The equation of motion of a particle started at* t *= 0 is given by *x* = 5 sin (20*t* + 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?

#### Solution

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* = *ω*^{2}*x *

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

*For a = 0,*

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

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

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

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

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

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

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