Question

A wave represented by equation $y=2\left(mm\right)sin\left[4\pi \right(se{c}^{-1})t-2\pi (m^{-1}\left)X\right]$ is superimposed with another wave $y=2\left(mm\right)sin\left[4\pi \right(se{c}^{-1})t+2\pi (m^{-1})x+\pi /3]$ on a tight string.
Phase difference between two particles with are located at $x_1=1/7$ and $x_2=5/12$ is :

• A. $0$
• B. $\frac{5\pi }{6}$
• C. $\pi$
• D. $\frac{5\pi }{3}$

Answer 1

The correct option is C $\pi$

$y_1=2\sin \left[4\pi t-2\pi x\right]$

$y_2=2\sin \left[4\pi t-2\pi x\right]$

$y=y_1+y_2=2\left[2\sin 4\pi t\cos 4\pi x\right]$

$=4\sin 4\pi t\cos 4\pi x$

$y=4\cos 4\pi \sin 4\pi t$

Amp pass$\left|4\cos 4\pi x\right|=4$

$\Rightarrow \cos 4\pi x=\pm 1$

$=0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},\frac{5}{4},\frac{6}{4}.......$

$\therefore for{x}_1=\frac{1}{9}and{x}_2=\frac{5}{12}$

hence,

phase difference is$\pi$