1. The problem statement, all variables and given/known data Consider 2 radio antennas placed on a 2 dimensional grid at points x1 = (300,0) and x2 = (-300, 0). If the antennas are out of phase such that ϕ2-ϕ1 = π/4 and both are emitting waves of frequency 2.59 MHz. What will the amplitude of the resultant wave be at the point (200, 800)? Radio waves being a form of light travel at a speed of 3x108 m/s. You may assume at the point of interference that the two waves have the same amplitude and you may leave your final answer in terms of this amplitude. 2. Relevant equations c = √(a2+b2) , Δφ = 2π(Δx/λ) , D(x,t) = A sin(kx – ωt + ϕ0). 3. The attempt at a solution I began this problem by getting the distance from x1 and x2 to the point (200, 800). The results I got were as follows; From x1 = 943.4 From x2 = 806.2 I would use these values as my displacement values. Following that I used the formula Δφ = 2π(Δx/λ) to determine the wavelength; Δφ = 2π(Δx/λ) → λ = 2π(Δx/Δφ) From x1 λ = 2π(500/(π/4)) = 4000 From x2 λ = 2π(100/(π/4)) = 800 This allowed me to get the wave number via k = 2π/λ. x1 wave number k = 2π/4000 = 0.0016 x2 wave number k = 2π/800 = 0.0079 To get the angular frequency ω I used the formula ω = 2π⋅f. x1 ω = 2π⋅f = 2π(2.59) = 16.27 x1 ω = 2π⋅f = 2π(2.59) = 16.27 My phase constants used were; φ1 = 0 φ2 = π/4 And this is as far as I was able to get. I still need to find the Amplitude of the resultant wave but am left with the time variable t still unsolved. My initial thought would be to take A sin(kx – ωt + ϕ0) of each wave, set the two sides equal to each other, drop the Amplitude because they will be the same and will therefore cancel, and solve for time in this formula but I am not sure if this is an appropriate way to approach this. Perhaps the time value t doesn't matter because the waves will intersect that point at the same time and have the same angular frequency so the values will be identical. I suppose what I'm looking for is a nudge in the right direction to ensure that I'm thinking about this in the right way.