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Simple Harmonic Motion of skydiver

  1. Jul 26, 2005 #1

    BDR

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    I can't seem to get the right answer, where is my mistake?

    A 90 kg skydiver hanging from a parachute bounces up and down with a period of 1.5 seconds. What is the new period of oscillation when a second skydiver, whose mass is 60 kg, hangs from the legs of the first?

    I am using the equation T = 2pie square root of m/k.

    I found k and then combined the 2 masses (m1+m2), and put that back into the equation and found an answer but not the right one.

    The right answer is suppose to be 1.94 s ....i'm getting 58
     
  2. jcsd
  3. Jul 26, 2005 #2

    LeonhardEuler

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    Your method is correct, you must be doing some calculation wrong. What did you get for k?
     
  4. Jul 26, 2005 #3

    BDR

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    I got k = 38.14 I don't know if its right though
     
  5. Jul 26, 2005 #4

    LeonhardEuler

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    There's your problem. k is much bigger than that. How did you find k?
     
  6. Jul 26, 2005 #5

    mukundpa

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    your value of k is not correct check the calculations

    You may also use T1/T2 = squrt of m1/(m1+m2)
     
  7. Jul 26, 2005 #6

    BDR

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    I'm not sure that is the correct number i got for K, i have so many numbers on my paper everything is mixed together. I took the period given which was 1.5 divided by 2 pie 1.5/2pie = .239. Then i squared to get rid of the square root. Which gave me .239 = m/k. Then multiplied by k and divided.
     
  8. Jul 26, 2005 #7

    BDR

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    I'm getting an answer of 2.5, is that correct? The correct answer is 1.94 s, or thats what the book says.
     
  9. Jul 26, 2005 #8

    mukundpa

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    T = 2p sqr(m/K)
    T^2= 4p^2(m/K)
    K = (4*P^2*m)/T^2
    put the values it gives 1579.14 N/m
    how you were calculating

    I suggested an easy method in the earlier posting
    As K is not changing you can eliminate it from the two cases by dividing the equations of the two time periods you will get the equation I gave in that posting and by substituting the values you can calculate T2

    It is always better to solve first in notations and finally calculating the numerical value.
     
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