Question

The equations of a travelling and stationary waves are $y_1=a\sin (\omega t-kx)$ and $y_2=a\sin kx\cos \omega t.$ The phase difference between two points $x_1=\frac{\pi }{4k}$ and $x_2=\frac{4\pi }{3k}$ are ${\varphi }_1$ and ${\varphi }_2$ respectively for two. waves, where $k$ is the wave number. The ratio of ${\varphi }_1/{\varphi }_2$ is:

• A. 6/7
• B. 16/3
• C. 12/13
• D. 13/12

Answer 1

Answer: (C)

$\Delta x=x_2-x_1=\left(\frac{4}{3}-\frac{1}{4}\right)\frac{\pi }{k}$
$=\frac{13}{12}\frac{\pi }{k}$
$\sin k{x}_1=\sin k\left(\frac{\pi }{4k}\right)=\sin \frac{\pi }{4}\ne 0$
$\sin k{x}_2=\sin k\left(\frac{4\pi }{3k}\right)$
$=\sin \left(\pi +\frac{\pi }{3}\right)\ne 0$
$x_1$ and $x_2$ are not the nodes
$\frac{2\pi }{k}>\Delta x>\frac{\pi }{k}$
$\Rightarrow \lambda >\Delta x>\frac{\lambda }{2}$
For ${\varphi }_1=\pi ,{\varphi }_2=k(\Delta x)$
$=k\left(\frac{13\pi }{12k}\right)=\frac{13\pi }{12}$
$\frac{{\varphi }_1}{{\varphi }_2}=\frac{\pi }{(13\pi /12)}=\frac{12}{13}$