Question

Equations of a stationary and a travelling waves are as follows, $y_1=a\sin kx\cos \omega t$ and $y_2=a\sin (\omega t-kx)$. The phase difference between two points $x_1=\frac{\pi }{3k}$ and $x_2=\frac{3\pi }{2k}$ are ${\varphi }_1$ and ${\varphi }_2$ respectively for two waves. The ratio $\frac{{\varphi }_1}{{\varphi }_2}$ is :

• A. $1$
• B. $5/6$
• C. $3/4$
• D. $6/7$

Answer 1

The correct option is D $6/7$

At $x_1=\frac{\pi }{3k}$ and $x_2=\frac{3\pi }{2k}$

$\sin k{x}_1$ or $\sin k{x}_2$ is not zero.

Therefore, neither of $x_1$ nor $x_2$ is a node.

$\Delta x=x_2-x_1=(\frac{3}{2}-\frac{1}{3})\frac{\pi }{k}=\frac{7\pi }{6k}$

Since, $\frac{2\pi }{k}>\Delta x>\frac{\pi }{k}$

$\lambda >\Delta x>\frac{\lambda }{2}$ $\{k=\frac{2\pi }{\lambda }\}$

Therefore, ${\varphi }_1=\pi$

and ${\varphi }_2=k$

$\Delta x=\frac{7\pi }{6}$

Therefore, $\frac{{\varphi }_1}{{\varphi }_2}=\frac{6}{7}$