Abstract. ;-) In this post, I assume my reader to be a confused intellect who swallowed the minus sign of Hooke's law ages ago. Bonus is the discussion of constructing the work-energy-potential things in section 1. Hookes law visits us in section 2.
1 Force and the happy terminologies
Consider a particle moving in one dimension, $x$. The force acting on it can depend on position, time, velocity etc. Let's first consider that we do not pay attention to the dependency of the force. So In this slightly general manner,
\begin{eqnarray}
\int F dx &=& \int _{x_1}^{x_2} ma dx = m \int _{x_1}^{x_2} \frac{d}{dt} \frac{dx}{dt} dx = m \int _{x_1}^{x_2} v dv = \frac{1}{2} m[v^2] _{x_1}^{x_2} \nonumber \\
\therefore \int F dx &=& \frac{1}{2} m v^2_2 - \frac{1}{2} m v_1^2
\label{eq:workEnergyTheorem}
\end{eqnarray}
We call equation \eqref{eq:workEnergyTheorem} as the Work-Energy theorem, where probably just for the sake of persuading ourselves that we understand the subject, we call the LHS by the name: work and the RHS as the change of kinetic energy.
Now, let's take a constant force $F$ which doesn't depend on position, time or velocity. Newton declares,
\begin{eqnarray}
F = ma
\label{eq:f=ma}
\end{eqnarray}
Integrating this force w.r.t position gives rise to another quantity which we convince to be called as the change in potential energy.
\begin{eqnarray}
\int F dx = \int_{x_1}^{x_2} ma \ dx = ma\int_{x_1}^{x_2} dx = ma [x]_{x_1}^{x_2} = max_2-max_1
\label{eq:potentialInConstantForce}
\end{eqnarray}
Equating equation \eqref{eq:workEnergyTheorem} and \eqref{eq:potentialInConstantForce},
\begin{eqnarray}
\frac{1}{2} m v^2_1 + max_1 &=& \frac{1}{2} m v^2_2 + max_2
\nonumber \\
&=& \ldots = \frac{1}{2} m v^2_n + mgx_n = constant
\label{eq:ConservationOfEnergyForF}
\end{eqnarray}
We play another name-game and call this constant as mechanical energy.
Now, we ask whether this integrating-over-position machinary works if the force depends on position. Let's take a leap of faith and assume that, the work-energy theorem will be valid at least for the position-dependence of the force $F(x)$. To rescue the potential, let's slightly broaden our imagination and think as the following: Since the difference of the potential energy was an integral of force over position, there must be a function of position whose derivative is force:
\begin{eqnarray}
\frac{dU(x)}{dx} = F(x)
\end{eqnarray}
Let's proceed by integrating:
\begin{eqnarray}
\int F(x) \ dx = \int _{x_1}^{x_2} \frac{dU(x)}{dx} dx = U(x_2) - U(x_2) =: U_2 - U_1
\label{eq:PotentialNively}
\end{eqnarray}
Remember that we have assumed the validity of the work-energy theorem,
\begin{eqnarray}
\int F(x)\ dx &=& \frac{1}{2} m v^2_2 - \frac{1}{2} m v_1^2
\label{eq:workEnergyTheoremForFx}
\end{eqnarray}
Equation \eqref{eq:PotentialNively} and \eqref{eq:workEnergyTheoremForFx} gives:
\begin{eqnarray}
\frac{1}{2} m v^2_1 - U_1 = \frac{1}{2} m v^2_2 - U_2
\end{eqnarray}
And we formally say, oops! There is an embarrassing minus. Let's glue it with $U$ in the first place and define the potential function:
\begin{eqnarray}
V(x) &=&: -U(x) \nonumber \\
F(x) &=& - \frac{d(V(x))}{dx} \
\label{eq:potentialFunction}
\end{eqnarray}
and thus we make our energy conservation law surviving in our thoughts:
\begin{eqnarray}
\frac{1}{2} m v^2_1 + V_1 = \frac{1}{2} m v^2_2 + V_2 = \ldots = constant
\end{eqnarray}
2 Making sense of Hooke's law
Let's consider a particle's motion to be around a stable equilibrium. Its position is described by $x$ w.r.t the origin $O$. Clearly the existence of the stable equilibrium manifests that, at the equilibrium point, the potential energy of the particle is minimum. If the equilibrium exists at $x=x_0$, then $V'(x)= 0$ (prime means the $x$-derivatives only).
Then the displacement from the equilibrium point, as shown in the figure, is $(x-x_0)$. Let's make a coordinate transformation such that we can analyze this motion in terms of the displacement $(x-x_0)$. Natural choice would be:
\begin{eqnarray}
X = x-x_0
\end{eqnarray}
 |
| Fig: A happy footballtron under a stable equilibrium. |
Back in the old cooridinate, the potential was $V(x)$. Let's Taylor-expand it near the point $x=x_0$.
\begin{eqnarray}
V(x) &=& V(x_0) + V'(x_0) (x-x_0) + \frac{1}{2!} V''(x_0) (x-x_0)^2 + \ldots \\
\implies V(x) &=& V(x_0) + V'(x_0) X + \frac{1}{2} V''(x_0) X^2 + \mathcal{O}(X^3) \nonumber \\
\implies \frac{dV}{dX} &\approx& 0 + 0 + V''(x_0)\ X
\nonumber \\
\implies - \frac{dV}{dX} &=& - V''(x_0) X \nonumber \\
\implies F(x) &=& - V''(x_0) X
\label{eq:TaylorExpnd}
\end{eqnarray}
Of course we assumed the displacement to be sufficiently small, and then, who cares about the higher order terms of taylor expansion? (special smile for our friend Advait)
However, in the RHS of \eqref{eq:TaylorExpnd}, the constant $V''(x_0)$ is really a positive constant near the equilibrium, because $V(x)$ is minimized at $x=x_0$. Notice that:
Of course we assumed the displacement to be sufficiently small, and then, who cares about the higher order terms of taylor expansion? (special smile for our friend Advait)
However, in the RHS of \eqref{eq:TaylorExpnd}, the constant $V''(x_0)$ is really a positive constant near the equilibrium, because $V(x)$ is minimized at $x=x_0$. Notice that:
- $X>0 \implies x>x_0$, i.e, a positive displacement yields a negative force,
- $X<0 \implies x<x_0$, i.e, a negative displacement yields a positive force by canceling the minus in \eqref{eq:TaylorExpnd}.
Let's choose $V''(x) = k$ and celebrate equation \eqref{eq:TaylorExpnd} as Hooke's law happily ever after with the mysterious minus sign which previously managed our embarrassment concerning the conservation of energy.
: -)
