Question

A particle executes simple harmonic motion on a straight line. The amplitude of oscillation is $2\text{cm}$. At the instant when the displacement of the particle from mean position is $1\text{cm}$, the magnitude of acceleration is equal to the magnitude of velocity. The frequency of the simple harmonic motion is

• A. $\frac{\sqrt{3}}{7\pi }\text{Hz}$
• B. $\frac{\sqrt{3}}{5\pi }\text{Hz}$
• C. $\frac{\sqrt{3}}{2\pi }\text{Hz}$
• D. $\frac{\sqrt{3}}{4\pi }\text{Hz}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2\pi }\text{Hz}$

Given:
Amplitude of the oscillation, $A=2\text{cm}$
Displacement of the particle from mean position, $r=1\text{cm}$
It is given that when displacement of the particle from mean position is $1\text{cm}$, the magnitude of acceleration is equal to the magnitude of velocity of the particle.
$\Rightarrow a=v$
$\Rightarrow {\omega }^{2}x=\omega \sqrt{A^2-x^2}$
$\Rightarrow \omega (\omega x-\sqrt{A^2-x^2})=0$
$\Rightarrow \omega =0$ or $\omega =\frac{\sqrt{A^2-x^2}}{x}$
Angular frequency cannot be zero.
So, $\omega =\frac{\sqrt{A^2-x^2}}{x}=\frac{\sqrt{2^2-1^2}}{1}=\sqrt{3}\text{rad/s}$
Hence, the frequency of oscillation,
$f=\frac{\omega }{2\pi }$
$\Rightarrow f=\frac{\sqrt{3}}{2\pi }\text{Hz}$