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Doppler effect and Beat frequency

  1. Apr 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A runner is running with speed V along the straight line connecting two identical speakers. Both speakers are playing a tone of the same frequency f.
    a)What is the beat frequency that the runner hears?
    b)If the standing wave forms between two speakers, how frequently will the runner hit a node of the standing wave?
    c)Compare your results from (a) and (b).

    2. Relevant equations
    fl = fs(v+Vl)/(v+Vs), where v is the velocity of the wave, Vs is the velocity of the source and Vl is the velocity of the listener.
    v = λƒ

    3. The attempt at a solution
    I solved part a using the doppler effect equation. Tbeat = Vf/2v
    For part b, I got the answer Vf/4v. Because it is two open ends, the start of the wave is an displacement antinode and thus the displacement node is at every λ/4.
    But according to the answer it seems that the node they are referring to is the pressure nodes which are the displacement antinodes, thus the displacement antinodes is at every λ/2 which gives the answer Vf/2v which is the given answer.
    So am I correct that the question refers the node as an pressure nodes? Or are they referring to the displacement nodes thus my concept is a bit shaky.
     
  2. jcsd
  3. Apr 22, 2015 #2
    What exactly is a "pressure" node?

    And why do you assume the wave is "open" at two ends? It actually doesn't matter.

    Also using "T" for frequency is a bit confusing since this is typically used for period.
     
    Last edited: Apr 22, 2015
  4. Apr 22, 2015 #3

    haruspex

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    Don't nodes and antinodes alternate? How can there be twice as many of one as of the other?
     
  5. Apr 22, 2015 #4
    Okay shall change it to fbeat = Vf/2v. Because a wave equation is y(x.t) = Acos(kx-wt). And y represents the displacement of a particle from the equilibirum point. However in longitudinal wave, like sound wave, we can represent the wave in terms of a particle displacement or the pressure variations in the fluid where ΔP = BkAsin(kx-wt), where B is the bulk modulus of the fluid and k is the wave number (2π/λ). Then when the particle displacement from the equilibrium is 0 (displacement node), the pressure variation is the highest (pressure antinodes). That 2 equation have a phase difference of λ/4. So I'm asking that the question refers the node as a displacement node, or a pressure node. In a pressure node I assume there is not variation in pressure, thus you can't hear anything in a pressure node? While if you are in a displacement node, the variation in pressure is the largest thus the sound will be the loudest?
    upload_2015-4-23_12-59-48.png
     
  6. Apr 23, 2015 #5
    Yeah they alternate. The wave formed between the 2 speakers is a standing wave with 2 open ends? So the displacement nodes are λ/4? Is kind of hard to explain so I'll use a picture. upload_2015-4-23_13-5-30.png
     
  7. Apr 23, 2015 #6

    haruspex

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    As per your picture, successive nodes are λ/2 apart, and successive antinodes are λ/2 apart.
    The question asks how frequently, not how long to the first one.
     
  8. Apr 23, 2015 #7
    Okay I get what you mean. So that is why it does not matter whether is an open end because each successive nodes/antinodes are always λ/2 apart?
     
  9. Apr 23, 2015 #8

    haruspex

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    Yes.
     
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