Two antennas located at points A and B are broadcasting radio waves of frequency

Question

Two antennas located at points A and B are broadcasting radio waves of frequency 95.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d= 12.40 m. An observer, P, is located on the x axis, a distance x= 80.0 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?

A) = .1.901 rad

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?

=4.790 * 10^1 m

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?

Need answer for this part.

Explanation / Answer

a>
phase difference will be only due to displacement difference.
=k|(AP-PB)|=k(x1-x2)
now AP=80 and PB =(12.42+802).5=80.955.
and k=2pi*f/c =1.9896

hence putting in the equation we get
phase difference=
1.579007*(80.955-80)=1.9007=1.901rad

b>we have to solve the equation
k(root(12.4*12.4+x*x)-x)=pi for destructive interference.

hence putting in the values we get
root(12.4*12.4+x*x)=x+1.579
x=47.9m

for part c>

we have to modify the equation
k(root(12.4*12.4+x*x)-x)= N*pi
N= ODD NUMBER.

hence putting N= 1 we get x= 47.9
(N=3) we get x=13.861
(N=5) we get x=5.79
(N=7) we get x=1.429
(N>=9) we get x=negative
it will be left to x axis hence not to be counted

hence other than that described in part (b) there will be destructive interference at 3 more points and in total 4 points

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