# Hooke's Law

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Hi,

I can't seem to understand the concept of Hooke's Law. My Mathematics Book tells that:

" The tension in an elastic string or spring is proportional to the extension. If a spring is compressed, the thrust is proportional to the decrease in length of the spring "

This implies:

" T = kx , where k is stiffness of the string "

Whereas my Physics Book states that:

" Hooke’s law states that the extension of a material is directly proportional to the applied force as long as its elastic limit is not exceeded. "

This implies:

" F = kx , where k is stiffness of the spring "

Now if I take both of them to be true simultaneously, there seems to be a contradiction as if " F = T ", the net resultant force is zero, therefore no acceleration and hence there will be no motion, indicating that there is no extension whatsoever.

Also I have read in calculus that:

" Work Done = Integral of F(x) dx from a to b "

" where F(x) is the net resultant force "

And if I put:

F(x) = F - T = 0

I cant derive the work done formula for elastic potential energy.

I can't seem to understand the concept of Hooke's Law. My Mathematics Book tells that:

" The tension in an elastic string or spring is proportional to the extension. If a spring is compressed, the thrust is proportional to the decrease in length of the spring "

This implies:

" T = kx , where k is stiffness of the string "

Whereas my Physics Book states that:

" Hooke’s law states that the extension of a material is directly proportional to the applied force as long as its elastic limit is not exceeded. "

This implies:

" F = kx , where k is stiffness of the spring "

Now if I take both of them to be true simultaneously, there seems to be a contradiction as if " F = T ", the net resultant force is zero, therefore no acceleration and hence there will be no motion, indicating that there is no extension whatsoever.

Also I have read in calculus that:

" Work Done = Integral of F(x) dx from a to b "

" where F(x) is the net resultant force "

And if I put:

F(x) = F - T = 0

I cant derive the work done formula for elastic potential energy.

Last edited by Tesla3; 6 months ago

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#2

(Original post by

**Tesla3**)**Hi,**

" The tension in an elastic string or spring is proportional to the extension. If a spring is compressed, the thrust is proportional to the decrease in length of the spring "

" T = kx , where k is stiffness of the string "" The tension in an elastic string or spring is proportional to the extension. If a spring is compressed, the thrust is proportional to the decrease in length of the spring "

" T = kx , where k is stiffness of the string "

**Whereas my Physics Book states that:**

**" Hooke’s law states that the extension of a material is directly proportional to the applied force as long as its elastic limit is not exceeded. "**

**This implies:**

**" F = kx , where k is stiffness of the spring "**

This is also correct. When a force is applied and the spring extends, then in equilibrium, the F = T and there is no movement so F = T = kx, though F and T are in opposite directions. To equate them, one should be negated so T+F=0

**Now if I take both of them to be true simultaneously, there seems to be a contradiction as if " F = T ", the net resultant force is zero, therefore no acceleration and hence there will be no motion, indicating that there is no extension whatsoever.**

That is correct apart from there is no contradiction and saying there is no extension. The applied force and the tension sum to zero so the particle is in equilibrium at displacement x.

**" Work Done = Integral of F(x) dx from a to b "**

**" where F(x) is the net resultant force "**

**And if I put:**

**F(x) = F - T = 0**

**I cant derive the work done formula for elastic potential energy.**

Work done is against the tension, so you'd integrate the tension (kx) from a to b which gives kx^2/2 (and sub the limits).

Last edited by mqb2766; 6 months ago

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(Original post by

This is correct. The tension (force) in the spring (Hooke's law) is proportional to the extension from the natural length or equilibrium state.

This is also correct. When a force is applied and the spring extends, then in equilibrium, the F = T and there is no movement so F = T = kx, though F and T are in opposite directions. To equate them, one should be negated so T+F=0

That is correct apart from there is no contradiction. The applied force and the tension sum to zero so the particle is in equilibrium at displacement x.

Work done is against the tension, so you'd integrate the tension (kx) from a to b which gives kx^2/2 (and sub the limits).

**mqb2766**)This is correct. The tension (force) in the spring (Hooke's law) is proportional to the extension from the natural length or equilibrium state.

**Whereas my Physics Book states that:****" Hooke’s law states that the extension of a material is directly proportional to the applied force as long as its elastic limit is not exceeded. "****This implies:****" F = kx , where k is stiffness of the spring "**This is also correct. When a force is applied and the spring extends, then in equilibrium, the F = T and there is no movement so F = T = kx, though F and T are in opposite directions. To equate them, one should be negated so T+F=0

**Now if I take both of them to be true simultaneously, there seems to be a contradiction as if " F = T ", the net resultant force is zero, therefore no acceleration and hence there will be no motion, indicating that there is no extension whatsoever.**That is correct apart from there is no contradiction. The applied force and the tension sum to zero so the particle is in equilibrium at displacement x.

**" Work Done = Integral of F(x) dx from a to b "****" where F(x) is the net resultant force "****And if I put:****F(x) = F - T = 0****I cant derive the work done formula for elastic potential energy.**Work done is against the tension, so you'd integrate the tension (kx) from a to b which gives kx^2/2 (and sub the limits).

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#4

(Original post by

So " F = T just for equilibrium , not before that. " ?

**Tesla3**)So " F = T just for equilibrium , not before that. " ?

ma = F - T (putting the sign in to denote opposite directions)

* If F > T then the spring will extend as a > 0

* If F < T, the spring will contract as a < 0

* If its equal, the particle will be in equilbrium as the net force is 0 (newton 1).

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(Original post by

Newton 2 gives

ma = F - T (putting the sign in to denote opposite directions)

* If F > T then the spring will extend as a > 0

* If F < T, the spring will contract as a < 0

* If its equal, the particle will be in equilbrium as the net force is 0 (newton 1).

**mqb2766**)Newton 2 gives

ma = F - T (putting the sign in to denote opposite directions)

* If F > T then the spring will extend as a > 0

* If F < T, the spring will contract as a < 0

* If its equal, the particle will be in equilbrium as the net force is 0 (newton 1).

Last edited by Tesla3; 6 months ago

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#6

(Original post by

Also, I can't understand one more thing. Lets suppose that we extend the spring from a to b and at b " F = T " . Just before b, we would have a certain velocity " v " as the spring is moving and some acceleration as well. And at b, the net force is zero, and therefore the acceleration is zero, so over here we will have velocity " v " as well. It certainly has some kinetic energy at this point. Then will the spring ever balance or this process continue till infinity? Can you elaborate your meaning of equilibrium, as I think equilibrium means that the net resultant force must be zero as well as the velocity must be zero as well perhaps.......

**Tesla3**)Also, I can't understand one more thing. Lets suppose that we extend the spring from a to b and at b " F = T " . Just before b, we would have a certain velocity " v " as the spring is moving and some acceleration as well. And at b, the net force is zero, and therefore the acceleration is zero, so over here we will have velocity " v " as well. It certainly has some kinetic energy at this point. Then will the spring ever balance or this process continue till infinity? Can you elaborate your meaning of equilibrium, as I think equilibrium means that the net resultant force must be zero as well as the velocity must be zero as well perhaps.......

**Can you elaborate your meaning of equilibrium, as I think equilibrium means that the net resultant force must be zero as well as the velocity must be zero as well perhaps.......**

For a spring, equilibrium would imply that both the velocity and acceleration (net force) are zero at that extension. Its fairly easy to imagine a thought experiment, where if the velocity is not zero, the spring extension will change and hence the net force (acceleration) will no longer be zero as the applied force would be unchanged, but the tension would have changed.

**Also, I can't understand one more thing. Lets suppose that we extend the spring from a to b and at b " F = T " . Just before b, we would have a certain velocity " v " as the spring is moving and some acceleration as well. And at b, the net force is zero, and therefore the acceleration is zero, so over here we will have velocity " v " as well. It certainly has some kinetic energy at this point. Then will the spring ever balance or this process continue till infinity?**

Good point. A better way to think about it is if the spring is perturbed from the equilibrium position, it will not return to the equilibrium but will exhibit simple harmonic motion (sinusoidal) about the equilibrium point forever (like a pendulum). You would generally assume that its in the equilibrium position to start with, not worry about how you get there.

Last edited by mqb2766; 6 months ago

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(Original post by

For a spring, equilibrium would imply that both the velocity and acceleration (net force) are zero at that extension. Its fairly easy to imagine a thought experiment, where if the velocity is not zero, the spring extension will change and hence the net force (acceleration) will no longer be zero.

Good point. A better way to think about it is if the spring is perturbed from the equilibrium position, it will not return to the equilibrium but will exhibit simple harmonic motion (sinusoidal) about the equilibrium point forever (like a pendulum). You would generally assume that its in the equilibrium position to start with, not worry about how you get there.

**mqb2766**)**Can you elaborate your meaning of equilibrium, as I think equilibrium means that the net resultant force must be zero as well as the velocity must be zero as well perhaps.......**For a spring, equilibrium would imply that both the velocity and acceleration (net force) are zero at that extension. Its fairly easy to imagine a thought experiment, where if the velocity is not zero, the spring extension will change and hence the net force (acceleration) will no longer be zero.

**Also, I can't understand one more thing. Lets suppose that we extend the spring from a to b and at b " F = T " . Just before b, we would have a certain velocity " v " as the spring is moving and some acceleration as well. And at b, the net force is zero, and therefore the acceleration is zero, so over here we will have velocity " v " as well. It certainly has some kinetic energy at this point. Then will the spring ever balance or this process continue till infinity?**Good point. A better way to think about it is if the spring is perturbed from the equilibrium position, it will not return to the equilibrium but will exhibit simple harmonic motion (sinusoidal) about the equilibrium point forever (like a pendulum). You would generally assume that its in the equilibrium position to start with, not worry about how you get there.

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#8

(Original post by

Also in real life, we would have some air resistance and some of the energy will be converted to heat energy. That might just help it to get to the equilibrium point and stop it from oscillating forever.

**Tesla3**)Also in real life, we would have some air resistance and some of the energy will be converted to heat energy. That might just help it to get to the equilibrium point and stop it from oscillating forever.

In real life you'd have some form of damping (mechanical, air resistance, ...) which would introduce a decaying term to the sinusoidal osciallations.

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(Original post by

Of course, but A levels generally are not real life.

In real life you'd have some form of damping (mechanical, air resistance, ...) which would introduce a decaying term to the sinusoidal osciallations.

**mqb2766**)Of course, but A levels generally are not real life.

In real life you'd have some form of damping (mechanical, air resistance, ...) which would introduce a decaying term to the sinusoidal osciallations.

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