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Pendulum Oscillations problem

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data
    A pendulum consists of a massless rigid rod with a mass at one end. The other end is pivoted on a frictionless pivot so that it can turn through a complete circle. The pendulum is inverted, so the mass is directly above the pivot point, and then released. The speed of the mass as it passes through the lowest point is 6.0 m/s. If the pendulum undergoes small amplitude oscillations at the bottom of the arc, what will be the frequency of the oscillations?

    2. Relevant equations
    F= 1/T

    3. The attempt at a solution
    I attempted to use Vmax to solve for w or A but neither w or A or given so im not sure how to work through this problem
  2. jcsd
  3. Dec 9, 2011 #2


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    Check in your textbook or class notes for a discussion of pendulums. What does it say about the period (or frequency) of a pendulum?
  4. Dec 9, 2011 #3
    w = 2(pie)f = sqroot(g/l) is the only thing i found
  5. Dec 9, 2011 #4
    You can use what you found but you need a bit more.

    Also think about the energy change in the mass as it drops from the top to the bottom and goes from a speed of zero to 6.0 m/sec. Draw a picture and see what you can calculate from this information.
  6. Dec 9, 2011 #5
    i got PE at the top = mgh = 9.8m(length of string) which is equal to 1/2mv^2 = 18m therefore L= 1.84m which would give a frequency of .367 which would give a period of 2.72? correct?
  7. Dec 9, 2011 #6
    Be careful here. How far does the mass fall relative to the length of the pendulum. You may be right as I haven't calculated it but you didn't mention an important point in this part of the problem. You have the right idea.
  8. Dec 9, 2011 #7
    the mass starts off one string length above the pivot of the pendulum and ends up one string length below the pivot, how does this affect the problem?
  9. Dec 9, 2011 #8
    so how many string length does it fall to achieve the final velocity of 6 m/sec?
  10. Dec 9, 2011 #9


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    Yes, so what is the change in height, in terms of number of string lengths?
  11. Dec 9, 2011 #10
    change in height is 2L, from 2L, so does this mean the PE should be double what i had since i used mgh and a height of only one L?
  12. Dec 9, 2011 #11
    That's what you didn't say - that your mgh is mg2L. That's correct
  13. Dec 9, 2011 #12
    so mgh= 9.8m(2L) which is equal to KE = 1/2m(6)^2 which gives L = .918m which gives a freq of .5199 which gives a period of 1.92s?
  14. Dec 9, 2011 #13
    Good job. You are using correct concepts. I haven't checked your math.
  15. Dec 9, 2011 #14
  16. Dec 9, 2011 #15
    at your service :-)
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