A wave of frequency 400 Hz has a phase velocity of 300 m/s The phase difference between two displacement at a certain point at time t= `10^(-3) `s is

Options

A. ` 72^(@) `
B. ` 102^(@) `
C. `144^(@)`
D. ` 180^(@) `

Answer

To find the phase difference between two displacements at a certain point at time \( t = 10^{-3} \) seconds for a wave of frequency 400 Hz and phase velocity 300 m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency \( f = 400 \) Hz
- Time \( t = 10^{-3} \) s

2. **Calculate Angular Frequency (\( \omega \)):**
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 400 \]
\[ \omega = 800\pi \text{ rad/s} \]

3. **Calculate the Phase Difference (\( \Delta \phi \)):**
The phase difference at time \( t \) can be calculated using the formula:
\[ \Delta \phi = \omega t \]
Substituting the values of \( \omega \) and \( t \):
\[ \Delta \phi = 800\pi \times 10^{-3} \]
\[ \Delta \phi = 0.8\pi \text{ radians} \]

4. **Convert Radians to Degrees:**
To convert radians to degrees, we use the conversion factor \( \frac{180}{\pi} \):
\[ \Delta \phi = 0.8\pi \times \frac{180}{\pi} \]
The \( \pi \) cancels out:
\[ \Delta \phi = 0.8 \times 180 = 144 \text{ degrees} \]

5. **Final Answer:**
The phase difference between the two displacements at time \( t = 10^{-3} \) seconds is \( 144 \) degrees.