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(A) function represents simple harmonic motion $(S.H.M.)$ if it satisfies the differential equation $\frac{d^2x}{dt^2} + \omega^2 x = 0$.Option $A$: $

Vedclass · Accepted Answer

$\sin \omega t - \cos \omega t$

Vedclass · Answer

(A) function represents simple harmonic motion $(S.H.M.)$ if it satisfies the differential equation $\frac{d^2x}{dt^2} + \omega^2 x = 0$.Option $A$: $x = \sin \omega t - \cos \omega t$. This can be written as $x = \sqrt{2} [\frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t] = \sqrt{2} \sin(\omega t - \frac{\pi}{4})$. This is a standard $S.H.M.$ equation.Option $B$: $x = \sin^2 \omega t = \frac{1 - \cos 2\omega t}{2}$. This represents motion with a constant shift and frequency $2\omega$,which is not simple harmonic.Options $C$ and $D$: These are superpositions of two different frequencies ($\omega$ and $2\omega$),which result in periodic motion but not simple harmonic motion.Therefore,the correct option is $A$.

Which of the following functions represents a simple harmonic oscillation?

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