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Humphrey the Singing Whale and his mate Matilda.

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Humphrey the Singing Whale and his mate Matilda swim toward each other, with Matilda swimming twice as fast as Humphrey. When Humphrey sings a note of a frequency 299 Hz. Maltida hears a frequency of 302 Hz. How fast is Humphrey swimming? (The speed of sound i sea water is 1533 m/s)

    2. Relevant equations

    f=fo(v+-vo/v-+vs)

    3. The attempt at a solution

    v0=(fo/f)*vs and i get 1526.4 m/s

    the answer is 5.11 m/s I'm lost , any help?
     
  2. jcsd
  3. Apr 15, 2012 #2
    Hi ScienceGeek24,

    I can confirm that 5.11m/s is the correct answer, and that you're starting out from the right Doppler effect equation - you're on the right track! Just be careful of your signs (the whales are moving in opposite directions, but I think you got that from your +- / -+ notation you wrote down), and start over on your algebra. It should start out like:

    302Hz = 299Hz*[(1533m/s - 2*v_Humphrey) / (1533m/s + v_Humphrey)]

    which is exactly what you wrote down under 'relevant equations'. If something is still unclear please let me know,

    Hope this helps,
    Bill Mills
     
    Last edited by a moderator: Apr 16, 2012
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