Simple harmonic motion of an ideal spring

[pb23me]

i get the equation kxi^2-kxf^2=mg(4A) so I am having the same problem again... if i plug in x=0 for one of the x values it cancels one of the potential energy equations...

 

[pb23me]

so then -k=mg/A

 

[Delzac]

You know initially there is no elastic potential energy. Why? Because there is no extension of the spring, x=0. But there is GPE of mg(2A). At the bottom, extension is maximum, EPE= 1/2 k (2A)^2. Does it make sense?

At the bottom, we can define GPE = 0, this can be done since energy is relative.

 

[Delzac]

We are now comparing absolute value in energy, so there shouldn't be any negative sign.

 

[pb23me]

yeah that makes sense but i was thinking about it as the spring is in motion... At the top it would be getting ready to "spring" back out... so i do think it would have elastic energy at the top as the spring is oscillating

 

[pb23me]

Also what is the value of x in the equation E=1/2mv²+1/2kx²
if i plug in zero at x=0 the equation becomes Zero? is it supposed to be?

 

[Delzac]

If x=0, then yes, there is no kinetic and elastic energy.

 

[pb23me]

Ok then say we had called the point of equillibrium zero... Then it would have elastic energy at the top.. How is that?

 

[Delzac]

It still wouldn't have any elastic energy. The x are talking about here is no longer position of the mass vertically.

The x we are talking about here in 1/2 k x^2 is the extension of the spring.

[STRIKE]Imagine if you were to call the centre point of your oscillation zero, the mass will still be accelerating at centre point! This is because, at centre of oscillation the spring is still stretched.[/STRIKE] EDIT: Mistake, at centre point there is no net acceleration, but there is still spring force exerted on mass.

x = -Acos wt + A, x is position. You can define this however you want.

But k=-kx and E = 1/2 kx^2, the x is extension, not position anymore, though there are times where it can be position.

 

[pb23me]

Just to clarify.. By extension you mean distance away from the center of oscillation?

 

[Delzac]

Nope. Amount of spring is stretch from its unstretched state.

 

[pb23me]

Ok so how would you go about calculating that?

 

[Delzac]

Well, the question will tell you that.

The oscillation is started by hanging and releasing a mass on an unstretched* spring. There you have it, that's your reference.

 

[pb23me]

Ok so when you hang a mass on the spring..and then let it go.. Theoretically isn't it supposed to rise back up to the original unstretched position? I have a hard time imagining it would do exactly that depending on the mass.

 

[Delzac]

Yes it will rise back up to the unstretched position, then go back down again.

I don't understand about the depending on mass part. What do you mean depend on mass? The period?

 

[pb23me]

no I mean if the mass is heavy enough(not so heavy that it deforms the spring)then it wouldn't quite rise back up to that same position..

 

[Mindscrape]

no I mean if the mass is heavy enough(not so heavy that it deforms the spring)then it wouldn't quite rise back up to that same position..

If it doesn't then it's because the spring deformed.

 

[Delzac]

It will rise back up to the original position, disregarding any dissipative force like air drag, and damping.

Conservation of energy demands that i rise back to the original position.

Edit: mindscape beat me to it...

 

[pb23me]

thanx for all the help...appreciatcha!