Question

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

Answer 1

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

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