JEE Advanced · 2006 · 09 Apr 2006 · Elasticity & SHM

Using the identities ${\sin }^2\omega t=\frac{1-\cos 2\omega t}{2}$, ${\cos }^2\omega t=\frac{1+\cos 2\omega t}{2}$, and $\sin \omega t\cos \omega t=\frac{\sin 2\omega t}{2}$, we get:
$x=\frac{A+B}{2}+\frac{B-A}{2}\cos 2\omega t+\frac{C}{2}\sin 2\omega t$

This is of the form $x=x_0+R\sin (2\omega t+\varphi )$, which represents SHM about mean position $x_0$ with angular frequency $2\omega$ and amplitude $R=\sqrt{{\left(\frac{B-A}{2}\right)}^2+{\left(\frac{C}{2}\right)}^2}$.

For any A, B, C (except when both $A=B$ and $C=0$), this represents SHM (option A correct).

For $A=-B,C=2B$: $x=B\cos 2\omega t+B\sin 2\omega t$, amplitude $=|B\sqrt{2}|$ (option B correct).

For $A=B,C=0$: $x=A$ (constant), not SHM (option C incorrect).

For $A=B,C=2B$: $x=B+B\sin 2\omega t$, amplitude $=|B|$ (option D correct).