Question

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

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