A travelling harmonic wave is given by `y = 5 cos(20t - 0.0070 x + 0.12)` where x, y are in cm and t is in seconds. Calculate the phase difference between two points separated by a distance of 0.5 m.

Answer

To calculate the phase difference between two points separated by a distance of 0.5 m for the given wave equation \( y = 5 \cos(20t - 0.0070x + 0.12) \), we can follow these steps:

### Step 1: Identify the wave number \( k \)
The general form of a traveling wave is given by:
\[ y = A \cos(\omega t - kx + \phi) \]
From the given wave equation, we can identify that:
\[ k = 0.0070 \, \text{cm}^{-1} \]

### Step 2: Convert \( k \) to meters
Since \( k \) is given in cm\(^{-1}\), we need to convert it to m\(^{-1}\):
\[ k = 0.0070 \, \text{cm}^{-1} = 0.0070 \times \frac{1}{100} \, \text{m}^{-1} = 0.00007 \, \text{m}^{-1} \]

### Step 3: Calculate the phase difference \( \Delta \phi \)
The phase difference \( \Delta \phi \) between two points separated by a distance \( \Delta x \) is given by:
\[ \Delta \phi = k \Delta x \]
Where \( \Delta x = 0.5 \, \text{m} \).

### Step 4: Substitute the values
Now substituting the values we have:
\[ \Delta \phi = 0.00007 \, \text{m}^{-1} \times 0.5 \, \text{m} = 0.000035 \, \text{radians} \]

### Step 5: Final result
Thus, the phase difference between the two points separated by a distance of 0.5 m is:
\[ \Delta \phi = 0.000035 \, \text{radians} \]