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

• labfiz3
In summary, the conversation discusses a problem involving a pendulum with a small body hanging from it. The motion of the pendulum is described by the equation \xi = A\sin\Omega t and the force acting on the body is F = (0, -mg). The problem is solved using Lagrange's equation of second kind, which yields a differential equation for the angle \varphi of the pendulum. After solving the equation, the initial conditions are used to find the value of the constant D. However, another initial condition related to the initial velocity is needed to fully solve the problem. The conversation concludes with a reminder to check the coefficients in the equation of motion and the solution for accuracy.
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 [tex]cos\varphi = 1[/itex] (Which I've already used to solve the differential equation\$. There is no other information given.

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.

## What is the Lagrange equation of second kind?

The Lagrange equation of second kind is a mathematical equation used to find the constants of a solution to a differential equation. It is derived from the Lagrange multiplier method and is commonly used in physics and engineering to solve problems involving constraints.

## What is the difference between the Lagrange equation of first and second kind?

The Lagrange equation of first kind is used to find the equations of motion for a system with constraints, while the Lagrange equation of second kind is used to find the constants of a solution to a differential equation. The first kind involves the use of Lagrange multipliers, while the second kind involves the use of undetermined coefficients.

## How is the constant of a solution determined using the Lagrange equation of second kind?

The constant of a solution is determined by substituting the solution into the differential equation and solving for the undetermined coefficients. These coefficients are then used to find the specific solution to the given problem.

## What are the applications of the Lagrange equation of second kind?

The Lagrange equation of second kind has various applications in physics and engineering, including in the study of mechanics, electromagnetism, and fluid dynamics. It is also used in optimization problems and in the analysis of systems with constraints.

## Are there any limitations to using the Lagrange equation of second kind?

One limitation of the Lagrange equation of second kind is that it can only be used for linear differential equations with constant coefficients. It also requires that the solution to the problem be known beforehand, as it only helps in determining the constants of the solution.

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