Question

A particle is performing a linear simple harmonic motion of amplitude ${}^{\prime }{A}^{\prime }$ . When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

• A. $\frac{2\pi }{\sqrt{3}}s$
• B. $\frac{\sqrt{3}}{2\pi }s$
• C. $2\pi \sqrt{3}s$
• D. $\frac{1}{2\pi \sqrt{3}}s$

Answer 1

Answer: (A)

In linear simple harmonic motion, the velocity of particle is given by
$v=\omega \sqrt{A^2-x^2}...(i)$
where, $\omega =$ angular frequency
$A=$ maximum displacement of amplitude
and $x=$ displacement from mean position.
The acceleration of a particle in simple harmonic motion, (SHM) is given by
$a={\omega }^{2}x...(ii)$
Here, $x=\frac{A}{2}$
Also, $,v=a$ (given)
$\omega \sqrt{\left(A^2-x^2\right)}={\omega }^{2}x[$ from Eqs. (i) and (ii), we get]
$\Rightarrow \sqrt{\left(A^2-\frac{A^2}{4}\right)}=\omega \times \frac{A}{2}\Rightarrow \frac{\sqrt{3}A}{2}=\omega \times \frac{A}{2}$
$\Rightarrow \frac{2\pi }{T}=\sqrt{3}\left[\because \omega =\frac{2\pi }{T}\right]$
$\Rightarrow T=\frac{2\pi }{\sqrt{3}}S$