Calculus-based derivation

From Newton's laws we know that F = ma where m is the mass of the weight, and a is its acceleration. Substituting this into Hooke's Law, we get

ma = -kx

Dividing through by m:

${\displaystyle a=-{\frac {k}{m}}x}$

The calculus definition of acceleration gives us

${\displaystyle x''=-{\frac {k}{m}}x}$

${\displaystyle x''+{\frac {k}{m}}x=0}$

Thus we have a second-order differential equation. Solving it gives us

${\displaystyle x=c_{1}\cos \left({\sqrt {\frac {k}{m}}}t\right)+c_{2}\sin \left({\sqrt {\frac {k}{m}}}t\right)}$ (2)

with an independent variable t for time.

We can change this equation into a simpler form. By lettting c₁ and c₂ be the legs of a right triangle, with angle φ adjacent to c₂, we get

${\displaystyle \sin \phi ={\frac {c_{1}}{\sqrt {c_{1}^{2}+c_{2}^{2}}}}}$

${\displaystyle \cos \phi ={\frac {c_{2}}{\sqrt {c_{1}^{2}+c_{2}^{2}}}}}$

and

${\displaystyle c_{1}={\sqrt {c_{1}^{2}+c_{2}^{2}}}\sin \phi }$

${\displaystyle c_{2}={\sqrt {c_{1}^{2}+c_{2}^{2}}}\cos \phi }$

Substituting into (2), we get

${\displaystyle x={\sqrt {c_{1}^{2}+c_{2}^{2}}}\sin \phi \cos \left({\sqrt {\frac {k}{m}}}t\right)+{\sqrt {c_{1}^{2}+c_{2}^{2}}}\cos \phi \sin \left({\sqrt {\frac {k}{m}}}t\right)}$

Using a trigonometric identity, we get:

${\displaystyle x={\sqrt {c_{1}^{2}+c_{2}^{2}}}\left[\sin \left(\phi +{\sqrt {\frac {k}{m}}}t\right)+\sin \left(\phi -{\sqrt {\frac {k}{m}}}t\right)\right]+{\sqrt {c_{1}^{2}+c_{2}^{2}}}\left[\sin \left({\sqrt {\frac {k}{m}}}t+\phi \right)+\sin \left({\sqrt {\frac {k}{m}}}t-\phi \right)\right]}$

${\displaystyle x={\sqrt {c_{1}^{2}+c_{2}^{2}}}\sin \left({\sqrt {\frac {k}{m}}}t+\phi \right)}$ (3)

Let ${\displaystyle A={\sqrt {c_{1}^{2}+c_{2}^{2}}}}$ and ${\displaystyle \omega ^{2}={\frac {k}{m}}}$. Substituting this into (3) gives

${\displaystyle x=A\sin \left(\omega t+\phi \right)}$

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