Question

If the displacement x and velocity v of a particle executing simple harmonic motion are related through the expression $4 v^{2}=25-x^{2}$, then its time period is

• A. $\pi$
• B. $2 \pi$
• C. $4 \pi$
• D. $6 \pi$

Answer 1

Answer: (C)

$4 v^{2}=25-x^{2}$
Differentiating it both sides, we get
$4\left(2 v \frac{d v}{d t}\right)=-2 x \frac{d x}{d t} \text { or } 4 \frac{d v}{d t}=-x \,\,\,\,\,\,\left(\because \frac{d x}{d t}=v\right)$
or $4 a=-x \,\,\,\,$ or $\,\,\,\,\,a=-\frac{1}{4} x \,\,\,\,\,\,\,\left(\because \frac{d v}{d t}=a\right)$
Comparing it with $a=-\omega^{2} x$, we get $\omega^{2}=\frac{1}{4}$ or $\omega=\frac{1}{2}$
$\therefore \,\,\,\,\,\,$ Time period, $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{1 / 2}=4 \pi$