AP Physics C: Mechanics · Topic 7.1

Topic 7.1: Defining Simple Harmonic Motion (SHM)

Q: What is the definition of simple harmonic motion in AP Physics C?

Essential knowledge 7.1.A.2 of AP Physics C: Mechanics says simple harmonic motion results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position, and 7.1.A.2.i adds that a restoring force is exerted in a direction opposite to that displacement. Statement 7.1.A.1 adds that it is a special case of periodic motion, so every simple harmonic motion repeats but not every repeating motion is simple harmonic. In this calculus-based course the same condition is usually written as a differential equation: the second derivative of position with respect to time equals a negative constant times the position, and the square root of that constant is the angular frequency.

Q: How do you tell whether a motion is simple harmonic motion?

Write the net force on the object as a function of its displacement from equilibrium, divide by the mass, and check whether the acceleration comes out as a negative constant times the displacement. If it does, the motion is simple harmonic and that constant is the angular frequency squared. If the force goes as any other power of the displacement, or has constant magnitude, or acts only during part of the cycle, the motion may still repeat but it is not simple harmonic. The tell that carries the most weight is that a simple harmonic period does not depend on amplitude, which is essential knowledge 7.3.A.5. Any oscillator whose period changes when you pull it further has failed the test.

Q: Is the equation ma equals minus kx on the AP Physics C Mechanics equation sheet?

Not as a single line. The sheet prints the spring force as minus the force constant times the displacement vector, and it prints Newton's second law as the time derivative of momentum, so you assemble the two yourself. The second-order differential equation that defines simple harmonic motion is also absent: it appears in the framework as the Derived equation of essential knowledge 7.3.A.2, and Derived is the course description's label for something you are expected to produce rather than look up. The acceleration relation that sets acceleration equal to minus the angular frequency squared times the position is likewise not printed. What the sheet does print is the spring force, the spring potential energy, the force as minus the derivative of potential energy, the spring period, and the cosine solution with a phase angle.

Q: Where is the equilibrium position of a block hanging on a spring?

At the position where the spring is already stretched by the block's weight divided by the force constant, not at the spring's natural length. Essential knowledge 7.1.A.2.ii defines an equilibrium position as a location at which the net force exerted on an object or system is zero, and at the hanging position two forces act and cancel. Measure displacement from there and the gravitational force cancels out of the equation of motion completely, leaving the same relation as a horizontal spring. That is why a vertical spring oscillator has the same period as the identical horizontal one, and why the period contains no gravitational field strength. It also gives you a second way to measure a force constant: the static stretch alone determines it, with no timing involved.

Q: Why is a pendulum only approximately simple harmonic motion?

Because the restoring torque on a pendulum is proportional to the sine of the angular displacement, not to the angular displacement itself. Essential knowledge 7.5.A.2.ii of AP Physics C: Mechanics fixes this by applying the small-angle approximation, that the sine of the angle is approximately the angle in radians, which is only accurate for small amplitudes. Since the sine of an angle is always smaller than the angle, the real restoring effect is weaker than the linear model predicts, so a real pendulum's period is slightly longer than the standard formula gives and it grows as the amplitude grows. At 10 degrees the angle in radians exceeds its own sine by about 0.5 percent; at 30 degrees the gap is about 5 percent.

Q: Can you identify simple harmonic motion from a potential energy graph?

Yes, and in AP Physics C: Mechanics this is a standard route, because that course's equation sheet prints the force as minus the derivative of the potential energy with respect to position. Equilibrium positions are where the graph has zero slope. A minimum is a stable equilibrium, since displacing the object gives a force back toward the minimum. The motion about that minimum is simple harmonic when the potential energy is quadratic in the displacement from it, because differentiating a quadratic gives a force proportional to displacement. The curvature sets the effective force constant. The AP Physics 1 equation sheet does not print that derivative relation at all, so this route belongs to the calculus-based course.

Q: Does AP Physics C expect you to solve the differential equation for SHM?

No, and the course description says so explicitly. The only boundary statement in Unit 7 sits under Topic 7.3 and reads that AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes simple harmonic motion, as well as be able to identify simple harmonic motion, and that the course does not expect students to mathematically prove that the solution is correct. So you need to recognise the equation's shape in an unfamiliar situation and to know that a cosine with an angular frequency and a phase angle solves it. You do not need to substitute the cosine back in and verify it, and you are not asked to integrate the equation from scratch.

Unit 7: Oscillations10-15% of the multiple-choice section

Simple harmonic motion happens when the restoring force is proportional to the displacement from equilibrium and points the opposite way. In AP Physics C that is a differential equation: if the acceleration comes out as a negative constant times the position, the motion is simple harmonic.

AP Physics: Unit 7 (topics 7.1 Defining Simple Harmonic Motion (SHM)). Topic 7.1 of the current AP Physics C: Mechanics course and exam description, inside Unit 7, which is weighted 10 to 15% of the multiple-choice section at about 12 to 17 class periods. One learning objective, 7.1.A (describe simple harmonic motion), and four essential-knowledge statements: 7.1.A.1 that SHM is a special case of periodic motion; 7.1.A.2 that SHM results when the magnitude of the restoring force is proportional to the displacement from equilibrium, with the Derived equation F_x = -k(delta x); 7.1.A.2.i defining a restoring force as one exerted opposite to the displacement from equilibrium; and 7.1.A.2.ii defining an equilibrium position as a location where the net force on an object or system is zero. Suggested skills 1.A, 2.C, 3.B, 3.C. Topic 7.1 prints no boundary statement; Unit 7 prints exactly one, under Topic 7.3. Calculus differentiator against the identically titled AP Physics 1 Topic 7.1: this course states the condition as the second-order differential equation d2x/dt2 = -omega^2 x (7.3.A.2), reads omega off the coefficient, and can identify SHM from a potential energy function because F_x = -dU/dx is printed on the C: Mechanics sheet and appears nowhere on the AP Physics 1 sheet. Equation-sheet facts verified against the AP Physics C: Mechanics Table of Information rendered as images: printed are F_s = -k(delta x) in vector form, U_s = k(delta x)^2/2, F_x = -dU/dx, F_net = dp/dt, T_s = 2*pi*sqrt(m/k), and x = x_max cos(omega t + phi) as the only position function; not printed are the differential equation itself, a = -omega^2 x, and ma = -kx as a single line. The three exam conventions printed are an inertial reference frame, negligible air resistance, and ideal springs and strings. Neither of the CED's two Unit 7 sample multiple-choice questions aligns to 7.1.A: question 5 aligns to 7.2.A and question 12 to 7.3.A.

What Topic 7.1 requires

Topic 7.1 of AP Physics C: Mechanics carries one learning objective and four essential-knowledge statements. That is all of the required content.

Learning objective 7.1.A: describe simple harmonic motion.

Statement	What it says
7.1.A.1	Simple harmonic motion is a special case of periodic motion
7.1.A.2	SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position. Derived equation: $F_x = -k\Delta x$
7.1.A.2.i	A restoring force is a force that is exerted in a direction opposite to the object's displacement from an equilibrium position
7.1.A.2.ii	An equilibrium position is a location at which the net force exerted on an object or system is zero

The suggested skills are 1.A (create diagrams, tables, charts, or schematics to represent physical situations), 2.C (compare physical quantities between two or more scenarios or at different times and locations within a single scenario), 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim) and 3.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws). Skill 2.A is not on that list, and neither is 2.B, so this topic is framed as a recognition and justification topic rather than a derivation one.

Topic 7.1 prints no boundary statement. Unit 7 prints exactly one across all five topics, and it sits under Topic 7.3. It is quoted in full further down this page, because it is the statement that decides how far the definition has to be taken.

Note what 7.1.A.2 does and does not say. It says the magnitude of the restoring force is proportional to the displacement. The direction is handled separately, by 7.1.A.2.i. Two conditions, stated in two places, and a motion has to satisfy both.

In this course the definition is a condition on a differential equation

This is the difference between AP Physics C: Mechanics and the algebra-based course, and it is worth stating before anything else.

Start from 7.1.A.2 and the spring force the equation sheet prints in vector form, $\vec{F}_s = -k\Delta\vec{x}$ . Apply Newton's second law, which this course's sheet gives in its primitive form as $\vec{F}_{\text{net}} = d\vec{p}/dt$ , so for constant mass $F_x = m\,d^2x/dt^2$ . Put the two together:

m\frac{d^2x}{dt^2} = -kx \quad \Longrightarrow \quad \frac{d^2x}{dt^2} = -\frac{k}{m}x

Compare that with the Derived equation of statement 7.3.A.2:

\frac{d^2x}{dt^2} = -\omega^2 x

The two match, with $\omega^2 = k/m$ . That comparison is the whole test. A motion is simple harmonic when its equation of motion has that shape, and the angular frequency is whatever constant sits in front of the position. You do not need to know in advance that the answer is a cosine, and you do not need a period formula.

In AP Physics 1 the definition stops at the proportionality of the force, because that course has no second derivative to write. Here the proportionality of the force and the shape of the differential equation are the same statement, and the second form is the one that generalises. It is why the same method handles a spring, a pendulum, a rigid body swinging on a pivot and a disk twisting on a wire, instead of four remembered results. Topic 7.5 uses the angular twin, $d^2\theta/dt^2 = -\omega^2\theta$ .

The practical consequence for a Topic 7.1 question: when you are asked whether a motion is SHM, do not look at the trajectory and do not look for a sine wave. Write the net force as a function of displacement from equilibrium, divide by the mass, and look at whether what multiplies the displacement is a negative constant.

The route this course has and the algebra-based one does not: start from the potential energy

The C: Mechanics equation sheet prints

F_x = -\frac{dU(x)}{dx}

and the AP Physics 1 sheet does not print it in any form. That single line opens a second, entirely equivalent way to identify simple harmonic motion, and it is a way the algebra-based course cannot use at all.

If the force has to be proportional to displacement, then the potential energy has to be quadratic in displacement, because differentiating a quadratic gives something linear. Run it forwards:

U(s) = \tfrac{1}{2}ks^2 \quad \Longrightarrow \quad F_s = -\frac{dU}{ds} = -ks

where $s$ is the displacement from equilibrium. The sheet's $U_s = \frac{1}{2}k(\Delta x)^2$ is exactly that function, so the spring case is the standard example rather than a special one.

So three descriptions of the same condition are available on this exam, and a question can hand you any one of them:

a force law, $F = -ks$ ,
a differential equation, $d^2s/dt^2 = -\omega^2 s$ ,
a potential energy function that is quadratic in the displacement from its minimum.

The third is the one to watch for on a potential energy curve. An equilibrium position is where $dU/dx = 0$ , which is 7.1.A.2.ii read through the printed relation, since zero slope means zero force. A minimum of $U$ is a stable equilibrium: push away from it and the force pushes back, so the force is restoring. A maximum is an equilibrium too, and it is not restoring, so nothing oscillates about it. Worked example 2 takes a quadratic $U(x)$ with a shifted minimum and reads a period straight off it, which is a Topic 7.1 question the algebra-based course has no equipment for.

One honest limit. The CED does not ask you to expand a general potential energy about its minimum, and it prints no series expansion anywhere. What it does do, in Topic 7.5, is apply a small-angle approximation to make one specific non-linear restoring torque linear. Treat that as the pattern the exam uses, and do not go further than a stated approximation.

Restoring force and equilibrium, in the framework's own words

Both definitions in this topic are sub-statements, and both are worth having exactly.

Statement 7.1.A.2.i: "A restoring force is a force that is exerted in a direction opposite to the object's displacement from an equilibrium position."

Statement 7.1.A.2.ii: "An equilibrium position is a location at which the net force exerted on an object or system is zero."

Read 7.1.A.2.ii carefully, because it says net force, not "no forces". A block hanging at rest on a vertical spring has two forces on it and neither is zero; the position is an equilibrium because they cancel. That is the case the next section works through, and it is where most of the sign errors in this unit start.

Read 7.1.A.2.i carefully too, because it constrains direction only. A force that always points back toward equilibrium makes the motion periodic but not necessarily simple harmonic. The proportionality in 7.1.A.2 is a separate requirement, and it is the one that fails in real systems.

A useful pair of glossary entries here: restoring force and equilibrium. The spring force itself is Topic 2.8, and the Topic 2.8 page covers where $-k\Delta x$ comes from and how a force against displacement graph gives you $k$ as a slope and stored energy as an area.

A vertical spring: gravity moves the equilibrium and leaves the period alone

This is the single most useful worked consequence of the two definitions, and it is the setup the framework's own sample multiple-choice question 12 uses.

Hang a block of mass $m$ from a spring of force constant $k$ . At rest, the spring is stretched by $x_0$ where $kx_0 = mg$ , so

x_0 = \frac{mg}{k}

That stretched position, not the spring's natural length, is the equilibrium position of 7.1.A.2.ii, because it is where the net force is zero. Now measure a displacement $y$ downward from there. The spring is stretched $x_0 + y$ , so the net downward force is

F = mg - k(x_0 + y) = mg - kx_0 - ky = -ky

The $mg$ cancelled against $kx_0$ . So

\frac{d^2y}{dt^2} = -\frac{k}{m}y

which is the identical differential equation to the horizontal case. Gravity relocates the equilibrium and does not appear in the equation of motion at all. The angular frequency is $\sqrt{k/m}$ , the period is the printed $T_s = 2\pi\sqrt{m/k}$ , and $g$ is nowhere in either.

Three things follow, and all three get tested:

Turning a horizontal spring oscillator vertical does not change its period. Taking it to the Moon does not change its period either, though it does change how far the spring hangs at rest.
Measuring the static stretch $x_0$ gives you $k$ without any timing, since $k = mg/x_0$ . That is a genuine second route to the spring constant and it is worth remembering for the experimental-design question.
The energy accounting still comes out as a single parabola. The sheet prints $U_s = \frac{1}{2}k(\Delta x)^2$ and $\Delta U_g = mg\Delta y$ separately, and if you add them for a vertical spring and measure from the hanging equilibrium, the cross terms cancel exactly as the forces did, leaving $\frac{1}{2}ky^2$ plus a constant. So $E_{\text{total}} = \frac{1}{2}kA^2$ from statement 7.4.A.4.ii still applies, provided $A$ is measured from the hanging equilibrium. Topic 7.4 carries that out.

One limit worth knowing for a vertical spring: the analysis assumes the spring stays stretched. It does, as long as the amplitude is smaller than $x_0$ . Since $a_{\max} = A\omega^2$ and $x_0 = g/\omega^2$ , that condition is exactly $a_{\max} < g$ .

Proportional is the word that does the work

Statement 7.1.A.2 requires the magnitude of the restoring force to be proportional to the displacement. Not merely increasing with it. Not roughly linear over the range you care about. Proportional.

The framework's first sample instructional activity for Unit 7 is built on exactly this. It asks students to predict whether a ball rolling back and forth inside a spherical bowl is simple harmonic motion, then take data to confirm the prediction, and it names the checks it wants: whether the period is independent of amplitude, whether the motion is a sine function, and whether the force is proportional to displacement. Those are three tests of the same condition, and any of them failing settles it.

Cases that are periodic and not simple harmonic, so worth being able to explain:

A ball in a bowl, or a pendulum, at large amplitude. The restoring force goes as $\sin\theta$ , not $\theta$ . Since $\sin\theta$ is always smaller than $\theta$ , the true restoring effect is weaker than the linear model, so the true period is longer than the small-angle value and it grows with amplitude. Topic 7.5 is where the approximation gets stated.
A force that goes as a higher power, say $F = -cs^3$ . The force is restoring, the motion repeats, and it is not SHM. Its period depends on amplitude, which is precisely what statement 7.3.A.5 rules out for SHM.
A ball bouncing elastically on the floor. Perfectly periodic; the force is zero for most of each cycle and enormous during contact, which is not proportional to anything.
A cart bouncing between two walls. Same objection.

And one case that is easy to dismiss and should not be. A rigid body swinging on a pivot, a disk twisting on a wire, and a cylinder rolling on a spring all satisfy 7.1.A.2 for the right variable, once you write Newton's second law in the correct form. The restoring quantity for a rotating system is a torque and the displacement is an angle. That is why Topic 7.5 exists and why the differential-equation test is more useful than a memorised force law.

What the equation sheet prints for Topic 7.1

Checked against the Table of Information in the AP Physics C: Mechanics course and exam description, rendered and read line by line, rather than recalled.

Relation	On the C: Mechanics sheet
$\vec{F}_s = -k\Delta\vec{x}$	yes, in vector form
$U_s = \frac{1}{2}k(\Delta x)^2$	yes
$F_x = -\dfrac{dU(x)}{dx}$	yes
$\vec{F}_{\text{net}} = \dfrac{d\vec{p}}{dt}$	yes
$\vec{a}_{\text{sys}} = \dfrac{\sum \vec{F}}{m_{\text{sys}}}$	yes
$T_s = 2\pi\sqrt{m/k}$	yes
$x = x_{\max}\cos(\omega t + \phi)$	yes, and it is the only position function printed
$\dfrac{d^2x}{dt^2} = -\omega^2 x$	no, it is the Derived equation of 7.3.A.2 only
$a = -\omega^2 x$	no
$ma = -kx$ as a single line	no, you assemble it

Note the form difference in the first row. Statement 7.1.A.2 gives the Derived equation as a component, $F_x = -k\Delta x$ ; the sheet prints the vector version. Same physics, and the component form is what you actually substitute into a one-dimensional differential equation.

The row that matters most is the second from last. The differential equation that defines simple harmonic motion in this course is not printed on this course's equation sheet. Neither is $a = -\omega^2 x$ . Both are labelled Derived equations in the framework, which is the CED's way of saying you are expected to produce them rather than look them up. What is printed is everything you build them from.

Also worth knowing what is printed elsewhere in the booklet, because `equations.ts` on this site transcribes the physics tables and not the rest. The Table of Information also carries a geometry and trigonometry table, a table of trigonometric values at 0, 30, 37, 45, 53, 60 and 90 degrees, a vectors table, a calculus table with the derivative and integral rules, and an identities table. There are three exam conventions listed, and one of them matters here: springs and strings are assumed to be ideal unless otherwise stated. That is the licence for using $-k\Delta x$ without arguing for it.

How Topic 7.1 is tested

Unit 7 is weighted 10 to 15% of the multiple-choice section of the AP Physics C: Mechanics exam, at about 12 to 17 class periods. The exam is 3 hours: 42 multiple-choice questions in 85 minutes for half the score, then 4 free-response questions in 95 minutes for the other half. A four-function, scientific, or graphing calculator is allowed on both sections.

On the multiple-choice section, skill 2.C is weighted 10 to 15%, 3.B is 15 to 25% and 3.C is 5 to 10%. Skill 1.A is one of this topic's suggested skills and Science Practice 1 is not assessed on the multiple-choice section at all, so the diagram work points at the free-response section, where Practice 1 carries 20 to 35%.

The framework's exam-weighting page adds a caution worth taking literally: required course content can be assessed with any skill. The suggested skills are a teaching signal, not a promise about the exam.

Neither of the CED's two Unit 7 sample multiple-choice questions is aligned to 7.1.A. Question 5 aligns to 7.2.A and question 12 to 7.3.A. What Topic 7.1 supplies is the vocabulary the other four topics are graded in, and the unit's Preparing for the AP Exam note shows where that vocabulary is cashed in. It ties Unit 7 to the second free-response question, Translation Between Representations, worth 12 points at a suggested 25 to 30 minutes, and gives a Unit 7 example in full: a student might be asked to sketch free-body diagrams of a block oscillating on a spring at maximum displacement and at equilibrium, then create energy bar charts for the block and spring system at those same two positions, then explain how the two representations are consistent with each other. The free-body diagram at equilibrium is where 7.1.A.2.ii gets tested: for a vertical spring the net force is zero there and neither individual force is.

The note closes by saying that while the Unit 7 content provides especially good practice for that question, content from any unit may be included in it on the AP Exam.

For the step-by-step routines rather than the framework, the simple harmonic motion guide owns the procedures, and the simple harmonic motion practice set is the drill.

If you are in AP Physics 1, this is not your page

AP Physics 1 has a Topic 7.1 with the same title, and its essential knowledge on this topic reads close to this one's. The honest difference is the one this page opened with.

The [AP Physics 1 Topic 7.1 page](/ap-physics-1/unit-7-oscillations/7-1-defining-simple-harmonic-motion-shm) is for students in the algebra-based course, where the definition of SHM stops at the proportionality of the restoring force. This page is for students in AP Physics C: Mechanics, who are expected to write the second-order differential equation, read the angular frequency off its coefficient, and identify SHM from a potential energy function as readily as from a force law. Neither page is a reading-level variant of the other. If you are in Physics 1, $F_x = -dU/dx$ is not on your equation sheet and no differential equation will be asked of you.

AP Physics C: Mechanics is a calculus-based, college-level course, equivalent to a first course in an introductory college sequence in calculus-based physics. Its stated prerequisite is that students have taken or are concurrently taking calculus. AP Physics 1 weights its own Unit 7 at 5 to 8% of its multiple-choice section against 10 to 15% here, and gives it four topics rather than five.

Where to go next: Topic 7.2 turns the coefficient into a period, Topic 7.3 solves the differential equation and carries the unit's only boundary statement, and Topic 7.5 applies the same test to rigid bodies and has no algebra-based counterpart at all. The Unit 7 hub lists all five topics and the unit's equations.

Three restoring forces, and which one is simple harmonic

A 0.25 kg object moves along the $x$ axis under a single net force, measured from the object's equilibrium position at $x = 0$ . Decide whether each of the following is simple harmonic motion, and where it is, find the angular frequency, the period and the frequency. (a) $F_x = -(36\ \mathrm{N/m})x$ . (b) $F_x = -(36\ \mathrm{N/m^3})x^3$ . (c) $F_x = -F_0$ for $x > 0$ and $F_x = +F_0$ for $x < 0$ , with $F_0$ a positive constant.

Set the convention once: $x$ is the displacement from equilibrium, positive in one fixed direction, and a negative $F_x$ at positive $x$ means the force is restoring in the sense of 7.1.A.2.i.
(a) Apply Newton's second law: $m\dfrac{d^2x}{dt^2} = -(36)x$ , so $\dfrac{d^2x}{dt^2} = -\dfrac{36}{0.25}x = -(144\ \mathrm{s^{-2}})x$ .
That has the shape of 7.3.A.2's $\dfrac{d^2x}{dt^2} = -\omega^2 x$ , so this is SHM and $\omega^2 = 144\ \mathrm{rad^2/s^2}$ , giving $\omega = 12\ \mathrm{rad/s}$ .
The sheet prints $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$ , so $T = \dfrac{2\pi}{12} = 0.5236 \approx 0.524$ s and $f = 1/T = 1.910 \approx 1.91$ Hz.
Cross-check against the printed spring period, since the force law is a spring force with $k = 36$ N/m: $T_s = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.25/36} = 2\pi\sqrt{6.944 \times 10^{-3}} = 2\pi(0.08333) = 0.5236$ s. Identical, as it must be, because the printed formula is what reading the coefficient produces.
(b) Newton's second law gives $\dfrac{d^2x}{dt^2} = -\dfrac{36}{0.25}x^3 = -(144)x^3$ . The coefficient of $x$ is not a constant here, it is $144x^2$ , so the equation does not have the required shape.
This force is restoring, by 7.1.A.2.i, and the motion does repeat. It is not SHM, because 7.1.A.2 requires the magnitude to be proportional to the displacement and a cube is not. There is no single $\omega$ to report: a larger amplitude makes the force disproportionately larger, so the motion is faster and the period is shorter. That amplitude dependence is exactly what statement 7.3.A.5 forbids for SHM.
(c) A constant-magnitude force that flips sign at the origin is restoring and periodic, and the motion is two parabolic arcs joined at the origin, not a sinusoid. $\dfrac{d^2x}{dt^2} = \mp F_0/m$ , a constant, so nothing multiplies $x$ at all. Not SHM.
Sanity check on (c) from the other direction: with a constant force magnitude the object is in uniform acceleration on each half, so the standard kinematics apply and the time for one half cycle goes as $\sqrt{A}$ . Amplitude-dependent period again, so not SHM.

(a) SHM, with $\omega = 12\ \mathrm{rad/s}$ , $T = 0.524$ s and $f = 1.91$ Hz, confirmed twice. (b) Not SHM: the force is restoring but goes as the cube of the displacement, so no constant $\omega$ exists and the period depends on amplitude. (c) Not SHM: the force magnitude is constant, so nothing is proportional to displacement.

A period read straight off a potential energy function

A 0.25 kg particle moves in one dimension with potential energy $U(x) = \alpha x^2 - \beta x$ , where $\alpha = 18\ \mathrm{J/m^2}$ and $\beta = 12\ \mathrm{J/m}$ . Find (a) the equilibrium position and the potential energy there, (b) the force as a function of displacement from equilibrium, (c) the angular frequency and period of small oscillations, and (d) what changes if the particle instead has $U(x) = \gamma x^4$ .

The sheet prints $F_x = -\dfrac{dU(x)}{dx}$ , which is the only tool needed. Differentiate: $F_x = -\dfrac{d}{dx}\left(\alpha x^2 - \beta x\right) = -(2\alpha x - \beta) = \beta - 2\alpha x$ .
(a) Equilibrium is where the net force is zero, per 7.1.A.2.ii, so $\beta - 2\alpha x_{\text{eq}} = 0$ and $x_{\text{eq}} = \dfrac{\beta}{2\alpha} = \dfrac{12}{36} = 0.3333\ \mathrm{m}$ .
The potential energy there is $U(x_{\text{eq}}) = 18(0.3333)^2 - 12(0.3333) = 18(0.11111) - 4.000 = 2.000 - 4.000 = -2.00\ \mathrm{J}$ . It is a minimum, not a maximum, because $\alpha$ is positive so the parabola opens upward. A minimum is what makes the force restoring.
(b) Let $s = x - x_{\text{eq}}$ be the displacement from equilibrium. Substituting $x = s + \beta/(2\alpha)$ into the force gives $F_s = \beta - 2\alpha\left(s + \dfrac{\beta}{2\alpha}\right) = -2\alpha s$ .
So $F_s = -(36\ \mathrm{N/m})s$ : proportional to the displacement and opposite to it, which is 7.1.A.2 and 7.1.A.2.i together. The effective force constant is $k = 2\alpha = 36$ N/m. Check the units: $\mathrm{J/m^2}$ is $\mathrm{N/m}$ , so $2\alpha$ has the units of a spring constant.
(c) $\dfrac{d^2s}{dt^2} = -\dfrac{2\alpha}{m}s = -\dfrac{36}{0.25}s = -(144\ \mathrm{s^{-2}})s$ , so $\omega^2 = 144\ \mathrm{rad^2/s^2}$ and $\omega = 12\ \mathrm{rad/s}$ .
$T = \dfrac{2\pi}{\omega} = 0.5236 \approx 0.524$ s, and $f = 1.91$ Hz.
Notice the $\beta$ term never reached the period. A term linear in $x$ shifts where the minimum sits and does nothing to the curvature, so it moves the equilibrium and leaves the period alone. That is the same structure as gravity on a vertical spring, where $mg$ is exactly such a linear term in the total potential energy.
(d) With $U = \gamma x^4$ , $F_x = -4\gamma x^3$ . Equilibrium is still at $x = 0$ and it is still a minimum, so the motion still repeats and the force is still restoring. It is not SHM, because the force is not proportional to the displacement, so no $\omega$ exists and the period depends on the amplitude. Quadratic in $U$ is the condition, not merely having a minimum.

(a) $x_{\text{eq}} = 0.333$ m, where $U = -2.00$ J and the curve has a minimum. (b) $F_s = -(36\ \mathrm{N/m})s$ measured from equilibrium, so the effective force constant is $2\alpha$ . (c) $\omega = 12\ \mathrm{rad/s}$ , $T = 0.524$ s, $f = 1.91$ Hz. (d) A quartic potential gives a restoring but non-proportional force, so the motion is periodic and not simple harmonic.

A block hanging on a spring, and where its equilibrium actually is

A 0.80 kg block hangs at rest from an ideal vertical spring of force constant $k = 50\ \mathrm{N/m}$ . Take $g = 9.8\ \mathrm{m/s^2}$ . Find (a) how far the spring is stretched at rest, (b) the net force on the block when it is 0.040 m below that position, (c) the differential equation of its motion and its period, and (d) the largest amplitude for which this analysis holds.

Declare the convention: let $y$ be measured downward from the hanging rest position, so $y = 0$ is the equilibrium of 7.1.A.2.ii and positive $y$ is a downward displacement.
(a) At rest the net force is zero, so the upward spring force balances the downward gravitational force: $kx_0 = mg$ , giving $x_0 = \dfrac{mg}{k} = \dfrac{(0.80)(9.8)}{50} = \dfrac{7.84}{50} = 0.1568 \approx 0.157\ \mathrm{m}$ .
Note what is true at $y = 0$ : two forces act, each of magnitude 7.84 N, and neither is zero. The position is an equilibrium because the net force is zero, which is precisely the wording of 7.1.A.2.ii.
(b) At displacement $y$ the spring is stretched $x_0 + y$ , so the net downward force is $F = mg - k(x_0 + y)$ . Since $kx_0 = mg$ by part (a), this collapses to $F = -ky$ .
At $y = 0.040$ m: $F = -(50)(0.040) = -2.0\ \mathrm{N}$ , that is 2.0 N upward, back toward equilibrium. Check it the long way: $mg - k(x_0 + y) = 7.84 - 50(0.1968) = 7.84 - 9.84 = -2.00\ \mathrm{N}$ . The two agree.
(c) Newton's second law gives $m\dfrac{d^2y}{dt^2} = -ky$ , so $\dfrac{d^2y}{dt^2} = -\dfrac{k}{m}y = -\dfrac{50}{0.80}y = -(62.5\ \mathrm{s^{-2}})y$ .
That is 7.3.A.2's equation with $\omega^2 = 62.5\ \mathrm{rad^2/s^2}$ , so $\omega = \sqrt{62.5} = 7.9057\ \mathrm{rad/s}$ and $T = \dfrac{2\pi}{7.9057} = 0.7948 \approx 0.795\ \mathrm{s}$ .
Confirm against the printed $T_s = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.80/50} = 2\pi\sqrt{0.016} = 2\pi(0.126491) = 0.7948\ \mathrm{s}$ . Identical. No $g$ appears in either. Gravity set where the block hangs and then dropped out of the equation of motion entirely.
(d) The derivation assumed the spring stays stretched, which needs $A < x_0 = 0.157$ m. Equivalently, since $a_{\max} = A\omega^2$ and $x_0 = g/\omega^2$ , the condition is $a_{\max} < g = 9.8\ \mathrm{m/s^2}$ . Beyond that the spring would have to push down on a block it is only hanging from, and the motion stops being simple harmonic.

(a) $x_0 = 0.157$ m. (b) 2.0 N directed back up toward equilibrium, confirmed two ways. (c) $d^2y/dt^2 = -(62.5\ \mathrm{s^{-2}})y$ , so $\omega = 7.91\ \mathrm{rad/s}$ and $T = 0.795$ s, with no $g$ in it. (d) Amplitudes below 0.157 m, which is the same condition as $a_{\max} < g$ .

Frequently asked questions

What is the definition of simple harmonic motion in AP Physics C?

Essential knowledge 7.1.A.2 of AP Physics C: Mechanics says simple harmonic motion results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position, and 7.1.A.2.i adds that a restoring force is exerted in a direction opposite to that displacement. Statement 7.1.A.1 adds that it is a special case of periodic motion, so every simple harmonic motion repeats but not every repeating motion is simple harmonic. In this calculus-based course the same condition is usually written as a differential equation: the second derivative of position with respect to time equals a negative constant times the position, and the square root of that constant is the angular frequency.

How do you tell whether a motion is simple harmonic motion?

Write the net force on the object as a function of its displacement from equilibrium, divide by the mass, and check whether the acceleration comes out as a negative constant times the displacement. If it does, the motion is simple harmonic and that constant is the angular frequency squared. If the force goes as any other power of the displacement, or has constant magnitude, or acts only during part of the cycle, the motion may still repeat but it is not simple harmonic. The tell that carries the most weight is that a simple harmonic period does not depend on amplitude, which is essential knowledge 7.3.A.5. Any oscillator whose period changes when you pull it further has failed the test.

Is the equation ma equals minus kx on the AP Physics C Mechanics equation sheet?

Not as a single line. The sheet prints the spring force as minus the force constant times the displacement vector, and it prints Newton's second law as the time derivative of momentum, so you assemble the two yourself. The second-order differential equation that defines simple harmonic motion is also absent: it appears in the framework as the Derived equation of essential knowledge 7.3.A.2, and Derived is the course description's label for something you are expected to produce rather than look up. The acceleration relation that sets acceleration equal to minus the angular frequency squared times the position is likewise not printed. What the sheet does print is the spring force, the spring potential energy, the force as minus the derivative of potential energy, the spring period, and the cosine solution with a phase angle.

Where is the equilibrium position of a block hanging on a spring?

At the position where the spring is already stretched by the block's weight divided by the force constant, not at the spring's natural length. Essential knowledge 7.1.A.2.ii defines an equilibrium position as a location at which the net force exerted on an object or system is zero, and at the hanging position two forces act and cancel. Measure displacement from there and the gravitational force cancels out of the equation of motion completely, leaving the same relation as a horizontal spring. That is why a vertical spring oscillator has the same period as the identical horizontal one, and why the period contains no gravitational field strength. It also gives you a second way to measure a force constant: the static stretch alone determines it, with no timing involved.

Why is a pendulum only approximately simple harmonic motion?

Because the restoring torque on a pendulum is proportional to the sine of the angular displacement, not to the angular displacement itself. Essential knowledge 7.5.A.2.ii of AP Physics C: Mechanics fixes this by applying the small-angle approximation, that the sine of the angle is approximately the angle in radians, which is only accurate for small amplitudes. Since the sine of an angle is always smaller than the angle, the real restoring effect is weaker than the linear model predicts, so a real pendulum's period is slightly longer than the standard formula gives and it grows as the amplitude grows. At 10 degrees the angle in radians exceeds its own sine by about 0.5 percent; at 30 degrees the gap is about 5 percent.

Can you identify simple harmonic motion from a potential energy graph?

Yes, and in AP Physics C: Mechanics this is a standard route, because that course's equation sheet prints the force as minus the derivative of the potential energy with respect to position. Equilibrium positions are where the graph has zero slope. A minimum is a stable equilibrium, since displacing the object gives a force back toward the minimum. The motion about that minimum is simple harmonic when the potential energy is quadratic in the displacement from it, because differentiating a quadratic gives a force proportional to displacement. The curvature sets the effective force constant. The AP Physics 1 equation sheet does not print that derivative relation at all, so this route belongs to the calculus-based course.

Does AP Physics C expect you to solve the differential equation for SHM?

No, and the course description says so explicitly. The only boundary statement in Unit 7 sits under Topic 7.3 and reads that AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes simple harmonic motion, as well as be able to identify simple harmonic motion, and that the course does not expect students to mathematically prove that the solution is correct. So you need to recognise the equation's shape in an unfamiliar situation and to know that a cosine with an angular frequency and a phase angle solves it. You do not need to substitute the cosine back in and verify it, and you are not asked to integrate the equation from scratch.