Question

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

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

Answer 1

Answer: (D)

Time period $T=\frac{2\pi }{{\omega }^{\prime }}$
$\frac{\pi }{\omega }=\frac{2\pi }{{\omega }^{\prime }}$
${\omega }^{\prime }=2\omega \to$ Angular frequency of SHM Option (3)
${\sin }^2\omega t=\frac{1}{2}\left(2{\sin }^2\omega t\right)=\frac{1}{2}(1-\cos 2\omega t)$
Angular frequency of $\left(\frac{1}{2}-\frac{1}{2}\cos 2\omega t\right)$ is $2\omega$
Option (4)
Angular frequency of $SHM$
$3\cos \left(\frac{\pi }{4}-2\omega t\right)$ is $2\omega$.
So option (3) & (4) both have angular frequency $2\omega$ but option (4) is direct answer