Constructive or destructive interference- Car radio

  • Thread starter sbayla31
  • Start date
  • #1
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Homework Statement



A car’s radio antenna is located at a distance of 56.983 m from a brick wall, which is
assumed to act like a dense barrier. The car is attempting to listen to an FM radio
station with a frequency of 89.5 MHz. These waves travel at the speed of light, 3.00
× 108 m/s. The waves are simultaneously received by the car’s antenna directly from
the distant broadcasting tower, as well as reflected from the brick wall. Describe the
type of interference at the antenna (constructive or destructive) and as such, the
radio reception in the car (fuzzy or clear), and prove your answer.

Homework Equations



I don't know the equations for interference... I know that the condition for constructive is m[tex]\lambda[/tex] and for destructive is (m+0.5)[tex]\lambda[/tex].
v=f[tex]\lambda[/tex] ?

The Attempt at a Solution



I'm totally stuck, as I don't know what equation to use.

Thanks for your help :)
 

Answers and Replies

  • #2
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Use c=νλ with the values given for the frequency and the speed of light.
(ν here is the greek letter "nu", denoting frequency)

From this, you'll find the wavelength of the wave.
Now, the waves will "hit" the wall and be reflected towards the car again, so the intereference will be costrunctive if distance/wavelength is an integer, whilst we would have destructive interference otherwise. With the maximum destructive interference occurring if the ratio distance/wavelength is half-integer.

A diagram of the wave wavelengths approaching the car from the tower and the wall would really help you "see" why this is the case.

R.
 
  • #3
gneill
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Do a search on "clapotis wave".
 
  • #4
SammyS
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The wall will reflect the radio waves, setting up standing waves caused by interference between the incident waves and the reflected waves.

One important fact is that the wall is a dense barrier. That should tell you whether the the wall position is a node or an anti-node for he standing waves. Use the wavelength to decide if the car antenna is near a node or an anti-node.
 
  • #5
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Thanks everyone :)

Use c=νλ with the values given for the frequency and the speed of light.
(ν here is the greek letter "nu", denoting frequency)

From this, you'll find the wavelength of the wave.
Now, the waves will "hit" the wall and be reflected towards the car again, so the intereference will be costrunctive if distance/wavelength is an integer, whilst we would have destructive interference otherwise. With the maximum destructive interference occurring if the ratio distance/wavelength is half-integer.

A diagram of the wave wavelengths approaching the car from the tower and the wall would really help you "see" why this is the case.

R.

I got 3351955.307 for the wavelength, so I know it is not an integer, and the interference is thus destructive. I still don't understand how that explains why it is destructive.. :confused:
 
  • #6
gneill
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I got 3351955.307 for the wavelength, so I know it is not an integer, and the interference is thus destructive. I still don't understand how that explains why it is destructive.. :confused:

3351955.307 what? Check your units. Show your work.
 
  • #7
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Sorry! I think I've figured it out.

Finding wavelength:

lambda=v/f
=(3)(10^8) / (8.95)(10^7)
=3.352 m

And then, to find out if the wave hits the wall at a node or an antinode:

56.983m / 3.352m = 16.9997
Which can be rounded to 17 full wavelengths from the antenna to the wall.
And because it is an integer, the interference is constructive.

Is that right? Or am I missing a step?
 
  • #8
lol im in the same physics class as you
your in mr jackson class right
 

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