# Doubling the energy of an oscillating mass on a spring

1. Apr 16, 2013

### Turion

1. The problem statement, all variables and given/known data

Attached.

2. Relevant equations

3. The attempt at a solution

From my calculations, option A is correct. What am I doing wrong? Also, I can't find a relationship between amplitude and energy.

$${ KE }_{ max }=\frac { 1 }{ 2 } m{ { v }_{ max } }^{ 2 }\\ =\frac { 1 }{ 2 } m{ (rω) }^{ 2 }\\ =\frac { 1 }{ 2 } mr^{ 2 }{ ω }^{ 2 }\\ { { KE }'_{ max } }=\frac { 1 }{ 2 } mr^{ 2 }{ (\sqrt { 2 } ω })^{ 2 }\\ =m{ r }^{ 2 }{ ω }^{ 2 }\\ =2{ KE }_{ max }$$

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2. Apr 16, 2013

### rock.freak667

By conservation of mechanical energy during SHM,

KE + PE = constant.

Total energy = constant = ½kA2

Where A is the amplitude, so option A should be correct.

3. Apr 16, 2013

### Turion

Option A doesn't involve amplitudes though. You mean option B?

4. Apr 16, 2013

### Turion

Okay, I understand now how I increasing the amplitude by √2 doubles total energy. However, it seems that increasing angular frequency by √2 also doubles total energy, so are both option A and option B correct?

5. Apr 16, 2013

### BruceW

yep. Nice work!

Edit: is the mark scheme saying that only B is correct? as you said, it looks like A and B are both true.

6. Apr 16, 2013

### Turion

Arghh... I'm not sure how my physics professor missed that. Now I'm a little nervous that she might mark my correct answers as wrong for the final tomorrow.

Thanks for the help! :)

7. Apr 16, 2013

### rock.freak667

I meant option B, sorry.

In your equations, you used v=rω which is for circular motion. Your motion happens to be a mass oscillation.

8. Apr 16, 2013

### Turion

Oh, I see. This is my fault for plugging and chugging. I have to understand what the equations mean.

So I can only use v=rω for circular motion (i.e. two-dimensional oscillations) only? It doesn't apply to one-dimensional oscillations?

9. Apr 16, 2013

### BruceW

still, even for mass oscillation, the 'angular frequency' is defined by the equation:
$$\omega = \sqrt{\frac{k}{m}}$$
Even though it is not related to a circular motion. So I think A and B should both be correct.

Edit: but yeah, as rockfreak says, it is not circular motion, so the calculation should be a bit different.

10. Apr 16, 2013

### BruceW

yeah. After all, the velocity in one-dimensional SHM should be varying, right?

11. Apr 16, 2013

### Turion

Yes, it does vary. But what does that have to do with it?

30 seconds of thinking...

Oh wow! Brilliant!

So the equation v=rω makes use of the fact that the magnitude of velocity in circular motion stays constant!

12. Apr 16, 2013

### BruceW

yes, for uniform circular motion.

edit: I mean yes, for uniform circular motion the magnitude of velocity is constant, so the angular frequency is constant. (i.e. a parameter of the system).