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 difference between two points in a stationary wave and a traveling wave, and then calculate the ratio of these phase differences.

### Step 1: Identify the wave equations
The stationary wave is given by:
\[ y_1 = \sin(kx) \cos(\omega t) \]
The traveling wave is given by:
\[ y_2 = a \sin(\omega t - kx) \]

### Step 2: Determine the positions
We have two positions:
- \( x_1 = \frac{\pi}{3k} \)
- \( x_2 = \frac{3\pi}{2k} \)

### Step 3: Calculate the phase difference in the stationary wave (\( \phi_1 \))
For the stationary wave, the phase at \( x_1 \) is:
\[ \phi_1(x_1) = kx_1 = k \left(\frac{\pi}{3k}\right) = \frac{\pi}{3} \]

For the stationary wave at \( x_2 \):
\[ \phi_1(x_2) = kx_2 = k \left(\frac{3\pi}{2k}\right) = \frac{3\pi}{2} \]

Now, the phase difference \( \phi_1 \) between these two points in the stationary wave is:
\[ \phi_1 = \phi_1(x_2) - \phi_1(x_1) = \frac{3\pi}{2} - \frac{\pi}{3} \]

To subtract these fractions, we need a common denominator:
- The common denominator of 2 and 3 is 6.

Converting the fractions:
\[ \frac{3\pi}{2} = \frac{9\pi}{6} \]
\[ \frac{\pi}{3} = \frac{2\pi}{6} \]

Now, substituting back:
\[ \phi_1 = \frac{9\pi}{6} - \frac{2\pi}{6} = \frac{7\pi}{6} \]

### Step 4: Calculate the phase difference in the traveling wave (\( \phi_2 \))
For the traveling wave, the phase difference \( \phi_2 \) is given by:
\[ \phi_2 = k \Delta x \]
where \( \Delta x = x_2 - x_1 \).

Calculating \( \Delta x \):
\[ \Delta x = \frac{3\pi}{2k} - \frac{\pi}{3k} \]

Finding a common denominator (which is 6k):
\[ \Delta x = \frac{9\pi}{6k} - \frac{2\pi}{6k} = \frac{7\pi}{6k} \]

Now substituting into the phase difference formula:
\[ \phi_2 = k \Delta x = k \left(\frac{7\pi}{6k}\right) = \frac{7\pi}{6} \]

### Step 5: Calculate the ratio \( \frac{\phi_1}{\phi_2} \)
Now we can find the ratio of the phase differences:
\[ \frac{\phi_1}{\phi_2} = \frac{\frac{7\pi}{6}}{\frac{7\pi}{6}} = 1 \]

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

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