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

Step-by-step explanation:

Check attachment for solution

Answer 2

a) Δφ = 1.51 rad , b) x = 21.17 m

Step-by-step 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