Question

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

• 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

Answer: (B)

$v=\omega \sqrt{A^{2}-x^{2}}$
$a=x \omega^{2}$
$v=a$
$\omega \sqrt{A^{2}-x^{2}}=x \omega^{2}$
$\sqrt{(3)^{2}-(2)^{2}}=2\left(\frac{2 \pi}{T}\right)$
$\sqrt{5}=\frac{4 \pi}{T}$
$T=\frac{4 \pi}{\sqrt{5}}$