Two antennas located at points A and B are broadcasting radio waves of frequency
ID: 1304992 • Letter: T
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