​ The function of time representing a simple harmonic motion with a period of π/ω is :

Question

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

(1) sin(ωt) + cos (ωt)

(2) cos(ωt) + cos (2ωt) + cos (3ωt)

(3) sin²(ωt)

(4) 3 cos(\(\frac{\pi}4-2\omega t\))

Answer 1

Correct option is (4) \(3 cos(\frac{\pi}4-\omega t)\)

Time period T = \(\frac{2\pi}{\omega'}\)

\(\frac{\pi}\omega=\frac{2\pi}{\omega'}\)

ω' = 2 ω → Angular frequency of SHM Option (3)

sin² ωt = \(\frac 12\)(2 sin² ωt ) = \(\frac 12\)(1 - cos 2ωt)

Angular frequency of (\(\frac 12-\frac 12\) cos 2ωt) is 2ω

Option (4)

Angular frequency of SHM

\(3cos(\frac{\pi}4-2\omega t)\) is 2ω.

So option (3) & (4) both have angular frequency 2w but option (4) is direct answer.