Q. If a simple harmonic is represented by $\frac{d^{2}x}{dt^{2}} + \alpha x = 0$, its time period is
A) $\frac{2\pi}{\alpha}$
B) $\frac{2\pi}{\sqrt{\alpha}}$
C) $2\pi\alpha$
D) $2\pi\sqrt{\alpha}$
Explanation
(b) $\frac{d^{2}x}{dt^{2}} = - \alpha x$ …(i)
We know,
$a = \frac{d^{2}y}{dt^{2}} = - \omega^{2}x$ …(ii)
From Eqs.(i) and (ii), we have
$\omega^{2} = \alpha$
$\omega = \sqrt{\alpha}$
or $\frac{2\pi}{T} = \sqrt{\alpha}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = \frac{2\pi}{\sqrt{\alpha}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.