Question

Prove the solutions to this simple harmonic oscillation a=w^2x

Answer 1

For this simple harmonic oscillation

$a=-\omega^2x$

$a=\frac{d^2x}{dt^2}$

$\frac{d^2x}{dt^2}=-\omega^2x$

Substitute $x(t)=e^{\lambda t}$

We have

$e^{\lambda t}(\omega^2+\lambda^2)=0$

$e^{\lambda t}\neq0$

$\lambda=i\omega$ or $\lambda=-i\omega$

$x_1(t)=C_1e^{-i\omega t}$ and $x_2(t)=C_2e^{i\omega t}$

The general solution is the sum of the above solutions

$x(t)=C_1e^{-i\omega t}+C_2e^{i\omega t}$

Using the Euler's identity we get

$x(t)=(C_1+C_2)\cos(\omega t)+i(-C_1+C_2)\sin(\omega t)$

In general

$x(t)=A_1\cos(\omega t)+A_2\sin(\omega t)$

or

$x(t)=A\cos(\omega t+\phi)$