Question

Three travelling waves in same direction are superimposed. The equations of wave are $y_1=A_0\sin (kx-\omega t),y_2=3\sqrt{2}{A}_0\sin (kx-\omega t+\varphi )$ and $y_3=4A_0\cos (kx-\omega t)$. If $0\le \varphi \le \frac{\pi }{2}$ and the phase difference between resultant-wave and. first wave is $\frac{\pi }{4}$, then $\varphi$ is________.

Answer 1

Answer: 0.26

$\tan \left(\frac{\pi }{4}\right)=\frac{BC}{AC}=\frac{A_0(4+3\sqrt{2}\sin \varphi )}{A_0(1+3\sqrt{2}\cos \varphi )}$

$\Rightarrow \cos \varphi -\sin \varphi =\frac{1}{\sqrt{2}}$
Squaring both sides,
$\Rightarrow {\cos }^2\varphi +{\sin }^2\varphi -2\cos \varphi \sin \varphi =\frac{1}{2}$
$\Rightarrow 2\sin \varphi \cos \varphi =\frac{1}{2}$
$\Rightarrow \sin 2\varphi =\frac{1}{2}$
$\Rightarrow \varphi =\frac{1}{2}{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{12}=0.26$