Finding $x(t)$ for the mechanical system $x'' = -x$

Question

Using energy conservation theorem and method of integration by quadrature, find $x(t)$ for the mechanical system $x'' = -x$, considering a spring with mass $1$ and elasticity constant $1$.

My attempt:

$E = \frac12 \dot{x}^{2} + U(x) = k$, then $\dot{x} = \sqrt{2(E-U(x)}$

Suppose we know $\displaystyle g(x) = \int \frac{1}{\sqrt{2(E-U(x)}} dx$, then, by chain rule:

$\displaystyle\frac{d}{dt} g(x(t)) = \frac{1}{\sqrt{2(E-U(x)}} \frac{dx}{dt}=1$

Then $g(x(t)) = t+c$

Is this correct? Thanks

EDIT I know how to solve this differential equation, I want to know specifically how to solve it with energy conservation and quadrature, because this is an exercise of my Mathematica-Physics/Classical Mechanics course.

Answer 1

The potential energy of the spring is $\frac{1}{2}x^2$ so,

$$\frac{1}{2}x^2+\frac{1}{2} \left(\frac{dx}{dt} \right)^2=E$$

It follows,

$$(\frac{dx}{dt})^2=2E-x^2$$

Supposing that $\frac{dx}{dt} \geq 0$ (ie spring is moving in positive direction), we get:

$$\frac{dx}{dt}=\sqrt{2E-x^2}$$

$$dt=\frac{dx}{\sqrt{2E-x^2}}$$

$$t+c=\arcsin(\frac{x}{\sqrt{2E}})$$

Answer 2

$$x(t) = A \cos (2 \pi \omega t + \phi)$$

where $\omega = 1$.

Answer 3

you could say

$x'' = kx\\ x = C_1 e^{(\sqrt k)t} + C_2 e^{-(\sqrt k)t}$

$k = -1\\ \sqrt k = i\\ x = C_1 e^{it} + C_2 e^{-it}\\ e^{it} = \cos t+ i\sin t\\ x = A\cos t + B\sin t$

Or you could say

let $v = x', v' =x'' = -x$

Giving us the system

$v' = -x\\ x' = v$

or

$\begin{bmatrix} x\\v \end{bmatrix}' = \begin{bmatrix} 0&1\\-1&0\end{bmatrix}\begin {bmatrix} x\\v \end{bmatrix}$

$\mathbf x' = A\mathbf x\\ \mathbf x = e^{At}\mathbf x_0$

and $e^{At} = \sum_{n=0}^{\infty} \frac {(At)^n}{n!}$

$A^2 = -I\\ A^3 = -A\\ A^4 = I$

$e^{At}=\sum_{n=0}^{\infty} \frac {(-1)^n(It)^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac {(-1)^n(At)^{2n+1}}{(2n+1)!} = \begin{bmatrix} \cos t&\sin t\\-\sin t&\cos t\end{bmatrix}$

$\begin{bmatrix} x\\v \end{bmatrix} = \begin{bmatrix} x_0\cos t + v_0\sin t\\-x_0 \sin t + v_0\cos t \end{bmatrix}$

finally

$v = \frac{dx}{dt}\\ x'' = \frac {dv}{dt}$

By the chain rule:

$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx} v$

$\frac{dv}{dx} v = -x$

This is a seprerable diff eq

$\frac 12 v^2 = -\frac12 x^2 + C\\ v = \sqrt {C-x^2}\\ x' =\sqrt {C-x^2}$

Which is also a separable diff eq

$\int \frac {1}{\sqrt{C-x^2}} \ dx = \int dt$

$\arcsin{\frac{x}{\sqrt C}} = t + \phi\\ x = \sqrt C \sin (t+\phi)$

Answer 4

HINT

Suppose $x(t) = e^{kt}$, where $k\in\mathbb{C}$: \begin{align*} x^{\prime\prime} + x = 0 \Longleftrightarrow k^{2}e^{kt} + e^{kt} = 0 \Longleftrightarrow k^{2} + 1 = 0 \Longleftrightarrow k = \pm i \end{align*} Therefore $x(t) = Ae^{it} + Be^{-it}$. Can you proceed from here?