A particle is exhibiting simple harmonic motion has its displacement $x$ and velocity $v$ related as $4 v^2=25-x^2$. The time period of SHM is

• A

  $\pi \mathrm{s}$
• B

  $2 \pi \mathrm{s}$
• C

  $3 \pi \mathrm{s}$
• Correct answer
  D

  $4 \pi \mathrm{s}$

Solution

Given that, $4 v^2=25-x^2 \Rightarrow v^2=\frac{1}{4}\left(25-x^2\right)$
So, on comparing Eqs. (i) and (ii), we get Angular frequency, $\omega=\frac{1}{2} \mathrm{rad} / \mathrm{s}$ Amplitude, $A=5 \mathrm{~m}$ So, time period, $T=\frac{2 \pi}{\omega} \Rightarrow T=\frac{2 \pi}{(1 / 2)} \Rightarrow T=4 \pi \mathrm{s}$

Hence, the correct answer is option: 'D'.