# Transposing Terms for an Isotropic Oscillator

• jsgoodfella
In summary, the problem involves finding the equation for an isotropic oscillator in the form of an ellipse. The equations for x and y are in the form of parametric equations for an ellipse, but the derivation process may be confusing. To simplify the equations, we can use the Pythagorean theorem and the equation for a circle. By substituting the equation for y into the equation for x, we can obtain the parametric equations for an ellipse.
jsgoodfella

## Homework Statement

The problem asks us to find the equation for an isotropic oscillator in the form of an equation for an ellipse. I'm in PHYS 212, Analytic Mechanics.

## Homework Equations

x=Acost(ωt)

y=5Acos(ωt-.6435)

The equations I found for x and y.

## The Attempt at a Solution

So far I have the equation at y=5[x*cos(-.6435)+(1-x^2)^(1/2)*sin(-.6435)]

I looked in the example in the text for the original derivation, and it basically skips through this part. I need to know how to transpose the equation I got to further simplify my work.

Last edited by a moderator:

Dear fellow student,

Thank you for sharing your progress on this problem. The equations you have found for x and y are correct and are in the form of parametric equations for an ellipse. However, I can understand your confusion about the derivation of this equation. Let me explain it in a bit more detail.

An isotropic oscillator is a system that oscillates with the same frequency in all directions. This means that the motion of the oscillator can be described by a circle or an ellipse. In order to find the equation for an ellipse, we need to use the Pythagorean theorem and the equation for a circle.

Let's start with the equation for a circle, x^2 + y^2 = A^2, where A is the radius of the circle. We can rewrite this equation as y = ±(A^2 - x^2)^(1/2). Now, we can substitute this into the equation for x that you have found, x = Acos(ωt), to get y = ±(A^2 - A^2cos^2(ωt))^(1/2). This simplifies to y = ±A(1 - cos^2(ωt))^(1/2).

Next, we can use the Pythagorean theorem to get rid of the square root. Recall that cos^2(ωt) + sin^2(ωt) = 1. Therefore, we can rewrite the equation as y = ±A(sin^2(ωt))^(1/2). This simplifies to y = ±Asin(ωt).

Finally, we can substitute this equation for y back into the equation for x to get the parametric equations for an ellipse: x = Acos(ωt) and y = 5Acos(ωt - .6435).

I hope this helps clarify the derivation process for you. Remember, when dealing with parametric equations, it is important to consider the relationships between the variables and use algebraic manipulations to simplify the equations. Keep up the good work!

## 1. What is an Isotropic Oscillator?

An isotropic oscillator is a physical system that undergoes oscillatory motion with equal frequency and amplitude in all directions.

## 2. How are terms transposed in an Isotropic Oscillator?

In an Isotropic Oscillator, terms are transposed by rearranging the equation to isolate the desired term on one side of the equation.

## 3. Why is it important to transpose terms in an Isotropic Oscillator?

Transposing terms in an Isotropic Oscillator allows for a better understanding of the system and its behavior, as well as the ability to manipulate the equation for further analysis and calculations.

## 4. What are some common techniques for transposing terms in an Isotropic Oscillator?

Some common techniques for transposing terms in an Isotropic Oscillator include using algebraic manipulation, substitution, and differentiation.

## 5. Are there any limitations to transposing terms in an Isotropic Oscillator?

One limitation to transposing terms in an Isotropic Oscillator is that it may not always be possible to isolate the desired term due to the complexity of the equation and the presence of multiple variables.

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