Restoring Force and One-Dimensional Simple Harmonic Oscillator
Simple Harmonic Motion1
Let’s consider the motion of an object hanging on a spring. It oscillates back and forth due to the restoring force of the spring. Such motion is called a harmonic oscillation. The functions representing harmonic oscillation, $\sin$ and $\cos$, were called harmonic functions a long time ago, which is why the motion is designated as such. Among harmonic oscillations, those with no friction or other external forces, moving solely by the spring’s restoring force, are called simple harmonic oscillations. Let’s first find out how the restoring force by the spring is expressed. Let $V(x)$ be the potential energy of a 1-dimensional simple harmonic oscillator. And let’s assume it is expressed as an infinite sum of polynomials, a series. Then it can be expressed as follows.
$$ V(x)=a_{0} + a_{1}x + a_2x^2+ a_{3}x^3 + \cdots $$
However, since only the difference between two potentials has physical meaning, it is irrelevant to consider the constant term as $0$. It’s akin to setting the equilibrium point at $0$. Furthermore, since $-\dfrac{dV}{dx}=F(x)$ and the restoring force is $F(0)=0$, $V^\prime(0)=0$ results. That is, the coefficient of the first order term must be $0$. Therefore, the potential of the restoring force is as follows.
$$ V(x)=a_2x^2+a_{3}x^3+\cdots $$
When $x$ is sufficiently small, terms of third order and above can be neglected. Ultimately, the restoring force can be determined as follows.
$$ F(x)=-\dfrac{dV}{dx}=-2a_2x=-kx \ \ (k=2a_2) $$
At this time, $k$ is called the modulus of elasticity or the spring constant. And $F(x)=-kx$ is called Hooke’s law2. Now let’s solve the equation of motion related to the restoring force. $F=ma=m\ddot{x}$ and the restoring force is $F=-kx$, so the following equation is obtained.
$$ \begin{align*} && m \ddot{x} & =-kx \\ \implies && m \ddot{x} + kx &= 0 \\ \implies && \ddot{x} + \dfrac{k}{m}x &= 0 \end{align*} $$
Here, let’s substitute ${\omega_{0}}^2 \equiv \dfrac{k}{m}$. The reason for squaring is to simplify the form of the final equation. $\omega_{0}$ is referred to as the angular frequency of the system. It is also called the natural angular frequency or natural frequency to distinguish from systems in damped and forced oscillation. Now the equation of motion is as follows.
$$ \ddot{x} + {\omega_{0}}^2x=0 $$
Solving a 2nd Order Differential Equation with Negative Coefficients
The solution to a 2nd order differential equation $$ \dfrac{d^{2}X}{dx^{2}} = -\alpha^{2}X $$ is as follows.
$$ X(x) = Ae^{i\alpha x} + B e^{-i \alpha x} $$
Thus, the following solution is obtained.
$$ \begin{align*} x(t) &=A_{1}e^{i\omega_{0} t}+A_2e^{-i\omega_{0} t} \\ &=A_{3}\cos \omega_{0} t+ A_{4}\sin \omega_{0} t \\ &=A \cos (\omega_{0} t + \phi) \end{align*} $$
Here, $A_{1}$, $A_{2}$, $A_{3}$, $A_{4}$, $A$ are arbitrary complex or real constants, respectively. Usually, it is often represented in the form of cosine or sine functions, like the third equation.