Question

If a simple harmonic motion 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. $2\pi\sqrt{\alpha}$

Answer 1

Answer: (B)

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

Answer 2

$\frac{d^{2}x}{dt^{2}}=-\alpha\,x\,...\left(i\right)$ We know $a=\frac{d^{2}x}{dt^{2}}=-\omega^{2}\,x\,...\left(ii\right)$ From Eqs. $\left(i\right)$ and $\left(ii\right)$, we have $\omega^{2}=\alpha$ $\omega=\sqrt{\alpha}$ or $\frac{2\pi }{T}=\sqrt{\alpha}$ $\therefore T=\frac{2\pi }{T}=\sqrt{\alpha}$