Question

The time period of the simple harmonic motion represented by the equation $\frac{d^{2}x}{d{t}^2}+\alpha x=0$ is

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

Answer 1

The correct option is D 2 $\pi /\sqrt{\alpha }$

In a simple harmonic motion $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$ where $\omega$ is the angular frequency. Comparing it with $\frac{d^{2}x}{d{t}^2}=-\alpha x$ we get ${\omega }^2=\alpha or\omega =\sqrt{\alpha }or\frac{2\pi }{T}=\sqrt{\alpha }orT=\frac{2\pi }{\sqrt{\alpha }}$, Hence the correct choice is (d)