Lagrangian of Pendulum: Calculation & Small Oscillations

[PhysicsGente]

Homework Statement

Consider a pendulum of mass m and length b in the gravitational field whose point of attachment moves horizontally $$x_0=A(t)$$ where $$A(t)$$ is a function of time.

a) Find the Lagrangian equation of motion.
b) Give the equation of motion in the case of small oscillations. What happens in that case when $$A(t)=cos\left(\sqrt{\frac {a} {b}}t\right)$$

Homework Equations

$${\cal L} = T - U$$

The Attempt at a Solution

a) The position of the pendulum would be given by:

$$x = A(t) + bsin\left(\theta\right)$$ $$\dot{x} = \dot{A}(t) + b\dot{\theta}cos\left(\theta\right)$$
$$y = bcos\left(\theta\right)$$ $$\dot{y} = -b\dot{\theta}sin\left(\theta\right)$$

The kinetic energy $$T$$ would be equal to:

$$T = \frac {m} {2} \left({\dot{A}(t)}^2 + 2\dot{A}(t) b \dot{\theta} cos\left(\theta\right) + b^2\dot{\theta}^2\right)$$

and taking the zero potential to be at $$x = 0$$ I get that the potential is equal to :

$$U = -mgy = -mgbcos\left(\theta\right)$$

And the Lagrangian would be:

$${\cal L} = T - U = \frac {m} {2} \left({\dot{A}(t)}^2 + 2\dot{A}(t) b \dot{\theta} cos\left(\theta\right) + b^2\dot{\theta}^2\right) + mgbcos\left(\theta\right)$$

I would like to know if I have represented the position of the pendulum the right way because I get a non-linear differential equation for part b and I doubt that's right. Thanks!

 

[clamtrox]

I'm sure the system will be nonlinear. What you should assume is that the oscillations are small, θ << 1. Then you can, by construction, drop all the nonlinear terms.

 

[rahul.6sept]

May i know what will be the phase plot for the same? I mean how should i proceed to get the phase plot for the same pendulum as above.