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Sound intensity increase as a result of wave interference?

  1. Dec 2, 2007 #1
    1. The problem statement, all variables and given/known data

    You're in an airplan whose two engines are running at 560rpm and 570 rpm. How often do you hear the sound intesity increase as a result of wave interference?

    Well, i know that the intensity will increase when the wave interference is constructive, but point me in the right direction in how to solve this.
  2. jcsd
  3. Dec 2, 2007 #2
    Have you ever sat in traffic in a turn lane, and watched the turn signals of the cars in front of you blink? Sometimes there's a garbage truck there, and its turn signal blinks at a slower rate than the Camry behind it. If you watch for a bit, the signals seem to alternate, back and forth, but slowly the Camry's turn signal slowly catches up, until for a few moments, they seemingly blink on and off at the same time. That is, until, the Camry's signal goes by, and they go back to blinking at different times.

    If you knew the periods of each turn signal, you could calculate when they would blink together. You have the same type of situation, except instead of blinking lights you have engines cycling many hundreds of times a minute.

    Here's hoping you've been as bored in traffic as I have :-)
  4. Dec 2, 2007 #3


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  5. Dec 2, 2007 #4

    I know exactly what you're saying because i always notice that! You're not the only one.
    I definitely understand the visualization of the what the question is asking, i just don't know how to set it up. That's always the hardest part!

    I'm going to look into it.
  6. Dec 2, 2007 #5
    If i convert the rpm to rad/s, is the rad/s the sound frequency? Sorry if that's a stupid question! I'm behind in this...
  7. Dec 2, 2007 #6
    so if i go with what i said previously, then --->

    560rpm = 58.62 rad/s
    570rpm = 59.69 rad/s

    then simply,

    59.69 - 58.62 = 1.07 rad/s

    T = 2[tex]\pi[/tex]/w --->

    5.87 seconds? how does this look?
  8. Dec 2, 2007 #7
    Not a stupid question, I don't know either. Maybe someone knows for sure.

    In any event, if it is, I think your answer is correct, although you've kind of done a roundabout thing-- typically you find the frequency and calculate beat frequency, like Astronuc said. I think you've done the equivalent here with the velocities instead-- perhaps a beat velocity, if you will. Dunno.

    I -do- know that:

    [tex] T = \frac{1}{\nu} [/tex]


    [tex] T_{beat} = \frac{1}{\nu_{beat}} = \frac{1}{\nu_2 - \nu_1}[/tex]

    and since

    [tex] \omega = 2\pi\nu \Rightarrow \nu = \frac{\omega}{2\pi} [/tex]


    [tex] T_{beat} = \frac{1}{\nu_2 - \nu_1} = \frac{1}{\frac{\omega_2}{2\pi} - \frac{\omega_1}{2\pi}} = \frac{2\pi}{\omega_2 - \omega_1} = \frac{2\pi}{\omega_{beat}}[/tex]

    which is what you've done, so yeah, I think you're good, assuming that the sound frequency is equivalent to the engine frequency.
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