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SHM- differential Equations. Finding the constants of Integration from exp. data

  1. Feb 18, 2005 #1


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    Hey guys,

    bit stuck here with this problem

    I have a differential equation for SHM here in the form

    x= A cos 10.7t + B sin 10.7t

    I need to find the constants of integration. w = 10.7 as above

    I dont have anything other than experimental data

    Time Second Distance / m
    0 0
    0.05 -0.008
    0.1 -0.008
    0.15 -0.008
    0.2 -0.009
    0.25 -0.014
    0.3 -0.019
    0.35 -0.029
    0.4 -0.036
    0.45 -0.045
    0.5 -0.051
    0.55 -0.056
    0.6 -0.054
    0.65 -0.029
    0.7 -0.012
    0.75 0.011
    0.8 0.031
    0.85 0.055
    0.9 0.049
    0.95 0.025
    1 0.002
    1.05 -0.032
    1.1 -0.047
    1.15 -0.055
    1.2 -0.047
    1.25 -0.021
    1.3 0.002
    1.35 0.035
    1.4 0.049
    1.45 0.054
    1.5 0.045
    1.55 0.016
    1.6 -0.006
    1.65 -0.038
    1.7 -0.051
    1.75 -0.053
    1.8 -0.042
    1.85 -0.012
    1.9 0.011
    1.95 0.04
    2 0.052

    Since x=0 when t = 0

    I get A = 0 ... A Cos 0 =0

    But then I also seem to get B=0 which makes no sense.

    If B is the amplitude of the function then sure it must 0.055m or thereabouts

    Any ideas on working out B

    I have tried other values but to no avail.

    thanks a lot


    Edit/ The first value of the data ist he time value and the second one is the displacement value
  2. jcsd
  3. Feb 18, 2005 #2


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    IF i write the equation in the format

    x= A sin (wt + e)

    then A is the amplitude and e is the phase shift factor. can I just write down A = 0.055m from the data??

    This is very urgent, please advise

  4. Feb 18, 2005 #3


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    What you wrote are ALL the possible solutions to the diff. equation. You need to look at your "boundary conditions" from your data to pick out which solution to use. At t=0, your data shows that you have ZERO amplitidue. Now look at both the sine function and the cosine function. Which one of these would fit your data more accurately?

  5. Feb 18, 2005 #4


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    Yes, x(0)= 0 means that the "A" in x= A cos 10.7t + B sin 10.7t must be 0.
    I think I would be inclined to look at t such that 10.7t= pi/2= 1.57 or 0.15: that's
    x(0.15)= -0.008 so x(t)= -0.008 sin 10.7t. Yes, that's awfully close to identically 0.

    If your data were exact, that should match all other values. If it is, rather, experimental measurements, you might want to take t as large as possible. Noticing that your t values go up to 2, 10.7t goes up to 21.4. 13 pi/2= 20.4 which corresponds to t= 1.9. x(1.9)= 0.011 which would give x(t)= 0.011 sin 10.7t.

    Yet another way to handle the fact that the data are not exact is to look at the largest value. The largest value of x that I see, with a quick scan is
    0.055. Since the largest possible value of A sin 10.7 is A, that would indicate that x(t)= 0.055 sin 10.7t.

    The best way to handle it would be to average the values: The integral of
    x(t)= A sin 10.7t between t= 0 and t= 2 is A/10.7 cos(21.4). Do a numerical integration of the given data and choose A to fit that. Of course, you don't HAVE to use x(0)= B= 0 (unless that is the one value you are SURE of). There are an infinite number of ways to fit a formula to approximate data.
    Last edited by a moderator: Feb 18, 2005
  6. Feb 18, 2005 #5


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    Thanks HallsofIvy

    The data itself goes all the way upto 25 seconds. Although this meant to be perfect, there is some damping as it progresses but nothing much.

    I dont think i completely understand your integration method, I dont know anything about numerical integration anyways(paralleogram rule i do know)

    I just choose A = 0.055 because it is the largest possible value.
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