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Differentiation and Simple Pendulums

  1. Nov 20, 2015 #1
    1. The problem statement, all variables and given/known data
    *according to my teacher, the problem can be solved either through differentials or quick algebra, but she prefers differentials as the answer in order for us to get used to Calculus, so that's the route I'd like to take it

    A pendulum made from a string and small metal ball swings back and forth. The formula for its period is given by the following: T = 2*pi*sqrt(L/g), where T is the period, pi is a constant, L is the length of the string and g is the gravitational force acting on it. In this situation, do not assume gravity is under the constant of 980 cm/s^2.

    Solve the following problems:

    A) Suppose the value of T is measured exactly (100%), but L is off by 1%. How much will g be off by?
    B) Suppose that the value of L is measured exactly (100%), but T is off by 1%. How much will g be off by?

    2. Relevant equations
    T = 2*pi*sqrt(L/g)
    Chain Rule
    Product Rule
    Quotient Rule
    Power Rule

    3. The attempt at a solution

    A) T = 2pi*(0.99L/g)^.5
    T' = 2pi * .5 * (0.99L/g)^-.5 * [(0.99g-0.99Lg')/g^2]
    Then after lots of cancelling out:
    T' = [pi(g - Lg')]/Lg
    g' = [(LgT'/pi) - g] / -L

    B) 0.99T = 2pi*(0.99L/g)^.5
    0.99 = [(pi)(gL'-Lg')] / Lg
    g' = -g[(0.99/pi) - (L'/L)]

    I have no idea if I did this right, or if I even answered my question correctly (since I don't even have a percentage... so I don't really know :/ ), so any help is appreciated, thank you!
     
  2. jcsd
  3. Nov 20, 2015 #2

    PeroK

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    I would start by re-arranging the original equation for ##g##.
     
  4. Nov 20, 2015 #3
    Thanks, so g = L / (T/2pi)^2 ... then what do I do? >.<
     
  5. Nov 20, 2015 #4

    PeroK

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    I thought you said you were supposed to use some calculus?
     
  6. Nov 20, 2015 #5
    Oh! Well that's kind of where I ended up in the answer I put in, except I rearranged everything at the end of the process instead.
     
  7. Nov 20, 2015 #6

    PeroK

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    You didn't use differentials.
     
  8. Nov 20, 2015 #7
    Isn't a differential the same as a derivative, though? That's as far as I know from my Calc class.
     
  9. Nov 20, 2015 #8

    PeroK

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    Not quite. Think ##\frac{dg}{dL}##
     
  10. Nov 20, 2015 #9
    So the derivative of g with respect to l?... But then what do I do with T?
     
  11. Nov 20, 2015 #10

    PeroK

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    ##T## is constant (for the time being). That's what the question tells you. Do ##L## first with ##T## constant, then try ##T## with ##L## constant.
     
  12. Nov 20, 2015 #11
    So... is this more like related rates or something like that?

    g = L / (T/2pi)^2 = L (T/2pi)^-2
    dg/dL = [dL/dL*(T/2pi)^-2] + [L*-2(T/2pi)*{(0*dT/dL - T*0)/(2pi)^2}]
    dg/dL = (T/2pi)^-2

    So if T is a constant and constants go to zero despite the dT/dL, is this undeterminable then?
     
  13. Nov 20, 2015 #12

    PeroK

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    First, let me show you the answer!

    ##g = \frac{4\pi^2}{T^2}L##

    If ##T## is constant, ##g## is proportional to ##L## and a 1% change in ##L## causes a 1% change in ##g##. That was easy! That's why I think your teacher wanted you to use calculus to get used to using differentials. You need to figure out how to use differentials to show this.

    You got as far as:

    ##\frac{dg}{dL} = \frac{4\pi^2}{T^2}##

    Another hint is how to express a percentage change in a variable using differentials. If you think of ##dL## as a small change in ##L##, then ##\frac{dL}{L}## is the proportional change in ##L##. Does that help?

    I'm going offline now. Maybe someone else can help if you need a bit more advice.
     
    Last edited: Nov 20, 2015
  14. Nov 22, 2015 #13
    Ah okay, I think I'm starting to see it. I think your hints enough have helped me. I'm going to brush up on differential skills after this (clearly, I'm lacking), but this is a good start. :-) Thank you!
     
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