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Problem with basic doppler effect question

  1. Aug 13, 2013 #1
    1. The problem statement, all variables and given/known data
    The source of a 1 kilohertz sound is getting closer to the listener at a speed of .9 times the speed of sound. What frequency does the listener hear?

    2. Relevant equations
    fl = ((v-vs) / (v + vl))fs

    3. The attempt at a solution
    ((343+343(.9)) / 343)*1000=1900 Hz
    this is the wrong answer, what did I do wrong? it's supposed to be 10,000 Hz
  2. jcsd
  3. Aug 13, 2013 #2
    Correct formula:
    [itex]f = \frac{c\ +\ v_{r}}{c\ +\ v_{s}} \cdot f_{0}[/itex]

    In your problem the source is getting closer, so [itex]v_{s}[/itex] is negative.

    As you didn't miss that, your error is in the formula. Use the one above.

    You also SUMMED values that you should have SUBTRACTED.


    [itex]v_{r}[/itex] is the velocity of the receiver relative to the medium; positive if the receiver is moving towards the source;

    [itex]v_{s}[/itex] is the velocity of the source relative to the medium; positive if the source is moving away from the receiver;
  4. Aug 13, 2013 #3


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    Homework Helper

    Use this equation : ##f = f_0( \frac{v}{v ± v_s} )##

    Where ##f## is the stationary frequency. ##f_2## is the apparent frequency detected. ##v## is the speed of sound in air and ##v_s## is the speed of the source.

    Since the source is moving closer to your listener, you want to use ##v - v_s## ( Doppler effect ). Don't forget to convert kHz to Hz otherwise this wont work!

    I got ##f = 10000 Hz##.
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