1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Power and Intensity for a Sound Wave

  1. Aug 29, 2008 #1
    The power for a sound wave is give by P = 1/2*p*A*(w*s)^2*v...in which p is density (rho), A is cross-sectional area, w is angular frequency, s is maximum displacement (amplitude), and v is speed of propagation. The intensity is given by P/A...the intensity for a spherical sound wave is supposed to decrease over time (according to experience)...but I'm not seeing it in the equations. The power is proportional to the area over some region and the intensity is inversely proportional to the area over some region. Therefore...the Intensity at a given point shouldn't depend on A (the area)...that would make the intensity constant throughout the wave. Where am I wrong in the reasoning?

    Oh wait...never mind...the source determines the power...I shift to a new question: the A (area) in the equations is the cross-sectional area immediately "touching" the source, right?
     
    Last edited: Aug 29, 2008
  2. jcsd
  3. Aug 29, 2008 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No, A is the area at the place where power or intensity is being measured.
     
  4. Sep 6, 2008 #3
    So in that case...if a drum is the source of a sound...then the area A would be the surface area of the oscillating surface of the drum, correct?
     
  5. Sep 6, 2008 #4
    ಠ_ಠ ??????
     
  6. Sep 6, 2008 #5

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    A is the area at wherever the sound power is being calculated or measured. It could be at the drum's surface, it could also be somewhere else.
     
  7. Sep 6, 2008 #6
    Wait...how could it also be somewhere else? Isn't it where the surface of the source intercepts the medium? Is it an arbitrary value?
     
  8. Sep 6, 2008 #7
    ಠ_x????...the area A I'm referring to is the one in the equation for sound wave power: P = 1/2*p*A*(w*s)^2*v
     
    Last edited: Sep 6, 2008
  9. Sep 7, 2008 #8

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Do sound waves have power only right at the source? Or do they also have power wherever they travel?
     
  10. Sep 7, 2008 #9
    They have power wherever they travel...but how does that fit in the equation if density, angular velocity, maximum displacement, and velocity are constant (it would only leave area, A, as variable)?
     
  11. Sep 8, 2008 #10

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Maximum displacement will be different, decreasing as you move farther from the source.

    edit:
    Another suggestion: is A defined in the book or article you got the equation from? Perhaps they are only concerned with power at the source, and that's all that you have to be concerned with ... even though we could also calculate/measure power at locations away from the source if we wish to.
     
    Last edited: Sep 8, 2008
  12. Sep 9, 2008 #11
    oh....that makes sense.....
     
  13. Sep 9, 2008 #12
    Does this mean that for spherical waves, the pressure and displacement graphs are damped?
    In the book, A is not explicitly defined. They come up with a proof for the energy and power equations using a model, in which the area is taken at the source...thing is that in this model, the area stays constant (sound propagating through a cylinder). They then defined intensity for this model, which would also stay constant due to the constant area...then they generalized the situation for spherical waves with I = P(avg)/A. Looks like I wasn't understanding the concept well.
     
    Last edited: Sep 9, 2008
  14. Sep 9, 2008 #13

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes. Intensity will follow an inverse-square law with distance from the source.
    Pressure and displacement amplitudes are proportional to square root of intensity, and hence are inversely proportional to distance from the source.

    Okay.
    "A" would be the surface area of a sphere, at a distance "r" from the source.
    So A = 4 pi r2, giving an inverse-square relation for I=P/A
     
  15. Sep 10, 2008 #14
    Alright...thanks for clarifying things
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Power and Intensity for a Sound Wave
Loading...