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Homogenity physics homework

  1. Sep 7, 2013 #1
    1. So for today's exam, I was given this equation, and I was required to get the value and units of K:

    T4 = (4K∏4)/g2 - (8d∏4)/g2

    I had to follow an experiment, plot the results, get the gradient. After getting the value of g, I was required to find the value of K, and its units.

    T4 has units s4, d has units m, g is gravitational acceleration, hence I believe it is m.s-2



    2. The problem lies in the fact that I don't believe I was given a homogeneous equation, hence I couldn't provide the units. Any help? Note that the units are enough for me, I managed to solve for K (correctly, I believe).

    Thanks in advance :)
     
  2. jcsd
  3. Sep 7, 2013 #2
    You can only add or subtract things with same units.
     
  4. Sep 7, 2013 #3
    So you confirm the equation is not homogeneous?
     
  5. Sep 7, 2013 #4
    I neither confirm nor deny, I merely point out the obvious. What then is ∏? Or did you mean [itex]\pi[/itex]?
     
    Last edited: Sep 7, 2013
  6. Sep 7, 2013 #5
    Well, after working it out, the left side is this: T4 has units s4.

    For the K part, I can't work it out without knowing whether it's homogeneous or not. So I go where there's d.

    (8d∏4)/g2 would then become:
    Numerator: m
    Denominator: m2.s-4

    Therefore this part would be s4.m-1

    For homogeneity with addition/subtraction, all parts have to be the same. These two parts aren't the same, evidently. Any confirmation would be very much appreciated, so I could e-mail my invigilators.
     
  7. Sep 7, 2013 #6
    What is the experiment?
     
  8. Sep 7, 2013 #7
    That should be a pi :) I couldn't find the symbol :P
     
  9. Sep 7, 2013 #8
    Nope, it doesn't look homogenous. What's the Experiment?
    You sure d is in meters and not m^2 ?
     
  10. Sep 7, 2013 #9
    The experiment isn't very important per-se, but for the record, there are two stand clamps, with a wire attached to both, forming a triangle. At the bottom, there's a pendulum, and d is the distance between the two stands. So yeah, definitely in meters. It's also given, but because of copyright issues, I'm afraid I'm not allowed to take a picture of it and upload it.
     
  11. Sep 7, 2013 #10

    gneill

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    Staff: Mentor

    Could be a typo in the given equation; I suspect that the 'd' in the second term should have been ##d^2##.

    Even if the second term has a typo, the terms should individually have the same units as the LHS of the equation. So you can still determine what k should be and even repair the typo!
     
  12. Sep 7, 2013 #11
    While I did as you said, gneill, I couldn't ascertain which was the right one - the 'd' part or the LHS.
     
  13. Sep 7, 2013 #12

    gneill

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    Staff: Mentor

    I think you can be pretty certain the LHS would consist of a single variable, and that as missed typos go, dropping a square on a variable in the midst of a term is more likely than dropping a variable on the LHS.
     
  14. Sep 7, 2013 #13
    Well, I haven't got the derivation of the formula yet but with off-hand information the formula's probably wrong. When you put d=0 the bifilar pendulum becomes a simple pendulum the formula should be reduced to a simple one.
    Making $$k= 4 l^2$$ from the standard formula of time period. Formula's probably missing a square factor on d, will confirm it and let you know.
    Whoops gneil got there before me....
     
    Last edited: Sep 7, 2013
  15. Sep 7, 2013 #14
    Thanks a lot for your help :) I really appreciate it!
     
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