Two radio antennas separated by d = 310 m, as shown in the figure below, simulta

Question

Two radio antennas separated by d = 310 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, 230 m from the center point between the antennas, and its radio receives the signals. If the car is at the position of the second maximum after that at point 0 when it has traveled a distance of y = 400 m northward, what is the wavelength of the signals?_______ m How much farther must the car travel from this position to encounter the next minimum in reception?________ m

Explanation / Answer

SOLUTION:
1. The distance from the "top" antenna to the car = [ (400 - 155)^2 + 1230^2 ]^0.5 = 1254.163068 m
2. The distance from the "bottom" antenna to the car = [ (400 + 155)^2 + 1230^2 ]^0.5 = 1349.416541 m
3. The path DIFFERENCE = 95.25347258 m
4. This is the position of the second maximum, therefore the path difference = 2*lambda ===>
lambda = 95.25347258 / 2 = 47.62673629 m

Question 2. The next minimum reception is the path difference is 2.5 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 - 155)^2 + 1230^2 =
1572925 + 490x + x^2
B) The distance from the "bottom" antenna to the car = (400 + x + 155)^2 + 1230^2 =
1820925 +1110x + x^2
C) The "path difference = 248000 + 620x
D) The path difference must be 2.5(lambda) = 2.5(47.62673629) = 119.0668407 m
We can find how far the car moves: 248,000 +620x = 119.0668407 ===> x = 399.80779567m

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