The equation of the progressive wave, where t is the time in second , x is the distance in metre is `y= A cos 240 (t-(x)/(12))` . The phase difference ( in SI units ) between two position 0.5 m apart is

Options

A. 40
B. 20
C. 10
D. 5

Answer

To solve the problem of finding the phase difference between two positions 0.5 m apart for the given wave equation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the wave equation**: The given wave equation is
\[ y = A \cos(240t - \frac{x}{12}). \]

2. **Rewrite the wave equation**: We can express the wave equation in the standard form \( y = A \cos(\omega t - kx) \). Here, we need to identify the angular frequency (\(\omega\)) and the wave number (\(k\)).
- From the equation, we have:
- \(\omega = 240\)
- \(\frac{x}{12}\) can be rewritten as \(kx\), where \(k = \frac{2\pi}{\lambda}\). Thus, we can find \(k\):
\[ k = \frac{240}{12} = 20. \]

3. **Calculate the phase difference**: The phase difference (\(\Delta \phi\)) between two points separated by a distance \(d\) is given by the formula:
\[ \Delta \phi = k \cdot d. \]
- Here, the path difference \(d = 0.5 \, \text{m}\).

4. **Substitute the values**: Now, substitute the value of \(k\) and the path difference \(d\) into the formula:
\[ \Delta \phi = 20 \cdot 0.5 = 10 \, \text{radians}. \]

5. **Conclusion**: The phase difference between the two positions 0.5 m apart is \(10 \, \text{radians}\).

### Final Answer:
The phase difference between two positions 0.5 m apart is \(10 \, \text{radians}\).

---