Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Out of the following functions, which represents SHM? I. $y = \sin \omega t - \cos \omega t$ II. $y = \sin^3 \omega t$ III. $y = 5 \cos\left(\frac{3\pi}{4} - 3\omega t\right)$ IV. $y = 1 + \omega t + \omega^2 t^2$

Only (IV) does not represent SHM

(I) and (III)

(I) and (II)

Only (I)

Step-by-Step Solution

Condition for SHM: A function represents Simple Harmonic Motion (SHM) if it can be written in the form $y = A \sin(\Omega t + \phi)$ or $y = A \cos(\Omega t + \phi)$ , involving a single frequency.
Analysis of Function I: $y = \sin \omega t - \cos \omega t$ . Multiply and divide by $\sqrt{2}$ : $y = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t \right) = \sqrt{2} \sin\left(\omega t - \frac{\pi}{4}\right)$ This is in the standard SHM form with amplitude $\sqrt{2}$ and frequency $\omega$ . So, I is SHM.
Analysis of Function II: $y = \sin^3 \omega t$ . Using the identity $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$ , we get: $y = \frac{1}{4} (3 \sin \omega t - \sin 3\omega t)$ This represents the superposition of two SHMs with different frequencies ( $\omega$ and $3\omega$ ). While the motion is periodic, it is not SHM because it involves more than one frequency.
Analysis of Function III: $y = 5 \cos\left(\frac{3\pi}{4} - 3\omega t\right)$ . Using $\cos(-\theta) = \cos(\theta)$ : $y = 5 \cos\left(3\omega t - \frac{3\pi}{4}\right)$ This is in the standard SHM form with amplitude $5$ and angular frequency $3\omega$ . So, III is SHM.
Analysis of Function IV: $y = 1 + \omega t + \omega^2 t^2$ . This function increases indefinitely with time (parabolic growth) and never repeats. It is non-periodic and thus not SHM.
Conclusion: Only functions I and III represent SHM.

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