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

The correct option is C

$4\pi$

Given that, $4v^2=25-x^2$
$\Rightarrow 8v\frac{dv}{dt}=0-2x\frac{dx}{dt}$
$\Rightarrow 8va=-2xv$ $(\because a=\frac{dv}{dt},v=\frac{dx}{dt})$
where a is the acceleration.
Thus, $a=-\left(\frac{1}{4}\right)x$ ..... (i)
For a simple harmonic motion, $a=-{\omega }^{2}x$ ...... (ii)
Comparing (i) and (ii), we have, $\omega =\frac{1}{2}$
$\Rightarrow \frac{2\pi }{T}=\frac{1}{2}$
$\Rightarrow T=4\pi$
Hence, the correct choice is (c).