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

The general equation of S..H.M. is give by,
$x=A\sin (\omega t+\varphi )$

(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.
$\therefore \mathrm{Displacement}$ x = 5, which is also the amplitude of the particle.
$$\begin{gathered} \Rightarrow 5=5\sin \left(20t+\frac{\mathrm{\pi }}{3}\right) \\ \mathrm{Now}, \\ \sin \left(20t+\frac{\mathrm{\pi }}{3}\right)=1=\sin \frac{\mathrm{\pi }}{2} \\ \Rightarrow 20t+\frac{\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{2} \\ \Rightarrow t=\frac{\mathrm{\pi }}{120}\mathrm{s} \end{gathered}$$

The particle will come to rest at $\frac{\mathrm{\pi }}{120}\mathrm{s}$

(b) Acceleration is given as,
a = ω²x
$={\mathrm{\omega }}^2\left[5\sin \left(20t+\frac{\mathrm{\pi }}{3}\right)\right]$

For a = 0,
$$\begin{gathered} 5\sin \left(20t+\frac{\mathrm{\pi }}{3}\right)=0 \\ \Rightarrow \sin \left(20t+\frac{\mathrm{\pi }}{3}\right)=\sin \pi \\ \Rightarrow 20t=\mathrm{\pi }-\frac{\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow t=\frac{\mathrm{\pi }}{30}\mathrm{s} \end{gathered}$$

(c) The maximum speed $\left(v\right)$ is given by,
$v=\mathrm{A}\omega \cos \left(\omega t+\frac{\mathrm{\pi }}{3}\right)$ (using $v=\frac{dx}{dt}$)
$=20\times 5\cos \left(20t+\frac{\mathrm{\pi }}{3}\right)$
For maximum velocity:
$$\begin{gathered} \cos \left(20t+\frac{\mathrm{\pi }}{3}\right)=-1=\cos \mathrm{\pi } \\ \Rightarrow 20t=\mathrm{\pi }-\frac{\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow t=\frac{\mathrm{\pi }}{30}\mathrm{s} \end{gathered}$$