Question

The velocity of a particle of mass m moving along a straight changes with time 't' as $\frac{d^{2}V}{d{t}^2}=-KV$ where $K$ is a positive constant. Which of the following statements are correct ?

• A. The particle will perform SHM.
• B. The particle willl have time period $\frac{2\pi }{\sqrt{K}}$
• C. The particle will have time period $2\pi \sqrt{\frac{m}{K}}$
• D. The particle will have angular frequency $\sqrt{K}.$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The particle will perform SHM.
B The particle willl have time period $\frac{2\pi }{\sqrt{K}}$
D The particle will have angular frequency $\sqrt{K}.$

If we integrate the given equation,
$\frac{d^{2}V}{d{t}^2}=-KV$
$\frac{d^{3}x}{d{t}^3}=-K\frac{dx}{dt}$
We get,
$\frac{d^{2}x}{d{t}^2}=-Kx$
and if we compare with basic relation of S.H.M,
$a=-{\omega }^{2}x$
we get,
$\omega =\sqrt{K}$ and $T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{K}}$