text{s}$

8 cm long line means amplitude is 4 cm (both sides of mean)
Velocity at mean position is the maximum velocity given by $A\omega$
Thus $16=4\omega$ or $\omega =4$
Thus, $T=\frac{2\pi }{\omega }=\frac{2\pi }{4}=\frac{\pi }{2}=1.57\text{s}$