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Tough question - tough ODE

  1. Jun 15, 2008 #1
    Tough question -- tough ODE

    1. I'm working over various texts this summer, one being Kibble's 'Classical Mechanics'. This being problem 15 on page 45. I'm stuck and your help would be much appreciated!!

    Q) A particle moves vertically under gravity and a retarding force proportional to the square of its velocity.

    2. The relevant equation of motion is this:

    [tex]\ddot{z} = -g -k\dot{z}^2[/tex] (k is a constant, g grav. acceleration)

    What I need to work out is its position at time, t, i.e. [tex]z(t)[/tex].

    3. If the force were proportional to the velocity, and not the square of it, I would integrate once and then the resulting ODE i'd solve by finding some integrating factor.

    However, I have issues with the [tex]\left(\frac{dz}{dt}\right)^2[/tex] term.

    One thing I thought would fix this would be the substitution,

    [tex]p = \left(\frac{dz}{dt}\right)[/tex]

    This yields the equation,

    [tex]pdp + (kp^2 + g)dz = 0 [/tex]

    which I made exact by multiplying through by


    which is then solved by integration -- yields,

    [tex] \frac{p^2e^{2kz}}{2} + \frac{g}{2k}e^{2kz} = c [/tex]

    where c is a constant.

    Rearranging for p then and since

    [tex]p = \left(\frac{dz}{dt}\right)[/tex]

    we can find z:

    [tex] \left(\frac{dz}{dt}\right) = \sqrt{2ce^{-2kz} - g/k}[/tex]

    The GIVEN ANSWER is:

    [tex]z = z_0 + \frac{1}{k}\ln\cos[\sqrt{gk}{(t_0 - t)}][/tex]

    It's quite possible that [tex]c = -z_0[/tex] if i remember rightly, but I still cannot see how I'm going to get to the answer from here; rather in fact, how i might adequately find z(t) at all.
    Last edited: Jun 15, 2008
  2. jcsd
  3. Jun 15, 2008 #2


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    Homework Helper

    How did you get this? In fact you don't have to resort to solving exact differential equations at all. Substituting as you have given yields:

    [tex]-\frac{dp}{dt} = g+kp^2[/tex] which is solvable by separation of variables. You need to make a trigo substitution in order to integrate wrt p. Once you get that, substitute in p = dz/dt and integrate again. You'll get the answer.
  4. Jun 15, 2008 #3
    Thankyou! A solution is borne.

    [tex] \ddot{z} = \frac{dp}{dt} = \frac{dp}{dz}\frac{dz}{dt}= p\frac{dp}{dz} [/tex]

    is where the equation in question comes from. I am not really sure why I took this that far -- I think I have that method in mind for if say the ODE contains the independent variable z but not t explicitly.

    Again, thanks!
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