13.5 Linearization Near Equilibrium Points

A Marble, a Bowl, and a Hill

A marble sits at the bottom of a bowl. Nudge it gently, and it rolls back and forth --- oscillating around the lowest point. The oscillation is smooth, rhythmic, and predictable.

Now place a bead at the very top of a hill. Nudge it even slightly, and it accelerates away. No oscillation. No return. It just leaves.

Both positions are equilibrium points --- places where the net force is zero and a particle at rest would stay at rest forever. But their responses to a small push could not be more different. One produces oscillation. The other produces escape.

[Video: Split screen. Left side: a marble in a curved bowl. A finger nudges it. It rocks back and forth, gradually settling back to the bottom. The potential energy curve $U(x)$ is overlaid, showing a minimum at the marble's rest position. Right side: a bead balanced on a smooth hill. A finger nudges it. It rolls away, accelerating. The potential energy curve is overlaid, showing a maximum at the bead's original position. Both curves are labeled at the equilibrium point with $F = -dU/dx = 0$.]

Why the difference? And for the case that does oscillate --- what determines the frequency? Does it depend on the shape of the bowl? Its depth? Its width? Something else entirely?

This section answers those questions. And the answer turns out to be one of the most powerful ideas in all of physics: near any stable equilibrium, every system behaves like a spring.

Prediction

Before you read on: Here is a potential energy curve $U(x)$ with a minimum. A particle sits at the bottom. You displace it slightly and release it.

The oscillation frequency will depend on some property of the curve near that minimum. Which property matters?

(a) The depth of the potential well (how far the minimum is below the surrounding energy)

(b) The width of the potential well (how far apart the walls are)

(c) The curvature of the potential at the minimum (how sharply it curves upward)

(d) Both the depth and the width together

Commit to your answer before continuing. Think about what you know from springs --- what property of a spring determines the oscillation frequency? Is there an analogy here?

The Guiding Question

How can complicated systems look simple when examined close to equilibrium?

You have already seen specific examples. A mass on a spring oscillates with $\omega = \sqrt{k/m}$. A pendulum at small angles oscillates with $\omega = \sqrt{g/L}$. In each case, you identified a restoring force proportional to displacement and derived the frequency from the proportionality constant.

But these feel like separate calculations for separate systems. The spring had its own derivation. The pendulum had its own. Is there a single principle that explains why both of them oscillate sinusoidally near equilibrium --- and that would also apply to a molecule vibrating in a chemical bond, or an atom trapped in a crystal lattice, or a satellite bobbing in a gravitational well?

There is. And it comes from looking at potential energy.

Exploration: Zooming In on a Minimum

[Interactive: Potential Energy Zoom. A smooth, general potential energy curve $U(x)$ is displayed. The curve has a single minimum (adjustable in position and shape via sliders that control the curve's asymmetry, steepness, and higher-order features --- the student can make the well narrow or wide, symmetric or lopsided, deep or shallow). A draggable zoom window surrounds the minimum. As the student zooms in (narrows the window), the portion of $U(x)$ visible in the zoom panel is rescaled to fill the display.

Three overlays are available as toggles:

Overlay 1: The best-fit parabola $U(x) \approx U_0 + \frac{1}{2}k_{\text{eff}}(x - x_0)^2$ is drawn in orange over the actual curve.

Overlay 2: A readout displays $k_{\text{eff}} = U''(x_0)$ and the resulting frequency $\omega = \sqrt{k_{\text{eff}}/m}$, with $m$ adjustable via a slider.

Overlay 3: A small particle is placed at a displaced position near the minimum. When "play" is pressed, the particle oscillates. Two motion traces appear side by side: one from the exact $U(x)$, one from the parabolic approximation. At small displacements, they are visually identical.

Guided prompts appear below:

Prompt 1: Zoom in as far as you can near the minimum. What shape does the curve approach?

Prompt 2: Toggle the parabolic overlay. How well does the parabola match the actual curve near the minimum? What happens to the match as you zoom out?

Prompt 3: Change the shape of the well using the sliders --- make it asymmetric, narrow, wide. Does the zoomed-in curve still look parabolic? Does $k_{\text{eff}}$ change?

Prompt 4: Now place the particle at a small displacement and press play. Compare the exact motion with the parabolic approximation. How far can you displace the particle before the two motions visibly disagree?]

Spend real time with this interactive. It illustrates the central idea of this section, and no amount of reading can substitute for watching it happen.

Here is what you should notice:

No matter what shape the well has --- symmetric or lopsided, narrow or wide --- when you zoom in close enough to the minimum, the curve looks like a parabola. Always.
The parabola that matches the curve near the minimum is not a guess. It is determined by a single number: the curvature of $U(x)$ at the minimum, which is $U''(x_0)$.
Different wells produce different curvatures, and therefore different effective spring constants. A narrow, steep well has a large $U''(x_0)$ and produces rapid oscillations. A broad, gentle well has a small $U''(x_0)$ and produces slow oscillations.
The depth of the well does not affect the frequency at all. You can make the well deeper or shallower without changing the curvature at the bottom. Depth determines how much total energy the particle can have while remaining trapped, but it says nothing about how fast it oscillates for small displacements.

Pause and think: Why does the depth of the well not matter for the oscillation frequency? Think about what "frequency" means physically. It depends on the restoring force for small displacements --- on how hard the curve pushes the particle back. A deep well with a flat bottom pushes gently. A shallow well with steep sides pushes hard. Depth and curvature are independent properties.

Concept Reveal: The Taylor Expansion Argument

Here is why every minimum looks like a parabola. The argument is short, elegant, and worth understanding deeply.

Setting up the expansion

Let $x_0$ be a stable equilibrium point --- a position where $U(x)$ has a minimum. We want to understand the motion of a particle near $x_0$, so we expand $U(x)$ in a Taylor series around $x_0$:

$$U(x) = U(x_0) + U'(x_0)(x - x_0) + \frac{1}{2}U''(x_0)(x - x_0)^2 + \frac{1}{6}U'''(x_0)(x - x_0)^3 + \cdots$$

Now we use what we know about equilibrium points.

The first derivative vanishes

At an equilibrium point, the force is zero. Since $F = -dU/dx$, this means:

$$U'(x_0) = 0$$

The linear term in the Taylor expansion drops out. This is what makes it an equilibrium point --- the slope of $U(x)$ is zero there. The potential has neither an uphill nor a downhill tendency.

For small displacements, the quadratic term dominates

With the linear term gone, the expansion becomes:

$$U(x) \approx U(x_0) + \frac{1}{2}U''(x_0)(x - x_0)^2 + \text{higher-order terms}$$

If the displacement $(x - x_0)$ is small, then $(x - x_0)^3$ is much smaller than $(x - x_0)^2$, and $(x - x_0)^4$ is smaller still. The cubic, quartic, and all higher terms become negligible. What remains is:

$$U(x) \approx U(x_0) + \frac{1}{2}U''(x_0)(x - x_0)^2$$

This is the potential energy of a spring. Compare it with $U = \frac{1}{2}k\,\delta x^2$ (plus a constant). The effective spring constant is:

$$\boxed{k_{\text{eff}} = U''(x_0)}$$

Stable vs. unstable

Here is where the sign of $U''(x_0)$ decides everything.

If $U''(x_0) > 0$: The curve bends upward --- it is concave up. The equilibrium is a minimum. The effective spring constant is positive, and the restoring force pulls the particle back toward $x_0$. The particle oscillates. This is a stable equilibrium.

If $U''(x_0) < 0$: The curve bends downward --- it is concave down. The equilibrium is a maximum. The effective "spring constant" is negative, and the force pushes the particle away from $x_0$. There is no oscillation --- the particle accelerates away. This is an unstable equilibrium.

If $U''(x_0) = 0$: The quadratic term vanishes, and the behavior is determined by higher-order terms. This case requires more careful analysis and arises at inflection points of $U(x)$.

This is why the marble in the bowl oscillates and the bead on the hill does not. The bowl is a minimum ($U'' > 0$). The hill is a maximum ($U'' < 0$). The mathematics of Taylor expansion makes the distinction automatic.

The oscillation frequency

For stable equilibrium ($U''(x_0) > 0$), the particle moves in a parabolic potential well with effective spring constant $k_{\text{eff}} = U''(x_0)$. From Section 13.1, a spring with constant $k$ produces SHM with angular frequency $\omega = \sqrt{k/m}$. Therefore:

$$\boxed{\omega = \sqrt{\frac{U''(x_0)}{m}}}$$

This is the central result of this section. It says: to find the oscillation frequency of a particle near a stable equilibrium, compute the second derivative of the potential energy at that point, divide by the mass, and take the square root. That is all.

Check your prediction: Look back at your earlier answer. The frequency depends on the curvature of the potential at the minimum --- option (c). Not the depth, not the width (though width and curvature are often correlated, they are not the same thing). The curvature $U''(x_0)$ is the single number that determines everything about the small-oscillation behavior.

Why This Is the Climax of Chapter 13

Step back for a moment and look at the arc of this chapter.

In Section 13.1, you derived simple harmonic motion from a specific system --- a mass on a spring obeying Hooke's law, $F = -kx$. That gave $\omega = \sqrt{k/m}$.

In Section 13.4, you analyzed the pendulum. By making the small-angle approximation $\sin\theta \approx \theta$, you turned the nonlinear pendulum equation into the SHM equation, with $\omega = \sqrt{g/L}$. That was a specific linearization of a specific system.

Now you see the general principle. Any smooth potential energy function with a minimum produces, for small displacements, a parabolic potential --- and therefore SHM. The spring was not special. The pendulum was not special. They were both instances of a universal phenomenon:

Every stable equilibrium is locally a spring.

The spring constant $k_{\text{eff}} = U''(x_0)$ is determined by the shape of the potential energy landscape at the equilibrium point. The frequency $\omega = \sqrt{U''(x_0)/m}$ follows immediately. This is why so many wildly different systems --- springs, pendulums, atoms in crystals, molecules in bonds, charges in traps, stars in galactic orbits --- all exhibit sinusoidal oscillations at small amplitudes. They are all living in the parabolic neighborhood of their equilibrium points.

Linearization is not a trick. It is a unifying principle.

The Connection to Energy Diagrams

In Section 7.5, you learned to read motion qualitatively from energy diagrams. You identified stable equilibria as positions where $U(x)$ has a minimum, and you saw that a particle near a minimum oscillates between turning points. But Section 7.5 could not tell you how fast that oscillation occurs, or why the motion is sinusoidal. It gave you the qualitative picture.

This section fills in the quantitative detail. The curvature of the potential well at its minimum determines the effective spring constant, which determines the frequency, which determines the period. The energy diagram is no longer just a qualitative tool --- with Taylor expansion, it becomes a quantitative predictor of oscillatory behavior.

And the connection to Section 4.4 is equally direct. There, you learned linearization as a technique --- replace a function with its tangent line to simplify a differential equation. Here, you are doing the same thing, but one order higher. You are replacing $U(x)$ with its best-fit parabola (the quadratic Taylor approximation) rather than its tangent line. The tangent line was not useful here because $U'(x_0) = 0$ --- the first derivative vanishes at equilibrium, so the linear term tells you nothing. The quadratic term is the leading non-trivial contribution, and it is the one that generates SHM.

A Worked Example: The Lennard-Jones Potential

To see the linearization procedure in action, let's apply it to a realistic system: two atoms interacting via the Lennard-Jones potential.

$$U(r) = 4\epsilon\left[\left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6}\right]$$

Here $r$ is the distance between the two atoms, $\epsilon$ is the depth of the potential well, and $\sigma$ is a length parameter related to the atomic size.

Step 1: Find the equilibrium

At equilibrium, $U'(r_0) = 0$. Taking the derivative and setting it to zero:

$$U'(r) = 4\epsilon\left[-12\frac{\sigma^{12}}{r^{13}} + 6\frac{\sigma^{6}}{r^{7}}\right] = 0$$

This gives $12\sigma^{12}/r_0^{13} = 6\sigma^6/r_0^7$, which simplifies to:

$$r_0^6 = 2\sigma^6 \quad \Longrightarrow \quad r_0 = 2^{1/6}\sigma \approx 1.122\,\sigma$$

This is the equilibrium bond length --- the distance at which the two atoms naturally sit.

Step 2: Find the effective spring constant

We need $U''(r_0)$. Taking the second derivative:

$$U''(r) = 4\epsilon\left[\frac{(12)(13)\sigma^{12}}{r^{14}} - \frac{(6)(7)\sigma^{6}}{r^{8}}\right]$$

Evaluating at $r_0 = 2^{1/6}\sigma$:

$$U''(r_0) = 4\epsilon\left[\frac{156\,\sigma^{12}}{(2^{1/6}\sigma)^{14}} - \frac{42\,\sigma^{6}}{(2^{1/6}\sigma)^{8}}\right]$$

Working through the exponents: $(2^{1/6})^{14} = 2^{14/6} = 2^{7/3}$ and $(2^{1/6})^{8} = 2^{8/6} = 2^{4/3}$.

$$U''(r_0) = 4\epsilon\left[\frac{156}{2^{7/3}\sigma^{2}} - \frac{42}{2^{4/3}\sigma^{2}}\right] = \frac{4\epsilon}{\sigma^2}\left[\frac{156}{2^{7/3}} - \frac{42}{2^{4/3}}\right]$$

Factoring out $1/2^{4/3}$:

$$U''(r_0) = \frac{4\epsilon}{2^{4/3}\sigma^2}\left[\frac{156}{2} - 42\right] = \frac{4\epsilon}{2^{4/3}\sigma^2}(78 - 42) = \frac{4\epsilon \cdot 36}{2^{4/3}\sigma^2} = \frac{144\epsilon}{2^{4/3}\sigma^2} = \frac{36 \cdot 2^{2/3}\epsilon}{\sigma^2}$$

So the effective spring constant is:

$$k_{\text{eff}} = U''(r_0) = \frac{36 \cdot 2^{2/3}\,\epsilon}{\sigma^2} \approx \frac{57.1\,\epsilon}{\sigma^2}$$

Step 3: Find the oscillation frequency

If $\mu$ is the reduced mass of the two-atom system (since both atoms can move), the vibration frequency is:

$$\omega = \sqrt{\frac{k_{\text{eff}}}{\mu}} = \sqrt{\frac{36 \cdot 2^{2/3}\,\epsilon}{\mu\,\sigma^2}}$$

This is exactly how physicists and chemists predict molecular vibration frequencies. The Lennard-Jones potential is complicated --- it has both attractive and repulsive parts, it is highly nonlinear, and it depends on two parameters. But near its minimum, it is just a spring. All of the complexity collapses into a single number, $k_{\text{eff}}$, and the motion is SHM.

Pause and think: The Lennard-Jones potential describes the interaction between noble gas atoms like argon. The parameter $\epsilon$ sets the energy scale, and $\sigma$ sets the length scale. Notice that the effective spring constant scales as $\epsilon/\sigma^2$ --- stiffer bonds (larger $\epsilon$) and shorter bonds (smaller $\sigma$) produce higher vibration frequencies. Does this match your physical intuition?

Three Systems, One Procedure

The power of linearization is that the procedure is always the same, regardless of the physical system. Here is the recipe:

Write down $U(x)$ --- the potential energy as a function of the relevant coordinate.
Find the equilibrium: solve $U'(x_0) = 0$.
Check stability: compute $U''(x_0)$. If it is positive, the equilibrium is stable.
Read off the spring constant: $k_{\text{eff}} = U''(x_0)$.
Compute the frequency: $\omega = \sqrt{k_{\text{eff}}/m}$.

Let's apply this to three different physical systems, side by side, to see the universality.

System 1: Small oscillations of a pendulum

The potential energy of a simple pendulum (measuring from the lowest point) is:

$$U(\theta) = mgL(1 - \cos\theta)$$

Equilibrium: $U'(\theta) = mgL\sin\theta = 0 \Rightarrow \theta_0 = 0$.

Second derivative: $U''(\theta) = mgL\cos\theta$, so $U''(0) = mgL$.

Frequency: The "mass" in angular coordinates is the moment of inertia $I = mL^2$, so:

$$\omega = \sqrt{\frac{U''(0)}{I}} = \sqrt{\frac{mgL}{mL^2}} = \sqrt{\frac{g}{L}}$$

This is the familiar pendulum frequency from Section 13.4 --- derived here without ever writing $\sin\theta \approx \theta$. The linearization happens automatically through the Taylor expansion of $U$.

System 2: A particle in a gravitational potential near a Lagrange point

Consider a particle of mass $m$ oscillating in a one-dimensional potential:

$$U(x) = -\frac{A}{x} + \frac{B}{x^2}$$

where $A, B > 0$. This form arises in effective gravitational potentials (such as the effective radial potential for orbital motion).

Equilibrium: $U'(x) = \frac{A}{x^2} - \frac{2B}{x^3} = 0 \Rightarrow x_0 = \frac{2B}{A}$.

Second derivative: $U''(x) = -\frac{2A}{x^3} + \frac{6B}{x^4}$.

At $x_0 = 2B/A$:

$$U''(x_0) = -\frac{2A}{(2B/A)^3} + \frac{6B}{(2B/A)^4} = -\frac{2A^4}{8B^3} + \frac{6A^4B}{16B^4} = -\frac{A^4}{4B^3} + \frac{6A^4}{16B^3} = \frac{A^4}{8B^3}$$

Since $U''(x_0) > 0$, the equilibrium is stable, and:

$$\omega = \sqrt{\frac{A^4}{8mB^3}}$$

System 3: A charge between two fixed charges

A positive charge $q$ is free to move along the line connecting two fixed positive charges $Q$, separated by distance $2d$. By symmetry, the midpoint is an equilibrium. But is it stable?

The potential energy of $q$ at distance $x$ from the midpoint (along the line) is:

$$U(x) = \frac{kQq}{d - x} + \frac{kQq}{d + x}$$

Equilibrium: By symmetry, $U'(0) = 0$. The midpoint is an equilibrium.

Second derivative:

$$U'(x) = \frac{kQq}{(d - x)^2} - \frac{kQq}{(d + x)^2}$$

$$U''(x) = \frac{2kQq}{(d - x)^3} + \frac{2kQq}{(d + x)^3}$$

At $x = 0$:

$$U''(0) = \frac{2kQq}{d^3} + \frac{2kQq}{d^3} = \frac{4kQq}{d^3}$$

Since $U''(0) > 0$, the equilibrium along this axis is stable, and the charge oscillates with:

$$\omega = \sqrt{\frac{4kQq}{md^3}}$$

(Note: Earnshaw's theorem tells us this equilibrium is unstable in at least one transverse direction. The charge is stable along the axis but unstable perpendicular to it. We are analyzing only the one-dimensional motion along the line.)

The pattern is unmistakable. Three completely different physical systems. Three different potential energy functions. But the procedure was identical in every case: find the equilibrium, compute $U''$, read off the frequency. Linearization does not care what system you are studying. It only asks: what is the curvature of the potential at the minimum?

Why Does Every Stable Equilibrium Produce SHM?

This question has a short answer and a deep answer. Both are worth knowing.

The short answer: At a minimum of $U(x)$, the first derivative vanishes and the second derivative is positive. The Taylor expansion to second order is therefore a parabola opening upward. A parabolic potential is the potential of a spring, and a spring produces SHM. So any minimum produces local SHM.

The deep answer: This is really a statement about smoothness. If $U(x)$ is a smooth function (meaning it has continuous derivatives of all orders), then Taylor's theorem guarantees that near any point, the function can be approximated by a polynomial. At an equilibrium point, the constant term is just a reference energy (irrelevant to the motion), and the linear term vanishes (because $U' = 0$). The first term that actually affects the dynamics is the quadratic one. And a quadratic potential is a harmonic oscillator, by definition.

The only way to avoid SHM near a stable equilibrium is if $U''(x_0) = 0$ as well --- making the quadratic term vanish and forcing you to look at cubic or higher terms. This is non-generic: it requires the potential to be very specially shaped (like $U(x) = (x - x_0)^4$ near a minimum). For virtually every potential you encounter in nature, $U''(x_0) \neq 0$ at a minimum, and the oscillations are locally simple harmonic.

This is why SHM is everywhere. It is not because nature has a fondness for springs. It is because every smooth minimum is locally parabolic, and parabolic means harmonic.

When the Approximation Breaks Down

The linearized model is excellent for small displacements. But "small" is relative --- small compared to what?

The answer: small compared to the length scale over which the higher-order terms in the Taylor expansion become significant. If the potential is very nearly parabolic over a wide range (like a true spring), then the SHM approximation works for large amplitudes. If the potential deviates from a parabola quickly (like the Lennard-Jones potential, which is highly asymmetric), then the SHM approximation is accurate only for very small displacements.

[Interactive: Breakdown Explorer. The same adjustable $U(x)$ curve from the earlier interactive, but now the student can set the particle's displacement amplitude with a slider. At small amplitudes, the period is constant (SHM). As the amplitude increases, the period begins to change. A graph of period vs. amplitude shows the deviation from the SHM prediction (a flat horizontal line). For an asymmetric well, the deviation starts earlier. For a nearly parabolic well, the deviation is delayed.

Guided prompts:

Prompt 1: Set the amplitude very small. Is the period constant? Does it match $T = 2\pi/\omega$ with $\omega = \sqrt{U''(x_0)/m}$?

Prompt 2: Increase the amplitude. At what point does the measured period differ from the SHM prediction by more than 1%? By 5%?

Prompt 3: Make the well asymmetric by adjusting the shape sliders. Does the breakdown happen sooner?

Prompt 4: Make the well as close to parabolic as possible. How far can you push the amplitude before the approximation breaks?]

The key takeaway: linearization is a local approximation. It captures the behavior near equilibrium with extraordinary accuracy, but it is not the whole story. For large-amplitude oscillations, the full nonlinear dynamics must be considered. The oscillation period becomes amplitude-dependent, the waveform is no longer purely sinusoidal, and new phenomena (like anharmonic effects in crystals or the amplitude-dependent period of a pendulum) emerge.

But here is the remarkable thing: for the vast majority of physical systems in their normal operating range, the displacements are small enough for the linearization to work. Atoms vibrate by fractions of an angstrom in bonds that are angstroms long. Pendulum clocks swing through angles of a few degrees. Building foundations oscillate by millimeters during earthquakes. In all these cases, the SHM approximation is not just qualitatively correct --- it is quantitatively precise.

Practice

Layer 1: Concrete

A particle of mass $m = 0.5 \, \text{kg}$ moves in the potential energy function:

$$U(x) = 3x^4 - 8x^3 + 6x^2 \quad \text{(in SI units)}$$

(a) Find all equilibrium points.

(b) Determine which equilibria are stable and which are unstable.

(c) For each stable equilibrium, find the effective spring constant and the angular frequency of small oscillations.

Check your answer

**(a) Find equilibria:** Set $U'(x) = 0$. $$U'(x) = 12x^3 - 24x^2 + 12x = 12x(x^2 - 2x + 1) = 12x(x - 1)^2$$ So $U'(x) = 0$ at $x = 0$ and $x = 1$. **(b) Classify stability:** Compute $U''(x)$. $$U''(x) = 36x^2 - 48x + 12$$ At $x = 0$: $U''(0) = 12 > 0$. This is a **stable equilibrium** (minimum). At $x = 1$: $U''(1) = 36 - 48 + 12 = 0$. The second derivative vanishes, so the standard test is inconclusive. Examining $U(x)$ near $x = 1$: since $U'(x) = 12x(x-1)^2$, the derivative does not change sign at $x = 1$ (it is positive on both sides for $x > 0$), so $x = 1$ is an **inflection point** of $U$, not a true minimum or maximum. This is a *marginally* stable equilibrium point (sometimes called a "neutral" point or a degenerate case). **(c) Effective spring constant at $x = 0$:** $$k_{\text{eff}} = U''(0) = 12 \, \text{N/m}$$ $$\omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{12}{0.5}} = \sqrt{24} = 2\sqrt{6} \approx 4.90 \, \text{rad/s}$$ The period of small oscillations is $T = 2\pi/\omega \approx 1.28 \, \text{s}$.

Layer 2: Pattern

Apply the linearization procedure to each of the following potential energy functions. For each one, find the equilibrium position $x_0$ (or $r_0$), the effective spring constant $k_{\text{eff}}$, and the angular frequency of small oscillations.

(a) Gravitational near a surface: $U(h) = mgh + \frac{1}{2}k h^2$ for a particle in a gravitational field with a confining spring (like a mass bouncing on a spring on a table). Here $h$ is measured upward from the natural length, $m = 0.2 \, \text{kg}$, $k = 50 \, \text{N/m}$.

(b) Molecular bond (Morse potential): $U(r) = D(1 - e^{-\alpha(r - r_0)})^2$ with $D = 5.0 \times 10^{-19} \, \text{J}$, $\alpha = 2.0 \times 10^{10} \, \text{m}^{-1}$, $r_0 = 1.5 \times 10^{-10} \, \text{m}$, and reduced mass $\mu = 1.0 \times 10^{-26} \, \text{kg}$.

(c) Electromagnetic trap: $U(x) = U_0 \cos^2\left(\frac{\pi x}{L}\right)$ near $x = L/2$, with $U_0 = 0.01 \, \text{J}$, $L = 0.1 \, \text{m}$, and $m = 10^{-6} \, \text{kg}$.

Check your answer

**(a)** $U'(h) = mg + kh = 0 \Rightarrow h_0 = -mg/k = -(0.2)(9.8)/50 = -0.0392 \, \text{m}$. $U''(h) = k = 50 \, \text{N/m}$ (constant, since $U$ is already a parabola plus a linear term). $\omega = \sqrt{k/m} = \sqrt{50/0.2} = \sqrt{250} \approx 15.8 \, \text{rad/s}$. The linear gravitational term shifts the equilibrium but does not affect the frequency --- this is consistent with what you learned in Section 13.1 about a vertical spring-mass system. **(b)** The Morse potential has its minimum at $r = r_0$ by construction (check: $U'(r_0) = 0$ since $e^{-\alpha(r_0 - r_0)} = 1$ and $(1 - 1) = 0$). $U'(r) = 2D\alpha(1 - e^{-\alpha(r - r_0)})e^{-\alpha(r - r_0)}$ $U''(r) = 2D\alpha^2[2e^{-2\alpha(r - r_0)} - e^{-\alpha(r - r_0)}]$ At $r = r_0$: $U''(r_0) = 2D\alpha^2[2(1) - 1] = 2D\alpha^2$. $k_{\text{eff}} = 2D\alpha^2 = 2(5.0 \times 10^{-19})(2.0 \times 10^{10})^2 = 2(5.0 \times 10^{-19})(4.0 \times 10^{20}) = 400 \, \text{N/m}$. $\omega = \sqrt{k_{\text{eff}}/\mu} = \sqrt{400 / (1.0 \times 10^{-26})} = \sqrt{4.0 \times 10^{28}} = 2.0 \times 10^{14} \, \text{rad/s}$. This corresponds to a vibration frequency $f = \omega/(2\pi) \approx 3.2 \times 10^{13} \, \text{Hz}$, which falls in the infrared --- consistent with molecular vibration frequencies measured by spectroscopy. **(c)** $U(x) = U_0\cos^2(\pi x/L)$. At $x = L/2$: $\cos(\pi/2) = 0$, so $U(L/2) = 0$, a minimum (since $\cos^2$ is non-negative and equals zero here). $U'(x) = -U_0 \cdot 2\cos(\pi x/L)\sin(\pi x/L) \cdot (\pi/L) = -U_0 \frac{\pi}{L}\sin(2\pi x/L)$ $U''(x) = -U_0 \frac{2\pi^2}{L^2}\cos(2\pi x/L)$ At $x = L/2$: $\cos(2\pi(L/2)/L) = \cos(\pi) = -1$, so $U''(L/2) = \frac{2\pi^2 U_0}{L^2}$. $k_{\text{eff}} = \frac{2\pi^2(0.01)}{(0.1)^2} = \frac{2\pi^2(0.01)}{0.01} = 2\pi^2 \approx 19.74 \, \text{N/m}$. $\omega = \sqrt{k_{\text{eff}}/m} = \sqrt{19.74 / 10^{-6}} = \sqrt{1.974 \times 10^{7}} \approx 4443 \, \text{rad/s}$.

Layer 3: Structure

Why does every stable equilibrium produce local simple harmonic motion? Answer this question in three parts:

(a) What mathematical fact about a minimum causes the linear term in the Taylor expansion to vanish?

(b) What mathematical fact about a minimum guarantees that the quadratic coefficient is positive?

(c) Why does a positive quadratic potential produce oscillation rather than some other type of motion?

Now address a follow-up: is there any stable equilibrium where the motion near the minimum is not SHM? If so, give an example. If not, explain why not.

Check your answer

**(a)** At a minimum, the first derivative $U'(x_0) = 0$. This is the condition for equilibrium --- no net force. Because $F = -U'(x)$, zero slope means zero force. This eliminates the linear term $U'(x_0)(x - x_0)$ from the Taylor expansion. **(b)** At a minimum, the function curves upward, meaning $U''(x_0) > 0$. This is the second-derivative test from calculus. A positive second derivative means the parabolic approximation opens upward, so the "effective spring constant" $k_{\text{eff}} = U''(x_0)$ is positive. **(c)** A positive quadratic potential $U \propto (x - x_0)^2$ produces a force $F = -U' \propto -(x - x_0)$, which is a restoring force proportional to displacement. Newton's second law then gives $ma = -k_{\text{eff}}(x - x_0)$, which is the SHM equation. Its solutions are sinusoidal by the mathematics of second-order linear ODEs with constant coefficients. **Follow-up:** Yes, there are edge cases. If $U''(x_0) = 0$ but the equilibrium is still stable (for example, $U(x) = (x - x_0)^4$ has a minimum at $x_0$ with $U'' = 0$), then the motion near the minimum is *not* SHM. The restoring force goes as $(x - x_0)^3$, not $(x - x_0)$, and the oscillation period depends on amplitude. However, such potentials are non-generic --- they require the second derivative to vanish at the minimum, which is a special condition that is not satisfied by "typical" physical potentials. For all practical purposes, stable equilibria in nature produce SHM.

Layer 4: Transfer

A chemical bond is modeled by the Morse potential:

$$U(r) = D\left(1 - e^{-\beta(r - r_0)}\right)^2$$

where $D$ is the dissociation energy, $\beta$ controls the width of the well, and $r_0$ is the equilibrium bond length.

(a) Verify that $r = r_0$ is an equilibrium and that it is stable.

(b) Find the effective spring constant $k_{\text{eff}}$ in terms of $D$ and $\beta$.

(c) Find the vibration frequency $\omega$ in terms of $D$, $\beta$, and the reduced mass $\mu$ of the two atoms.

(d) For HCl: $D = 7.4 \times 10^{-19} \, \text{J}$, $\beta = 1.81 \times 10^{10} \, \text{m}^{-1}$, and $\mu = 1.63 \times 10^{-27} \, \text{kg}$. Compute the vibration frequency in Hz and compare it to the measured infrared absorption frequency of HCl ($f \approx 8.66 \times 10^{13} \, \text{Hz}$).

Check your answer

**(a)** At $r = r_0$: the exponential $e^{-\beta(r_0 - r_0)} = e^0 = 1$, so $U(r_0) = D(1 - 1)^2 = 0$. To check equilibrium, compute: $$U'(r) = 2D\beta(1 - e^{-\beta(r - r_0)})e^{-\beta(r - r_0)}$$ At $r = r_0$: $U'(r_0) = 2D\beta(1 - 1)(1) = 0$. So $r_0$ is indeed an equilibrium point. For stability, note that $U(r) = D(\cdots)^2 \geq 0$ everywhere, and $U(r_0) = 0$ is the minimum value. So $r_0$ is a minimum --- a stable equilibrium. **(b)** Compute $U''(r)$: $$U''(r) = 2D\beta^2\left[2e^{-2\beta(r-r_0)} - e^{-\beta(r-r_0)}\right]$$ At $r = r_0$: $$U''(r_0) = 2D\beta^2[2(1) - 1] = 2D\beta^2$$ So $k_{\text{eff}} = 2D\beta^2$. **(c)** The vibration frequency is: $$\omega = \sqrt{\frac{k_{\text{eff}}}{\mu}} = \sqrt{\frac{2D\beta^2}{\mu}} = \beta\sqrt{\frac{2D}{\mu}}$$ **(d)** Plugging in the values: $$k_{\text{eff}} = 2(7.4 \times 10^{-19})(1.81 \times 10^{10})^2 = 2(7.4 \times 10^{-19})(3.276 \times 10^{20}) = 485 \, \text{N/m}$$ $$\omega = \sqrt{\frac{485}{1.63 \times 10^{-27}}} = \sqrt{2.98 \times 10^{29}} = 5.45 \times 10^{14} \, \text{rad/s}$$ $$f = \frac{\omega}{2\pi} = \frac{5.45 \times 10^{14}}{6.28} = 8.68 \times 10^{13} \, \text{Hz}$$ This matches the measured infrared absorption frequency of HCl ($8.66 \times 10^{13} \, \text{Hz}$) to within 0.2%. The linearization procedure, applied to a two-parameter empirical potential, reproduces a spectroscopic measurement with remarkable accuracy. This is the power of $\omega = \sqrt{U''(x_0)/m}$ applied to real molecular data.

Reflection

Think back over this section.

You started the chapter with specific systems: a mass on a spring, a pendulum. Each had its own derivation, its own formula for $\omega$. Now you have a single procedure that unifies all of them --- and extends to systems you have never seen before.

How does linearization connect the specific case (spring, pendulum) to the general principle?

You might also consider: linearization throws away the cubic, quartic, and higher terms in the Taylor expansion. What kind of physical behavior do those terms capture? What would you miss if you stopped at the parabolic approximation and never looked further? (Think about what happens to a pendulum at large angles, or a molecule stretched far from equilibrium.)

Looking Ahead

You have just seen the climax of Chapter 13's narrative arc. Simple harmonic motion is not a special phenomenon limited to springs and pendulums. It is the universal local behavior near any stable equilibrium. The mathematical reason is elegant: every smooth minimum is locally parabolic, and a parabolic potential is a harmonic oscillator. The physical consequence is profound: the same sinusoidal oscillation appears in atoms, molecules, structures, circuits, and stars.

In the next section, we step back and look at SHM from a completely different angle --- literally. You will discover that simple harmonic motion is the shadow of uniform circular motion. A ball moving steadily around a circle, viewed from the side, oscillates sinusoidally. The radius of the circle is the amplitude, the angular speed is the angular frequency, and the starting angle on the circle is the phase. This geometric connection will give you a powerful new way to visualize and reason about oscillatory motion --- one that ties together circular kinematics from Section 3.7 with everything you have built in this chapter.