A wave whose frequency is 500 Hz -1 has a speed of 400 m s-1' What is the minimum distance between twopoints where the phase difference of this wave is /4@

The phase difference of π/4 corresponds to a quarter of the wavelength.

The formula for the speed of a wave is v = fλ, where v is the speed, f is the frequency, and λ is the wavelength.

We can rearrange this formula to solve for the wavelength: λ = v/f.

Given that the speed of the wave is 400 m/s and the frequency is 500 Hz, we can substitute these values into the formula:

λ = 400 m/s / 500 Hz = 0.8 m.

Since the phase difference of π/4 corresponds to a quarter of the wavelength, the minimum distance between two points where the phase difference of this wave is π/4 would be λ/4:

0.8 m / 4 = 0.2 m.

So, the minimum distance between two points where the phase difference of this wave is π/4 is 0.2 meters.