The distance between two consecutive points with phase difference of 60 ° in a wave of frequency 500 Hz is 6 . 0 m . The velocity with which wave is travelling is ______ km s - 1 .

Step-by-step solution

Distance between two consecutive points 60 ° out of phase is given by Δ x = λ 2 π Δ ϕ Putting the values, we have 6 = λ 2 π π 3 ⇒ λ = 36   m Now, velocity of wave is v = f λ = ( 500   Hz ) ( 36   m ) = 18000   m   s - 1 = 18   km   s - 1