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loonychune
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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
[tex]e^{2kz}[/tex]
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.
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
[tex]e^{2kz}[/tex]
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.
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