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Second order Differential (Simple)

  1. Aug 4, 2008 #1
    1. The problem statement, all variables and given/known data
    I have no idea about how to solve the second order differential equation. Please someone help me solve this. Or provie me some articles for beginnrs. I have only basic knowldge of calculus.

    While solving a question at physics, I got this:

    g and l are constants. Plase help me solve this.
  2. jcsd
  3. Aug 4, 2008 #2


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    Are you sure it isn't

  4. Aug 4, 2008 #3
    I cant be very sure. Actually its a chain falling off the table. as one part of it decreases, the other increases. Can you teach me how to integrate this?
    Last edited: Aug 4, 2008
  5. Aug 4, 2008 #4
    I dont ask for the whole process of solving second ordr differential. Just this particular case.
  6. Aug 4, 2008 #5


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    Well, we'll look at both cases.

    For, [tex]\frac{d^{2}x}{dt^{2}}=\frac{xg}{l}[/tex]
    the solution is
    [tex]x(t)=Ae^{\omega t}+Be^{-\omega t}[/tex]where A and B are arbitrary constants, that you solve for using the initial condition.

    For [tex]\frac{d^{2}x}{dt^{2}}=-\frac{xg}{l}[/tex] ,
    the solution is
    [tex]x(t)=A\cos(\omega t)+B\sin(\omega t)[/tex]

    In each case [itex]\omega=\sqrt{\frac{g}{l}}[/itex]
    You should verify that these are, in both cases, indeed solutions
  7. Aug 4, 2008 #6
    At t=0; x=h
    I wanted to know t, when x=L ????
  8. Aug 4, 2008 #7
    What I want now is just the value of AB
    and A separately.

    I know that A+B=h constant.

    See the diagram I have attached. I was trying to calculate the time taken by th chain to fall off the table.

    Attached Files:

  9. Aug 4, 2008 #8


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    One initial condition isn't enough for a second order differential equation, you need two. Presumably the other is at t=0, v=0, if the chain is falling from rest. (v for velocity)

    Now, you have two equations with two unknowns (A and B), so you should be able to solve it for x(t). Then all you have to do is solve for t in x(t)=L.

    Note that it might be easier if you write the solution as [tex]x(t)=A\cosh(\omega t) + B\sin(\omega t)[/tex], and use the properties that the derivative of coshx is sinhx and the derivative of sinhx is coshx, and that cosh(0)=1 and sinh(0)=0. See http://en.wikipedia.org/wiki/Hyperbolic_trigonometric_function

    Why don't you show an attempt at the solution?
  10. Aug 4, 2008 #9
    I have attempted at the solution:
    First of all, I want to say thanks a lot for the whole thing.
    My answer will match the required answer, only if A and B are equal. An you are right dexactly thqat at t=0, v=0. the body starts from rest
    but v= dx/dt ?? isnt it?
    How can I use this condition to prove that A an B are equl?
  11. Aug 4, 2008 #10
    Thank you very very much. I have got my solution. Thanks a lot.
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