Question

Equations of a stationary and a travelling waves are as follows, $y_{1}=a \sin k x \cos \omega t$ and $y_{2}=a \sin (\omega t-k x)$. The phase difference between two points $x_{1}=\frac{\pi}{3 k}$ and $x_{2}=\frac{3 \pi}{2 k}$ are $\phi_{1}$ and $\phi_{2}$ respectively for two waves. The ratio $\frac{\phi_{1}}{\phi_{2}}$ is

• A. 1
• B. 5/6
• C. 3/4
• D. 6/7

Answer 1

Answer: (D)

At $x_{1}=\frac{\pi}{3 k}$ and $x_{2}=\frac{3 \pi}{2 k}$
$\sin k x_{1}$ or $\sin k x_{2}$ is not zero.
Therefore, neither of $x_{1}$ nor $x_{2}$ is a node.
$\Delta x=x_{2}-x_{1}=\left(\frac{3}{2}-\frac{1}{3}\right) \frac{\pi}{k}=\frac{7 \pi}{6 k}$
Since,
$\frac{2 \pi}{k} >\Delta x>\frac{\pi}{k} $
$\lambda >\Delta x>\frac{\lambda}{2} \left\{k=\frac{2 \pi}{\lambda}\right\}$
Therefore, $\phi_{1}=\pi$
and $\phi_{2}=k $
$\Delta x=\frac{7 \pi}{6}$
Therefore, $ \frac{\phi_{1}}{\phi_{2}}=\frac{6}{7}$