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Physics frequency and interference question?
Your firm has been hired to design a system that allows airplane pilots to make instrument landings in rain or fog. You've decided to place two radio transmitters 50 m apart on either side of the runway. These two transmitters will broadcast the same frequency, but out of phase with each other. This will cause a nodal line to extend straight off the end of the runway. As long as the airplane's receiver is silent, the pilot knows she's directly in line with the runway. If she drifts to one side or the other, the radio will pick up a signal and sound a warning beep. To have sufficient accuracy, the first intensity maxima need to be 60 m on either side of the nodal line at a distance of 3.0 km. What frequency should you specify for the transmitters?
I can't seem to figure out what the 3.0km represents. Help please? How would you solve this?
Signal emitted from point A: sin(2πf(t - sqrt((y-50)^2 + x^2))/c)
Signal emitted from point B: - sin(2πf(t - sqrt((y+50)^2 + x^2))/c)
For y = 0, these two signals will be equal in magnitude and opposite in sign: they will cancel.
In order to achieve the first antinodes, the difference in path must introduce an additional phase difference of π:
2πf(t - sqrt((y-50)^2 + x^2))/c = (2πf(t - sqrt((y+50)^2 + x^2))/c ± π
sqrt((y-50)^2 + x^2) = sqrt((y+50)^2 + x^2)) ± c/(2f)
(y-50)^2 + x^2) = (y+50)^2 + x^2) ± c/(2f)*sqrt((y+50)^2 + x^2)) +(c/(2f))^2
-200*y - (c/(2f))^2 = ± c/(2f)*sqrt((y+50)^2 + x^2))
40000y^2+ (c/(2f))^4 + 400(c/(2f))^2 = (c/(2f))^2 * ((y+50)^2 + x^2)
40000(2f/c)^2 + (c/(2f))^2 + 400 = (y+50)^2 + x^2
The point is that the above equation defines the antinodal curve, where the maxima occur: for the given distance x ( = 3000) from the pair of antennae, the value y should be ± 60
40000(2f/c)^2 + (c/(2f))^2 + 400 = (y+50)^2 + x^2
= y^2 + 100y + x^2
0 = y^2 + 100y + x^2 - (40000(2f/c)^2 + (c/(2f))^2 + 400)
0 = y^2 + 100y - (40000(2f/c)^2 + (c/(2f))^2 + 400 - x^2)
y = [-100 ± sqrt(10000 + 4(40000(2f/c)^2 + (c/(2f))^2 + 400 - x^2))]/2
= -50 ± sqrt(2500 + 40000(2f/c)^2 + (c/(2f))^2 + 400 - x^2))
Now we specify that we are interested in the value of y for x = 3000:
y = -50 ± sqrt(2500 + 40000(2f/c)^2 + (c/(2f))^2 + 400 - 3000^2))
Since y should be -60, the sqrt value should be 10:
10 = sqrt(2500 + 40000(2f/c)^2 + (c/(2f))^2 + 400 - 3000^2))
100 = 2500 + 40000(2f/c)^2 + (c/(2f))^2 + 400 - 3000^2)
0 = 2300-3000^2 + 40000(2f/c)^2 + (c/(2f))^2
0 = (c/(2f))^4 + (2300-3000^2)*c/(2f))^2 + 40000
0 = (c/(2f))^4 - (-2300 + 3000^2)*c/(2f))^2 + 40000
=>
(c/(2f))^2 = ((-2300 + 3000^2) ± sqrt((-2300 + 3000^2)^2 - 4*40000))/2
= 9e6, 4.45e-3
=>
f = c/(2sqrt(X)) = (3/2)e8/(3e3, 6.67e-2) = 5e4 or e7
OK, there are a few loose ends:
- Need to figure out which which of the +/- signs relates to the left-hand side and the right-hand side of the center line. When I squared the square roots initially, we made a decision about which one we were looking at.
- Also need to figure out which of the two quadratic solutions is right; maybe they both are, but I doubt it.
But to answer your basic question: The point is that the "phase = 0" nodal line is the only constant-phase curve that is a line. The rest of them are hyperbolas: the curves defined by the fact that the distance to any point of the curve from the two antennae differ by a constant length, which needs to be 1 wavelength. Actually, that's probably a better way to do the problem:
Ra = distance to A = sqrt(x^2 + (y - d)^2)
Rb = distance to B = sqrt(x^2 + (y + d)^2)
Ra = Rb + Wavelength = Rb + w
x^2 + (y - d)^2 = Ra^2 = x^2 + (y + d)^2 + w^2 + 2w*sqrt(x^2 + (y + d)^2)
-2dy = w^2 + 2w*sqrt(x^2 + (y + d)^2)
- (w^2 + 2yd) = 2w*sqrt(x^2 + (y + d)^2)
((w^2 + 2yd)/(2w))^2 = x^2 + (y + d)^2 = x^2 + y^2 + 2yd + d^2
(w^4 + 4w^2*yd + 4(yd)^2)/(2w)^2 = x^2 + y^2 + 2yd + d^2
OK, it's ugly any way you do it. It's quite doable, but easier with pencil & paper than on keyboard!
The Power Hour: Nila Sagadevan (6-6-05)[1of6]
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