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Sound waves propagation and the doppler effect

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data

    A lighthouse emits a noise with a frequency of 2000 Hz in the direction of the sea, with a power of 100W. Consider the speed of sound, vs, to be 341 m/s.

    A ship approaches the coast with a speed of 30 km/h and produces noise with an intensity of 40dB. Consider it is possible to distinguish two sounds with equal levels of intensity. Find the maximum distance from the coast where the ship can distinguish the sound of the lighthouse.

    2. Relevant equations

    IDB = 20log10(P/P0) , with P0 = 2*10-5 N m-2
    ψ(x,t) = ψ0cos(wt-kx)
    ρ(x,t) = kρ0ψ0cos(wt-kx+pi/2)
    P(x,t) = vsωρ0ψ0cos(wt-kx+pi/2)
    I = 0.5 * vsωρ02ψ02 =
    = 0.5 * Ps02 / (vsρ0) , with Ps0 equal to the amplitude of P(x,t), vsωρ0ψ0

    3. The attempt at a solution

    It will be impossible to hear the sound of the lighthouse when the intensity of the sound is equal to 0 decibels. Therefore:

    IDB = 20log10(P/P0) = 0 ⇔ P/P0 = 1 ⇔ P = P0

    I also know that the intensity is equal to the power divided by the area:
    I = Power / Area

    Therefore, assuming the lighthouse emits in every direction I have:
    I = Power / (4∏r^2) , where r will be the distance from the lighthouse.

    Equaling this equation with the expression of intensity that depends on pressure we get:
    Power / (4∏r^2) = 0.5 * Ps02 / (vsρ0) ⇔
    ⇔ r^2 = 2*Power*(vsρ0) / [4∏*Ps02]

    If I solve this equation I will not get to the correct result tho.
    If anyone could point me in the right direction I'd really appreciate.
    Thanks!
     
  2. jcsd
  3. Mar 14, 2013 #2

    rude man

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    Homework Helper
    Gold Member

    For what it's worth:

    The units are confusing to me. I suppose I at the ship due to the lighthouse sound is 100W/4πd2 or 100W/2πd2 depending on whether the ground sound is absorbed or reflected.

    The intensity at the boat is said to be +40dB, but I don't know what the intensity reference level is. In one webpage I found I_ref = 1e-12 W/m^2 so by that +40dB would be 1e-8 W/m^2.

    So you can equate that level of intensity with what you get from the lighthouse at a distance d.

    Maybe you can fill in some of the blanks.

    I don't know how to relate sound pressure to either power or intensity but I don't see why that should be necessary here.
     
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