Radio station KPHY uses two antennae to transmit its signal at an AM frequency o

Question

Radio station KPHY uses two antennae to transmit its signal at an AM frequency of 1, 260 kHz. While driving by on a semicircular road around these antennae, you notice that there is only one location where the signal from both antennae is totally destructive. At this location (X), you are 114.0 meters from antenna 1 and 82.0 meters from antenna 2. What is the difference in the phase constant delta phi (in radians; note that this is little phi) between antenna 1 and antenna 2? 0.84 radians 2.30 radians 3.14 radians 2.72 radians 5.44 radians Compared to red light, blue light has a wavelength frequency and (in air at 20 degree C).

Explanation / Answer

Path difference between the two waves = L2 -L1
= 114.0 -82.0
= 32 m
Phase difference = 2 * pi * path difference/lambda

Frequency = 1260 kHz

Radio waves have a speed of 3 * 10^8m/s so
we know c =f * wavelength

Hence ;

lambda = 238.1 m

Hence;

Phase Differecnce = (2 * pi * 32)/238.1
= 0.844 radians
= 0.84 radians

Therefore option a is correct

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