Question

Equations of a stationary wave and a travelling wave are $y_1=asinkxcos\omega t$ and $y_2=asin(\omega t-kx)$. The phase difference between two points $x_1=\frac{\pi }{3k}and{x}_2=\frac{3\pi }{2k}is{\varphi }_1$ for the first wave and ${\varphi }_2$ for the second wave. The ratio $\frac{{\varphi }_1}{{\varphi }_2}$ is :

• A. $1$
• B. $\frac{5}{6}$
• C. $\frac{3}{4}$
• D. $\frac{6}{7}$

Answer 1

The correct option is D $\frac{6}{7}$

Phase difference between two points in a standing wave $=n\pi$

where $n$ is the no of nodes between two points.

Given points are $x_1=\pi /3K=60/K$

$x_2=3\pi /2K=210/K$

Equation of the standing wave.

$y_1=asinKxcoswt$

at node points, $Kx=n\pi$

$x=\frac{n\pi }{K}$ $(n=0,1,2,\mathrm{3......})$

so nodes are $=\frac{\pi }{K},\frac{2\pi }{K},\frac{3\pi }{K}.....$

$\frac{180}{K},\frac{360}{K},\frac{540}{K}$

Since there is only one node in b/w phase difference ${\varphi }_1=\pi$

for travelling wave

${\varphi }_2=\frac{2\pi }{\lambda }△x$

from the equation, $y_2=a\sin (wt-Kx)$

$K=\frac{2\pi }{\lambda }$

$\therefore {\varphi }_2=K[x_2-x_1]=K[\frac{3\pi }{2K}-\frac{\pi }{3K}]$

$=\pi [\frac{3}{2}-\frac{1}{3}]=\frac{7}{6}\pi$

$\therefore \frac{{\varphi }_1}{{\varphi }_2}=\frac{\pi }{\frac{7}{6}\pi }$

$=\frac{6}{7}$