Question

A particle is performing SHM along a straight line. Its amplitude is A and time period T.The minimum time interval between two instants, when magnitude of its acceleration become $\frac{2\sqrt{3}{\pi }^{2}A}{T^2}$ is :

• A. T/6
• B. T/12
• C. T/3
• D. 2T/3

Answer 1

The correct option is C T/3

Hence, option $C$ is correct answer.

$a=w^{2}x$

$\Rightarrow \frac{2\sqrt{3}{\pi }^{2}A}{7^2}=w^{2}x$

$\Rightarrow \frac{2\sqrt{3}{\pi }^{2}A}{7^2}=\frac{4{\pi }^2}{7^2}x$

$\Rightarrow x=\frac{\sqrt{3}A}{2}$

Refer image.

$\sin \theta =\frac{\sqrt{3}A}{2A}$

$=\frac{\sqrt{3}}{2}$

$=\theta ={60}^0$

$\therefore$ required time is $t=\frac{2\theta }{w}=\frac{2\pi T}{3\times 2\pi }=\frac{T}{3}$