Does Hooke's Law Explain Asymmetric Spring Oscillations?

  • Thread starter Thread starter qazxsw11111
  • Start date Start date
  • Tags Tags
    Hooke's law Law
AI Thread Summary
Hooke's Law can explain asymmetric spring oscillations when considering gravitational potential energy (GPE) and elastic potential energy (EPE). The discussion highlights that at the top of the oscillation, both GPE and EPE are present, leading to different extensions at the top and bottom of the oscillation. The equilibrium position shifts due to gravity, affecting the amplitude of oscillation. The reasoning that energy conservation implies different extensions is validated, as the equilibrium point is adjusted by the weight of the load. Overall, the analysis confirms that while the spring exhibits simple harmonic motion, the presence of gravity alters the effective equilibrium position.
qazxsw11111
Messages
95
Reaction score
0
If I have a spring with a load and I oscillate it freely, applying hooke's law,

TOP: GPE+EPE(Given by area under F-x graph)
Equilibrium:KE (No EPE since x=0)
Bottom:EPE Only

Since energy is conserved, I have to assume that the extension (actually compression) at the top is less than extension at bottom as there is also GPE also at the top.

Is this reasoning correct?

But what about an oscillating spring being a example of simple harmonic motion? Isnt the amplitude from the equilibrium position supposed to be the same?

Anyone help?
 
Physics news on Phys.org
qazxsw11111 said:
… But what about an oscillating spring being a example of simple harmonic motion? Isnt the amplitude from the equilibrium position supposed to be the same?

Hi qazxsw11111! :smile:

Everything is shifted slightly.

The equation of motion is x'' = -kx - mg,

so put y = x + mg/k, then y'' = -ky, which is SHM, with equal amplitude either side of the equilibrium position, but with the equilibrium position where the mass would remain at rest. :wink:
 
Last edited:
Hi Tiny-Tim =),

How do you get a = -kx - mg ?

What is actually the loophole in my reasoning:

Quoted: "Since energy is conserved, I have to assume that the extension (actually compression) at the top is less than extension at bottom as there is also GPE also at the top."

At the top of the oscillation, both GPE and EPE are present right?

I appreciate the help but still I can't help feeling confused.

Thank you for your time.
 
(ooh, i left out an m … it should be mx'' = -kx - mg and my'' = - ky :redface:)
qazxsw11111 said:
How do you get a = -kx - mg ?

Hi qazxsw11111! :smile:

Good ol' Newton's second law … F = ma

the F is -kx for the spring and -mg for gravity.
What is actually the loophole in my reasoning:

Quoted: "Since energy is conserved, I have to assume that the extension (actually compression) at the top is less than extension at bottom as there is also GPE also at the top."

At the top of the oscillation, both GPE and EPE are present right?

Your reasoning is correct, but it doesn't contradict the actual result …

everything is displaced slightly, by a distance mg/k.

The extension (actually compression) at the top is less than the extension at bottom, if you measure it from the equilibrium-position-without-gravity (eg, if the spring is horizontal), but they are the same if you measure from a position mg/k lower. :wink:
 
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 '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