Homework Help: My Challenge Question

1. Mar 14, 2012

domyy

So, I am supposed to answer this question for my class and I found some answers already. I would like someone to confirm them for me or if not, guide me into getting the right ones. This is physics introduction.

The question:

Suppose that a 0.20kg is oscillating at the end of spring upon a frictionless surface. The spring can be both stretched and compressed and has a spring constant of 240N/m. It was originally stretched a distance of 12cm from its equilibrium.

1) What´s its initial potential PEmax? My answer is 1.728 J.
2) Where is the oscillation maximum reached? My answer is equilibrium position.
3) What´s the maximum velocity that the mass will reach in its oscillation? My answer is 4.157 m/s.

Last edited: Mar 15, 2012
2. Mar 14, 2012

Staff: Mentor

Hi domyy! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Can you show your working for part 1?

Incidently, I think homework questions should be placed in the homework forum.

Last edited by a moderator: May 5, 2017
3. Mar 14, 2012

Staff: Mentor

Please provide the full question. Your problem statement is missing key information.

4. Mar 14, 2012

domyy

Thanks! Where should I post my question?

Last edited: Mar 14, 2012
5. Mar 14, 2012

willem2

1) The question seems to be incomplete, you'll need the inital position and velocity of the mass.
2) If oscillation maximum means, maximum velocity of the mass, then yes, if you mean maximum amplitude, then no.
3) see 1.

6. Mar 14, 2012

domyy

Oh yes. I am so sorry. I am missing one last line:

IT WAS ORIGINALLY STRETCHED A DISTANCE OF 12 CM FROM ITS EQUILIBRIUM.

Last edited: Mar 14, 2012
7. Mar 14, 2012

domyy

What I did to calculate the initial potential energy was:

PE = 1/2 (240N/m)(0.12)^2 = 1.728.

To calculate max. velocity that the mass will reach in its oscillation:

Vmax = aw, being w= square root of k/m = square root of 240/0.2 = 34.64.
Vmax = (34.64)(0.12) = 4.157 m/s.

Last edited: Mar 15, 2012
8. Mar 14, 2012

domyy

anyone?

9. Mar 14, 2012

Staff: Mentor

Good.
That's fine.

Another way is just to set the Max PE (which you just calculated) equal to the Max KE and then solve for the velocity that way.

10. Mar 14, 2012

domyy

Oh thanks a lot.

About the question where oscillation is maximum reached, I answered Equilibrium Position. Is that ok, as well?

I also have part two for the same problem:

When the mass is 6cm from the equilibrium position, what are the values of the elastic potential energy, kinetic energy and velocity:

PE = (240N/m)(0.06)^2 divided by 2 = 0.432.
KE = (240N/m)(0.12)^2 divided by 2 - 0.432 = 1.296
V= ? I am not sure. I have one formula: (2KE/M)^1/2 Is that right?

11. Mar 14, 2012

Staff: Mentor

I don't understand the meaning of "oscillation is maximum". What about the oscillation is maximum?

This is good.

That equation doesn't quite make sense, but the answer does. To find the KE at that point, subtract the PE from the total energy.

That should work fine.

12. Mar 14, 2012

domyy

The question is exactly like this on my paper: "Where in the oscillation is this maximum reached?"

The question relates to the first one I answered in the beginning about the initial potential energy.

_____________________________________________________________________

I missed the PE value. That´s why the answer is correct! KE = 1.728 - 0.432(PE) = 1.296.

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I just would like to thank you very very much for taking your time to assist me!

Last edited: Mar 15, 2012
13. Mar 14, 2012

Staff: Mentor

Presumably they mean at which point is the speed of the mass greatest. Yes, at the equilibrium position because at that point there is zero energy in the spring, so the system's energy must all be as KE of the body.

14. Mar 14, 2012

Staff: Mentor

As you can see, your post was moved to the correct sub-forum. You can see it listed at the top of this screen, Homework & Coursework Questions.

Visit again!

15. Mar 15, 2012

domyy

Oh thank you so much! You guys were amazing! I feel more confident about my answers now! Thanks again!

Last edited: Mar 15, 2012