Question

A particle moves so that its acceleration $a$ is given by $a=-bx$ , where $x$ is displacement from equilibrium position and $b$ is a non-negative real constant. The time period of oscillation of the particle is

• A. $2\pi \sqrt{b}$
• B. $\frac{2 \pi }{b}$
• C. $\frac{2 \pi }{\sqrt{b}}$
• D. $2\sqrt{\frac{\pi }{b}}$

Answer 1

Answer: (C)

The relation between acceleration ( $a$ ) and displacement ( $x$ ) for a body in SHM is
$a=-\omega ^{2}x$
Given, $a=-bx$
On comparing the two equations, we get
$\omega ^{2}=b$
$\therefore $ $\omega =\sqrt{b}$
Since, $\omega =\frac{2 \pi }{T}$
$\therefore $ $\frac{2 \pi }{T}=\sqrt{b}$
$\Rightarrow T=\frac{2 \pi }{\sqrt{b}}$