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Sound Beats

  1. Jan 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440 Hz. One of the strings is retuned by adjusting its tension. When this is done, 2 beats per second are heard when both strings are plucked simultaneously.

    a. What is the highest possible fundamental frequency of the retuned string?
    b. What is the lowest possible fundamental frequency of the retuned string?
    c. By what fractional amount was the string tension changed if it was increased?


    2. Relevant equations

    F_beat = T_b - T_a / Ta*Tb => 1/Ta - 1/Tb

    F_beat = f_a - f_b


    3. The attempt at a solution

    I'm only given 2 values and I'm not sure where to go with this problem
     
  2. jcsd
  3. Jan 20, 2007 #2
    for a and b--you know that if two pitches are sounded two hertz apart, two beats per second will be heard (your formula). you can figure this out prett easily from there (there are only two possibilities--one high and one low.
     
  4. Jan 22, 2007 #3
    If they both have a frequency of 440hz, and it's 2 beats per second and the beat frequency is just the difference between them, is it just 442 and 440 ?
     
  5. Jan 22, 2007 #4

    berkeman

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    442 is the correct answer for a). What about b) and c) ?
     
  6. Jan 22, 2007 #5
    b) 438

    c) I'm not sure how to start it.
     
  7. Jan 22, 2007 #6

    berkeman

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    What is the relationship between the tension in a string and the velocity of propagation of a travelling wave in the string? And what is the relationship between the propagation velocity of a travelling wave in a string, versus the fundamental oscillation frequency of a string that is fixed at both ends and plucked?
     
  8. Jan 22, 2007 #7
    For the 1st relationship v = sqrt(Force of Tension / Mass Density)

    For the 2nd relationship fundamental oscillation frequency is Fn = nv / 2L
     
  9. Jan 22, 2007 #8

    berkeman

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    So that gives you all you need for c). What answer do you get?
     
  10. Jan 22, 2007 #9
    F_1 = nv / 2L

    442 = 1*v / 2L

    442 = 1 * (F_t / Mass Density) / 2L

    Though the given's don't provide the mass density of the string or the length of the string. This is where I'm a little confused on.
     
  11. Jan 22, 2007 #10

    berkeman

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    The mass density and the length do not change. So when you calculate the percent change in the tension based on the percent change in the frequency, those other variables should cancel out. Oh, they asked for the "fractional" change, so maybe I should have said fractional instead of percent. Same concept, though.
     
  12. Jan 22, 2007 #11
    So the fractional change would be

    fn = n *Force of tension / 2

    f1 = 1 * Force of tension / 2

    f1 = 442

    442 = 1*Force of Tension / 2

    ? If this is right so far, what is the original force of tension, I know it gives the original frequency at 440hz at the same force of tension.
     
  13. Jan 22, 2007 #12

    berkeman

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    This is what you should use. The frequency ratios directly with the velocity of propagation. And the velocity ratios with the sqrt of the tension. So if the frequency goes up to 442 Hz from 440 Hz (fractional amount 2/440), what does that say about the fractional change of the tension required to make that change?
     
  14. Jan 22, 2007 #13
    So basically the fractional increase is 2/240 and the fractional decrease is 2/338 ? Thank for taking the time to explain this, it's been a few years since I've taken physics, jumping back in to finish the final course.
     
  15. Jan 22, 2007 #14

    berkeman

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    Hold on.... Those are the fractional changes of the frequency. You are asked in c) to find the corresponding fractional change in the tension to cause the positive fractional frequency change. The tesion change is not 2/440.
     
  16. Jan 22, 2007 #15
    Okay so the fractional amount from 442hz to 440hz is 2/440. Now the question ask for the fractional change in tension. The thing I'm confused about is finding out the tension force. You said that the Mass Density and the Lenght is constant so they should cancel out.

    From that

    F = 1 * Force of Tension / 2

    I guess I'm still not following on what is the current Force of Tension.
     
  17. Jan 22, 2007 #16

    berkeman

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    Okay, I'm just trying to get you to see how you involve the square root thing.

    Take this analogy to see if it helps. The length of the side of a square is equal to the sqrt(Area), right? Now, if I increase the length of a side of a square by 2/440, what is the fractional change in the Area?
     
  18. Jan 22, 2007 #17
    Fractional change in Area would be Sqrt(2/440) ?
     
  19. Jan 22, 2007 #18

    berkeman

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    In the square Area analogy:

    [tex]\frac{\Delta L}{L} = \sqrt{\frac{\Delta A}{A}}[/tex]

    What mathematical operation do you perform on both sides in order to isolate the answer that is desired?

    [tex]\frac{\Delta A}{A} = ??[/tex]
     
  20. Jan 22, 2007 #19
    You have to square both sides.

    EDIT:

    Would the fractional change in tension be the sqrt(442/440)?
     
  21. Jan 22, 2007 #20

    berkeman

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    Woo-hoo! So what is (2/440)^2?

    Note how the tension goes up, but not as much as the frequency goes up (since the fractional change in frequency is less than one).
     
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