Question

A function of time given by $(\sin \omega t-\cos \omega t)$ represents

• A. simple harmonic motion
• B. non-periodic motion
• C. periodic but not simple harmonic motion
• D. oscillatory but not simple harmonic motion

Answer 1

The correct option is A simple harmonic motion

$\begin{array}{c}\sin \omega t-\cos \omega t\\ =\sqrt{2}\left[\frac{1}{\sqrt{2}}\sin \omega t-\frac{1}{\sqrt{2}}\cos \omega t\right]\\ =\sqrt{2}\left[\sin \omega t\times \cos \frac{\pi }{4}-\cos \omega t\times \sin \frac{\pi }{4}\right]\\ =\sqrt{2}\sin \left(\omega t-\frac{\pi }{4}\right)\\ thisfunctionrepresentsSHMasitcanbewrittenintheform:\\ a\sin \left(\omega t+\varphi \right)\\ itsperiodis,\frac{2\pi }{\omega }\end{array}$

Hence,

option $\left(A\right)$ is correct answer.