Question

A particle is performing a linear simple harmonic motion of amplitude $'A'$ . 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 $