Question

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

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

Answer 1

The correct option is C $\frac{2\pi }{\sqrt{\alpha }}$

The equation of SHM,
$\frac{d^{2}x}{d{t}^2}+\alpha x=0$ or $\frac{d^{2}x}{d{t}^2}=-\alpha x$
Comparing it with the equation of SHM
$\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x;{\omega }^2=\alpha or\omega =\sqrt{\alpha }\therefore T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\alpha }}$.