Why does the spring constant go down after a specific mass?

Click For Summary

Homework Help Overview

The discussion revolves around the behavior of a spring's constant in relation to added mass during an oscillatory motion experiment. The original poster notes an unexpected decrease in the spring constant after reaching a specific mass, prompting inquiries into the underlying reasons.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand the relationship between mass and spring constant, questioning whether their observations indicate an error in their methodology. Some participants suggest that real-life springs may not adhere strictly to Hooke's law, introducing the concept of nonlinearity in spring behavior.

Discussion Status

Participants are exploring the implications of nonlinearity in spring behavior and its effect on the results of the experiment. Guidance has been offered regarding how to document these findings in a lab report, but there is no explicit consensus on the interpretation of the results.

Contextual Notes

There is a mention of potential limitations in the application of Hooke's law due to the nature of real-world springs, which may not behave linearly at larger displacements. The original poster's observations may be influenced by these factors.

boii
Messages
15
Reaction score
0
I have been conducting a lab to find the force of a spring by calculating the oscillatory motion of the period. Each time i added a mass, i gained a higher spring constant, but on the 250g i ended up losing spring constant. I was wondering if anyone know why, or if i have done something wrong? The thumbnail i posted will have a better description of what i am trying to say.
 

Attachments

  • physicsfor.png
    physicsfor.png
    2.8 KB · Views: 431
Physics news on Phys.org
That's just because in practice, real-life springs behave nonlinearly (don't perfectly obey Hooke's law). In other words, the spring potential energy is not just ##V=\frac{1}{2}k(x-x_0)^2## but also contains higher order terms in ##(x-x_0)##. Hooke's law works well only for small displacements.
 
hilbert2 said:
That's just because in practice, real-life springs behave nonlinearly (don't perfectly obey Hooke's law). In other words, the spring potential energy is not just ##V=\frac{1}{2}k(x-x_0)^2## but also contains higher order terms in ##(x-x_0)##. Hooke's law works well only for small displacements.

So for this instance on my lab report, should i record down because it doesn't perfectly obey hooke's law?
 
Yes, that's what I would write in the report if I were you.
 

Similar threads

Replies
3
Views
983
Replies
3
Views
2K
Replies
9
Views
4K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
5
Views
2K
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K