Question

Three travelling waves in same direction are superimposed. The equation 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 $0\le \varphi \le \pi /2$ and the phase difference between resultant wave and first wave is $\pi /4$, then $\varphi$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{12}$
• D. none of these

Answer 1

The correct option is C $\frac{\pi }{12}$

Given, $y_1=A_{0}sin(kx-\omega t)$
$y_2=3\sqrt{2}{A}_{0}cos(kx-\omega t+\varphi )$
and $y_3=4A_{0}cos(kx-\omega t)$,
These waves can be represented by phase diagram as shown.
From phase diagram,
tan $\left(\frac{\pi }{4}\right)=\frac{BC}{AC}=\frac{A_0(4+3\sqrt{2}sin\varphi )}{A_0(1+3\sqrt{2}cos\varphi )}$
$⟹4+3\sqrt{2}sin\varphi =1+3\sqrt{2}cos\varphi$
$⟹cos\varphi -sin\varphi =dfrac1\sqrt{2}$
squaring both sides, we get
$co{s}^2\varphi +si{n}^2\varphi -2sin\varphi cos\varphi =\frac{1}{2}$
$1-2sin\varphi cos\varphi =\frac{1}{2}$ or $2sin\varphi cos\varphi =\frac{1}{2}$
or 2 $sin2\varphi =\frac{1}{2}=sin\frac{\pi }{6}$ or $2\varphi =\frac{\pi }{6}⟹\varphi =\frac{\pi }{12}$