What is the correct way to integrate this equation: $m{\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}} = -kx$

Question

After some hours without a solution I came here to ask how to integrate directly this equation:

$$ m{\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}} = -kx $$

The conditions are: $$ x(0) = 0 $$

and

$$ \frac{\mathrm {d}x}{\mathrm {d}t}\Big|_{t=0} = 0 $$

This is about physics, but I what I can not handle is the mathematical part, so I came to ask here.

Would you help me?

Answer 1

Let's multiply both sides by $x'$, then

$$x'(t)x''(t) = -\frac{k}{m} x(t)x'(t).$$

This can be realized as

$$ \frac{1}{2}\frac{d}{dt} (x'(t))^2 = -\frac{k}{2m} \frac{d}{dt}(x(t))^2. $$

Integrating, using your initial conditions, and canceling the factor of $\frac{1}{2}$ this becomes

$$ (x'(t))^2 = -\frac{k}{m} (x(t))^2. $$

Moving the $(x(t))^2$ to the other side, we get

$$\left(\frac{x'(t)}{x(t)}\right)^2 = -\frac{k}{m}.$$

Square rooting,

$$\frac{x'(t)}{x(t)} = \pm i\sqrt{\frac{k}{m}}.$$

The left hand side can be recognized as $\frac{d}{dx} \log(x(t))$ so we can rewrite this as

$$ \frac{d}{dx} \log(x(t)) = \pm i \sqrt{\frac{k}{m}}.$$

Integrating both sides, then exponentiating we get

$$ x(t) = A\exp\left(\pm i \sqrt{\frac{k}{m}}\right).$$

There are details left out ensuring that things are well-defined in places which you are more than welcome to work out. This is meant as a guide, not a full foolproof proof.

Answer 2

hint

Put $\omega^2=\frac {k}{m}$.

observe that

$\sin (\omega t) $ and $\cos (\omega t) $ are independent solutions.

Answer 3

The differential equation: $$ ax'' + bx'+ cx = 0 $$ is solved first by solving its characteristic equation: $$ ar^2 + br + c = 0$$ which (most often in these problems) produced two roots $r_1, r_2$.

The general solution to the differential equation is then: $$ x(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} $$ The initial conditions for $x(0)$ and $x'(0)$ can be used to solve for the constants $c_1$ and $c_2$ corresponding to the particular solution with the given initial conditions.

Going back to your problem, the characteristic equations has two imaginary roots: $$ r = \pm \sqrt{\frac{k}{m}} \, i$$ By using complex exponentials, we can show that, when we have two imaginary roots to the characteristic equation $\pm i\omega$, the general solution to the differential equation is: $$ x(t) = c_1 \cos \omega t + c_2 \sin \omega t $$ which in this case reduces to: $$ x(t) = c_1 \cos \sqrt{\frac{k}{m}} t + c_2 \sin \sqrt{\frac{k}{m}} t $$ Often in physics, this is usually rewritten using an amplitude and phase on a single sinusoid, rather than two constants for a linear combination of sinusoids. The above form is easier for solving for the particular solution given the initial conditions.

The differential equation you have provided describes the periodic motion of a mass $m$ attached to a spring with spring constant $k$. Solving the differential equation shows that motion is periodic with angular frequency $\omega = \sqrt{\frac{k}{m}}$, or actual frequency $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$.

However, the initial conditions you have provided means that you start with the mass at rest position ($x(0) = 0$) and with no initial velocity ($x'(0) = 0$). So the particular solution is $x(t) = 0$ because the mass is not moving and will not be moved. (You can verify mathematically by solving for $c_1, c_2$).