Question

A particle moves such that its acceleration is given by

$a=-\beta \left(x - 2\right)$

Here : $\beta $ is a positive constant and $x$ is the position from origin. Time period of oscillations is

• A. $2\pi \sqrt{\beta }$
• B. $2\pi \sqrt{\frac{1}{\beta }}$
• C. $2\pi \sqrt{\beta + 2}$
• D. $2\pi \sqrt{\frac{1}{\beta + 2}}$

Answer 1

Answer: (B)

$a=-\beta \left(x - 2\right)$
as $a=-\omega^{2}\left(x - x_{0}\right)$
by comparing
$\omega ^{2}=\beta $
$\Rightarrow T=2\pi \sqrt{\frac{1}{\beta }}$