Why is the spring constant W′′_p(0)?

Damon Campos

Damon Campos

Answered question

2022-08-12

According to my physics book, the spring constant can be calculated from knowing the potential energy, with the formula k = W p ( 0 )
I don't really understand why, and the book doesn't explain it any further. How do I know the two are equal?

Answer & Explanation

vibrerentb

vibrerentb

Beginner2022-08-13Added 21 answers

The key concept is small oscilation. It is very easy to see this via Lagrangian Formalism, so we write the Lagrangian for our problem
L ( x , v ) = 1 2 m v 2 W ( x )
We expand the potential in series about the equilibrium point x 0 so that x x 0 is a small oscillation in some way, say x x 0 x << 1
W ( x ) = W ( x 0 ) + W ( x 0 ) ( x x 0 ) + 1 2 W ( x x 0 ) 2 + O ( ( x x 0 ) 3 )
Why truncate the series at ( x x 0 ) 3 . Because ( x x 0 ) 2 gives the first non trivial contribution: W ( x 0 ) is a constant, and it does not contribute to the equations of motion, because the motion is governed by the derivatives of the Lagrangian. On the other hand W ( x 0 ) = 0 by hypothesis so we can write
W ( x ) 1 2 W ( x 0 ) ( x x 0 ) 2
and hence
L ( x , v ) = 1 2 m v 2 1 2 W ( x 0 ) ( x x 0 ) 2
this is the Lagrangian for a spring with constant W ( x 0 ). Remember the notation meaning, W ( x 0 ) it is a number, the second derivative evaluated at x 0 , it is not a function.
Silvina2b

Silvina2b

Beginner2022-08-14Added 5 answers

The oscillator potential is the one quadratic in the degree of freedom, let's call it x. If you expand a potential energy term W via Taylor series, you get
W ( x ) = W ( 0 ) + W ( 0 ) x + 1 2 W ( 0 ) x 2 +  
and here you can identify W ( 0 ) as the spring constant k.
Remark: The first term is just a constant energy value and the second one must be taken zero, assuming x=0 is a stable position for your system. So effectvely W ¯ ( x ) = k 2 x 2 + O ( x 3 )

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