Question

A particle executes linear simple harmonic motion with an amplitude of $3\text{cm}$. When the particle is at $2\text{cm}$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Find the time period in seconds.

• A. $\frac{\sqrt{5}}{2\pi }$
• B. $\frac{4\pi }{\sqrt{5}}$
• C. $\frac{2\pi }{\sqrt{3}}$
• D. $\frac{\sqrt{5}}{\pi }$

Answer 1

The correct option is B $\frac{4\pi }{\sqrt{5}}$

Velocity-displacement relationship is given by
$v=\omega \sqrt{A^2-x^2}$
Acceleration-displacement relationship is given by
$|a|={\omega }^{2}x$
Given that,
$|\overrightarrow{v}|=|\overrightarrow{a}|$
$\Rightarrow \omega \sqrt{A^2-x^2}={\omega }^{2}x$
$\Rightarrow {\omega }^2(A^2-x^2)={\omega }^4{x}^2$
Given that amplitude $A=3\text{cm}$ & required distance $x=2\text{cm}$
$\Rightarrow {\omega }^2=\frac{A^2-x^2}{x^2}=\frac{9-4}{4}=\frac{5}{4}$
$\Rightarrow \omega =\frac{\sqrt{5}}{2}$
$\Rightarrow T=\frac{4\pi }{\sqrt{5}}\text{seconds}$
$(\because T=\frac{2\pi }{\omega })$