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concepts/mechanics/u2/m2-6-springs.tex

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+\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.