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Beats from Different Strings

  1. Jul 6, 2011 #1
    Two strings which are fixed at both ends are identical except that one is 0.65 cm longer than the other. Waves on both of these string propagate with a speed of 34.9 m/sec and the fundamental frequency of the shorter string is 220 Hz.

    a) What is frequency of the beat that would result if these two strings were plucked at the same time?

    For this, I did 34.9 / 220 to get the length of the string, which was 0.158636m. Then I added 0.065m to it to get 0.2236m. Then I did, 34.9 / 0.2236 to get the new frequency which is 156.0569 Hz, but thats not right. It makes sense since the new frequency should be greater. What am I doing wrong?

    c) What is the beat frequency if the length difference is now 0.8 cm?

    For this one, I would pretty much follow the same steps for part a, but use 0.08m and then subtract 220 from this new frequency.
     
  2. jcsd
  3. Jul 6, 2011 #2

    rl.bhat

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    For this, I did 34.9 / 220 to get the length of the string,
    This in not correct.
    v = fλ.
    In the fundamental mode of vibration, length of the stretched string is λ/2.
     
  4. Jul 6, 2011 #3
    So then I did 34.9 / 220 and got 0.15863 for the wavelength. Then I divided that by 2 and got 0.0793. After that, do I follow the same steps as I desctibed in the problem?
     
  5. Jul 6, 2011 #4
    Ok, so I used this new answer and followed the same steps as I described earlier and its still not the right answer. My answer was 241.927. Its greater than the original frequency, but still not the right answer.
     
  6. Jul 6, 2011 #5

    rl.bhat

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    0.65 cm = 0.0065 m.
     
  7. Jul 6, 2011 #6
    Ok, so I added 0.0065 to 0.079318m and got 0.085818m. Then I did 34.9/ 0.085818 and got 406.6737 Hz, but thats not right either.
     
  8. Jul 7, 2011 #7

    rl.bhat

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    Length of the wire is 0.085818 = λ/2.
    Find λ and then frequency.
     
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