A progressive wave of frequency 400 Hz is travelling with velocity 336 ~m / s . How far apart are the two points on a wave which are 60^ out of phase?

Step-by-step solution

The wavelength is determined from the wave velocity and frequency using v = f . Given f = 400 , Hz and v = 336 , m/s , the wavelength is = v f = 336 400 = 0.84 , m . The phase difference and path difference x are related by = 2 x . A phase difference of 60^ corresponds to = 3 , rad . Solving for the path difference yields x = 2 = 0.84 2 3 = 0.84 6 = 0.14 , m . Final answer: 0.14 , m