Question

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

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

Answer 1

Answer: (D)

$\frac{4\pi }{\sqrt{5}}$

Amplitude, $A=3cm$

In Harmonic Motion,

$x=A\sin wt$, where $w=\frac{2\pi }{T}$

$v=Aw\cos wt$

Acceleration, $a=-A{w}^2\sin wt$

At, $x=2cm$

$|v|=|a|$

$|Aw\cos wt|=|-A{w}^2\sin wt|$ ...(1)

$A\sin wt=2cm$

$3\sin wt=2$

$\sin wt=\frac{2}{3}$

$\cos wt=\sqrt{1-(\frac{2}{3}{)}^2}=\frac{\sqrt{5}}{3}$

From eqn. (1)

$|3w\frac{\sqrt{5}}{3}|=|-3w^2\times \frac{2}{3}|$

$\Rightarrow w=\frac{\sqrt{5}}{2}$

$T=\frac{4\pi }{\sqrt{5}}$