Question

The velocity of a particle of mass $m$ moving along a straight line varies with time $t$ as $\frac{d^{2}v}{d{t}^2}=-kv$, the particle's acceleration was zero when it started its motion from $x=0$. Here $k$ is a positive constant. Choose the incorrect statement.

• A. The particle is performing $SHM$
• B. The particle has a time period of $\frac{\pi }{\sqrt{k}}$
• C. The particle has angular frequency of $\sqrt{k}$
  
• D. The particle has a time period of $\frac{2\pi }{\sqrt{k}}$

Answer 1

The correct option is B The particle has a time period of $\frac{\pi }{\sqrt{k}}$

Given,
$\frac{d^{2}v}{d{t}^2}=-kv$
$\Rightarrow \frac{d}{dt}\left(\frac{dv}{dt}\right)=-k\frac{dx}{dt}$
$\Rightarrow \frac{da}{dt}=-k\frac{dx}{dt}$
$\Rightarrow da=-kdx$
Integrating both sides, we get,
${\int }_0^{a}da=-k{\int }_0^{x}dx$
$[\because a=0\text{at}x=0]$
$\Rightarrow a+kx=0$
$\Rightarrow \frac{d^{2}x}{d{t}^2}+kx=0$
This equation is similar to the equation of free oscillations.
$\therefore$ Particle performs SHM.
On comparing with the equation ,
$\frac{d^{2}x}{d{t}^2}+{\omega }_o^{2}x=0$, we get,
Angular frequency, ${\omega }_o=\sqrt{k}$
Now, Time period of oscillation,
$T=\frac{2\pi }{{\omega }_o}=\frac{2\pi }{\sqrt{k}}$
Thus, option (b) is the required answer.