Four simple harmonic vibrations $x_1=8$  sin wt, $x_2=6\sin \left(wt+\pi /2\right),x_3=4\sin \left(wt+\pi \right)$  and $x_4=2\sin \left(\omega t+\frac{3\pi }{2}\right)$  are superimposed on each other. The resulting amplitude is

Given

$x_1=8\sin \omega t$ , $x_2=6\sin \left(\omega t+\frac{\pi }{2}\right)$

$x_3=4\sin \left(\omega t+\pi \right)$ and $x_4=2\sin \left(\omega t+\frac{3\pi }{2}\right)$

From super position principle application displacement $x=x_1+x_2+x_3+x_4$

$x=8\sin \omega t+6\sin \left(\omega t+\frac{\pi }{2}\right)+4\sin \left(\omega t+\pi \right)+2\sin \left(\omega t+\frac{3\pi }{2}\right)$

$x=8\sin \omega t+6\cos \omega t-4\sin \omega t-2\cos \omega t$

$x=4\sin \omega t+4\cos \omega t$

$x=4(\sin \omega t+\cos \omega t)$

$x=4\left(\left(\sqrt{2}\sin \omega t\cos \frac{\pi }{4}\right)+\left(\sqrt{2}\cos \omega t\sin \frac{\pi }{4}\right)\right)$

$x=4\sqrt{2}\left(\sin \omega t\cos \frac{\pi }{4}\right)+\left(\cos \omega t\sin \frac{\pi }{4}\right)$

$\mathrm{Sin}A\mathrm{Cos}B+\mathrm{Cos}A\mathrm{Sin}B=\mathrm{Sin}(A+B)$

$\therefore x=4\sqrt{2}\sin (\omega t+\frac{\pi }{4})$

$\therefore$ The resulting amplitude =$4\sqrt{2}$