Two antennas located at points A and B are broadcasting radio waves of frequency
ID: 1388376 • Letter: T
Question
Two antennas located at points A and B are broadcasting radio waves of frequency 98.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d=12.40m. An observer, P, is located on the x axis, a distance x=79.0m 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?
Explanation / Answer
A)
lambda=C/f
=3*10^8/98*10^6
=3.06 m
phase difference =(2pi/lambda)*path difference
=(2pi/lambda)*(x2-x1)
here x2=sqrt(d^2+x1^2)
=sqrt(12.4^2+79^2)
=79.97 m
now
phase difference=(2pi/lambda)*(x2-x1)
=(2*3.14/3.06)*(79.97-79)
=1.99 rad <------------
B)
to get destructive interference ,
path difference =lambda/2
x2-x1=lambda/2
and
x2=sqrt(d^2+x1^2)
now
sqrt(d^2+x1^2)-x1=lambda/2
(d^2+x1^2)=(x1+lambda/2)^2
d^2=(x1*lambda)+(lambda^2/4)
====>
x1=(d^2/lambda)-(lambda/4)
=(12.4^2/3.06)-(3.06/4)
=49.48 m.....is answer