Question

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

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

Given,

$\frac{d^{2}x}{d{t}^2}+\alpha x=0$

$a+\alpha x=0$

$a=-\alpha x$. . . . ..(1)

The equation of motion in S.H.M is given by

$a=-{\omega }^{2}x$. . . . . .(2)

By comparaing equation (1) and (2), we get

${\omega }^2=\alpha$

$\omega =\sqrt{\alpha }$

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

Time period, $T=\frac{2\pi }{\sqrt{\alpha }}$

The correct option is B.