Why is simple harmonic motion considered to be a sinusoidal function? [closed]

Question

Motion of a pendulum is said to be an harmonic motion. What do we exactly mean by harmonic and why is that the displacement and velocity etc. Represented by a sine function?

Answer 1

The harmonic oscillator

The harmonic oscillator is a system described by the following equation:

$$\vec{F}=-k\vec{x},$$

where $\vec{x}$ is the displacement from the equilibrium position and $\vec{F}$ is the force that opposes this displacement. If writing Newton's second law for a system gives such an equation, then the force $\vec{F}$ is said to be elastic. By going a step further and expressing $\vec{F}$ in terms of the second derivative of $x$, we encounter a second order linear differential equation ($\hat{x}$ is the unit vector of the Ox axis):

$$\vec{F}=m\ddot{x}\hat{x}=-kx\hat{x}\iff \ddot x=-\frac{k}{m}x=-\omega^2x,\text{ where we use }\omega=\sqrt{k/m}.$$

Solving the differential equation of the harmonic oscillator

We are aiming to solve $$\ddot x=-\frac{k}{m}x=-\omega^2x.$$ Recall that the number $e$ has the nice property that $\frac{d}{dx}e^x=e^x$, and if we multiply the exponent by a constant, we get $\frac{d}{dx}e^{ax}=ae^{ax}$. This may sound silly, but notice that this is exactly what we need! Let $x=Ce^{\alpha t}$, for instance. Plugging this into our original equation, we get $$C\alpha^2e^{\alpha t}=-\omega^2Ce^{\alpha t}\iff \alpha^2=-\omega^2\iff a=\pm i\omega$$ We have found two solutions of our differential equation, namely $C_1e^{i\omega t}$ and $C_2e^{-i\omega t}$. The most general solution is a sum of these two, and therefore

$$x(t)=C_1e^{i\omega t}+C_2e^{-i\omega t}.$$

It is easy to notice that the sum of two solutions will also be a solution to a differential equation. And don't worry, there won't be "other" solutions, in the sense that a linear differential equation ($x$ and its derivatives only show up with exponent $1$) of order $n$ ($n$ is the maximum order of a derivative that appears) has exactly $n$ linearly independent solutions (i.e. that can't be expressed as linear combinations of each other).

What does this all have to do with sines and cosines?

According to Euler's formula, $e^{i\alpha}=\cos\alpha+i\sin\alpha$, so the "expanded" solution becomes:

$$\begin{align}\tilde{x}(t)&=C_1\cos(\omega t)+iC_1\sin(\omega t)+C_2\cos(-\omega t)+iC_2\sin(-i\omega t)\\&=(C_1+C_2)\cos(\omega t)+i(C_1-C_2)\sin(\omega t)\end{align}.$$

I've put a tilde on top of $x$ because the $x$ we are working with describes the position of a body, so it must be real, not complex. Therefore, we actually use either $x=\Re(\tilde{x})$ or $x=\Im(\tilde{x})$, the real part or the imaginary part. Let's choose $\sin$ this time:

$$x(t)=(C_1-C_2)\sin(\omega t)=A\sin(\omega t) \text{ (where }A\text{ is called amplitude})$$

And this is why harmonic oscillators are described by sinusoidal functions. We can also introduce an initial phase ($\varphi$), by, say, using $ce^\varphi$ (where $c$ is real) instead of using $C$ (which is a complex number), and the equation becomes $x(t)=A\sin(\omega t+\varphi)$ (I didn't do this from the beginning in order to reduce clutter). The other constants involved, $A$ and $\varphi$ can only be deduced from initial conditions.

The pendulum

Consider the following pendulum diagram:

Let $\vec{g}$ be the gravitational acceleration and $\vec{g_r}$ and $\vec{g_\tau}$ be its components along the radial and tangential axes. Newton's second law states that $$m\vec{g_\tau}=m\vec{a}\implies \vec{a}=\vec{g_\tau}\implies a=-g\sin\theta \text{ (because it opposes the growth of }\theta).$$

This acceleration, $\vec{a}$, is related to the angular acceleration, $\vec{\varepsilon}$, by $a=\ell\varepsilon$, where $\ell$ is the length of the pendulum and $\varepsilon$ is the speed of the rate of change the angle $\theta$ (that is, $\varepsilon=d^2\theta/dt^2=\ddot\theta$). This leads us to the differential equation of the pendulum:

$$g\sin\theta=-\ell \ddot \theta$$

This equation is hard to solve for an arbitrary angle $\theta$. But we know that for small angles, $\sin\theta\approx\theta$ (when $x\ll 1 \text{ radian}$). Therefore, for small angles, the equation of the pendulum can be approximated by:

$$g\theta=-\ell \ddot \theta$$

And we arrive at what we just deduced, the differential equation of the harmonic oscillator, but in this case $\omega^2=g/l$ instead of $k/m$. Therefore:

$$\theta(t)=\alpha \sin\left(\sqrt{\frac{g}{l}}t+\varphi\right)$$

Angular velocity and velocity are just given by $\omega=\dot\theta=\alpha\sqrt{\frac{g}{l}}\cos\left(\sqrt{\frac{g}{l}}t+\varphi\right)$ ($\alpha$ is the angular amplitude) and $v=\ell \omega$. Remember that this is all valid only within the small oscillations approximation.

As an example, you can now state some initial conditions to solve for $\alpha$ and $\varphi$. Say, someone displaces the pendulum from its vertical equilibrium position by a small angle $\beta$ and lets it drop freely (so $v(0)=0$ and $\theta(0)=\beta$), then you get $\cos\varphi=1\implies \varphi=\pi/2$ and $\alpha=\beta$, which leads you to $\theta(t)=\beta\cos\left(\sqrt{\frac{g}{l}}t\right)$.

Answer 2

They result from physical systems with a peculiar type of restoring force: the acceleration is proportional to the displacement, but directed to the opposite side.

$x^{(2)}(t) = -k^2x(t)$

One way to examine what types of function have that property is testing as a solution a generic function expanded in a Taylor series. Depending on the boundary conditions it will be a sin, a cos, or combination of both:

$x(t) = x(0) + x^{(1)}(0)t + (1/2!)x^{(2)}(0)t^2 + (1/3!)x^{(3)}(0)t^3 + (1/4!)x^{(4)}(0)t^4 + (1/5!)x^{(5)}(0)t^5 + \dots$

For boundary conditions: $x(0) = 0\;$ and $\;x^{(1)}(0) = k$

$x^{(2)}(0) = -k^2x(0) = 0$
$x^{(3)}(0) = -k^2x^{(1)}(0) = -k^3$
$x^{(4)}(0) = -k^2x^{(2)}(0) = 0$
$x^{(5)}(0) = -k^2x^{(3)}(0) = k^5$

$x(t) = 0 + kt + 0 - (1/3!)k^3t^3 + 0 + (1/5!)k^5t^5 + \dots = \sin(kt)$

for boundary conditions: $x(0) = 1\;$ and $\;x^{(1)}(0) = 0$

$x^{(2)}(0) = -k^2x(0) = -k^2$
$x^{(3)}(0) = -k^2x^{(1)}(0) = 0$
$x^{(4)}(0) = -k^2x^{(2)}(0) = k^4$
$x^{(5)}(0) = -k^2x^{(3)}(0) = 0$

$x(t) = 1 + 0 - (1/2!)k^2t^2 + 0 + (1/4!)k^4t^4 + 0 + \dots = \cos(kt)$