Question

A particle performs SHM with amplitude 25 cm and period 3s. The minimum time required for it to move between two mean positions is :

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

Answer 1

Answer: (B)

$0.5$ s

Here, amplitude of particle, $A=25$ cm
and time period, $T=3$ s
If the particle at $t=0$ is at mean position, its displacement equation will be
$X=Asin\omega t$
$X=25sin\frac{2\pi t}{3}$ $[\because \omega =2\pi v=\frac{2\pi }{T}=\frac{2\pi }{3}]$
If it takes time $t_1$ to move a distance $x=12.5cm$ to one side of its mean position , then
$12.5=25sin\frac{2\pi t_1}{3}$

$\frac{1}{2}=sin\frac{2\pi t_1}{3}$

$sin\frac{\pi }{6}=sin\frac{2\pi t_1}{3}$
$\therefore \frac{\pi }{6}=\frac{2\pi t_1}{3}\Rightarrow t_1=\frac{1}{4}s$
The same will be the time to move 12.5 cm to the other side of its mean position, therefore, total time
$t=t_1+t_2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}=0.5s$