Question

A particle executes simple harmonic motion with an amplitude of $5\text{cm}$. When the particle is at $4\text{cm}$ from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is

• A. $\frac{4\pi }{3}$
• B. $\frac{3}{8}\pi$
• C. $\frac{8\pi }{3}$
• D. $\frac{7}{3}\pi$

Answer 1

The correct option is C $\frac{8\pi }{3}$

For a particle executing SHM,
Velocity $v=\omega \sqrt{A^2-x^2}$..........(i)
Accelaration $a=-{\omega }^{2}x$..........(ii)
and according to question,
$|v|=|a|$
$\Rightarrow \omega \sqrt{A^2-x^2}={\omega }^{2}x$
$\Rightarrow A^2-x^2={\omega }^2{x}^2$
$\Rightarrow 5^2-4^2={\omega }^2\left(4^2\right)$
$\Rightarrow 3=\omega \times 4\Rightarrow \omega =\frac{3}{4}$

$\therefore T=\frac{2\pi }{\omega }=\frac{2\pi }{\frac{3}{4}}=\frac{8\pi }{3}\text{s}$