A wave of frequency 500 Hz has a phase velocity of $360~m/s$. The phase difference between two displacements at a certain point at a time $10^{-3}$ sec apart is:

Question

Accepted Answer

To determine the phase difference between two displacements at a certain point at a time $10^{-3}$ seconds apart for a wave with a frequency of 500 Hz and a phase velocity of 360 m/s, we can use the relationship between frequency, phase velocity, and phase difference.

1. **Calculate the wavelength ($\lambda$):**

The phase velocity ($v$) is related to the frequency ($f$) and the wavelength ($\lambda$) by the equation:
   $$
   v = f \cdot \lambda
   $$
   Given $v = 360 \, \text{m/s}$ and $f = 500 \, \text{Hz}$, we can solve for $\lambda$:
   $$
   \lambda = \frac{v}{f} = \frac{360}{500} = 0.72 \, \text{m}
   $$

2. **Calculate the angular frequency ($\omega$):**

The angular frequency is given by:
   $$
   \omega = 2\pi f = 2\pi \times 500 = 1000\pi \, \text{rad/s}
   $$

3. **Determine the phase difference ($\Delta \phi$):**

The phase difference $\Delta \phi$ between two points in time separated by $\Delta t = 10^{-3}$ seconds is given by:
   $$
   \Delta \phi = \omega \Delta t = 1000\pi \times 10^{-3} = \pi
   $$

Thus, the phase difference is $\pi$, which corresponds to Option 4.

**Explanation for why other options are incorrect:**

- **Option 1 (2π):** This option suggests a full cycle phase difference, which would occur if the time difference were equal to the period of the wave. However, the calculated phase difference is $\pi$, not $2\pi$.

- **Option 2 ($\pi/2$):** This option suggests a quarter cycle phase difference. The calculated phase difference is $\pi$, which is half a cycle, not a quarter cycle.

- **Option 3 ($\pi/3$):** This option suggests a phase difference of one-sixth of a cycle. The calculated phase difference is $\pi$, which is half a cycle, not one-sixth of a cycle.

Therefore, the correct answer is Option 4, $\pi$, as it accurately represents the phase difference calculated based on the given parameters.