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Solving differential equation x'' = ax

  1. Feb 20, 2007 #1

    JK423

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    1. The problem statement, all variables and given/known data

    We have the differential equation x''=a*x. If a=0 then the motion is:
    X(t)=Xo+Vo*t [1]
    a)Prove (by solving the dif. equation) for a<>0 (both a>0 and a<0) that when we take the limit a->0, motion will be given by [1].
    b)BUT "a" has units [T^-2] and 0 is just a number, so the limit a->0 is not well defined. What is the correct form of the limit?



    2. Relevant equations



    3. The attempt at a solution
    I have some problem with question b.
    If 0 is just a number then we must have a simple number on the left side of the limit. So i assume that i have to write it this way:
    a[T^2]-->0
    so that the units of "a" (T^-2) and T^2 will cansel its other.
    What do you think?

    PS. Sorry for my bad english.. I translated it from greek to english o:)
     
  2. jcsd
  3. Feb 20, 2007 #2

    Hootenanny

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    Not quite. Consider you expression; x''=ax . Now, on the left you have units of m.s-2 (x'' is acceleration) and therefore on the RHS you must also have units of m.s-2. So, what can you do to;

    [tex]\lim_{a\to 0} a[/tex]

    To make it have units of m.s-2?
     
  4. Feb 20, 2007 #3

    JK423

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    lim(a) with a->0, must have units s^-2.
    But i still dont get the problem! :S
    We have a quantity with units 1/s^2 that goes to zero. so x''=0 m/s^2.
    Why is there any problem?
    Help me a little more :S
     
  5. Feb 20, 2007 #4

    Hootenanny

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    The problem is that if you only take the limit of a, the units on the LHS (m.s-2) don't match the units on the RHS (s-2). This cannot happen, you must have matching units. Try taking the limit [itex]a\to0[/itex] of the whole of the RHS, what units do you get now?
     
  6. Feb 20, 2007 #5

    JK423

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    We must take the limit of the whole RHS -lim(a*x)- because x depends on a (and t). Right?
    limx''=lim(a*x)

    (I must be very stupid :@)
     
  7. Feb 20, 2007 #6

    Hootenanny

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    Correct! :approve:
    Not at all. To be quite honest, at first I wasn't sure myself , I've never seen this type of quetion come up before.
     
  8. Feb 20, 2007 #7

    JK423

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    But i still dont get it!
    The exercise sais that the limit a->0 is not well defined because a has units while 0 has none (just a number). What does that mean..?
    It looks to me that it doesnt have to do with the "limx''=lim(a*x)" part ...
    Im sooo confused,
    confused to death :P
     
  9. Feb 20, 2007 #8

    JK423

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    Any help guys? ..
     
  10. Feb 21, 2007 #9

    JK423

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    I asked my proffessor and i`ll write his answer here for those who are curious about it:
    A quantity with units lacks of meaning to be considered very small because it depends on the units it has. For example, earths length is very small according to astronomical units (like light-year), but is earth small?? It`s size must be compared with something in order to say that its very small. So, by multiplying the differential equation with a spesific unit of time, the quantity a T^2 will be adimensional so we can move on and consider it very small. Moreover when we find the solution and substitute the initial conditions we will have:

    x(t)=(v0/sqrt(a))sin(sqrt(a)t)+x0 cos(sqrt(a)t)=
    (v0 T /sqrt(a T^2))sin(sqrt(a T^2)t/T)+
    x0 cos(sqrt(a T^2)t/T).

    In that way, the coefficient of sin has units of lenght (v0 T) divided by a number which can be considered very small.



    Hope that you all enjoyed the answer :D
     
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