Question

Equations of a stationary and a travelling waves are as follows ${y_1} = a\sin kx\cos \omega t$ and ${y_2} = a\sin \left( {\omega t - kx} \right)$. The phase difference between two points ${x_1} = \dfrac{\pi }{{3k}}$ and ${x_2} = \dfrac{{3\pi }}{{2k}}$ is ${\phi _1}$ in the standing waves $({y_1})$ and is ${\phi _2}$ in travelling wave (${y_2}$), then ratio $\dfrac{{{\phi _1}}}{{{\phi _2}}}$ is
A. $1$
B. $\dfrac{5}{6}$
C. $\dfrac{3}{4}$
D. $\dfrac{6}{7}$

Answer 1

Hint:Here, it is given in the question, there are standing waves and travelling waves. The phase difference between the two points of the standing waves is ${\phi _1}$ and the phase difference between the two points of the transverse wave is ${\phi _2}$. Here, we will consider a node at the one point of the standing wave.

Complete step by step answer:
As we know that the phase difference between the two points of a standing wave $ = n\pi $
Here, $n$ is the number of nodes between the two points and $\pi $ is the difference between the phases.
Also, we know that the equation of a standing wave is given by
$y = a\sin kx\cos \omega t$
Now we know that there are two positions of a standing wave that are node positions and antinode positions.
Therefore, at node positions, $kx = n\pi $
$ \Rightarrow \,x = \dfrac{{n\pi }}{k}$
Where, $n = 0,1,2,3.....$
Therefore, nodes at $x$ is given by
$x = \dfrac{\pi }{K},\dfrac{{2\pi }}{K},\dfrac{{3\pi }}{K},.........$
Now, as given in the question, there are two points ${x_1}$ and ${x_2}$ , therefore, the phase difference between the two points is ${\phi _1} = \pi $
Now, if we consider a travelling wave, the phase difference is given by
${\phi _2} = \dfrac{{2\pi }}{\lambda }\Delta x$
The second wave equation is given by
${y_2} = a\sin \left( {kx - \omega t} \right)$
Therefore, the nodes of this wave is given by
$K = \dfrac{{2\pi }}{\lambda }$
Therefore, the phase difference of the second wave is given by
${\phi _2} = K\left[ {{x_2} - {x_1}} \right]$
$ \Rightarrow \,{\phi _2} = K\left[ {\dfrac{{3\pi }}{{2K}} - \dfrac{\pi }{{3K}}} \right]$
$ \Rightarrow \,{\phi _2} = \pi \left[ {\dfrac{3}{2} - \dfrac{1}{3}} \right]$
$ \Rightarrow \,{\phi _2} = \dfrac{7}{6}\pi $
Now, the ratio of both the phase difference is given by
$\Rightarrow \,\dfrac{{{\phi _1}}}{{{\phi _2}}} = \dfrac{\pi }{{\dfrac{7}{6}\pi }}$
$ \therefore \,\dfrac{{{\phi _1}}}{{{\phi _2}}} = \dfrac{6}{7}$
Therefore, the ratio $\dfrac{{{\phi _1}}}{{{\phi _2}}}$ is $\dfrac{6}{7}$.

Hence, option D is the correct option.

Note:Here, we have taken the nodes of waves instead of the antinode of a wave. This is because node is the position where the standing wave remains at a fixed point. Here, the nodes of standing waves and the travelling waves are different.