I Why do we use Taylor expansion expressing potential energy

My textbook doesn’t go into it, can someone tell me why Taylor expansion is used to express spring potential energy? A lot of the questions I do I think I can just use F=-Kx and relate it to U(x) being F=-Gradiant U(x) but I see most answers using the Taylor expansion instead to get 1/2 kx^2. Usually it has to do with x being a small value but I never understood why the Taylor expansion is used
 

Chandra Prayaga

Science Advisor
648
147
The Hooke's law force, F = -Kx, defines an ideal spring. In a real spring, Hooke's law is valid only for small stretches (small x) of the spring. The Taylor expansion of the potential energy about the equilibrium position (in this case, x = 0), is used to express more general potential energies, not necessaily of an ideal spring. Since any function (inifinitely differentiable) can be expanded in a Taylor series, even the potential energy can be expanded in a Taylor series. If you are expanding the potential energy about its minimum (at x = 0), then
U(x) = U(0) + x (dU/dx)0 + (x2 / 2!) (d2U/dx2)0 + .......(higher order terms)
where the subscript 0 on the derivatives means that they are evaluated at x = 0. If x = 0 is a minimum, then the first derivative in the above expansion is zero, and that leaves the quadratic term as the dominant term in the Taylor expansion of the potential energy, and is identified as the "spring" potential energy.
 
The Hooke's law force, F = -Kx, defines an ideal spring. In a real spring, Hooke's law is valid only for small stretches (small x) of the spring. The Taylor expansion of the potential energy about the equilibrium position (in this case, x = 0), is used to express more general potential energies, not necessaily of an ideal spring. Since any function (inifinitely differentiable) can be expanded in a Taylor series, even the potential energy can be expanded in a Taylor series. If you are expanding the potential energy about its minimum (at x = 0), then
U(x) = U(0) + x (dU/dx)0 + (x2 / 2!) (d2U/dx2)0 + .......(higher order terms)
where the subscript 0 on the derivatives means that they are evaluated at x = 0. If x = 0 is a minimum, then the first derivative in the above expansion is zero, and that leaves the quadratic term as the dominant term in the Taylor expansion of the potential energy, and is identified as the "spring" potential energy.
Thank you that explained it in great detail!
 

Want to reply to this thread?

"Why do we use Taylor expansion expressing potential energy" You must log in or register to reply here.

Related Threads for: Why do we use Taylor expansion expressing potential energy

Replies
6
Views
1K
Replies
3
Views
4K
Replies
5
Views
2K
Replies
2
Views
1K
Replies
2
Views
4K
Replies
3
Views
2K
Replies
5
Views
714

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top