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

AI Thread Summary
The discussion centers on the observation that the spring constant increases with added mass until a specific point (250g), after which it decreases. This phenomenon occurs because real-life springs do not perfectly follow Hooke's law and exhibit nonlinear behavior at larger displacements. The potential energy of a spring includes higher-order terms beyond the standard formula, indicating that Hooke's law is only applicable for small displacements. Participants suggest documenting this deviation in the lab report, acknowledging the limitations of Hooke's law in practical scenarios. Understanding these nonlinear effects is crucial for accurate analysis in experiments involving springs.
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: 410
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top