# 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. - Physics

Sum

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 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 = ω2

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

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 :

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

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

$\Rightarrow t = \frac{\pi}{30} s$

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 12 Simple Harmonics Motion
Q 6 | Page 252