Phase Difference between two waves from antennas

Click For Summary
SUMMARY

The discussion focuses on calculating the phase difference between radio waves emitted from two antennas at a frequency of 96.0 MHz, separated by 12.40 meters. The observer is positioned 55.0 meters from antenna A, forming a right triangle with the path lengths AP and BP. The calculated path length difference is 1.4 meters, but the conversion to radians using the wavelength of 3.125 meters resulted in incorrect phase difference values. The key equations involved include the relationship between wavelength, frequency, and path length differences.

PREREQUISITES
  • Understanding of wave frequency and wavelength calculations
  • Knowledge of trigonometry and right triangle properties
  • Familiarity with phase difference concepts in wave physics
  • Ability to apply the formula for converting path length differences to radians
NEXT STEPS
  • Study the derivation of phase difference formulas in wave mechanics
  • Learn about the implications of path length differences in interference patterns
  • Explore the application of the wave equation: c/f = λ
  • Investigate the effects of antenna spacing on signal reception and phase alignment
USEFUL FOR

Students in physics, particularly those studying wave mechanics, antenna theory, and signal processing, will benefit from this discussion.

skibum143
Messages
112
Reaction score
0

Homework Statement


Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 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=55.0m 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? Use units of "rad" for the answer. (If you are stuck, read the hint.)


Homework Equations


1 wavelength = 2pi radians
m * lambda / distance between slits = distance between maxima (y) / Distance to screen
c/f = lambda

The Attempt at a Solution


First, I converted 96MHz to wavelengths, = 3.125 m
Then, I wanted to find the pathlength of AP and the pathlength of BP, but I don't know how to do this without knowing the distance between maxima (y). If I had y, I could find the value of the order (m), and then I would know the pathlength difference. How do I solve for this?

Finally, I will convert the pathlength distance to radians with the above formula. I just don't understand how to get the pathlength distance.
 

Attachments

  • triangle.gif
    triangle.gif
    714 bytes · Views: 885
Physics news on Phys.org
I know that AP = 55m and BP = 56.4m, I just don't know how to find the wavelength of pathlength difference
 
So I know the difference in the paths is 1.4m, but when I multiply that by the wavelength, (3.125m), and convert to radians it is wrong.
 

Similar threads

Replies
7
Views
1K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
13
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
8
Views
6K
  • · Replies 6 ·
Replies
6
Views
5K