The function of time representing a simple harmonic motion with a period of (π)/(ω) is: (a) cos( ω t) + cos(2 ω t) + cos(3 ω t) (b) sin2( ω t) (c) sin( ω t) + cos( ω t) (d) 3cos (π/4 - 2ω t)

Answer: (d). SHM needs a single sine/cosine term; period T = (2π)/(ω'). (a) has three frequencies and (c) has period 2π/ω. (d) 3cos (π/4 - 2ω t) has ω' = 2ω, so T = (2π)/(2ω) = (π)/(ω). So the answer is (d).

The function of time representing a simple harmonic motion with a… | JEE Main 2021 (18 Mar, Shift 2)

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Periodic Motion & SHM Basics

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

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Answer: (d)

SHM needs a single sine/cosine term; period $T = \frac{2\pi}{\omega'}$ .
(a) has three frequencies and (c) has period $2\pi/\omega$ .
(d) $3\cos\left(\frac{\pi}{4} - 2\omega t\right)$ has $\omega' = 2\omega$ , so $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$ .
So the answer is (d).

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