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Sound waves question - vibrations and amplitudes

  1. Dec 16, 2012 #1
    A tuning fork vibrates 235 times in 3.1 seconds with an amplitude of 0.59mm. Determine..
    i) its frequency
    ii) its period
    iii) the distance the end of one of the fork tines travels in one second

    for i) The answer I got is f=75.81Hz
    ii) The answer I got is 0.13s
    iii) ???
     
  2. jcsd
  3. Dec 16, 2012 #2
    hz is 1/s so 235/3.1s

    period is 1/frequency s
     
  4. Dec 16, 2012 #3
    You may want to check your calculation for part (ii) again! I think it is 0.013s
     
  5. Dec 16, 2012 #4
    I think a little more info is needed for part (iii) because you don't know where it is starting from. Do we want to find the distance it travels after 1 second from it's equilibrium position or from its maximum displacement?
     
  6. Dec 16, 2012 #5
    I'm going to guess equilibrium, since this is only gr. 11 physics. The supply teacher said there are 4 amplitudes in 1 vibration, so the distance is 4*0.59*frequency. I tried it, but I got an unreasonable answer and I don't think there are 4 amplitudes in a vibration.
     
  7. Dec 16, 2012 #6

    haruspex

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    Another reasonable guess is that it means on average (which is fact what you computed).
    Starting from equilibrium, how far will it travel in the first quarter period? How far in the second quarter?
    What answer did you get?
     
  8. Dec 16, 2012 #7
    First quarter = 44.73 mm
    Second quarter = 89.4 6mm
     
  9. Dec 16, 2012 #8

    haruspex

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    How did you get those numbers? The amplitude is only 0.59 mm.
     
  10. Dec 16, 2012 #9
    For the first quarter, there is 1 amplitude, so (1*0.00059*75.81)
    For the second quarter, there are 2 amplitude, so (2*0.00059*75.81)

    I multiplied by the frequency to get the answer for 1 second.
     
  11. Dec 16, 2012 #10

    ehild

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    Displacement is not the same as the distance travelled. If you go to school 500 m far, you walk 500 m and your displacement is also 500 m. If you go there and back, your displacement is zero, but you walked 1000 m, your "distance travelled" or "distance covered" is 1000 m. The tip of the fork travels from equilibrium to maximum displacement (amplitude) during a quarter of the period. Then it moves backwards, reaching zero displacement in quarter period again, but travelling A distance again, so the distance travelled in half period is 2A. The whole distance travelled in one period is 4A. The number of periods in one second is equal to the frequency. So the total distance travelled in one second is?


    ehild
     

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  12. Dec 16, 2012 #11
    so the answer is 4*0.00059*75.81 = .1789m or 178.9mm
     
  13. Dec 17, 2012 #12

    ehild

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    Yes it is, but round it to three significant digits.

    ehild
     
  14. Dec 17, 2012 #13

    haruspex

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    I had asked how far in one quarter period, i.e. one quarter of a complete oscillation.
    You had expressed doubts that it travelled four amplitudes in each period; I was trying to get you to see that it does.
     
  15. Dec 17, 2012 #14
    I finally understand it now. Thank you so much ehild and haruspex.
     
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