Show that the motion of a particle represented by $y = \sin \omega t - \cos \omega t$ is simple harmonic with a period of $\frac{2\pi}{\omega}$.

The given displacement equation is $y = \sin \omega t - \cos \omega t$.
To express this in the standard form of $SHM$,$y = A \sin(\omega t + \phi)$,we multiply and divide by $\sqrt{1^2 + (-1)^2} = \sqrt{2}$:
$y = \sqrt{2} \left[ \frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t \right]$
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,where $\cos(\pi/4) = 1/\sqrt{2}$ and $\sin(\pi/4) = 1/\sqrt{2}$:
$y = \sqrt{2} \left[ \sin \omega t \cos \frac{\pi}{4} - \cos \omega t \sin \frac{\pi}{4} \right]$
$y = \sqrt{2} \sin \left( \omega t - \frac{\pi}{4} \right)$
This equation is in the form $y = A \sin(\omega t + \phi)$,where the amplitude $A = \sqrt{2}$ and the phase constant $\phi = -\pi/4$.
The angular frequency is $\omega$. The time period $T$ is given by the relation $T = \frac{2\pi}{\omega}$.
Since the motion can be expressed as a sinusoidal function of time,it represents $SHM$ with a time period of $T = \frac{2\pi}{\omega}$.