# Derivation of Simple Harmonic motion equation [closed]

I don't seem to be getting anywhere. The differential equation is $$\frac{d^2x}{dt^2}=- \omega ^2x.$$

So, $$\frac{1}{x}dx^2=- \omega ^2dt^2$$

I integrated this equation twice but I'm not getting the general solution $x=A(\sin{( \omega t+\phi)})$. Please help.

## closed as off-topic by JMac, ZeroTheHero, sammy gerbil, stafusa, GertDec 27 '17 at 22:39

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• Have you had a course in differential equations? – Chet Miller Dec 27 '17 at 15:53
• Try to think intuitively what this differential equation implies: $x$ is a solution where its second derivative is equal to itself (scaled by a factor $-w^2$). Which functions have this property? – nluigi Dec 27 '17 at 15:54
• @nluigi I can think of $e^{iwt}$ which is not the sinusoid solution given in my book. Also I'd like a way to derive it instead of guessing. – Dove Dec 27 '17 at 16:17
• $d^2 x\ne dx^2$. – ZeroTheHero Dec 27 '17 at 16:22
• @Dove I don't think there is any way to "derive" the solution to the differential equation: there is going to be guesswork, or more politely, experience at play at some stage of solving a differential equation in closed form, unless you have a first order separable or exact equation. I would put this is an answer, but I'm afraid I might be wrong and there actually are special cases besides separable and exact equations that don't involve trying known solutions at all. – Styg Dec 27 '17 at 16:29

## 4 Answers

You are misinterpreting the equation. It is

$$\frac{{\rm d}}{{\rm d}t}\left( \frac{{\rm d}x}{{\rm d}t}\right) = -\omega^2 \,x$$

so you cannot separate the variables. You can however use the substitution ${\rm d}x = v\, {\rm d}t$ together with the chain rule $\frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} \frac{{\rm d}x}{{\rm d}t} =\frac{{\rm d}v}{{\rm d}x} v$

$$\left. \frac{{\rm d}}{{\rm d}t} (v) = -\omega^2 \,x \right\} v\, \frac{{\rm d}v}{{\rm d}x} = -\omega^2 \,x$$

And now the separation of variables can take place, by moving the ${\rm d}x$ to the right-hand side

$$\int v\,{\rm d}v =\int (-\omega^2 \,x) {\rm d}x +\mathtt{C}$$

$\Rightarrow$

$$\left. \frac{v^2}{2} = \mathtt{C} - \frac{\omega^2 x^2}{2} \right\} v = \sqrt{2 \mathtt{C}-\omega^2 x^2}$$

Consider the initial condition $x=0$ and $v=v_0$ and use it to find $\mathtt{C}=\frac{v_0^2}{2}$.

$$v = \sqrt{v_0^2-\omega^2 x^2}$$

The next integration finds the time dependency

$$t = \int {\rm d}t = \int \frac{{\rm d}t}{{\rm d}x} {\rm d}x = \int \frac{1}{v}\,{\rm d}x =\int \frac{1}{\sqrt{v_0^2-\omega^2 x^2}}\,{\rm d}x$$

You can use an integration technique like substitution, or an integration table to find

$$\left. t =\frac{1}{\omega}\, \sin^{-1} \left( \frac{\omega x}{v_0} \right) \right\} x = \frac{v_0}{\omega} \sin(\omega t)$$

The general method for solving 2nd order equations requires you to make an ansatz (or a guess) as to the form of the function, and refine this guess so it matches the details of the equation and the boundary conditions.

The equation $$\ddot{x}(t)=-\omega^2 x(t) \tag{1}$$ implies that the second derivative is proportional to the function itself, and this proportionality factor is negative. There are two types of functions that do this: the exponentials of the for $C_\pm e^{\pm i\lambda t}$ and the trigonometric $A\sin(\lambda t+\phi)$ or $B\cos(\lambda t+\phi)$. Here, $\lambda$ is to be determined, as are $C_\pm, A, B$ and $\phi$.

Insert these in turn into (1) to find the connection between $\lambda$, the other constants and $\omega$. The general solution will be a sum of all those that fit the bill. Since this is a 2nd order equation, you ought to be able to manipulate your general solution so that only two unknown constants remain.

Separation of variables is not the way to go here. Instead, use the auxiliary equation method, so that you'll have $m^2 = - \omega ^2$, where $m$ is the number of derivatives. That will get you to the solution $A\sin( \omega t) + B\cos( \omega t)$.

As being mentioned in the comments $d^2x \neq dx^2$,
$d^2x$ itself doesn't mean anything, and $dx^2$ means $(\Delta x)^2$, where $\Delta x$ is a very small segment.

A hand-waving "proof":

$$\frac{d^2x}{dt^2} = \frac{d}{dt}[\frac{x(t+\Delta t)-x(t)}{\Delta t}]\approx \frac{x(t+2\Delta t)-2x(t+\Delta t)+x(t)}{(\Delta t)^2}$$

All the $x$-terms are all linear, hence there is no $(\Delta x)^2$

Now return to the ODE, a usually good way to solve a ODE is to guess.

We guess the solution is $e^{at}$, then plug in and find out the value of $a$:

$$\frac{d^2}{dt^2}e^{at}=-\omega^2 e^{at} \implies a^2=-\omega^2 \implies a=\pm i\omega$$

hence, the solutions are either $x=Ae^{i\omega t}$or $x=Be^{-i\omega t}$

where $i=\sqrt {-1}$, that pretty much just means oscillation occuring.

Now if we want a complete solution, then we have: $$x(t)=Ae^{i\omega t} + Be^{-i\omega t}$$

• "The best way to solve a ODE is to guess." I'll have to disagree on that one. I repressed a lot of what I learned in DE and the courses following; but I'm pretty sure if guessing were the ideal method we would spend a lot less time covering them. – JMac Dec 27 '17 at 19:39