# Energy conversion in a spring SHM

1. May 20, 2016

1. The problem statement, all variables and given/known data
Looking for some feedback

a) Describe the energy conversions in a spring undergoing simple harmonic motion as it moves from the point of maximum compression to maximum stretch in a frictionless envoirnment. Be sure to point out the points at which there will be
I) maximum speed
ii) minimum speed
iii) minimum acceleration

2. Relevant equations

3. The attempt at a solution

When a spring is undergoing SHM energy is being conserved in the oscillating spring. This energy is being transformed between one type of energy and another. 1) At the extremes when the spring is fully stretched/compressed it has achieved the minimum speed and instantaneous velocity of 0. The reason for this is because the spring has lost its kinetic energy. When at this minimum speed it has achieved the greatest stretch Δx from its equilibrium. All the E_k at this point has been transferred over to E_e 2) At the equilibrium point or rest position of the spring the energy has all been transferred away from E_e into E_k and it has achieved the maximum speed in the system. 3) The acceleration when the spring is at the mean is minimum as well since Δx=0. Between the two extremes and the equilibrium the spring has its total energy distributed between E_e and E_k. The total energy in the system does not change during this process, it only transfers from one energy type to another.

2. May 20, 2016

### whit3r0se-

Are you assuming this spring has been placed on a horizontal plane?

3. May 20, 2016

I'm assuming the spring isn't in any plane. A fictional spring oscillating without any other influences on it i.e gravity.

4. May 21, 2016

### whit3r0se-

Well that isn't very helpful, why not attempt to factor in gravity?

5. May 21, 2016

I believe it would complicate the matter beyond the scope of the course, it seems like they want just a general test for knowledge of basic relationships with an ideal spring in SHM forever.

6. May 21, 2016

### lychette

I agree!....the properties of a spring do not depend on gravity. A vertical spring with a hanging weight may be the most familiar situation but it is not the general case.
Can you sketch graphs of PE and KE with displacement for SHM?

7. May 21, 2016

### whit3r0se-

If you factor gravity in, the point of equilibrium of the oscillations will be shifted

8. May 21, 2016

The way I visualize these factious springs for conceptual purposes is as a spring in an empty background, I don't really consider it on an x or y plane.

With regards to the sketching the graphs, wouldn't it depend on which axis the spring was oscillating? As an example if I choose the spring to be oscillating parallel to the floor and gravity pulling towards the center of the earth I would have no PE because Δh=0 since h is on the y axis.

9. May 22, 2016

### lychette

All of the equations relating to SHM are given in terms of displacement from the equilibrium position, x Displacement is a vector.
eg F = -kx or a = -k/m (x) if you prefer the acceleration equation.
The PE in SHM is ELASTIC PE of the spring system, not gravitational PE

10. May 23, 2016

I was doing some thinking about these graphs, how would I setup the restrictions for the graph because wouldn't displacement continue on to ±∞ and what definition would I use for k? -1? I can graph 1 variable equations but with 2 i'm not very familiar with the process in setting it up.

11. May 23, 2016

### whit3r0se-

Restrictions would be +/- A (amplitude)

12. May 23, 2016

### Nidum

A spring not connected to anything else is literally a free body . You need to either anchor it in space or define its actions in terms of nodal forces and displacements .

13. May 26, 2016

If we assume the spring is anchored to a surface with a negligible mass could you not disregard the direction which is it oscillating? I assumed it was anchored by default, my fault for forgetting something as obvious as it requiring an anchor.

14. May 28, 2016

### haruspex

I think you mean a surface with infinite mass, but that is not necessary. What you do need is that the acceleration of each endpoint of the spring is an affine function of the tension in the spring. Thus, it could be two masses that are otherwise free to move.
Typically you ignore the mass of the spring itself, since this creates complications. The motion would no longer be 'simple'.
... which has no bearing on any of the questions posed, provided the answer is phrased in terms of the reference points mentioned (min and max extension).
That said, the question is a bit wrong in suggesting it is necessarily sometimes under compression and sometimes under tension. It might always be compressed or always be stretched. Min and max extension is a better way to express it because each could be a negative value.