Question

The position-time $(x-t)$ equation of a particle moving along $x-$ axis is given by $x=A+A(1-\cos \omega t)$. Select the correct statement(s):

(Consider S.I units everywhere)

• A. The particle oscillates simple harmonically between points $x=2A$ and $x=A$
• B. Velocity of the particle is maximum at $x=2A$
• C. Minimum time taken by the particle in travelling from $x=A$ to $x=3A$ is $\frac{\pi }{\omega }$
• D. Minimum time taken by the particle in travelling from $x=A$ to $x=2A$ is $\frac{\pi }{2\omega }$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Velocity of the particle is maximum at $x=2A$
C Minimum time taken by the particle in travelling from $x=A$ to $x=3A$ is $\frac{\pi }{\omega }$
D Minimum time taken by the particle in travelling from $x=A$ to $x=2A$ is $\frac{\pi }{2\omega }$

Given,
$x=A+A(1-\cos \omega t)...\left(1\right)$
From $\left(1\right)$ we can say that,
$x_{\text{max}}=3A$ (when $\cos \omega t=-1)$
$x_{\text{min}}=A$ (when $\cos \omega t=1)$

Therefore, particle oscilates between $x=3A$ to $x=A$

Differentiating $\left(1\right)$ with respect to ${}^{\prime }{t}^{\prime }$ we get,

$v=\frac{dx}{dt}=0+0-(-A\omega \sin \omega t)$
$\Rightarrow v=A\omega \sin \omega t$

From the above equation we get,
$v_{\text{max}}=A\omega$, (when $\sin \omega t=1$)

When $\sin \omega t=1$, the value of $\cos \omega t=0$
Substituting, $\cos \omega t=0$ in $\left(1\right)$ we get,
$x=A+A(1-0)=2A$

Thus the velocity of particle is maximum at $x=2A$

Time period of oscillation , $T=\frac{2\pi }{\omega }$
When the particle in travelling from
$x=A$ to $x=3A$ . Particle is travelling from one extreme position to another extreme position , the time taken will be:
$t=\frac{T}{2}=\frac{\pi }{\omega }$

when the particle is travelling from $x=A$ to $x=2A$. Particle is travelling from one extreme position to its mean position, the time taken will be:
$t=\frac{T}{4}=\frac{\pi }{2\omega }$

Thus, options (b) , (c) and (d) are the correct answers.