Help on Superposition and Interference

  • Thread starter andrew410
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  • #1
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Ok...so I don't know how to do this problem...

Two speakers are driven by the same oscillator whose frequency is 200Hz. They are located on a vertical pole a distance of 4.00 m from each other. A man walks straight toward the lower speaker in a direction perpendicular to the pole. a) How many times will he hear a minimum in sound intensity? b) how far is he from the pole at these moments? Take speed of sound to be 330 m/s and ignore any sound reflections coming off the ground.

I don't know how to start this problem so any help will be much appreciated!
 
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Answers and Replies

  • #2
Tide
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This might get you started. First, you're talking about sound waves which mean they are longitudinal. Therefore, you can express each wave as something like

[tex]\vec A(r) \sin (\vec k \cdot \vec r)[/tex]

where A is the (vector) amplitude and is parallel with the direction of propagation for sound waves. To do it correctly you might note that the amplitude falls of inversely with the square of r which is the distance from the source to the detector (the man's ears). [itex]\vec k[/itex] is the (vector) wavenumber.

Write expressions like that for both waves (one from each speaker) paying particular attention to the directions of wave numbers, amplitudes and displacements between the source and observer.

Then you will need to add the vector quantities to obtain the resultant sound field at the observer's location. Finally, work out the magnitude of that vector and, after applying a trig identity or two it should be evident that the intensity of the sound varies periodically with distance walked.

Incidentally, there is a time dependence to the wave functions but since both speakers are driven by the same oscillator we can ignore it.
 
  • #3
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How would you get the amplitude or "r" ?
 
  • #4
Tide
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Unless you are very close to the source you should be able to write the magnitude of the amplitude of each wave as

[tex]\frac {constant}{r^2}[/tex]

where r is the distance from the source to the detector. Don't forget, though, the A that appears in the original equation I wrote is a vector quantity!
 
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  • #5
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So its like...
[tex]\frac{1}{r^2}*r\sin({\frac{2\pi}{\lambda}*r)}[/tex]
where lambda equals
[tex]\frac{v}{f}[/tex]

right?
 
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  • #6
Tide
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andrew410 said:
So its like...
[tex]\frac{1}{r^2}*r\sin({\frac{2\pi}{\lambda}*r)}[/tex]
where lambda equals
[tex]\frac{v}{f}[/tex]

right?
Yes - just be sure to remember there are two different r's in your waves!
 

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