# Elastic Potential Energy and SHM test

1. Jun 1, 2005

### PhysicsinCalifornia

Hello, I took a test on Simple Harmonic Motion today and the very last problem looked something like this:

A simple harmonic oscillator has a total energy of E. (a) Detemine the kinetic and potential energies when the displacement is one half the amplitude. (b) For what value of the displacement does the kinetic energy equal the potential energy?

I had no clue how to solve this one. I'm sure I got it wrong. Can anyone help me to start with this thing??

All I know is that I'm supposed to use the energy equation:
$$E_i = E_f$$
$$KE_i + PE_i + PE_{si} = KE_f + PE_f + PE_{sf}$$
$$KE = \frac{1}{2}mv^2$$
$$PE = mgh$$
$$PE_s = \frac{1}{2}kx^2$$

I seriously do NOT know how to do this problem.

2. Jun 1, 2005

### OlderDan

The usual treatment of this problem combines the gravitational and spring potential energies into one term. If you measure displacement from the equilibrium position as x, then the combined potentail energy is

$$PE = \frac{1}{2}kx^2$$

and the total energy is

$$PE = \frac{1}{2}kA^2$$

where A is the amplitude. At x = A/2, the PE is 1/4 of the total energy, so the KE is 3/4. for part b) you need to find where the potential energy is half the total.

3. Jun 1, 2005

### PhysicsinCalifornia

I don't think I'm getting this or not, but i just replace $$x^2$$ with $$(\frac{A}{2})^2$$ ?
or just directly replace $$x^2 \Rightarrow (\frac{A}{2})]/tex]? 4. Jun 1, 2005 ### PhysicsinCalifornia I don't think I'm getting this or not, but i just replace [tex]x^2$$ with $$(\frac{A}{2})^2$$ ?
or just directly replace $$x^2 \Rightarrow (\frac{A}{2})$$?

And if so, how can I show that the PE is 1/4 of E and KE is 3/4 E?

5. Jun 1, 2005

### Dr.Brain

For simplicity the harmonic oscillator can be thought like a spring in SHM .Its the same.

The net energy for a particle in SHM is given by: $\frac {kA^2}{2}$

Assuming Kinetic energy at any time t is : $\frac {kx^2}{2}$

And potential energy at t is: $\frac {k(A^2 - x^2)}{2}$

The above two relations can be easily though of as PE +KE = Total energy which is the case in above problems.

6. Jun 1, 2005

### PhysicsinCalifornia

The Kinetic energy can be written like that? I thought that the kinetic energy has to do with velocity.

So I can replace that kinetic energy with that elastic potential energy equation??

Last edited by a moderator: Jun 1, 2005
7. Jun 1, 2005

### Staff: Mentor

I think you mixed up the terms "potential" and "kinetic".

8. Jun 1, 2005

### Dr.Brain

In the above equation $\frac {Kx^2}{2}$, Remember that "x" is not the "x" for extension length but this is some variable . Here "x" has units of v/sec .

You can replace "x" by any other variable like a,b,c,d ...but it should have units of velocity ....once you have assumed a KE using some variable , now you can deduce the PE using the fact that their sum is $\frac {kA^2}{2}$

And int he Total Energy $\frac {kA^2}{2}$ A is the amplitude .

9. Jun 1, 2005

### Staff: Mentor

No.
The total energy of the system is given by:
$$E = \frac{1}{2}kA^2$$

At any position x from equilibrium, the potential energy is given by:
$$\mbox{PE} = \frac{1}{2}kx^2$$

To find the KE at any point, realize that $$E = PE + KE$$

10. Jun 1, 2005

### PhysicsinCalifornia

Since the problem is one half the amplitude, I thought the x would be simply replaced with A/2

Can you explain the concept of doing this?

Also, I realized that E = KE + PE, but in this case, the PE would be the elastic potential? $$PE_s = \frac{1}{2}kx^2$$

11. Jun 1, 2005

### Staff: Mentor

For part a, everything you say is correct. To find the PE, just plug in the displacement from equilibrium: x = A/2. (If you were saying this all along, I apologize if I misled you by saying "No". I responded to post #3, when I should have responded to #4.)

Now figure out what the PE equals in terms of E. (Hint: Write both E and PE in terms of A and compare.)