Question

The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t + π/3), where x is in centimetre and t in second. When does the particle
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed?

Answer 1

Given:
The equation of motion of a particle executing S.H.M. is,

\[x = 5 \sin \left( 20t + \frac{\pi}{3} \right)\]

The general equation of S..H.M. is give by,

\[x = A \sin (\omega t + \phi)\]

(a) Maximum displacement from the mean position is equal to the amplitude of the particle. 
     As the velocity of the particle is zero at extreme position, it is at rest.

\[\therefore \text { Displacement }\]  x = 5, which is also the amplitude of the particle.

\[\Rightarrow 5 = 5 \sin \left( 20t + \frac{\pi}{3} \right)\]

\[\text { Now }, \]

\[ \sin \left( 20t + \frac{\pi}{3} \right) = 1 = \sin\frac{\pi}{2}\]

\[ \Rightarrow 20t + \frac{\pi}{3} = \frac{\pi}{2}\]

\[ \Rightarrow t = \frac{\pi}{120} s\]

The particle will come to rest at \[\frac{\pi}{120} s\]

(b)  Acceleration is given as,
     a = ω²x 

\[= \omega^2 \left[ 5 \sin \left( 20t + \frac{\pi}{3} \right) \right]\]

For a = 0,

\[5 \sin \left( 20t + \frac{\pi}{3} \right) = 0\]

\[ \Rightarrow \sin \left( 20t + \frac{\pi}{3} \right) = \sin \pi\]

\[ \Rightarrow 20t = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\]

\[ \Rightarrow t = \frac{\pi}{30} s\]

(c) The maximum speed \[\left( v \right)\] is given by,

\[v = A\omega\cos \left( \omega t + \frac{\pi}{3} \right)\]

 (using \[v = \frac{\text {dx}}{\text{dt}}\])

\[= 20 \times 5 \cos \left( 20t + \frac{\pi}{3} \right)\]

 For maximum velocity : 

\[\cos \left( 20t + \frac{\pi}{3} \right) = - 1 = \cos \pi\]

\[ \Rightarrow 20t = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\]

\[ \Rightarrow t = \frac{\pi}{30} s\]