The ratio of `omega//k` for a travelling wave (where `omega` is the angular frequency and k is the angular wave number) is .........

Answer

To solve the problem of finding the ratio of \( \frac{\omega}{k} \) for a traveling wave, we can follow these steps:

### Step 1: Understand the Definitions
- The angular frequency \( \omega \) is defined as:
\[ \omega = 2\pi \nu \]
where \( \nu \) is the frequency of the wave.

- The angular wave number \( k \) is defined as:
\[ k = \frac{2\pi}{\lambda} \]
where \( \lambda \) is the wavelength of the wave.

### Step 2: Write the Ratio
We need to find the ratio \( \frac{\omega}{k} \):
\[ \frac{\omega}{k} = \frac{2\pi \nu}{\frac{2\pi}{\lambda}} \]

### Step 3: Simplify the Expression
When we simplify the expression, we can multiply by the reciprocal of \( k \):
\[ \frac{\omega}{k} = 2\pi \nu \cdot \frac{\lambda}{2\pi} \]
The \( 2\pi \) terms cancel out:
\[ \frac{\omega}{k} = \nu \lambda \]

### Step 4: Relate to Wave Speed
Now, we know that the wave speed \( v \) is given by the product of frequency and wavelength:
\[ v = \nu \lambda \]
Thus, we can conclude:
\[ \frac{\omega}{k} = v \]

### Final Answer
The ratio \( \frac{\omega}{k} \) for a traveling wave is equal to the speed of the wave \( v \).

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