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Homework Help: Ask help about the pendulum~ THX

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data

    angle [0,2Pi] from bottom most position, find the 2nd order nonlinear ODE for evolution of the pendulum in terms of angle
    then assume angle is small, do the linear approximation to the equation

    2. Relevant equations

    use balance of forces and newton's 2nd law

    3. The attempt at a solution

    y''+g*siny\L = 0 but then how to find the linear approximation about y'' + g*y\L=0
    i tried, by the formula, y = C1*cos(i*sqrt(g\L)t) + C2*sin(i*sqrt(g\L)t)
    Is that correct? Thanks a lot~
  2. jcsd
  3. Sep 18, 2008 #2


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    This is posted in Precalculus Mathematics but you talk about differential equations? I have no idea how much to expect you to know! Do you know the Taylor's series for sin(x) around x=0? Do you know that [itex]\lim_{x\rightarrow 0}sin(x)/x= 1[/itex]? Either of those should give you an idea of the linear approximation for sin(y) when y is close to 0.

    No, that formula is not correct. eix= cos(x)+ i sin(x). You should not have "i" inside the sine and cosine.
  4. Sep 18, 2008 #3
    well, there maybe some misunderstandings here between us. Like when x approach to 0, what I know is siny = y, so I change y''+g*siny\L = 0 to y'' + g*y\L=0. And your formula is not what I used. Whatever, thanks anyway~
  5. Sep 18, 2008 #4
    btw, I'm a rookie here, so I clicked wrong title just now. Please help me to move this to Calculus & Beyond or just delete it? Thanks.
  6. Sep 19, 2008 #5


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    "when x approach to 0, what I know is siny = y". No, you don't know that- it makes no sense. Perhaps you meant "when x approach 0, sin x= x" but that still is not true. For values of x very close to 0, sin x is very close to x. That results in [itex]\lim_{x\rightarrow 0} sin x/x= 1[/itex]

    What you originally wrote was "y''+g*siny\L = 0 but then how to find the linear approximation about y'' + g*y\L=0". As both you and I have said now, for small y, sin y is close to y so approximately, replacing sin y by y, y"+ gy/L= 0. That's what you wanted.

    Now, do you know how to find the characteristic equation for y"+ (g/L)y= 0? What are its roots?

    You need to know that if [itex]a\pm ib[/itex] are roots of the characteristic equation of a "linear homogenouse differential equation with constant coefficients", then
    [tex]y(x)= e^{at}(C_1 cos(bt)+ C_2 sin(bt)[/tex]
    is the general solution to the differential equation.

    We could write
    [tex]y(t)= D_1e^{(a+ bi)t}+ D_2e^{(a+bi)t}= e^{at}(D_1e^{bit}+ D_2e^{-bit}[/tex]
    but [itex]e^{bit}= cos(bt)+ i sin(bt)[/itex] and [itex]e^{-bit}= cos(bt)- i sin(bt)[/itex]
    We can combine those and then absorb the "i" into the constant. We would expect that if the original problem involved only real numbers, the solution will involve only real numbers.
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