Physics - SHM Question with Solution | TestHub

$\mathrm{x}={\mathrm{x}}_0{\cos }^2\mathrm{\omega t}$

$\therefore \mathrm{x}={\mathrm{x}}_0\left(\frac{\cos 2\mathrm{\omega t}+1}{2}\right)$

$\Rightarrow \mathrm{x}=\frac{{\mathrm{x}}_0}{2}+\frac{{\mathrm{x}}_0}{2}\cos 2\mathrm{\omega t}$
$\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=–\frac{{\mathrm{x}}_0}{2}\sin 2\mathrm{\omega t}\mathrm{ }.\mathrm{ }2\mathrm{\omega }$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=–{\mathrm{\omega x}}_0\sin 2\mathrm{\omega t}$
So, the speed at mean position is ${\mathrm{v}}_0={\mathrm{\omega x}}_0$