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Wave Interference and wavelength

  • Thread starter satchmo05
  • Start date
  • #1
114
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Homework Statement



In Fig. 17-35, sound with a 15.6 cm wavelength travels rightward from a source and through a tube that consists of a straight portion and a half-circle. Part of the sound wave travels through the half-circle and then rejoins the rest of the wave, which goes directly through the straight portion. This rejoining results in interference. What is the smallest radius r (cm) that results in an intensity minimum at the detector? (Take the speed of sound to be 343 m/s.)

Homework Equations



If it wants the minimum intensity, we have to minimize the following equation: I = P/(4pi*r^(2)), where I is intensity, P equals power, and r equals radius. But then again, I also think of derivative when I hear maximum/minimum values...

The Attempt at a Solution



343/15.6 = 21.99 Hz = frequency
2pi * frequency = angular velocity = 138.15
k = wave number = 0.403

So, these are the values we know, but I do not know how to manipulate them into the formula above...maybe I'm using the wrong formula?! Please help! I appreciate all the help. Thanks a bunch!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
114
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Oh, another piece of info. that I know, (but after I enter in the problem, it tells me I'm wrong) is that P = .5(4*pi^(2))*frequency^(2)*A^(2)*v, and A = 4*pi*r^(2). They also say on this one website that I = f^(2)*A^(2). Therefore, the equation simplifies down to:

r^(2) = .5*pi*velocity --> r = 23.2cm = WRONG!!!!

Please help me figure out this problem! Thanks again!
 
  • #3
114
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am I on the right track, or do I need to use a different formula?! Thanks for the help!
 

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