Question

The equations of three waves are given by $y_1=A_0\sin (kx-\omega t),y_2=3\sqrt{2}{A}_0\sin (kx-\omega t+\varphi )$ and $y_3=4A_0\cos (kx-\omega t)$. 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 $\varphi =\frac{\pi }{n}\le \frac{\pi }{2}$, then what is the value of $n$ ?

Answer 1

Answer: 12

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

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