Lagrange equation of second kind - find solution's constant?

[labfiz3]

Homework Statement

This could be a more general question about pendulums but I'll show it on an example.

We have a small body (mass m) hanging from a pendulum of length l.
The point where pendulum is hanged moves like this:
\xi = A\sin\Omega t, where A, \Omega = const. We have to find motion of the body. The force acting on the body is F = (0, -mg).

Homework Equations

Lagrange's equation of second kind for a generalized coordinate \varphi:
\frac{d}{dt}(\frac{dL}{d\varphi'}) - \frac{dL}{d\varphi} = 0

The Attempt at a Solution

[/B]
So, after making angle \varphi a generalized coordinate, we have:
x = A\sin \Omega t + lsin \varphi, \ \ y = -lcos\varphi.

After calculations, Lagrangian is: L = \frac{m}{2}\Big( \Omega^2 A^2 \cos^2 \Omega t + 2\Omega A l \frac{d\varphi}{dt}\cos \Omega t \cos \varphi + (\frac{d \varphi}{dt})^2l^2 + 2gl\cos \varphi \Big).

Lagrange's equation of second kind gives:
\frac{d}{dt}(\frac{dL}{d\varphi'}) - \frac{dL}{d\varphi} = 0
(where \varphi' = \frac{d\varphi}{dt}).

After calculations:
\varphi'' l + g\varphi = \Omega A^2 sin \Omega t

Solving it gives:

\varphi(t) = C\cos (\sqrt{\frac{g}{l}}t) + D\sin (\sqrt{\frac{g}{l}}t) + \frac{\Omega^2 A^2}{g-l\Omega^2}\sin \Omega t, where C, D are constants.

Initial conditions x(t=0)=0, \ \ y(t=0) = -l both yield the same:
C = 0

But how to find the D constant?
The only other information given is that we can assume \sin \varphi = \varphi and cos\varphi = 1[/itex] (Which I&#039;ve already used to solve the differential equation$. There is no other information given.<br /> <br /> Thank you for your answers.

 

[TSny]

Welcome to PF!

I believe you are overlooking an initial condition for $\varphi'$ which is related to the initial condition on $x'$.

Also, check your coefficients of $\sin \Omega t$ in your differential equation of motion and in your solution. The coefficients don't have the correct dimensions.