Physics - SHM Question with Solution | TestHub

$\because x=A\sin \omega t \Rightarrow sin \omega t=\frac{x}{A}$

Also,$cos \omega t= \sqrt{1-{\sin }^2\omega t}= \sqrt{1-\frac{x^2}{A^2}}$

$\Rightarrow \cos \omega t= \frac{\sqrt{A^2-x^2}}{A}$

As,$y=2A\sin \omega t\cos \omega t$

$\Rightarrow y=2 A\frac{x}{A} \frac{\sqrt{A^2-x^2}}{A}$

$\Rightarrow y=\frac{2}{A} x \sqrt{A^2-x^2}$

$\Rightarrow y=0 at x=0 and x= \pm A$

Which is possible only in option (C)