Given, frequency of second wave, f2=200Hz Speed of sound in air, v=340ms−1 Distance between two point, ∆x=85cm As we know that, ∆φ=k∆x . . . (i) where, ∆φ is phase difference, k is propagation constant and ∆x is path difference. ∴ωt=k∆x where, ω is angular frequency and t is time taken. k=
ω
v
(v=
∆x
t
) where, v is wave velocity. Substituting the value in Eq. (i), we get ∆φ=