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So how to solve the differential equation $\:\frac{d^2x}{dt^2}=-\frac{k}{m}x$.?

I know how to solve normal differential equation such as:$\:\frac{dx}{dt}=x$, but the above is new to me. I suppose how can I integrate for $x$ when we have $\:\frac{d^2x}{dt^2}$

Note I am trying to derive the equation of simple harmonic motion.

https://en.wikipedia.org/wiki/Simple_harmonic_motion#:~:text=In%20mechanics%20and%20physics%2C%20simple,towards%20the%20object's%20equilibrium%20position.

CountDOOKU
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  • This may be of help https://www.khanacademy.org/math/differential-equations/second-order-differential-equations – Ralph Clausen Aug 23 '20 at 13:20
  • Let $x=r \cdot e^{i\theta}$. And then it is a simple equation for which you can solve for $\theta$. Also use the identity $\omega^2$ to make a simplification. – Popular Power Aug 23 '20 at 13:21
  • Multiply both sides by $x'$ and integrate to reduce the order.of the differential equation. – user577215664 Aug 23 '20 at 13:36
  • @Aryadeva Hello sir, thanks for commenting. Do you mind showing how its done, a sneak peak will do? I tried it, but my working seems off. Many thanks. – CountDOOKU Aug 23 '20 at 13:49
  • yes $x''x'+k/m x'x=0 $ $ \implies (x'^2)'+k/m (x^2)'=0 $ integrate Nina. $x'^2+k/mx^2=c_1$ Then it's a first order differential equation. It's separable....separate x and t and integrate again and you're done. – user577215664 Aug 23 '20 at 13:51
  • I added some line sin my answer Nina I hope it's more clear now – user577215664 Aug 23 '20 at 13:59
  • The thing I have done is in spirit the same thing. It is just a more special case to solve if any initial position is already given. – Popular Power Aug 23 '20 at 14:04

2 Answers2

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$$\:\frac{d^2x}{dt^2}=-\frac{k}{m}x$$ Is a second order linear differential equation. You need the characteristic polynomial. I suppose $\dfrac km >0$: $$r^2+\dfrac k m =0$$ Solve that equation then the solution is : $$x(t)=c_1e^{r_1t}+c_2e^{r_2t}$$ That you can write as: $$x(t)=c_1\cos{(r_1t)}+c_2\sin{(r_2t)}$$ where $r_1,r_2 (r_1 \ne r_2)$ are solutions of the characteristic polynomial.

You can aslo multiply by $x'$ both sides and reduce the order of the DE: $$x''x'+\frac km xx'=0$$ $$(x'^2)'+\frac{k}{m}(x^2)'=0$$ Integrate: $$x'^2+\frac{k}{m}x^2=c_1$$ it's a first order differential equation that is separable. $$\dfrac {dx}{dt}=\pm \sqrt {c_1-\frac{k}{m}x^2}$$ Separate and integrate again.

user577215664
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Given that $a=\dfrac{d^2x}{dt^2}=-\dfrac{k}{m}x$

Let $\frac{k}{m}=\omega^2$. Here $\omega$ is the angular frequency.

$$a=-\omega^2 x \implies \frac{dv}{dt}=-\omega^2x \implies \dfrac{dv}{dx} \cdot \dfrac{dx}{dt}=-\omega^2x \implies \int_{0}^{v} v \cdot dv=\int_{A}^{x}-\omega^2 x \cdot dx \\ \implies v= \pm \omega \cdot \sqrt{A^2-x^2}$$

Where $A$ is amplitude or the maximum displacement of the block or object from the initial position.

$$\dfrac{dx}{dt}=\pm \omega \sqrt{A^2-x^2} \implies \dfrac{dx}{\sqrt{A^2-x^2}}=\pm \omega \cdot dt$$

The rest is trivial. The above integral is a standard one. Or you can solve it using the substitution $x=A \sin u$

Popular Power
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