Diffraction from a loud speaker

  • Thread starter pmd28
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  • #1
pmd28
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Homework Statement


Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted outside on a pole. In winter, when the temperature is 273 K, the diffraction angle θ has a value of 11°. What is the diffraction angle for the same sound on a summer day when the temperature is 307 K?


Homework Equations


v=sqrt([itex]\gamma[/itex]kT/m)
v=[itex]\lambda[/itex]f
sin[itex]\theta[/itex]=[itex]\lambda[/itex]/D


The Attempt at a Solution


I can't seem to relate these equations. I know that [itex]\gamma[/itex]k/m is constant for both T and I also know that v increases with temperature but I don't understand how that affects lambda and frequency.
 

Answers and Replies

  • #2
TSny
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The frequency of a sound wave is determined by the frequency of vibration of the source of sound (the speaker in this case). So, changing the temperature does not change the frequency of the sound if you assume the speaker vibrates at the same frequency for both temperatures.

A good way to approach this problem is to consider the ratio [itex]\frac{sin\theta_f}{sin\theta_i}[/itex] where the subscripts refer to the initial and final temperatures.
 
  • #3
pmd28
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So I set that ratio equal to the ratio of vf/vi and then canceled out sqrt([itex]\gamma[/itex]k/m) because it is constant in both cases therefore the ratio of Sinfθ/Siniθ is equal to sqrt(Tf)/sqrt(Ti) It was correct.
 

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