[CONFUSED]Question about Hooke's Law

In summary, the equilibrium position of a spring with a load is shifted slightly due to the presence of GPE and EPE. Everything is displaced slightly, by a distance mg/k.
  • #1
qazxsw11111
95
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?
 
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  • #2
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:
  • #3
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.
 
  • #4
(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:
 

What is Hooke's Law?

Hooke's Law is a principle in physics that describes the relationship between the force applied to an elastic object and the resulting deformation or change in shape. It states that the force applied is directly proportional to the amount of deformation, as long as the object is within its elastic limit.

Who discovered Hooke's Law?

Hooke's Law was first proposed by the English scientist Robert Hooke in the 17th century.

What is the formula for Hooke's Law?

The formula for Hooke's Law is F = -kx, where F is the force applied, k is the spring constant (a measure of the stiffness of the object), and x is the amount of deformation.

What type of materials does Hooke's Law apply to?

Hooke's Law applies to elastic materials, meaning those that can return to their original shape after being deformed. This includes materials such as rubber, metal springs, and some types of plastics.

What is the significance of Hooke's Law?

Hooke's Law is important in understanding the behavior of elastic materials and is used in various fields such as engineering, mechanics, and materials science. It also serves as the basis for many other laws and principles in physics.

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