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Out of the following functions representing motion of a particle, which represents SHM?
(A) $y = \sin\omega t - \cos\omega t$
(B) $y = \sin^3\omega t$
(C) $y = 5\cos\left(\frac{3\pi}{4} - 3\omega t\right)$
(D) $y = 1 + \omega t + \omega^2t^2$
(A) $y = \sin\omega t - \cos\omega t$
(B) $y = \sin^3\omega t$
(C) $y = 5\cos\left(\frac{3\pi}{4} - 3\omega t\right)$
(D) $y = 1 + \omega t + \omega^2t^2$
A
Only (A) and (B)
B
Only (A)
C
Only (D) does not represent SHM
D
Only (A) and (C)
Detailed Solution
A motion is SHM if the displacement is a single sine or cosine function of time, i.e., if $\frac{d^2y}{dt^2} \propto -y$.
(A) $y = \sin\omega t - \cos\omega t = \sqrt{2}\sin\left(\omega t - \frac{\pi}{4}\right)$: a single sine function, so it is SHM with angular frequency $\omega$.
(B) $y = \sin^3\omega t = \frac{1}{4}(3\sin\omega t - \sin3\omega t)$: a superposition of two SHMs of different frequencies; it is periodic but not SHM.
(C) $y = 5\cos\left(\frac{3\pi}{4} - 3\omega t\right) = 5\cos\left(3\omega t - \frac{3\pi}{4}\right)$: a single cosine function, so it is SHM with angular frequency $3\omega$.
(D) $y = 1 + \omega t + \omega^2t^2$: y keeps increasing with time, so the motion is not even periodic.
Hence only (A) and (C) represent SHM.
(A) $y = \sin\omega t - \cos\omega t = \sqrt{2}\sin\left(\omega t - \frac{\pi}{4}\right)$: a single sine function, so it is SHM with angular frequency $\omega$.
(B) $y = \sin^3\omega t = \frac{1}{4}(3\sin\omega t - \sin3\omega t)$: a superposition of two SHMs of different frequencies; it is periodic but not SHM.
(C) $y = 5\cos\left(\frac{3\pi}{4} - 3\omega t\right) = 5\cos\left(3\omega t - \frac{3\pi}{4}\right)$: a single cosine function, so it is SHM with angular frequency $3\omega$.
(D) $y = 1 + \omega t + \omega^2t^2$: y keeps increasing with time, so the motion is not even periodic.
Hence only (A) and (C) represent SHM.