Why do we use Taylor expansion expressing potential energy

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SUMMARY

Taylor expansion is utilized to express potential energy in the context of spring mechanics, particularly when analyzing Hooke's Law, defined by the equation F = -Kx. This method is essential for deriving the quadratic form of potential energy, U(x) = (1/2) kx², which is valid for small displacements (x) around the equilibrium position (x = 0). The Taylor series allows for the expansion of potential energy functions, where the first derivative at the equilibrium point is zero, leaving the quadratic term as the primary contributor to the potential energy in ideal springs.

PREREQUISITES
  • Understanding of Hooke's Law and its application in spring mechanics.
  • Familiarity with Taylor series and its mathematical implications.
  • Knowledge of potential energy concepts in physics.
  • Basic calculus, particularly differentiation and series expansion.
NEXT STEPS
  • Study the derivation of potential energy using Taylor expansion in various physical systems.
  • Explore the implications of non-ideal springs and how they deviate from Hooke's Law.
  • Learn about higher-order derivatives and their significance in Taylor series expansions.
  • Investigate applications of Taylor expansion in other areas of physics, such as thermodynamics and wave mechanics.
USEFUL FOR

Students of physics, mechanical engineers, and anyone interested in the mathematical modeling of physical systems, particularly in the context of spring dynamics and potential energy analysis.

Ahmed Abdalla
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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
 
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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.
 
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Chandra Prayaga said:
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!
 

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