Physics - SHM Question with Solution | TestHub

For $A=-B$ and $C=2B$ $$X=B\cos 2\omega t+B\sin 2\omega t=\sqrt{2B}\sin \left(2\omega t+\frac{\pi }{4}\right)$$ This is equation of SHM of amplitude $\sqrt{2}B$ If $A=B$ and $C=2B$, then $X=B+B\sin 2\omega t$ This is also equation of SHM about the point $X=B$. Function oscillates between $X=0$ and $X=2B$ with amplitude $B$.