\subsection{Hooke's Law and Spring Models}

This subsection uses the local displacement-from-equilibrium viewpoint for spring forces. That viewpoint makes the restoring nature of the force and the modeling of combined springs especially clear.

\dfn{Spring constant and displacement from equilibrium}{Let an ideal spring act along a line with positive direction given by the unit vector $\hat{u}$. Let $x$ denote the signed displacement of the attached object from the spring's equilibrium position, measured along that line, so $x>0$ means displacement in the $+\hat{u}$ direction and $x<0$ means displacement in the opposite direction.

Let $k>0$ denote the \emph{spring constant}. It measures the stiffness of the spring: a larger $k$ means a larger restoring force for the same displacement. The SI units of $k$ are $\mathrm{N/m}$.}

\thm{Hooke's law and equivalent spring constants}{Let $x$ denote the signed displacement from equilibrium for an ideal spring with spring constant $k$, measured along the spring's line of action with unit vector $\hat{u}$. Then the spring force on the attached object is
\[
\vec{F}_s=-kx\hat{u}.
\]
In one-dimensional scalar form,
\[
F_s=-kx.
\]
The negative sign means the spring force opposes the displacement from equilibrium.

If several ideal springs are modeled by one equivalent spring with constant $k_{\mathrm{eq}}$, then:
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item \textbf{Parallel:} if all springs undergo the same displacement $x$, then
\[
k_{\mathrm{eq}}=k_1+k_2+\cdots+k_n.
\]

\item \textbf{Series:} if all springs carry the same force magnitude, then
\[
\frac{1}{k_{\mathrm{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}.
\]
\end{enumerate}}

\pf{Why the sign and combination rules are correct}{If $x>0$, the object is displaced in the $+\hat{u}$ direction, so the restoring spring force must point in the $-\hat{u}$ direction. If $x<0$, the restoring force must point in the $+\hat{u}$ direction. Both cases are captured by the single vector formula
\[
\vec{F}_s=-kx\hat{u}.
\]

For springs in parallel, each spring has the same displacement $x$, so the forces add:
\[
\vec{F}_{\mathrm{tot}}=-(k_1+k_2+\cdots+k_n)x\hat{u}.
\]
Therefore $k_{\mathrm{eq}}=k_1+k_2+\cdots+k_n$.

For springs in series, let $F$ denote the common force magnitude through each spring, and let $x_i$ denote the magnitude of the displacement of spring $i$. Since $F=k_ix_i$, we have
\[
x_i=\frac{F}{k_i}.
\]
The total displacement magnitude is then
\[
x=x_1+x_2+\cdots+x_n=F\left(\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}\right).
\]
If the combination is replaced by one equivalent spring, then $F=k_{\mathrm{eq}}x$, so
\[
\frac{1}{k_{\mathrm{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}.
\]}

\cor{Vertical equilibrium shift and the local spring equation}{Let a mass $m$ hang from a vertical spring of spring constant $k$. Choose downward as positive. Let $y$ denote the downward displacement from the spring's unstretched length, let $y_{\mathrm{eq}}$ denote the equilibrium value of $y$, and let
\[
x=y-y_{\mathrm{eq}}
\]
denote the displacement from equilibrium.

At static equilibrium,
\[
mg-ky_{\mathrm{eq}}=0,
\]
so
\[
y_{\mathrm{eq}}=\frac{mg}{k}.
\]
Then the net force on the mass is
\[
F_{\mathrm{net}}=mg-ky=mg-k(y_{\mathrm{eq}}+x)=-kx.
\]
Thus, when motion is measured from equilibrium, gravity has already been accounted for, and the net force again has Hooke form.}

\qs{Worked example}{A block of mass $m=0.50\,\mathrm{kg}$ hangs at rest from a vertical spring with spring constant $k=200\,\mathrm{N/m}$. Choose downward as positive. Let $y$ denote the block's downward displacement from the spring's unstretched length, let $y_{\mathrm{eq}}$ denote the equilibrium displacement, and let $x=y-y_{\mathrm{eq}}$ denote the displacement from equilibrium.

\begin{enumerate}[label=(\alph*)]
\item Find $y_{\mathrm{eq}}$.
\item The block is pulled downward $0.030\,\mathrm{m}$ from equilibrium and released from rest. Find the spring force and the net force at the instant of release.
\item Find the acceleration at the instant of release.
\end{enumerate}}

\sol At equilibrium, the acceleration is zero, so the net force is zero:
\[
mg-ky_{\mathrm{eq}}=0.
\]
Therefore,
\[
y_{\mathrm{eq}}=\frac{mg}{k}=\frac{(0.50\,\mathrm{kg})(9.8\,\mathrm{m/s^2})}{200\,\mathrm{N/m}}=0.0245\,\mathrm{m}.
\]
So the equilibrium position is
\[
2.45\times 10^{-2}\,\mathrm{m}
\]
below the unstretched length.

At the instant of release, the block is displaced
\[
x=+0.030\,\mathrm{m}
\]
from equilibrium, because downward is positive. The actual spring displacement from the unstretched length is
\[
y=y_{\mathrm{eq}}+x=0.0245\,\mathrm{m}+0.030\,\mathrm{m}=0.0545\,\mathrm{m}.
\]

The spring force along the chosen vertical axis is
\[
F_s=-ky=-(200\,\mathrm{N/m})(0.0545\,\mathrm{m})=-10.9\,\mathrm{N}.
\]
The negative sign means the spring force is upward. Its magnitude is therefore
\[
10.9\,\mathrm{N}.
\]

The weight is
\[
mg=(0.50\,\mathrm{kg})(9.8\,\mathrm{m/s^2})=4.90\,\mathrm{N}
\]
in the positive direction. Hence the net force is
\[
F_{\mathrm{net}}=mg-ky=4.90\,\mathrm{N}-10.9\,\mathrm{N}=-6.0\,\mathrm{N}.
\]
Equivalently, using the local displacement-from-equilibrium form,
\[
F_{\mathrm{net}}=-kx=-(200\,\mathrm{N/m})(0.030\,\mathrm{m})=-6.0\,\mathrm{N},
\]
which agrees.

Now apply Newton's second law:
\[
F_{\mathrm{net}}=ma.
\]
Thus,
\[
a=\frac{F_{\mathrm{net}}}{m}=\frac{-6.0\,\mathrm{N}}{0.50\,\mathrm{kg}}=-12\,\mathrm{m/s^2}.
\]
The negative sign means the acceleration is upward, so the block's acceleration at release has magnitude
\[
12\,\mathrm{m/s^2}.
\]

Therefore,
\[
y_{\mathrm{eq}}=0.0245\,\mathrm{m},
\]
the spring force at release is
\[
10.9\,\mathrm{N}
\]
upward, the net force is
\[
6.0\,\mathrm{N}
\]
upward, and the acceleration is
\[
12\,\mathrm{m/s^2}
\]
upward.
