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Diving board

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A diving board oscillates with simple harmonic mothion of frequency 3.0 cycles per second.?
    What is the maximum amplitude with which the end of the board can vibrate in order that a pebble placed there to not lose contact with the board during oscillation.


    2. Relevant equations



    3. The attempt at a solution
    Since the intial velocity at the bottom of a cycle is zero,
    Distance = 1/2 a * t^2 where i got t by determining that quarter cycle takes
    1/(4*3.0) sec. Since the accelleration exceeds that of gravity, the pebble will jump off the board. So a = 9.8 m/sec^2
    Distance = 1/2 * 9.8 * (1/(4*3.0))^2
    But this is the motion from the bottom to the center so the Amplitude will be twice this.
    I get Amplitude = 9.8 * (1/(4*3.0))^2 = .068 meters but this answer is not correct
     
  2. jcsd
  3. Nov 11, 2008 #2

    Dick

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    Distance IS NOT (1/2)*a*t^2. That's for constant acceleration problems. Oscillatory motion is NOT constant acceleration. You are using completely wrong formulas. On the good side, your idea is correct. The acceleration of the board should not exceed 9.8m/sec^2. Write the displacement of the board as x(t)=A*sin(w*t) where A is the amplitude. Find w so you get a frequency of 3 cycles per second. Now figure out x''(t), since that's acceleration, correct? Equate the maximum of that to 9.8m/sec^2.
     
  4. Nov 11, 2008 #3
    How do I go about finding w from 3 cycles / sec? and are you saying that x(t) = 9.8 m/s^2
     
  5. Nov 11, 2008 #4

    Dick

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    What's the period of the wave sin(wt)? And no, I said x''(t)=9.8m/s^2, x''(t) is supposed to mean the second derivative of the displacement x(t). Isn't that acceleration?
     
  6. Nov 11, 2008 #5
    oh so i multiply the frequency by 2pi to get w i believe.
    Im sitll confused wht do i input for "t"
     
  7. Nov 11, 2008 #6

    Dick

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    Yes, I think you multiply the frequency by 2pi to get w. Your final expression for the acceleration will only have t in the form of sin(wt). You want the MAXIMUM acceleration. What's the MAXIMUM of sin(wt)?
     
  8. Nov 11, 2008 #7
    sorry im just confused...is it 9.8 m/s^2?
     
  9. Nov 11, 2008 #8

    Dick

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    What's the maximum of sin(wt) over all t? Look at a graph of sin. Tell me what the max is. Forget this specific problem.
     
  10. Nov 11, 2008 #9
    is it 1?
     
  11. Nov 11, 2008 #10

    Dick

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    What do you mean 'is it 1?'. It IS 1. That's the correct answer. 'is it 1?' is not even an answer. It's another question. Sorry, but people have doing this to me all night. Answering an obvious question with a tentative question. Be ASSERTIVE when you know you are right.
     
  12. Nov 11, 2008 #11
    I appreciate all your help. Just take me a little further please if you can.

    so i found w and set t=1

    now i have 9.8= A(-sin6pi)

    Is this correct?
     
  13. Nov 11, 2008 #12

    Dick

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    I'll take you as far as you want until I fall asleep, which might not be that long. The displacement is A*sin(wt) where w=6*pi/sec. I thought we had agreed to set the maximum of the second derivative of the displacement=9.8m/sec^2. What's the second derivative of sin(wt)? And whatever happened to replacing sin with it's maximum of 1? It's still there.
     
  14. Nov 11, 2008 #13
    hmmm so since the max of sin is 1 , do i substitute this value right away or after i have taken the derivate of sin(wt) ?

    1st derivative of sin(wt) = cos(wt)
    2nd derivative = -sin(wt)
     
  15. Nov 12, 2008 #14

    Dick

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    That's one problem. You aren't using the chain rule. d/dt of sin(wt) is not cos(wt). What is it? And there's a second problem. I thought we had figured out that |sin(wt)|<=1. You can set the max of the sin(wt) to 1. Don't put t=1.
     
  16. Nov 12, 2008 #15
    hmm i thought the w and t were numbers so i wudnt have to use the chain rule :S

    and you keep saying Max of sin(wt) =1
    so why are we taking teh derivative?
     
  17. Nov 12, 2008 #16

    Dick

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    Ack! What is d/dt of sin(2t). If you say cos(2t), I'm going to bed right now. Look up the chain rule.
     
  18. Nov 12, 2008 #17
    cos(2t)*(2) so cos4t
     
  19. Nov 12, 2008 #18

    Dick

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    cos(2t)*2 is not equal to cos(4t). It's equal to 2*cos(2t). They aren't the same. You might be more tired than I am.
     
  20. Nov 12, 2008 #19
    yes sorry i have been doing physics since the past 5-6 hrs
     
  21. Nov 12, 2008 #20

    Dick

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    d/dt of sin(wt). What is it? You can nap on it, if you want to. Until we get that this will go nowhere.
     
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