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?

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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&#039;t really understand why, and the book doesn&#039;t explain it any further. How do I know the two are equal?

vibrerentb · Accepted Answer

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        &amp;lt;&amp;lt;    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 · Accepted Answer

The oscillator potential is the one quadratic in the degree of freedom, let&#039;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    )

Why is the spring constant W′′_p(0)?

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