# How Do You Determine Normal Mode Frequencies in a Coupled Oscillator System?

• Yosty22
In summary: The determinant of the coefficients must be zero.In summary, we have a particle of mass m moving in a plane with a Lagrangian that can be expressed as a sum of kinetic and potential energy terms. To find the normal mode frequencies, we use the Euler-Lagrange Equation to solve for the equations of motion in both the x and y directions. We then equate the two equations and solve for the frequencies by assuming a solution in the form of x = x0e^(iωt) and y = y0e^(iωt). The frequencies can be found by taking the determinant of the coefficients of the equations and setting it equal to zero.
Yosty22

## Homework Statement

We have a particle of mass m moving in a plane described by the following Lagrangian:
\frac{1}{2}m((\dot{x}^2)+(\dot{y}^2)+2(\alpha)(\dot{x})(\dot{y}))-\frac{1}{2}k(x^2+y^2+(\beta)xy) for k>0 is a spring constant and \alpha and \beta are time-independent.

Find the normal mode frequencies, \omega_1,2

## Homework Equations

Euler-Lagrange Equation

## The Attempt at a Solution

I think I'm just missing here. There was a lot of math, so I won't explicitly write out all of it, but I will have my final answers. I used the Euler-Lagrange Equation twice: once for x, once for y. This yielded:

m*ddot{x}+m\alpha(\ddot{y}=-kx+\beta(y)
and
m*\ddot{y}+m\alpha(\ddot{x}=-ky+\beta(x)

I solved each for \ddot{x} and equated them, giving me:

\ddot{y} = \frac{\beta+k/(\alpha)}{m(\alpha-1/(\alpha))}*y + \frac{-k-\beta/(\alpha)}{m(\alpha-1/(\alpha))}*x

Am I approaching this the right way to find the frequencies? I know usually in 1D for example, you solve the Euler-Lagrange equation to yield something of the form: \ddot{x}=\omega^2*x, but it is a little more unclear to me as to what to do here. Would I find two separate frequencies, once in x and once in y and they are two separate answers?

Update: I should probably transform to coordinates like (r,theta) and find theta dot, right?

Yosty22 said:

## Homework Statement

We have a particle of mass m moving in a plane described by the following Lagrangian:
\frac{1}{2}m((\dot{x}^2)+(\dot{y}^2)+2(\alpha)(\dot{x})(\dot{y}))-\frac{1}{2}k(x^2+y^2+(\beta)xy) for k>0 is a spring constant and \alpha and \beta are time-independent.

Find the normal mode frequencies, \omega_1,2

## Homework Equations

Euler-Lagrange Equation

## The Attempt at a Solution

I think I'm just missing here. There was a lot of math, so I won't explicitly write out all of it, but I will have my final answers. I used the Euler-Lagrange Equation twice: once for x, once for y. This yielded:

m*ddot{x}+m\alpha(\ddot{y}=-kx+\beta(y)
and
m*\ddot{y}+m\alpha(\ddot{x}=-ky+\beta(x)

I solved each for \ddot{x} and equated them, giving me:

\ddot{y} = \frac{\beta+k/(\alpha)}{m(\alpha-1/(\alpha))}*y + \frac{-k-\beta/(\alpha)}{m(\alpha-1/(\alpha))}*x

Am I approaching this the right way to find the frequencies? I know usually in 1D for example, you solve the Euler-Lagrange equation to yield something of the form: \ddot{x}=\omega^2*x, but it is a little more unclear to me as to what to do here. Would I find two separate frequencies, once in x and once in y and they are two separate answers?

Perhaps if you presented the TeX/LaTeX expressions properly you would receive more responses. I edited the first of your equations above. The first one below is exactly what you wrote, but inserting the appropriate controls to make the PF processor understand you want to use LaTeX. As you can see, it looks ugly, and defeats the whole purpose of using LaTeX in the first place. The second one is the properly-edited version, in which all needed control characters are used. Right-click on it to see the TeX commands.

$$m*ddot{x}+m\alpha(\ddot{y}=-kx+\beta(y)$$

$$m \ddot{x}+m\alpha \ddot{y}=-kx+\beta y$$

Yosty22 said:

## Homework Statement

We have a particle of mass m moving in a plane described by the following Lagrangian:
##\frac{1}{2}m((\dot{x}^2)+(\dot{y}^2)+2(\alpha)(\dot{x})(\dot{y}))-\frac{1}{2}k(x^2+y^2+(\beta)xy)##
for k>0 is a spring constant and \alpha and \beta are time-independent.

Find the normal mode frequencies, \omega_1,2

## Homework Equations

Euler-Lagrange Equation

## The Attempt at a Solution

I think I'm just missing here. There was a lot of math, so I won't explicitly write out all of it, but I will have my final answers. I used the Euler-Lagrange Equation twice: once for x, once for y. This yielded:

##m\ddot{x}+m\alpha\ddot{y}=-k(x+\beta y)##
and
##m\ddot{y}+m\alpha \ddot{y}=-k(y +\beta x) ##

I solved each for \ddot{x} and equated them, giving me:

##\ddot{y} = \frac{\beta+k/(\alpha)}{m(\alpha-1/(\alpha))}*y + \frac{-k-\beta/(\alpha)}{m(\alpha-1/(\alpha))}*x##

Am I approaching this the right way to find the frequencies? I know usually in 1D for example, you solve the Euler-Lagrange equation to yield something of the form: \ddot{x}=\omega^2*x, but it is a little more unclear to me as to what to do here. Would I find two separate frequencies, once in x and once in y and they are two separate answers?
Check your equations. I corrected your TeX codes. Is not a factor of 2 in front of beta in the Lagrangian?
Find the solution in form ##x=x_0 e^{i\omega t}##, ##y=y_0 e^{i\omega t} ##. You get a system of linear homogeneous equations for xo and yo that has nonzero solution for certain ω-as only.

Last edited:
Delta2

## 1. What is Lagrangian Mechanics?

Lagrangian Mechanics is a mathematical framework used to describe the motion of a system of particles or a continuous system. It is based on the principle of least action, which states that the motion of a system can be described by minimizing the difference between the kinetic and potential energies.

## 2. What is the difference between Lagrangian Mechanics and Newtonian Mechanics?

The main difference between Lagrangian Mechanics and Newtonian Mechanics is the approach used to describe the motion of a system. While Newtonian Mechanics uses forces and accelerations to describe motion, Lagrangian Mechanics uses the concept of energy and minimization of the action.

## 3. What are the advantages of using Lagrangian Mechanics?

One of the main advantages of using Lagrangian Mechanics is that it provides a more elegant and concise way of describing the motion of a system compared to Newtonian Mechanics. It also allows for the use of generalized coordinates, which simplifies the equations of motion for complex systems.

## 4. What are the applications of Lagrangian Mechanics?

Lagrangian Mechanics has many applications in physics, engineering, and other fields. It is commonly used in celestial mechanics to study the motion of planets and satellites, in classical mechanics to analyze the motion of particles, and in quantum mechanics to describe the dynamics of quantum systems.

## 5. What is the Lagrangian function in Lagrangian Mechanics?

The Lagrangian function, denoted as L, is a mathematical function that represents the difference between the kinetic and potential energies of a system. It is defined as L = T - V, where T is the kinetic energy and V is the potential energy. The equations of motion in Lagrangian Mechanics are derived by minimizing the action integral of the Lagrangian function.

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