Question

Equations of a stationary wave and a travelling wave are $y_1 = a\;sin(kx)cos(\omega t)$ and $y_2 = a\;sin(\omega t-kx)$. The phase difference between two points $x_1 = \dfrac{\pi}{3k}$ and $x_2 =\dfrac{3\pi}{2k}$ is $\phi_1$ for the first wave and $\phi_2$ for the second wave. The ratio $\dfrac{\phi_1}{\phi_2}$ is:
A. 1
B. $\dfrac{5}{6}$
C. $\dfrac{3}{4}$
D. $\dfrac{6}{7}$

Answer 1

Hint: Begin by first finding the number of nodes that lie between the two points in a stationary wave. From this you can deduce the phase difference of the stationary wave by looking at which node region lies across both the points. After this determine the phase difference between the two points in the travelling wave by calculating the difference between the two positions. Then divide the two to get the final ratio.

Formula Used: For a travelling wave, phase difference between two points: $\phi = k\Delta x$, where k is the wave vector and $\Delta x$ is the difference between the position of two points.
For a stationary wave, phase difference between two points: $\phi = n\pi$, where n is the number of nodes.

Complete answer:
Let us first consider the stationary wave whose equation is given by $y_1 = a\;sin(kx)cos(\omega t)$.
Now, we know that successive nodes in a stationary wave are found at $n\pi$ phase difference.
This means that, at node points : $sin(kx) = n\pi \Rightarrow x = \dfrac{n\pi}{k}$, where n= 0,1,2, so on.
Therefore, the nodes are found at $\dfrac{\pi}{k}$, $\dfrac{2\pi}{k}$, $\dfrac{3\pi}{k}$, and so on.
Now, the two points are given as $x_1 = \dfrac{\pi}{3k}$ and $x_2 =\dfrac{3\pi}{2k}$.
We see that $x_1 = \dfrac{\pi}{3k}$<$\dfrac{\pi}{k}$, and $\dfrac{\pi}{k}$ < $x_2 = \dfrac{3\pi}{2k}$<$\dfrac{2\pi}{k}$
This means that there is only one node between the two points, given by
Now, from $kx = n\pi \Rightarrow k\left(\dfrac{\pi}{k}\right) = n\pi \Rightarrow n = 1$
The, $kx = \pi \Rightarrow x = \dfrac{\pi}{k}$
And this gives a phase difference
 $\phi_1 = kx = k\left(\dfrac{\pi}{k}\right) \Rightarrow \phi_1 = \pi$ for the stationary wave.
For the travelling wave, the phase difference $\phi_2$ is given as:
$\phi_2 = k(x_2)-k(x_1) = k(x_2-x_1) = k\left(\dfrac{3\pi}{2k}-\dfrac{\pi}{3k}\right) = \dfrac{3\pi}{2} - \dfrac{\pi}{3} = \dfrac{9\pi-2\pi}{6} \Rightarrow \phi_2 = \dfrac{7\pi}{6}$
Therefore, the ratio of the two phase differences:
$\dfrac{\phi_1}{\phi_2} = \dfrac{\pi}{\left(\dfrac{7\pi}{6}\right)} = \dfrac{6\pi}{7\pi} = \dfrac{6}{7}$

Therefore, the correct answer will be D. $\dfrac{6}{7}$.

Note:
It is important to determine the number of nodes that lie between the two points on the stationary wave as this is the way to ultimately find the phase difference depending on whether the points are present in successive loops of the wave or not. Note that the two points do not have any nodes at their position since the nodes are formed at $\pi$, $2\pi$, and so on. However, $x_2$ lies at an antinodes lie at $\dfrac{\pi}{2}$, $\dfrac{3\pi}{2}$ and so on.