Question

Radio station KBOB broadcasts at a frequency of 85.7 MHz on your dial using radio waves that travel at 3.00 × 108 m/s. Since most of the station's audience is due south of the transmitter, the managers of KBOB don't want to waste any energy broadcasting to the east and west. They decide to build two towers, transmitting in phase at exactly the same frequency, aligned on an east-west axis. For engineering reasons, the two towers must be AT LEAST 10.0 m apart. What is the shortest distance between the towers that will eliminate all broadcast power to the east and west?

Answer 1

12.5 m

Step-by-step explanation:

The first thing we would do is to calculate the wavelength. To do this, we use the formula

v = fλ, where

v = wave speed

f = frequency

λ = wavelength

If we make wavelength the formula, we have

wavelength = speed / frequency

Now, we substitute the values we had been given and we have

wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m

half of this said wavelength will be

= 3.50 / 2

= 1.75 m

As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,

(10 / 1.75) = 5.7

So the separation will have to be 7 half wavelengths

= (7 * 1.75) = 12.5 m