Out of the following functions representing motion of a particle which represents SHM :
(i) $\mathrm{y}= \sin \mathrm{\omega t}-\cos \mathrm{\omega t}$
(ii) $\mathrm{y}= {\sin }^3 \mathrm{\omega t}$
(iii) $\mathrm{y}= 5 \cos \left(\frac{3\mathrm{\pi }}{4}-3\mathrm{\omega t}\right)$
(iv) $\mathrm{y}=1+\mathrm{\omega t}+{\mathrm{\omega }}^2{\mathrm{t}}^2$

 $$\begin{gathered} \mathrm{For} \left(\mathrm{i}\right), \mathrm{y}=\sin \mathrm{\omega t}-\cos \mathrm{\omega t}=\sqrt{2}\left[\frac{1}{\sqrt{2}}\sin \mathrm{\omega t}-\frac{1}{\sqrt{2}}\cos \mathrm{\omega t}\right] \\ =\sqrt{2}\left[\sin \mathrm{\omega t} \cos \frac{\mathrm{\pi }}{4}- \cos \mathrm{\omega t} \sin \frac{\mathrm{\pi }}{4}\right] \\ =\sqrt{2}\sin \left(\mathrm{\omega t}-\frac{\mathrm{\pi }}{4}\right)\mathrm{it} \mathrm{is} \mathrm{SHM} \\ \mathrm{For} \left(\mathrm{ii}\right), \mathrm{y}={\sin }^3 \mathrm{\omega t} \left(\therefore \sin 3 \mathrm{A}=3 \sin \mathrm{A}-4 {\sin }^3 \mathrm{A}\right) \\ \mathrm{y}=\frac{1}{4}\left(3 \sin \mathrm{\omega t}-\sin 3\mathrm{\omega t}\right) \\ =\frac{3}{4}\sin \mathrm{\omega t}-\frac{1}{4}\sin 3\mathrm{\omega t} \end{gathered}$$
It is superposition of two SHMs of different frequencies. So it is not SHM.
$\mathrm{For} \left(\mathrm{iii}\right), \mathrm{y}=5 \cos \left(\frac{3\mathrm{\pi }}{4}-3\mathrm{\omega t}\right)$