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Simple pendulum graph

  1. Apr 27, 2014 #1
    As part of a Physics experiment I have to investigate how the amplitude of a pendulum bob (attached to a string) varies with the number of oscillations it undergoes. The equation I have to work with is:

    cvpv.png


    (where t = the number of swings, A = amplitude after t swings, A0 = initial amplitude and k = the damping constant)

    sssssqcq.png

    Am I correct in saying that ln(A/A0) would be the label on the y-axis and t would be the label of the x-axis?

    IF yes, this would suggest that at the y-intercept , ln(A/A0) = lnA0 which suggests that at the y-intercept (where t=0) A = (A0^2). Is this correct?

    Lastly, I am required to find the uncertainty in k by sketching the above graph and using error bars. I am unsure of how to go about this?

    Briefing sheet:

    http://photouploads.com/images/dzvsv.png
    (PS:in the above document, 'n' is used to represent the number of swings rather than 't')
     
  2. jcsd
  3. Apr 27, 2014 #2

    Simon Bridge

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    ... this would be a damped pendulum - not a simple pendulum.

    If you plot ##\ln (A/A_0)## vs ##t## then yes.
    Presumably you want to get a straight line?
    Did you try it and see what you get?

    (Experiments are about dealing with the data you have and not about trying to conform to an expected result: do it and see.)

    No.
    Check your algebra. How does the RHS go from the ##\ln A## above to ##\ln (A/A_0)##?

    You may be better to just plot ##\ln(A)## vs ##t## instead.

    You know the errors for the measurements you made - use them to calculate the errors on the numbers you are plotting.
    (Hint: if the error in y is small compared to the error in x, you can eave it off.)

    Your course will probably have a specific method they want you to use to convert error bars into an overall uncertainty - check your notes.
     
  4. Apr 27, 2014 #3
    Ah sorry. That's what I meant :)!

    Yes I want to get a straight line. I was thinking of plotting lnA against t but then the y-axis (lnA) would have units and I thought log graphs are not meant to have units? Hence why I decided to plot ##\ln (A/A_0)## vs ##t##.

    Well this is what I did (I can't see what I wrong?):

    SwoPKD2l.jpg


    Thanks
     
  5. Apr 27, 2014 #4

    Simon Bridge

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    when you change what you plot, you have to redo the y=mx+c step.
    off y=mx+c, show me how you decided
    y=
    x=
    m=
    c=

    ... this is missing from your calculation pictured.

    note: A/Ao is the size of A measured in units of Ao ... i.e. you still have units.
    I don't think there is anything wrong with a log plot having units/dimensions on both axis, but it is tidier.
     
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