Out of the following functions representing motion of a particle which represents SHM (1) y = sin ω t - cos ω t (2) y = sin3 ω t (3) y = 5cos (3π/4 - 3ω t) (4) y = 1 + ω t + ω 2t2 (a) Only (1) (b) Only (4) does not represent SHM (c) Only (1) and (3) (d) Only (1) and (2)

Answer: (c). (1) sin ω t - cos ω t = √(2)sin (ω t - π/4): a single sine term, so SHM. (2) sin ³ω t = (3sin ω t - sin 3ω t)/(4): two different frequencies, periodic but not SHM. (3) 5cos (3π/4 - 3ω t): a single cosine term, so SHM. (4) 1 + ω t + ω² t² grows without limit, not even periodic. So the answer is (c) only (1) and (3).

Out of the following functions representing motion of a particle… | NEET 2011 (AIPMT, Prelims)

Q68NEET 2011 · AIPMT · PRELIMSEasy

Periodic Motion & SHM Basics

Out of the following functions representing motion of a particle which represents SHM (1) y = sin $\omega$ t $-$ cos $\omega$ t

(2) y = sin3 $\omega$ t

(3) y = 5cos $\left(\frac{3\pi}{4} - 3\omega t\right)$

(4) y = 1 + $\omega$ t + $\omega$ 2t2

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Answer: (c)

(1) $\sin\omega t - \cos\omega t = \sqrt{2}\sin(\omega t - \pi/4)$ : a single sine term, so SHM.
(2) $\sin^3\omega t = \frac{3\sin\omega t - \sin 3\omega t}{4}$ : two different frequencies, periodic but not SHM.
(3) $5\cos\left(\frac{3\pi}{4} - 3\omega t\right)$ : a single cosine term, so SHM.
(4) $1 + \omega t + \omega^2 t^2$ grows without limit, not even periodic.
So the answer is (c) only (1) and (3).

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Periodic Motion & SHM Basics question from NEET 2011 (AIPMT, Prelims)

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