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Frequency and Beat Frequency

  1. Oct 19, 2008 #1
    Two-out-of-tune flutes play the same note. One produces a tone that has a frequency of 260 Hz, while the other produces 266 Hz. When a tuning fork is sounded together with the 260 Hz tone, a beat frequency of 1 Hz is produced. When the same tuning fork is sounded together with the 266 Hz tone, a beat frequency of 5 Hz is produced. What is the frequency of the tuning fork?

    I have no idea where to start. I know that Fbeat= f-Ffork, but I don't think that helps me any. Any ideas?
     
  2. jcsd
  3. Oct 19, 2008 #2

    cepheid

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    Really? Well, for each flute, you are given two out of three of those frequencies, and are trying to find a third one. Of course, each flute could be out of tune in either direction (too high or too low), so it should really be Fbeat = |F - Ffork|. Which is the reason why you need both flutes in order to ascertain what frequency they are supposed to be playing, not just one.
     
  4. Oct 19, 2008 #3
    I am still lost. If I solve the equation for the first flute: 1Hz= 260-Ffork then Ffork=259. If I solve for the 2nd flute 5Hz=266-Ffork then Ffork=261 which is an entirely different answer. Any suggestions. I know I am a little dense when it comes to physics!
     
  5. Oct 19, 2008 #4

    cepheid

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    NOT NECESSARILY. Read my first reply to you again. If we get a beat frequency of 1 hertz between the first flute and the tuning fork, then that means that the difference between the flute's frequency and the tuning fork's frequency is 1 hertz *in absolute value.* That's why I added the absolute value signs to your equation. Because (read this carefully)

    If we get a beat frequency of 1 hertz, we DON'T know whether that means that the flute is 1 hertz too HIGH or whether it is 1 hertz too LOW. Either situation is possible. Again, repeating what I said in my first reply, that's why we need the second flute to make the determination. So the proper way to do this problem is to say that:

    1 Hz = |260 - Ffork|

    case 1: (260 - Ffork) is positive

    then |260 - Ffork| = (260 - Ffork) = 1 Hz

    ==> Ffork = 259 Hz

    case 2: (260 - Ffork) is negative

    then |260 - Ffork| = -(260 - Ffork) = 1 Hz

    ==> Ffork = 261 Hz

    To determine which of these cases is true in this situation, you have to do the same thing for the second flute and pick the answer that is consistent between both flutes.
     
  6. Oct 19, 2008 #5
    The light bulb came on! Thanks you so much! I get it now!!
     
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