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labfiz3

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## 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 [itex]m[/itex]) hanging from a pendulum of length [itex]l[/itex].

The point where pendulum is hanged moves like this:

[tex]\xi = A\sin\Omega t,[/tex] where [itex]A, \Omega = const[/itex]. We have to find motion of the body. The force acting on the body is [itex]F = (0, -mg)[/itex].

## Homework Equations

Lagrange's equation of second kind for a generalized coordinate [itex]\varphi[/itex]:

[tex]\frac{d}{dt}(\frac{dL}{d\varphi'}) - \frac{dL}{d\varphi} = 0[/tex]

## The Attempt at a Solution

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So, after making angle [itex]\varphi[/itex] a generalized coordinate, we have:

[tex]x = A\sin \Omega t + lsin \varphi, \ \ y = -lcos\varphi.[/tex]

After calculations, Lagrangian is: [tex]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).[/tex]

Lagrange's equation of second kind gives:

[tex]\frac{d}{dt}(\frac{dL}{d\varphi'}) - \frac{dL}{d\varphi} = 0[/tex]

(where [itex]\varphi' = \frac{d\varphi}{dt}[/itex]).

After calculations:

[tex]\varphi'' l + g\varphi = \Omega A^2 sin \Omega t[/tex]

Solving it gives:

[tex]\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,[/tex] where [itex]C, D[/itex] are constants.

Initial conditions [itex]x(t=0)=0, \ \ y(t=0) = -l[/itex] both yield the same:

[tex]C = 0[/tex]

But how to find the [itex]D[/itex] constant?

The only other information given is that we can assume [itex]\sin \varphi = \varphi[/itex] and [tex]cos\varphi = 1[/itex] (Which I've already used to solve the differential equation$. There is no other information given.

Thank you for your answers.