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Graphing Potential and Kinetic Energy of an Oscillating Particle!

  • Thread starter sousou_88
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  • #1
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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!
 

Answers and Replies

  • #2
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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"
 
  • #3
Redbelly98
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Hint for potential energy graph: what is the potential energy of a mass that is at a height h?
 
  • #4
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U = mgh?
 
  • #5
Redbelly98
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Yes. So if you had h(x), could you graph U(x)?
 
  • #6
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yes of course!
 
  • #7
nrqed
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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.

Hint: read the question carefully.
 
  • #8
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I'm so confused!
 
  • #9
nrqed
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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.
 
  • #10
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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
 
  • #11
nrqed
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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
 

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