Created At: [[2024-11-11]] Refer to [[Princeton Lectures in Analysis I - Fourier Analysis]] - Chapter 1 on [Simple Harmonic Motion](x-devonthink-item://D058741C-540E-40F0-9EA9-5A13ABDEACF0?page=18&annotation=Highlight&x=152&y=501). ![[20241111155048.png]] Let $y(t)$ denote the displacement of the mass at time $t$. By Hooke's law, $F$ exerted by the spring on the mass is given by: $F = -ky(t).$ - $k > 0$ is the spring constant. By Newton's law $f = ma = m y''$, where $y''$ is the second derivative of the position of the object, we have: $ \begin{aligned} -ky(t) &= ma \\&= my''(t).\\ \end{aligned} $ Simplifying it, we have: $ \begin{aligned} my''(t) + ky(t) &= 0\\ &\Downarrow\\ y''(t) + \dfrac{k}{m} y(t)&=0. \end{aligned} $ If we then set $c = \sqrt{k/m}$, we have: $ y''(t) + c^2 y(t) = 0. $