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Homework Help: Fundamental Frequency

  1. Jan 6, 2005 #1
    Which of the following could be the fundamental freq. for a vibration that has an overtone freq. of 990 Hz?

    a. 330
    b. 660
    c. 148
    d. 1980
    e. 1990
    (All in Hz.)

    The formula that I think you have to use is f_n = n*f_1, where f_1 is fud. freq.

    Then, I thought

    f_2 = 2*f_1, where f_2 is the overtone freq. (is these a correct assumption?)

    but 990/2 is 495.

    Could someone direct me to the correct path of getting the solution?

    I'm confused over the definitions of overtone and fund. freq.

    Thanks.
     
    Last edited: Jan 6, 2005
  2. jcsd
  3. Jan 6, 2005 #2

    Gokul43201

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    There's no reason to assume than n=2, so you just leave it as n.

    So you have [itex]f_n = nf_1 [/itex]
    Or [itex]f_1 = \frac {990}{n} [/itex]

    Which of the choices satisfies this condition for some integer n ?

    Can you say if [itex]f_1[/itex] will be greater than or less than 990 Hz, from this ? Does this rule out any choices ?

    The alternative is simply to notice that [itex]f_n/f_1 = n [/itex], so try the remaining choices of [itex]f_1[/itex] to see which one gives you an integer ratio.
     
    Last edited: Jan 6, 2005
  4. Jan 6, 2005 #3
    Is the answer A?

    The fud freq. would be less than 990, would it not? That leaves A and B, and 990/3 is 330.
     
  5. Jan 6, 2005 #4

    Gokul43201

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    That's right.
     
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