Equations of a stationary and a travelling waves are as follows `y_(1) = sin kx cos omega t` and `y_(2) = a sin ( omega t - kx)`. The phase difference between two points `x_(1) = pi//3k` and `x_(2) = 3 pi// 2k is phi_(1)` in the standing wave `(y_(1))` and is `phi_(2)` in the travelling wave `(y_(2))` then ratio `phi_(1)//phi_(2)` is

Options

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

Explanation

To solve the problem, we need to find the phase differences \(\phi_1\) and \(\phi_2\) for the standing wave \(y_1\) and the travelling wave \(y_2\) at the specified positions \(x_1\) and \(x_2\). Then, we will calculate the ratio \(\frac{\phi_1}{\phi_2}\).

### Step 1: Identify the equations of the waves
The equations of the waves are given as:
- Standing wave: \(y_1 = \sin(kx) \cos(\omega t)\)
- Travelling wave: \(y_2 = a \sin(\omega t - kx)\)

### Step 2: Determine the phase of the standing wave at \(x_1\) and \(x_2\)
The phase of the standing wave \(y_1\) at position \(x\) is given by:
\[ \phi_1 = kx \]
For \(x_1 = \frac{\pi}{3k}\):
\[ \phi_1(x_1) = k \left(\frac{\pi}{3k}\right) = \frac{\pi}{3} \]
For \(x_2 = \frac{3\pi}{2k}\):
\[ \phi_1(x_2) = k \left(\frac{3\pi}{2k}\right) = \frac{3\pi}{2} \]

### Step 3: Calculate the phase difference \(\phi_1\)
The phase difference \(\phi_1\) between the two points in the standing wave is:
\[ \Delta \phi_1 = \phi_1(x_2) - \phi_1(x_1) = \frac{3\pi}{2} - \frac{\pi}{3} \]
To subtract these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6:
\[ \Delta \phi_1 = \frac{9\pi}{6} - \frac{2\pi}{6} = \frac{7\pi}{6} \]

### Step 4: Determine the phase of the travelling wave at \(x_1\) and \(x_2\)
The phase of the travelling wave \(y_2\) at position \(x\) is given by:
\[ \phi_2 = \omega t - kx \]
At \(x_1\):
\[ \phi_2(x_1) = \omega t - k\left(\frac{\pi}{3k}\right) = \omega t - \frac{\pi}{3} \]
At \(x_2\):
\[ \phi_2(x_2) = \omega t - k\left(\frac{3\pi}{2k}\right) = \omega t - \frac{3\pi}{2} \]

### Step 5: Calculate the phase difference \(\phi_2\)
The phase difference \(\phi_2\) between the two points in the travelling wave is:
\[ \Delta \phi_2 = \phi_2(x_2) - \phi_2(x_1) = \left(\omega t - \frac{3\pi}{2}\right) - \left(\omega t - \frac{\pi}{3}\right) \]
This simplifies to:
\[ \Delta \phi_2 = -\frac{3\pi}{2} + \frac{\pi}{3} \]
Again, we need a common denominator (6):
\[ \Delta \phi_2 = -\frac{9\pi}{6} + \frac{2\pi}{6} = -\frac{7\pi}{6} \]

### Step 6: Calculate the ratio \(\frac{\phi_1}{\phi_2}\)
Now we can find the ratio of the phase differences:
\[ \frac{\Delta \phi_1}{\Delta \phi_2} = \frac{\frac{7\pi}{6}}{-\frac{7\pi}{6}} = -1 \]

### Final Answer
Thus, the ratio \(\frac{\phi_1}{\phi_2}\) is:
\[ \frac{\phi_1}{\phi_2} = 1 \]