Question

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

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

Answer 1

Answer: (A)

0.5 s

Answer 2

In order to find the time taken by the particle from-12.5 cm to
+ 12.5 cm on either side of mean position, we will find the
time taken by particle to go from x = - 12. 5 cm to x = 0 and to
go from x = 0 to x = + 12.5 cm.
Let the equation of motion be x $=A sin \omega t$
First, the particle moves from x = -125 cm to x = 0
$\therefore \, \, \, \, \, \, \, \, \, 12.5 =25 sin \omega t \, \, \, \, \, \, \, (\because A=25 cm)$
$\Rightarrow \, \, \, \, \, \, \, \, \frac{1}{2}=sin \omega t$
$\Rightarrow \, \, \, \, \, \, \, \, \, \omega t =\frac{\pi}{6}$
$\therefore \, \, \, \, \, \, \, \, t=\frac{\pi}{6\omega}$
Similarly to go from x = 0 to x = 12. 5 cm
$\, \, \, \, \, \, \, \, \, \, \, \, \, \omega =\frac{\pi}{6} \, \, \Rightarrow \, \, \, t=\frac{\pi}{6 \omega}$
$\therefore \, \,$Total time taken from x = -12.5 cm to x = 12.5 cm
$\, \, \, \, \, \, \, \, \, t=\frac{\pi}{6\omega}+\frac{\pi}{6\omega}=\frac{\pi}{3\omega}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{\pi}{3\bigg(\frac{2\pi}{T}\bigg)}=\frac{T}{6}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{3}{6} =0.5 s$