(A)Two antennas located at points A and B are broadcasting radio waves of freque

Question

(A)Two antennas located at points A and B are broadcasting radio waves of frequency 94.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d= 9.30 m. An observer, P, is located on the x axis, a distance x= 72.0 m from antenna A, so that APB forms a right triangle with PB as hypotenuse. What is the phase difference between the waves arriving at P from antennas A and B?

(B)Now observer P walks along the x axis toward antenna A. What is P's distance from A when he first observes fully destructive interference between the two waves?

(C)If observer P continues walking until he reaches antenna A, at how many places along the x axis (including the place you found in the previous problem) will he detect minima in the radio signal, due to destructive interference?

I have taken so many appraoches to this I don't even know what I'm doing anymore. I've looked at multple other similiar questions on chegg, but no one seems to get the right answer...I am just lost.

I will love you forever if you can solve this before before the next hour and 40 minutes

Explanation / Answer

You will get destructive interference when the signal from antenna B arrives at point P half a wavelength out of phase with the signal from A. In other words, when the signal from B has travelled half a wavelength further than the signal from A. BP = AP + ?/2 Using Pythagoras: (26.3 + ?/2)² = 26.3² + 9.30² You should be able to solve that for ?. Then it's just using c = f? to find frequency (f) { I get ? = 3.192 m, so f = (3 * 10^8 / 3.192) = 94 MHz }

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