Question

The equation of displacement of a particle executing simple harmonic motion is x = (5m) $\sin \left[\right(\pi s^{-1})t+\frac{\pi }{3}]$. Write down the amplitude, time period and maximum speed. Also find the velocity at t = 1s.

• A. $-\frac{5\pi }{2}m/s$
• B. $-\frac{\pi }{2}m/s$
• C. $-\frac{3\pi }{2}m/s$
• D. $-\frac{\pi }{3}m/s$

Answer 1

The correct option is A $-\frac{5\pi }{2}m/s$

Comparing with equation $x=A\sin (\omega t+\delta )$ we see that the amplitude = 5 m
and time period = $\frac{2\pi }{\omega }=\frac{2\pi }{\pi s^{-1}}=2s$
The maximum speed = $A\omega =5m\times \pi s^{-1}=5\pi m/s$
The velocity at time $t=\frac{dx}{dt}=A\omega \cos (\omega t+\delta )$
At t = 1 s
$v=\left(5m\right)\left(\pi s^{-1}\right)\cos (\pi +\frac{\pi }{3})=-\frac{5\pi }{2}m/s$