# Graphing Potential and Kinetic Energy of an Oscillating Particle!

1. A particle oscillates back and forth in a frictionless bowl whose height is given by h(x) = 0.22x2 where h and x are meters. (a) Show graphically how the potential and kinetic energies of the particle vary with x.

2. I completed the (b) and (c) part of the problem, but I'm not sure as to how I graph to illustrate how the potential nad kinetic energies of the particle vary with x. Any assitance would be very much appreciated!

3. I really don't even know where to start or how to start, and how I would label my axis, etc. So if anybody can lead me in the right direction, that would be really great, becuase I'm lost!

I know that the particle has maximum kinetic energy at the bottom of the bowl in which h=0 and x=0 and that IF the maximum speed of the particle was 0.4ms-1, x=0.19 and -0.19 and h=0.008. Will this information help me in drawing the graph? Because this is what I had to determine for parts (b) and (c): "(b) Where does the particle have maximum kinetic energy? (c) If the maximum speed of the particle is 0.4 ms-1, find the x-coordinates at which the particle has maximum potential energy"

Redbelly98
Staff Emeritus
Homework Helper
Hint for potential energy graph: what is the potential energy of a mass that is at a height h?

U = mgh?

Redbelly98
Staff Emeritus
Homework Helper
Yes. So if you had h(x), could you graph U(x)?

yes of course!

nrqed
Homework Helper
Gold Member
As you said, the potential energy is mgh. All you need is to relate h to x, so you need the function h(x). If you know that, then you know th epotential energy as a function of x and you can plot that.

I'm so confused!

nrqed
Homework Helper
Gold Member
I'm so confused!

The potential energy is equal to mg h.

We need to know how the height h depends on the horizontal distance x of the ball. This given to you in the question.

right! U = mgh
h(x) = 0.22x2 (where x2 = x squared)
As x increases in the postive and negative direction, h increases exponentially. For instance, here is a small table I constructed:
x h
-10 22
-5 5.5
-3 1.98
0 0
3 1.98
5 5.5
10 22
*If plotted, it would look like a parabola

nrqed
Homework Helper
Gold Member
right! U = mgh
h(x) = 0.22x2 (where x2 = x squared)
As x increases in the postive and negative direction, h increases exponentially. For instance, here is a small table I constructed:
x h
-10 22
-5 5.5
-3 1.98
0 0
3 1.98
5 5.5
10 22
*If plotted, it would look like a parabola

Good!
It's indeed a parabola.
Note that it does not increase exponentially. Exponential growth is when x is in the exponent.

Now, for the kinetic energy, you need to express 1/2 mv^2 in tersms of x also. You already know the potential energy in terms of x. The trick is to use conservation of energy to
relate kinetic energy to potential energy