# Solving 2D Harmonic Motion Homework

• theuniverse
In summary, Homework Statement discusses a particle undergoing simple harmonic motion. The x and y coordinates of the particle are given by x=asin(wt) and y=bcos(wt). The particle's x and y coordinates are an ellipse, and the total energy is constant. The ellipse is a path of constant total energy, and the kinetic and potential energy at a point on the particle's trajectory are given by the sum of the separate energies of the x and y oscillations.
theuniverse

## Homework Statement

A particle undergoes simple harmonic motion in both the x and y directions simultaneously. Its x and y coordinates are given by
x=asin(wt)
y=bcos(wt)

a. Eliminate t from these equations to show that the path of the particle in the x-y plane is an ellipse.
b. Calculate the kinetic and potential energy at a point on the particle’s trajectory. Show that the ellipse is a path of constant total energy, and show that the total energy is given by the sum of the separate energies of the x and y oscillations.

## Homework Equations

Equation of an ellipse: (x/a)^2+(y/b)^2=1
KE = 1/2=mv^2
PE = ??

## The Attempt at a Solution

a) I rearranged the parametric eqns. like so: x/a=sinwt and y/b=coswt, squared both sides and using trig identities eliminated the term (sin^2wt+cos^2wt) leaving 1 on the right side and therefore getting the equation of an ellipse: (x/a)^2+(y/b)^2=1
b)This is where I'm having some troubles... I am pretty sure that I can calculate the KE of the particle by looking at:
KE(x)= 1/2*mv(t)^2
= (1/2)mw^2(acoswt)^2

KE(y)= -(1/2)mw^2(bsinwt)^2

but I don't know how to come up with the equations for the potential energy.

Regrading the fact that the ellipse is a path of constant total energy, I am thinking of showing that when the particle is on the x-axis it means that y=0 thus eliminating the y component of the energy, same goes for when its on the y axis. which essentially can be modeled with a cosine function where energy is always constant but changes from component to component.

I'd appreciate any help or some input on my attempt at the solution.
Thanks!

You can find the acceleration and the force. And remember the definition of potential energy. ehild

Your Ek calc looks good, except the Eky should be positive. I combined Vx and Vy with the Pythagorean theorem, then found the energy and got the sum of your Ekx and Eky - so between us we have the last part of (b). Time should be eliminated to give Ek as a function of x and y:
Ek = ½mw²(a²y²/b² + b²x²/a²)

The potential energy is mysterious. What causes it? There apparently is no charge so must it be gravitational potential energy?
If so, any calculation of it will have a factor of "g" in it. So it will not combine with the Ek expression to make a constant. Looks like a dead end to me.

@ehild: you mind expanding a bit about the "acceleration and the force" you talk of?
@Delphi51: Yea I got to the same conclusion regarding the Ek. It seems that the Ep is due to gravity but I'm really not sure how to write the equation.. been trying to work it out with a circle, which seems to work (U=mgl(theta)^2/2) but it's not really related to an ellipse or what the question asks for...

Any more ideas?

Ehild has a brilliant insight! Forget gravity, you're given the acceleration so you can find the force directly without knowing what caused it.
I have it almost working, just out by a sign from getting a constant Ek + Ep. Find ax and ay, Fx and Fy. Integrate each to get work in each direction, add them to get total Ep. I got an expression with constants and x², y² just like the Ek. Totalling them and then using the original expressions for x and y as functions of time to eliminate x and y, I get a function of time that is some constants times cos²(wt) - sin²(wt). If you can do all that and get rid of that minus sign, it adds up to 1 times the constants.

You know the particle is undergoing simple harmonic motion. What type of force causes simple harmonic motion, i.e., can an arbitrary force do it, or does it have to vary in a specific way?

I completely forgot to look at the relation between work done and the potential energy!
so I found Fx= m*ax= -maw^2sin(wt) and Fy= m*ay=-mbw^2cos(wt).
But how do I do the integration?

You can follow two ways: Make a vector equation:

$$\vec F = m \vec a =-m\omega^2\vec r$$

How do you get the potential energy if you know the force in terms of the position vector?

ehild

so I make my acceleration vector a = sqrt((ax)^2+(ay)^2) and after some rearranging and factoring I get a = w^2 sqrt(x^2+y^2). But what do I do with that result now? How do I reach the expression of PE?

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I think it is better to keep your x and y accelerations separate.
Figure out the work done in each direction, then add them.
To find work you must integrate dW = Fx*dx and dW = Fy*dy.
If Fx is a function of x only, it is easy integration. Strictly speaking, you should integrate from x = 0 to some value of x, but the evaluation at x = 0 is zero, so the integral of
Fx*dx = -maw²*sin(wt)*dx = -mw²x*dx is -mw²x²/2
Check this very carefully; remember I lost a minus sign at least.
The PE is the total of the work done in the x direction and the work done in the y direction.

The vector of acceleration is

$$\vec a =-\omega^2(a\sin(\omega t) \vec i +b\cos(\omega t )\vec j ) = -\omega^2\vec r.$$

The potential energy is equal to the work done by the force when the body moves from the selected point (x,y) to the one where the potential energy is 0. Let be this point at r=0, and the integration path a radius from point P(x,y) to (0,0). The work done is the integral of (-w^2 m rdr)=w^2 m r^2/2.

I wanted to write it the formulae in tex, but I did not succeed...

ehild

Last edited:
There is an other way to handle the problem. The motion is the superposition of two independent oscillation, one along x, the other along y. The energy in this case is the sum of the energies assigned to both motions. What is the energy of a body performing simple harmonic motion, in terms of its amplitude and angular frequency?

ehild

## 1. What is 2D harmonic motion and how is it different from 1D harmonic motion?

2D harmonic motion refers to the movement of an object in two dimensions, with both horizontal and vertical components. This is in contrast to 1D harmonic motion, which only has movement along a single axis. In 2D harmonic motion, the object's position is described by both x and y coordinates, while in 1D harmonic motion, it is only described by one coordinate.

## 2. How is the equation for 2D harmonic motion derived?

The equation for 2D harmonic motion is derived from the principles of Newton's second law of motion and Hooke's law. It takes into account both the horizontal and vertical forces acting on the object, as well as the object's mass and the spring constant of the system.

## 3. What are the key variables in the equation for 2D harmonic motion?

The key variables in the equation for 2D harmonic motion are the object's position (x and y), velocity (v), acceleration (a), mass (m), and the spring constant (k) of the system. These variables are used to calculate the displacement, velocity, and acceleration of the object at any given time.

## 4. How do you solve a 2D harmonic motion problem?

To solve a 2D harmonic motion problem, you will need to use the equation for 2D harmonic motion and the given values for the variables. First, identify the initial position, velocity, and acceleration of the object. Then, use the equation to calculate the displacement, velocity, and acceleration at any given time. Finally, plot the object's position over time to visualize its motion.

## 5. What are some real-life examples of 2D harmonic motion?

Some real-life examples of 2D harmonic motion include a pendulum swinging back and forth, a mass attached to a spring bouncing up and down, and a roller coaster moving along a track. These systems exhibit 2D harmonic motion because they have both horizontal and vertical components of movement, and are subject to the forces of gravity and elasticity.

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