Understanding SHM and Finding k in a Vertical Spring Lab

  • Thread starter Thread starter TN17
  • Start date Start date
  • Tags Tags
    Shm
Click For Summary

Homework Help Overview

The discussion revolves around understanding simple harmonic motion (SHM) in the context of a vertical mass-spring system, specifically focusing on determining the spring constant, k. Participants are exploring the relationship between forces acting on the system at equilibrium and during oscillation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equilibrium condition where the force due to gravity equals the spring force, and they question how this relationship changes when the system is oscillating. There is uncertainty about how to apply the force equation F = kx in the context of oscillation, particularly regarding the variables involved.

Discussion Status

Some participants are clarifying the forces acting on the mass during oscillation and how to relate these to the spring constant k. There is a recognition of the need to use both initial displacement and additional displacement in their calculations, but confusion remains about how to isolate k with the given information.

Contextual Notes

Participants are working under the constraints of a lab experiment where only certain measurements (displacement and mass) are available, and they are trying to reconcile these with theoretical equations. There is an emphasis on understanding the dynamics of the system rather than simply applying formulas.

TN17
Messages
45
Reaction score
0

Homework Statement



This is another question related to the vertical spring lab where we try to determine k, the spring constant.

When the vertical mass-spring system is at equilibrium, then the F in F = kx is equal to F = mg, or 9.8.

What would the F be in F = kx when the system is oscillating?
In that case, the only information we know from the experiment is x (the displacement) and mass of the object.
I'm not sure how to move on from there.
 
Physics news on Phys.org
TN17 said:

Homework Statement



This is another question related to the vertical spring lab where we try to determine k, the spring constant.

When the vertical mass-spring system is at equilibrium, then the F in F = kx is equal to F = mg, or 9.8.

What would the F be in F = kx when the system is oscillating?
In that case, the only information we know from the experiment is x (the displacement) and mass of the object.
I'm not sure how to move on from there.

When the mass is initial hung, there is an initial displacement δ, so you have kδ=mg

then you displace the mass a distance 'x' such that the spring force acting is k(δ+x).
 
rock.freak667 said:
When the mass is initial hung, there is an initial displacement δ, so you have kδ=mg

then you displace the mass a distance 'x' such that the spring force acting is k(δ+x).

I understand the first part that you said because that is when it's still.
And the second part, do you mean that F = k(δ+x)? Is that when the system is oscillating?
 
TN17 said:
I understand the first part that you said.
And the second part, do you mean that F = k(δ+x)?

right, that F is the spring force which is just one of the forces acting when you displace the mass an additional distance x. You will still have the weight mg acting downwards.
 
rock.freak667 said:
right, that F is the spring force which is just one of the forces acting when you displace the mass an additional distance x. You will still have the weight mg acting downwards.
Okay.
But if the purpose was to find k, the spring constant, how would I do so when I only know δ and x?
That's the part I'm confused about from your equation.
 
TN17 said:
Okay.
But if the purpose was to find k, the spring constant, how would I do so when I only know δ and x?
That's the part I'm confused about from your equation.

If you need to get 'k' and you have 'δ', then you just need to use kδ=mg or k=mg/δ.
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
5
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K