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

Answer: (D)

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.
$\Delta x=x_2-x_1=\left(\frac{3}{2}-\frac{1}{3}\right)\frac{\pi }{k}=\frac{7\pi }{6k}$
Since,
$\frac{2\pi }{k}>\Delta x>\frac{\pi }{k}$
$\lambda >\Delta x>\frac{\lambda }{2}\left\{k=\frac{2\pi }{\lambda }\right\}$
Therefore, ${\varphi }_1=\pi$
and ${\varphi }_2=k$
$\Delta x=\frac{7\pi }{6}$
Therefore, $\frac{{\varphi }_1}{{\varphi }_2}=\frac{6}{7}$