Question
Show that the function of time $\text{y}=(\sin\omega\text{t}-\cos\omega\text{t})$ represents simple harmonic motion.

Answer

We have $\text{y}=\sin\omega\text{t}-\cos\omega\text{t}$

$=\sqrt{2}\Big(\sin\omega\text{t}\times\frac{1}{\sqrt{2}}-\cos\omega\text{t}\times\frac{1}{\sqrt{2}}\Big)$

$=\sqrt{2}\Big(\sin\omega\text{t}\cos\frac{\pi}{4}-\cos\omega\text{t}\sin\frac{\pi}{4}\Big)$ $\Big[\because\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}\text{ and}\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\Big]$

$=\sqrt{2}\sin\Big(\omega\text{t}+\frac{\pi}{4}\Big)$ $[\because\sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}]$

Moreover, $\text{y}\Big(\text{t}+\frac{2\pi}{\omega}\Big)=\sqrt{2}\Big(\omega\text{t}+2\pi-\frac{\pi}{4}\Big)=\text{y(t)}$

Hence. it represents simple harmonic motion.

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