A progressive wave of frequency 500 ~Hz is travelling with a velocity of 360 ~ms ⁻¹ . The distance between the two points, having a phase difference of 60^ is .............

Step-by-step solution

gathered f=500 ~Hz , v=360 ~ms ⁻¹ = v f = 360 500 gathered Now, a phase difference of 60^ corresponds to a path difference of 60 360 . So, distance between 2 particles is d= 60 360 = 60 360 360 500 =0.12 ~m