Question

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

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

Answer 1

Answer: (C)

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