Equation of Simple Harmonic Motion of Spring Derivation

Hooke's Law states that:

$\begin{aligned} F&=-kx\\ F+kx&=0\\ \frac{F}{m}+\frac{kx}{m}&=0\\ a+\frac{kx}{m}&=0\\ \frac{d^2x(t)}{dt^2}+\frac{kx(t)}{m}&=0\\ \end{aligned}$

Assume $x(t)\propto e^{\lambda t}$ solves the differential equation:

$\begin{aligned} \frac{d^2e^{\lambda t}}{dt^2}+\frac{ke^{\lambda t}}{m}&=0\\ \lambda^2e^{\lambda t}+\frac{ke^{\lambda t}}{m}&=0\\ e^{\lambda t}\left(\lambda^2+\frac{k}{m}\right)&=0\\ \lambda^2+\frac{k}{m}&=0\\ \lambda^2&=-\frac{k}{m}\\ \lambda&=\pm i\sqrt{\frac{k}{m}} \end{aligned}$

$\begin{cases} x_1(t)=c_1e^{i\sqrt{\frac{k}{m}} t}\\ x_2(t)=c_2e^{-i\sqrt{\frac{k}{m}} t} \end{cases}$

$\begin{aligned} x(t)&=x_1(t)+x_2(t)\\ &=c_1e^{i\sqrt{\frac{k}{m}} t}+c_2e^{-i\sqrt{\frac{k}{m}} t}\\ &=c_1\left[\cos\left(\sqrt{\frac{k}{m}}t\right)+i\sin\left(\sqrt{\frac{k}{m}}t\right)\right]+c_2\left[\cos\left(\sqrt{\frac{k}{m}}t\right)-i\sin\left(\sqrt{\frac{k}{m}}t\right)\right]\\ &=c_1\cos\left(\sqrt{\frac{k}{m}}t\right)+c_1i\sin\left(\sqrt{\frac{k}{m}}t\right)+c_2\cos\left(\sqrt{\frac{k}{m}}t\right)-c_2i\sin\left(\sqrt{\frac{k}{m}}t\right)\\ &=(c_1+c_2)\cos\left(\sqrt{\frac{k}{m}}t\right)+(c_1-c_2)i\sin\left(\sqrt{\frac{k}{m}}t\right)\\ &=C_1\cos\left(\sqrt{\frac{k}{m}}t\right)+C_2\sin\left(\sqrt{\frac{k}{m}}t\right)\\ &=\sqrt{C_1+C_2}\cos\left[\sqrt{\frac{k}{m}}t-\tan^{-1}\left(\frac{C_2}{C_1}\right)\right]\\ \end{aligned}$

Let $A=\sqrt{C_1+C_2}$, $\omega^2=\frac{k}{m}$ and $\tan(\varphi)=\frac{C_2}{C_1}$:

$\boxed{x(t)=A\cos\left(\omega t-\varphi\right)}$

Proof by differentiation:

$\begin{aligned} x(t)&=A\cos\left(\omega t-\varphi\right)\\ \frac{dx(t)}{dt}&=-A\omega\sin\left(\omega t-\varphi\right)\\ \frac{d^2x(t)}{dt^2}&=-A\omega^2\cos\left(\omega t-\varphi\right)\\ \frac{d^2x(t)}{dt^2}&=-\omega^2[A\cos\left(\omega t-\varphi\right)]\\ \frac{d^2x(t)}{dt^2}&=-\frac{k}{m}x(t)\\ \frac{d^2x(t)}{dt^2}+\frac{kx(t)}{m}&=0\quad\blacksquare \end{aligned}$