Question

One radio transmitter A operating at 60.0 MHz is 10.0 m from another similar transmitter B that is 180° out of phase with transmitter A. How far must an observer move from transmitter A toward transmitter B along the line connecting A and B to reach the nearest point where the two beams are in phase?

Answer 1

1.25 m

Answer 2

Let the position of transmitter A be the origin (x=0) and the position of transmitter B be x = 10.0 m. The frequency of the radio waves is f = 60.0 MHz = 60.0 × 10⁶ Hz. The speed of radio waves is the speed of light, c = 3.00 × 10⁸ m/s. The wavelength of the radio waves is λ = c / f = (3.00 × 10⁸ m/s) / (60.0 × 10⁶ Hz) = 5.0 m.

Let the observer be at a point P located at a distance x from transmitter A along the line connecting A and B. The observer moves from A towards B, so 0 ≤ x ≤ 10.0 m. The distance of P from A is r_A = x. The distance of P from B is r_B = 10.0 - x.

The two transmitters are operating 180° out of phase. This means there is an initial phase difference of π radians between the waves emitted by A and B.

The phase difference between the waves from A and B arriving at point P is due to two factors: the path difference and the initial phase difference. The path difference is Δr = r_B - r_A = (10 - x) - x = 10 - 2x. The phase difference due to the path difference is ΔΦ_path = k * Δr = (2π/λ) * (10 - 2x).

The initial phase difference between the sources is ΔΦ_initial = π.

The total phase difference between the waves at point P is ΔΦ = ΔΦ_path + ΔΦ_initial

The condition for the two beams to be in phase at P is that the total phase difference is an integer multiple of 2π.

k(2x - 10) + π = n * 2π

Substitute k = 2π/λ:

(2π/λ)(2x - 10) + π = n * 2π

Substitute λ = 5.0 m:

(4/5)(x - 5) = 2n - 1

0.8(x - 5) = 2n - 1

0.8x - 4 = 2n - 1

x = (2n + 3) / 0.8

x = (2n + 3) / (4/5)

x = 5(2n + 3) / 4

x = (10n + 15) / 4

We are looking for the nearest point to A where the beams are in phase. This means we need to find the smallest positive value of x in the range [0, 10.0 m] that satisfies this equation for some integer n.

The possible locations within the range [0, 10] where the beams are in phase are x = 1.25 m, 3.75 m, 6.25 m, and 8.75 m. The nearest point to A is the smallest positive value of x, which is 1.25 m.

Therefore, the final answer is 1.25 m.