Wave of frequency 500 Hz has a phase velocity `360 m//s`. The phase difference between two displacement at a certain point at time `10^(-3)s` apart will be

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `pi`
D. `(3pi)/(2)`

Answer

To find the phase difference between two displacements at a certain point that are \(10^{-3}\) seconds apart, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Frequency (\(f\)) = 500 Hz
- Phase velocity (\(c\)) = 360 m/s
- Time difference (\(\Delta t\)) = \(10^{-3}\) s

2. **Calculate the wavelength (\(\lambda\)):**
The wavelength can be calculated using the formula:
\[ \lambda = \frac{c}{f} \]
Substituting the values:
\[ \lambda = \frac{360 \, \text{m/s}}{500 \, \text{Hz}} = 0.72 \, \text{m} \]

3. **Determine the phase difference (\(\Delta \phi\)):**
The phase difference can be calculated using the formula:
\[ \Delta \phi = 2\pi \frac{\Delta t}{\lambda} \]
However, we can also use the simplified formula:
\[ \Delta \phi = 2\pi f \Delta t \]
Substituting the values:
\[ \Delta \phi = 2\pi \times 500 \, \text{Hz} \times 10^{-3} \, \text{s} \]

4. **Calculate the phase difference:**
\[ \Delta \phi = 2\pi \times 500 \times 10^{-3} = 2\pi \times 0.5 = \pi \, \text{radians} \]

5. **Final Result:**
The phase difference between the two displacements is:
\[ \Delta \phi = \pi \, \text{radians} \]

### Summary:
The phase difference between two displacements at a certain point at time \(10^{-3}\) seconds apart is \(\pi\) radians.

---