Question

A particle performs S.H.M. with amplitude $25cm$ and period $3s$. The minimum time required for it to move between two points $12.5cm$ on either side of the mean position is

• A. $0.6s$
• B. $0.5s$
• C. $0.4s$
• D. $0.2s$

Answer 1

The correct option is B $0.5s$

$sin\theta =\frac{12.5}{25}\theta =\frac{\pi }{6}$

So, total angle to move$=\frac{2\pi }{6}=\frac{\pi }{3}$

It takes $3s$ to rotate ${360}^{\circ }$. So, to rotate ${60}^{\circ }$

$t=\frac{3}{360}\times 60t=0.5s$