Two radio antennas separated by d = 273 m, as shown in the figure below, simulta

Question

Two radio antennas separated by d = 273 m, as shown in the figure below, simultaneously broadcast identical signals at the same wavelength. A car travels due north along a straight line at position x = 1,180 m from the center point between the antennas, and its radio receives the signals. Hint: Do not use the small-angle approximation in this problem. (a) If the car is at the position of the fourth maximum after that at point O when it has traveled a distance of y = 400 m northward, what is the wavelength of the signals? (b) How much farther must the car travel to encounter the next minimum in reception?

Explanation / Answer

SOLUTION:
1. The distance from the "top" antenna to the car =sqrt{ (400 - 136.5)^2 + 1180^2} = 1209.06 m
2. The distance from the "bottom" antenna to the car = sqrt{(400 + 136.5)^2 + 1180^2} = 1296.23 m
3. The path DIFFERENCE = 87.17 m
4. This is the position of the fourth maximum, therefore the path difference = 4*lambda ===>
lambda = 87.17 / 4 = 21.79 m

Question 2. The next minimum reception is the path difference is 4.5 wavelengths.
Let's assume the car moves x meters north and encounters "the next minimum reception".
A) the distance from the "top" antenna to the car = (400 + x - 136.5)^2 + 1180^2 =
1,461,832 + 527x + x^2
B) The distance from the "bottom" antenna to the car = (400 + x + 136.5) + 1180^2 =
1,680,232 +1073x + x^2
C) The "path difference = 218,400 +546x
D) The path difference must be 4.5(lambda) = 4(21.79) = 87.16 m
We can find how far the car moves: 218,400 +546x = 87.16 ===> x = 400 m

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