solving simple harmonic oscillator

Question

I am not understanding the general solution of the simple harmonic oscillator.

How can it get from here:

$\frac{\mathrm{d}^2x }{\mathrm{d} t^2}=-\omega _{0}^2x$

to here:

$x(t)=x_{0}cos(\omega_{0}t) +\frac{v_{0}}{\omega_{0}}sin(\omega_{0}t)$

I do understand that: $x(t)=Asin(\omega_{0}t-\delta)$ and $x(t)=Acos(\omega_{0}t-\phi)$ are solutions. And in this case, i can summ all the possible solutions to have the general one. But i didn't undersand where $x_{0}$ and $\frac{v_{0}}{\omega_{0}}$ came from.

Answer 1

Consider the Cauchy problem $$\begin{cases}x''=-\omega^2 x \\x(0)=x_0 \\\dot{x}(0)=v_0 \end{cases}$$ The first equation is linear and has characteristic equation $$\lambda^2=-\omega_0^2$$ with solutions $\lambda_{1,2}=\pm i\omega_0$. This shows that the general solution is $$x(t)=A\cos(\omega t)+B \sin (\omega t)$$ for some $A,B\in \mathbb R$. Imposing the conditions of the Cauchy problem yields $$\begin{cases}x(0)=A \cos(0)+B \sin (0)=A=x_0 \\ \dot x(0)=-\omega A \sin(0)+\omega B \cos(0)=\omega B=v_0 \end{cases}$$ so $A=x_0$, $B=\frac{v_0}{\omega}$. Hence, $$x(t)=x_0 \cos (\omega t)+\frac{v_0}{\omega}\sin(\omega t)$$

Answer 2

To solve:

$$x''(t)=-\omega_0^2x(t)$$

Use Laplace transform:

$$s^2\text{X}(s)-sx(0)-x'(0)=-\omega_0^2\text{X}(s)$$

Solving this for $\text{X}(s)$ gives us:

$$\text{X}(s)=\frac{sx(0)+x'(0)}{s^2+\omega_0^2}$$

Now, usig inverse Laplace transform, we get:

$$x(t)=x(0)\cos(\omega_0t)+\frac{x'(0)\sin(\omega_0t)}{\omega_0}$$

Answer 3

$$\frac{d^2 x}{dt^2} + \omega_0^2x = 0$$

$$x(t)=e^{\lambda t} \implies e^{\lambda t}(\lambda^2 + \omega_o^2)=0\implies \lambda = \pm \omega_0 i$$

$$x(t) = A e^{\omega_0 i t} + B e^{-\omega_0 it}$$

$$e^{it} = \cos(t) + i\sin(t)$$

$$x(t) = A (\cos(\omega_0 t) + i\sin(\omega_0t)) + B (\cos(-\omega_0 t) + i\sin(-\omega_0t))$$

$$\cos(t) = \cos(-t) \quad\quad \sin (t) = - \sin (-t)$$

$$x(t) = A (\cos(\omega_0 t) + i\sin(\omega_0t)) + B (\cos(\omega_0 t) - i\sin(\omega_0t))$$

$$x(t) = (A+B) \cos \omega_0 t + (Ai - Bi) \sin \omega_0 t$$

$$x(t) = c_1 \cos \omega_0 t + c_2 \sin \omega_0 t$$

$$x(0) = x_0 \quad \quad x'(0) = v_0$$

$$x(0) = c_1 \cos 0 + c_2 \sin 0 \implies c_1 = x_0$$

$$x'(0) = -x_0 \omega_0 \sin 0 + c_2 \omega_0 \cos 0 \implies c_2 \omega_0 = v_0$$

$$x(t) = x_0 \cos \omega_0 t + \frac{v_0}{\omega_0} \sin \omega_0 t$$