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Finding Amplitude

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data
    An air-track glider attached to a spring oscillates between the 9.23 cm mark and the 50.23 cm mark on the track. The glider completes 27 oscillations in 44.7 s. angular frequency of the glider? (in rad/s)
    3.7933
    what is the amplitude of the glider?


    2. Relevant equations

    x=Acos(wt+ro)

    3. The attempt at a solution
    I considered that at x=9.23 the amplitude was zero and so is ro.
    Hence, x=50.23 would equal just 50.23 cm=Acos(3.79(44.7)). But A turned out to be negative. What am I doing wrong? How do I solve this problem? Thank you in advance.
     
  2. jcsd
  3. Jun 16, 2009 #2

    dx

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    What is the center of oscillation here? The amplitude is the maximum distance from the center that the glider reaches.
     
  4. Jun 16, 2009 #3
    Would it be half the distance the glider travels which is 20.5 cm?
     
  5. Jun 16, 2009 #4

    dx

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    Correct.
     
  6. Jun 16, 2009 #5
    It still gives me a negative number if I do .205/((cos(3.79*44.7)).
     
  7. Jun 16, 2009 #6

    dx

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    Negative number for what?
     
  8. Jun 16, 2009 #7
    For the amplitude A.
     
  9. Jun 16, 2009 #8

    dx

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    :confused: You just said the amplitude was 20.5 cm, which is correct.
     
  10. Jun 16, 2009 #9
    OH! Ok I get it. Thank you very much.
     
  11. Jun 16, 2009 #10
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