Question

Three travelling waves in same direction are superimposed. The equations of wave are $y_{1}=A_{0} \sin (k x-\omega t), y_{2}=3 \sqrt{2} A_{0} \sin (k x-\omega t+\phi)$ and $y_{3}=4 A_{0} \cos (k x-\omega t)$. If $0 \leq \phi \leq \frac{\pi}{2}$ and the phase difference between resultant-wave and. first wave is $\frac{\pi}{4}$, then $\phi$ is________.

Answer 1

$\tan \left(\frac{\pi}{4}\right)=\frac{ BC }{ AC }=\frac{ A _{0}(4+3 \sqrt{2} \sin \phi)}{ A _{0}(1+3 \sqrt{2} \cos \phi)}$

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