What is the Lagrangian of a Pendulum with Oscillating Top Point?

[the keck]

Homework Statement

Consider a pendulum the top point of which is oscillating vertically as y=a*cos*(gamma*t)

Find its Lagrangian and the equation of emotion

The Attempt at a Solution

I can do most of the question and obtain the Lagrangian, but when I derive the equation, I achieve the same one as that of pendulum which is fixed at the top point i.e. angular velocity or ("theta dot") = -g*sin(theta)/l

This seems strange to me, as I expected the oscillation at the top to vary the equation of motion

Regards,
The Keck

 

[variation]

Take ordinary x-y plane coordinate (x-right;y-up) and the origin is the oscillating center of the top point. The horizontal plane of the origin is set to be the zero gravitational potential.
The Lagrange of the pendulum is
L=\frac{m}{2}\left[\left(\dot{y}-l\dot{\theta}\sin\theta\right)^2+\left(l\dot{\theta}\cos\theta\right)^2\right]-\left[-mg\left(l\cos\theta-y\right)\right]
=\frac{m}{2}\left[\dot{y}^2-2\dot{y}l\dot{\theta}\sin\theta+l^2\dot{\theta}^2+2g(l\cos\theta-y)\right]
The equation of motion is
\frac{\partial L}{\partial\theta}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{\theta}}\right)=0
\Rightarrow-2\dot{y}l\dot{\theta}\cos\theta-2gl\sin\theta-\frac{d}{dt}\left(-2\dot{y}l\sin\theta+2l^2\dot{\theta}\right)=0
\Rightarrow-\underline{2\dot{y}l\dot{\theta}\cos\theta}-2gl\sin\theta-\left(-2\ddot{y}l\sin\theta-\underline{2\dot{y}l\dot{\theta}\cos\theta}+2l^2\ddot{\theta}\right)=0
\Rightarrow-2gl\sin\theta-\left(-2\ddot{y}l\sin\theta+2l^2\ddot{\theta}\right)=0
\Rightarrow-2gl\sin\theta+2\ddot{y}l\sin\theta-2l^2\ddot{\theta}=0
\Rightarrow\ddot{\theta}+\frac{g}{l}\sin\theta-\frac{\ddot{y}\sin\theta}{l}=0
recall y=a\cos(\gamma t)
the last term vanishes when the top point without oscillating.

i hope there is no mistake in my derivation and the result is your answer.

 

[the keck]

Thanks a lot, mate! I used a different reference system, which sort of stuffed me upRegards,
The Keck