Question

Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d=9.30m. An observer, P, is located on the x axis, a distance x=57.0m from antenna A, so that APB forms a right triangle with PB as hypotenuse. What is the phase difference between the waves arriving at P from antennas A and B?

Now observer P walks along the x axis toward antenna A. What is P's distance from A when he first observes fully destructive interference between the two waves?

Answer 1

Answer:

a)   Δφ = 1.51 rad , b)  x = 21.17 m

Explanation:

This is an interference problem, as they indicate that the distance AP is on the x-axis the antennas must be on the y-axis, the phase difference is

          Δr /λ = Δfi / 2π

          Δfi = Δr /λ 2π

          Δr = r₂-r₁

let's look the distances

         r₁ = 57.0 m

We use Pythagoras' theorem for the other distance

         r₂ = √ (x² + y²)

         r₂ = √(57² + 9.3²)

         r₂ = 57.75 m

The  difference is

         Δr = 57.75 - 57.0

         Δr = 0.75 m

Let's look for the wavelength

        c =  λ f

          λ = c / f

          λ = 3 10⁸ / 96.0 10⁶

          λ = 3.12 m

Let's calculate

         Δφ = 0.75 / 3.12 2π

         Δφ = 1.51 rad

b) for destructive interference the path difference must be λ/2, the equation for destructive interference with φ = π remains

           Δr = (2n + 1) λ / 2

           

For the first interference n = 0

           Δr = λ / 2

           Δr = r₂ - r₁

We substitute the values

        √ (x² + y²) - x = 3.12 / 2

Let's solve for distance x

          √ (x² + y²) = 1.56 + x

          x² + y² = (1.56 + x)²

          x² + y² = 1.56² + 2 1.56 x + x²

          y2 = 20.4336 +3.12 x

          x = (y² -20.4336) /3.12

          x = (9.3² -20.4336) /3.12

          x = 21.17 m

This is the distance for the first minimum

Answer 2

Answer:

Explanation:

Check attachment for solution