# Taking the general solution for simple harmonic motion and turning it into the conventional equation? [closed]

I was reading a question on the general solution to a simple harmonic oscillator and I never realized I derived it that way myself, so I decided to try. I have some questions on the process. After my derivation, my question is two-fold (I've also decided to make it in a quote box for aesthetic purposes and clarity):

To find an expression for the position of an object undergoing simple harmonic motion, we start with Hooke's Law:

$$F = -kx$$

In which $k$ is the spring constant.

Now, we begin to solve the second-order ordinary differential equation:

$$x'' + \frac{k}{m}x = 0$$

$$e^{\lambda t}\Big(\lambda^2 + \frac{k}{m}\Big)= 0$$

Thus:

$$\bbox{ \lambda^2 = -\frac{k}{m} }$$

$$\lambda = \mp i\sqrt\frac{k}{m}$$

Giving us a general solution of:

$$C_1e^{\alpha t}\cos(\beta t) + C_2e^{\alpha t}\sin(\beta t)$$

Since $\alpha = 0$ and $\beta = \sqrt\frac{k}{m}$ , we have:

$$\bbox[5px,border:2px solid red] { x(t) = C_1\cos\bigg(\sqrt\frac{k}{m} t\bigg) + C_2\sin\bigg(\sqrt\frac{k}{m} t\Bigg) }$$

As a general solution. Hopefully I did that right.

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## 1. How do we get to $x(t) = x_0cos(\omega t + \phi)$ from my derivation?

• And what do we do with the constants to get there?
• I've heard by making $c_1 = x_0$ and $c_2 = v$ as boundary conditions, we can arrive to it, since if we make $x_0 = x_{max}$ and $v_0 = 0$ we arrive at that formula, more or less, but why can we just say "I'm making $c_1$ and $c_2$, which are constants, equal to variables"? Huh?

## 2. Why is $\omega = \sqrt\frac{k}{m}$?

• It just seems to be a frequency scalar. Why and how has it been defined as the angular velocity?

## closed as off-topic by Bill N, David Hammen, Jon Custer, peterh, Qmechanic♦Jun 20 '17 at 10:57

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• Note that $c_2$ must have dimensions of length and so $c_2$ cannot be equal to the initial velocity on dimensional grounds. – Alfred Centauri Jun 19 '17 at 20:37

Start with $$x(t)=c_1\cos(\omega t)+c_2\sin(\omega t) \tag{1}$$ and take the second derivative to get $$\ddot{x}=-\omega^2 \left(c_1\cos(\omega t)+c_2\sin(\omega t)\right)=-\omega^2 x.$$ Thus inserting (1) in the differential equation \begin{align} \ddot{x}+\frac{k}{m}x&=0\, \\ \left(-\omega^2 +\frac{k}{m}\right)x&=0\, . \end{align} Since this must hold for any $x$ it follows that $\omega^2=k/m$. Note that by expanding \begin{align} x(t)=x_0\cos(\omega t+\phi)&=x_0\left(\cos\omega t)\cos\phi-\sin(\omega t)\sin\phi\right)\, ,\\ &= x_0\cos\phi \cos\omega t-x_0\sin\phi\sin\omega t \tag{2} \end{align} you recover (1) with $c_1=x_0\cos\phi$ and $c_2=-x_0\sin\phi$.

• I see. Two questions with that: 1) Can you explain the first bit again, with how you got $-\omega^2 x+\frac{k}{m}x=0$? And 2), do we define $c_1$ and $c_2$ as such because they'll be a percentage, for lack of a better word, at best 100% of a maximum initial displacement based on $\phi$? – sangstar Jun 19 '17 at 18:30
• @sangstar I've added a bit. $c_1$ and $c_2$ are not "defined" in this way: this just follows from comparing (2) with (1). Indeed $c_1$ and $c_2$ are arbitrary constants determined by the initial conditions: $x(0)=c_1$ and $\dot{x}(0)=\omega c_2$. – ZeroTheHero Jun 19 '17 at 18:58
• @sangstar A way that always helped me to think about differential equations is that you are trying to find a solution for your original equation. For the case of simple harmonic oscillators, we need something where the second derivative of a quantity is equal to the quantity itself. You can then consider the cos and sin equations, because we know that $\frac {d^2}{dt^2} sin (t) = sin (t)$ (same with cos) so we can start from there and using the conditions we know we have, we can solve for what constants will make our $x(t)$ satisfy our $\frac {-d^2 x}{dt^2} = \frac km x$. – JMac Jun 19 '17 at 20:44
• @JMac indeed this is the best way to justify the ansatz $x=Ae^{\lambda t}$ since the exponential is self-reproducing under derivative. – ZeroTheHero Jun 19 '17 at 23:58

I'll try to answer both of your questions be showing how I would derive the general solution. As is typical for this type of differential equation, we assume a solution of the form

$$x(t) = Ae^{rt}$$

Then

$$\dot x(t) = rAe^{rt} = rx(t)$$

$$\ddot x(t) = r^2Ae^{rt} = r^2x(t)$$

Inserting into the differential equation yields

$$\ddot x(t) + \frac{k}{m}x(t) = \left(r^2 + \frac{k}{m}\right)x(t) = 0$$

So, our assumption that $Ae^{rt}$ is a solution is true for $r = \pm i\sqrt{\frac{k}{m}}$. The general solution is then the weighted sum of the two independent solutions

$$x(t) = A^+e^{+i\omega t} + A^-e^{-i\omega t},\qquad \omega \equiv \sqrt{\frac{k}{m}}$$

where we define the angular frequency $\omega$. Now, for $x(t)$ to be real, the coefficients must be complex conjugates, $A^- = (A^+)^*$, and so let

$$A^+ = \frac{\alpha}{2}e^{i\phi},\qquad A^- = \frac{\alpha}{2}e^{-i\phi}$$

for some $\alpha$ and $\phi$ that are real numbers and then we can rewrite the general solution as

$$x(t) = \frac{\alpha}{2}e^{i\phi}e^{+i\omega t} + \frac{\alpha}{2}e^{-i\phi}e^{-i\omega t} = \alpha\frac{e^{i(\omega t + \phi)} + e^{-i(\omega t + \phi)}}{2}= \alpha \cos(\omega t + \phi)$$

where we have used the identity $\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$. Setting $t = 0$ yields

$$x(0) = \alpha \cos\phi \rightarrow \alpha = \frac{x(0)}{\cos\phi}$$

Taking the time derivative and evaluating at $t = 0$ yields

$$\dot x(0) = -\omega \alpha \sin\phi = -\omega \tan\phi \; x(0) = \rightarrow \phi = -\tan^{-1}\left(\frac{\dot x(0) }{\omega x(0)}\right)$$

• +1 for your answer, Alfred. Few questions. 1) Why is it $\frac{\alpha}{2}$ for the constants, and 2) I'm confused by the signage of $A^-$ and $A^+$. What is that indicating? – sangstar Jun 20 '17 at 7:22
• @sangstar, (1) in anticipation of the factor of $\frac{1}{2}$ in the identity $\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and (2) I chose plus and minus superscripts to remind me that, e.g, $A^+$ goes with the $r = +\sqrt{\frac{k}{m}}$ solution. – Alfred Centauri Jun 20 '17 at 11:38