Question

A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean position is 1 cm, the numerical value of the magnitude of acceleration is equal to the numerical value of the magnitude of the velocity. The frequency of SHM is:

• A. $2\pi \sqrt{3}$
• B. $\frac{2\pi }{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2\pi }$
• D. $\frac{1}{2\pi \sqrt{3}}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2\pi }$

Let y is the displacement $y=a\sin \omega t$

$\omega =\frac{2\pi }{T}$

It is given that,

Displacement y = 1

Amplitude a = 2

So $1=3\sin \frac{2\pi t}{T}$

Since, acceleration in SHM is $a={\omega }^{2}y$

And velocity in SHM is $v=\omega \sqrt{a^2-y^2}....\left(1\right)$

$\because v=\omega$

Put all the values in equation (1)

$\frac{4{\pi }^2}{T^2}=\frac{2\pi }{T}\left(\sqrt{a^2-1}\right)$

$\frac{2\pi }{T}=\left(\sqrt{3}\right)$

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