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Mathematical Pendulum

Plain Pendulum

We will illustrate the application of Lagrangian mechanics by analyzing the movement of a mathematical pendulum - a problem we have already discussed in the Physics 4 Teens - Mechanics - Pendulum, Spring - Pendulum using the Newtonian approach.

We recommend reviewing the lecture mentioned above and refresh the Newtonian method of deriving the main equation of motion of the pendulum:
α"(t) = −(g/l)·sin(α(t))
which was obtained from properly determining the force F that moves a pendulum as a vector sum of the gravity force directed vertically down and the tension of an unstretchable thread that keeps an object at the free end of a thread on a constant distance from the fixed end of a thread.

Let's apply the Lagrangian mechanics to this problem using an angle of a thread with a vertical α as the one and only parameter that determines a position of an object.

The Lagrangian is the difference between kinetic and potential energies.
L(α(t),α'(t)) = Ekin(α'(t)) − Epot(α(t))

Kinetic energy depends on mass M and linear speed of an object along its circular trajectory v=l·α'(t)
Ekin = ½M·v² = ½M·l²·[α'(t)]²

Potential energy depends on a mass of an object M, its height over the ground h(t) and an acceleration of free fall g.
Epot(t) = M·g·h(t)
If the origin of our coordinates, the fixed end of a thread, is at height H over the ground,
h = H − l·cos(α)
and, therefore,
Epot(t) = M·g·[H−l·cos(α(t))]

Now we can construct the Euler-Lagrange equation
(∂/∂α)L(α(t),α'(t)) = (d/dt)(∂/∂α')L(α(t),α'(t))

Spring Pendulum

A weightless spring replaces an unstretchable thread of the previous problem.
The spring and an object on its end are in a weightless frictionless tube that maintains a straight form, so an object has two degrees of freedom - radial inside a tube stretching and squeezing a spring and pseudo-circular as it moves together with a tube in a pendulum like motion.

The problem of specifying the motion of an object is much more complex here because the spring tension is changing not only with an angle α(t) but also because of the movement of an object within a tube.

However, using the Langrangian mechanics, this problem can be analyzed with much less efforts and the corresponding differential equation can be constructed relatively easy.