Confirming Answers for Oscillating Mass on a Spring

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In summary, the conversation revolved around a physics introduction question involving a mass oscillating on a spring. The question asked for the initial potential energy, the maximum reached in the oscillation, and the maximum velocity of the mass. The answers provided were 1.728 J for PE, equilibrium position for the maximum reached, and 4.157 m/s for the maximum velocity. The conversation also included a discussion about posting homework questions in the correct forum and clarification on the meaning of "maximum reached" in the question. In the end, the poster thanked the expert for their assistance.
  • #1
domyy
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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.
 
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  • #2
Hi domyy! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Can you show your working for part 1?

Incidently, I think homework questions should be placed in the homework forum.
 
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  • #3
Please provide the full question. Your problem statement is missing key information.

Show your work.
 
  • #4
Thanks! Where should I post my question?
 
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  • #5
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
Oh yes. I am so sorry. I am missing one last line:

IT WAS ORIGINALLY STRETCHED A DISTANCE OF 12 CM FROM ITS EQUILIBRIUM.
 
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  • #7
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.
 
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  • #8
anyone?
 
  • #9
domyy said:
What I did to calculate the initial potential energy was:

PE = 1/2 (240N/m)(0.12)^2 = 1.728.
Good.
To calculate max. velocity that the mass will reach in its oscilation:

I answered this one with some direction because I found it hard to solve:

Vmax = aw, being w= square root of k/m = square root of 240/0.2 = 34.64.
Vmax = (34.64)(0.12) = 4.15m/s.
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
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
domyy said:
Oh thanks a lot.

About the question where oscillation is maximum reached, I answered Equilibrium Position. Is that ok, as well?
I don't understand the meaning of "oscillation is maximum". What about the oscillation is maximum?

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.
This is good.

KE = (240N/m)(0.12)^2 divided by 2 = 1.296
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.

V= ? I am not sure. I have one formula: (2KE/M)^1/2 Is that right?
That should work fine.
 
  • #12
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.

_____________________________________________________________________

I just would like to thank you very very much for taking your time to assist me!
 
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  • #13
domyy said:
The question is exactly like this on my paper: "Where in the oscillation is this maximum reached?"
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
domyy said:
Thanks! Where should I post my question?
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! :wink:
 
  • #15
Oh thank you so much! You guys were amazing! I feel more confident about my answers now! Thanks again!
 
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