Doubling the energy of an oscillating mass on a spring

[Turion]

Homework Statement

Attached.

Homework Equations

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 }

 

[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.

 

[Turion]

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.

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

 

[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?

 

[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.

 

[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! :)

 

[rock.freak667]

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?

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.

 

[Turion]

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

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?

 

[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.

 

[BruceW]

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

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

 

[Turion]

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

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!

 

[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).