Question

A particle executes $SHM$ of amplitude $25cm$ and time period $3s$. What is the minimum time required for the particle to move between two points $12.5cm$ on either side of the mean position?

• A. $0.5s$
• B. $1.0s$
• C. $1.5s$
• D. $2.0s$

Answer 1

Answer: (A)

$0.5s$

$y=rsin\omega t$
$12.5=25sin\frac{2\pi }{3}\times t$ $(\because \omega =\frac{2\pi }{T})$
$\frac{\pi }{6}=\frac{2\pi }{3}t$
$t=\frac{1}{4}sec=0.25sec$

$t^{{}^{\prime }}=2t$$=2\times 0.25=0.5sec$ for either side