Question

The equation of motion of a particle started at $t=0$ is given by $x=5$ sin $(20t+\frac{\pi }{3}),$ 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

$x=5sin(20t+\frac{\pi }{3})$

(a) Max displacement from the mean position = Amplitude of the particle.

At the extreme position, the velocity becomes '0'.

$\therefore x=5=$ Amplitude

$\therefore 5=5sin(20t+\frac{\pi }{3})$

$sin(20t+\frac{\pi }{3})=1=sin\frac{\pi }{2}$

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

$\Rightarrow t=\frac{\pi }{120}sec$

So, at $\frac{\pi }{120}sec$ it first comes to rest.

$\left(b\right)a={\omega }^{2}x$

$={\omega }^2\left[5sin(205+\frac{\pi }{3})\right]$

For $a=0$,

$5sin(20t+\frac{\pi }{3})=0$

$\Rightarrow sin(20t+\frac{\pi }{3})=sinx$

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

$\Rightarrow t=\frac{\pi }{30}sec.$

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

$=20\times 5cos(20t+\frac{\pi }{3})$

when, v is maximum

$i.e.cos(20t+\frac{\pi }{3})=-1=cos\pi$

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

$\Rightarrow t=\frac{\pi }{30}sec.$