Question

The function of time representing a simple harmonic motion with a period of $\frac{\pi }{\omega }$ is :

• A. $\cos \left(\omega t\right)+\cos \left(2\omega t\right)+\cos \left(3\omega t\right)$
• B. $3\cos (\frac{\pi }{4}-2\omega t)$
• C. ${\sin }^2\left(2\omega t\right)$
• D. $\sin \left(\omega t\right)+\cos \left(\omega t\right)$

Answer 1

The correct option is B $3\cos (\frac{\pi }{4}-2\omega t)$

Given :
Time period of SHM,
$T=\frac{\pi }{\omega }$

We know that for SHM,
$T=\frac{2\pi }{{\omega }^{\prime }}$

Where, ${\omega }^{\prime }$ is the angular frequency of the SHM.

$\Rightarrow \frac{\pi }{\omega }=\frac{2\pi }{{\omega }^{\prime }}$

$\Rightarrow {\omega }^{\prime }=2\omega$

From the given functions, only $3\cos (\frac{\pi }{4}-2\omega t)$ has angular frequency of $2\omega$.

Therefore, option $\left(B\right)$ is the correct answer.