Question

A particle is subjected to SHM $x_1=A_1\sin \omega t$ and $x_2=A_2\sin (\omega t+\frac{\pi }{4})$ . The resultant SHM will have an amplitude of :

Answer 1

Two SHM’s are

$x_1=A_1\sin \omega t$

$x_2=A_2\sin \left(\omega t+\frac{\pi }{4}\right)$

Resultant amplitude is,

$A=\sqrt{\left({A_1}^2+{A_2}^2+2A_1{A}_2\cos \frac{\pi }{4}\right)}$

$A=\sqrt{\left({A_1}^2+{A_2}^2+2A_1{A}_2\times \frac{1}{\sqrt{2}}\right)}$

$A=\sqrt{\left({A_1}^2+{A_2}^2+\sqrt{2}{A}_1{A}_2\right)}$