Question

A particle is executing $S.H.M.$ If the amplitude is $2m$ and periodic time $2seconds$, then the max. velocity of the particle will be

• A. $\pi m/s$
• B. $\sqrt{2}\pi m/s$
• C. $2\pi m/s$
• D. $4\pi m/s$

Answer 1

The correct option is C $2\pi m/s$

In an $S.H.M$

$A=2m,T=2s$

So, $\frac{2\pi }{\omega }=2⟹\omega =\pi$

$V_{max}=A\omega$

$=2\pi m/s$