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Homework Help: Calculating phase difference of sound waves

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    An observer stands 3 m from speaker A and 5 m from speaker B. Both speakers, oscillating in phase, produce waves with a frequency of 250 Hz. The speed of sound in air is 340 m/s. What is the phase difference between the waves from A and B at the observer's location?

    2. Relevant equations

    3. The attempt at a solution

    $$Δr=|{ r }_{ 2 }-{ r }_{ 1 }|\\ =2$$

    I have no idea what to do. What the textbook says is attached. This was a test question. :(

    Attached Files:

  2. jcsd
  3. Apr 16, 2013 #2

    rude man

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    OK so the difference in distance is 2m as you state.

    What is the wavelength of the sound?

    And what relates wavelength to phase angle?
  4. Apr 16, 2013 #3
    Think of the sound waves produced from the speakers as being composed of multiple wavelengths, whose length corresponds to the frequencies given.

    When both sound waves hit the observer's ears, their wavelengths are not going to be aligned, given by the difference in their wavelengths. This difference is in fact the phase difference of the two sound waves.
  5. Apr 17, 2013 #4
    I've calculated the wavelength to be 1.36m. But I have no equation that relates wavelength to phase angle.

    Isn't the wavelength for both the same?


    v and f are the same, so λ must be the same.
  6. Apr 17, 2013 #5
    Any help guys? I have a physics exam in 8 hours from this post.

    Thank you in advance. The textbook has only 1 example and it is a simpler problem. :(
  7. Apr 17, 2013 #6

    rude man

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    If I separate two waves of the same wavelength by one wavelength the phase difference is defined to be 2pi. Take it from there.

    Did I say it wasn't? I hope not ...

    Oops, someone else did, and that was wrong. They're the same, as you say.
  8. Apr 17, 2013 #7
    2m/1.36m*2pi = 9.24 rad = 2.96 rad

    Is this right?

    I didn't realize Δr had anything to do with wavelength.
    Last edited: Apr 17, 2013
  9. Apr 17, 2013 #8

    rude man

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    Exactly right. The phase difference is actually 9.24 but you would measure 2.96.
  10. Apr 17, 2013 #9
    Aren't they the same thing?
  11. Apr 17, 2013 #10

    rude man

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    No. It's like a clock that went round from noon to noon. Difference = 0 but it's a day later!

    But OK - if you're asked to find the phase difference between two monochromatic (one and only one frequency) signals, always give an answer either between 0 and 2 pi or between -pi and +pi, just like you did. You can't measure any differently (unless we're dealing with signals of finite-width spectra, e.g. a signal with a mixture of between 500 and 501 nm. Then than interferogram (beating the two signals aginst each other) would look like a standard monochromatic interferogram except the peaks would diminish with the fringe number. And each fringe = 2 pi. If this sounds like gobbledygook, don't worry about it.)
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