Question

If a simple harmonic motion is represented by $\frac{d^{2}x}{\text{d}\text{t}^{2}} + \alpha x = 0$, its time period is

• A. $2\pi\sqrt{\alpha}$
• B. 2$\pi\alpha$
• C. $\frac{2\pi}{\sqrt{\alpha}}$
• D. $\frac{2\pi}{\alpha}$

Answer 1

Answer: (C)

$\frac{2\pi}{\sqrt{\alpha}}$

Answer 2

The equation of SHM,

$\frac{d^{2}x}{dt^{2}} + \alpha x = 0$ or $\frac{d^{2}x}{dt^{2}} = - \alpha x$

Comparing it with the equation of SHM

$\frac{d^{2}x}{dt^{2}} = - \omega^{2}x$

$\omega^{2} = \alpha or\omega = \sqrt{\alpha}$

$\therefore T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\alpha}}$