AP Physics C: Mechanics · Topic 7.4

Topic 7.4: Energy of Simple Harmonic Oscillators

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

The total energy of a simple harmonic oscillator is constant and set by the amplitude alone. For a spring and object it is one half the force constant times the amplitude squared. Kinetic energy peaks at equilibrium, potential energy at the turning points, and each cycles twice per period.

AP Physics: Unit 7 (topics 7.4 Energy of Simple Harmonic Oscillators). Topic 7.4 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, 7.4.A (describe the mechanical energy of a system exhibiting SHM), and four numbered essential-knowledge statements: 7.4.A.1 that the total energy is the sum of kinetic and potential, with the Relevant equation E_total = U + K; 7.4.A.2 that conservation of energy makes that total constant; 7.4.A.3 that kinetic energy is maximum when potential energy is minimum; 7.4.A.4 the converse, with 7.4.A.4.i that the minimum kinetic energy is zero and 7.4.A.4.ii that changing the amplitude changes the maximum potential energy and therefore the total energy, with the Relevant equation for a spring and object system E_total = kA^2/2. Suggested skills 1.C, 2.B, 2.D, 3.C. Topic 7.4 prints no boundary statement; Unit 7 prints exactly one, under Topic 7.3. Note that both equations here carry the framework's Relevant equation label rather than Derived equation, and neither is printed on the equation sheet, so the two labels do not track what is printed. Calculus differentiator against the identically titled AP Physics 1 Topic 7.4: the qualitative statements are close to identical in the two frameworks, and the C: Mechanics sheet additionally prints delta-U as minus the integral of a conservative force over displacement and F_x = -dU/dx, neither of which is on the AP Physics 1 sheet, so the spring potential energy is derived here rather than handed over, SHM can be identified from the curvature of a potential energy curve, and the energies can be written as functions of time and differentiated. Equation-sheet facts verified against the AP Physics C: Mechanics Table of Information rendered as images: printed are K = mv^2/2, U_s = k(delta x)^2/2, delta U_g = mg(delta y), K_rot = I omega^2/2, delta U = -integral F_cf dot dr, F_x = -dU/dx, W = integral F dot dr and delta K = sum W_i; not printed are E_total = U + K, E_total = kA^2/2, v = omega sqrt(A^2 - x^2) and v_max = A omega. The booklet's Identities table prints sin^2 + cos^2 = 1, which is what makes the total energy constant algebraically; it does not print the double-angle form for cosine. Neither of the CED's two Unit 7 sample multiple-choice questions aligns to 7.4.A, and no sample free-response question aligns to Unit 7. Sample instructional activity 4, on Topic 7.3, asks for kinetic, potential and total energy against time as well as energy bar charts.

What Topic 7.4 requires

Topic 7.4 of AP Physics C: Mechanics carries one learning objective and four numbered essential-knowledge statements, the last of which carries two sub-statements.

Learning objective 7.4.A: describe the mechanical energy of a system exhibiting SHM.

Statement	What it says
7.4.A.1	The total energy of a system exhibiting SHM is the sum of the system's kinetic and potential energies. Relevant equation: $E_{\text{total}} = U + K$
7.4.A.2	Conservation of energy indicates that the total energy of a system exhibiting SHM is constant
7.4.A.3	The kinetic energy of a system exhibiting SHM is at a maximum when the system's potential energy is at a minimum
7.4.A.4	The potential energy of a system exhibiting SHM is at a maximum when the system's kinetic energy is at a minimum
7.4.A.4.i	The minimum kinetic energy of a system exhibiting SHM is zero
7.4.A.4.ii	Changing the amplitude of a system exhibiting SHM will change the maximum potential energy of the system and, therefore, the total energy of the system. Relevant equation for a spring and object system: $E_{\text{total}} = \frac{1}{2}kA^2$

The suggested skills are 1.C (create qualitative sketches of graphs), 2.B (calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway), 2.D (predict new values or factors of change of physical quantities using functional dependence between variables) and 3.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws).

Skill 2.B appears here and nowhere else in Unit 7 except Topic 7.5. This is the numerical topic of the unit.

Topic 7.4 prints no boundary statement. Unit 7 prints exactly one across all five topics and it sits under Topic 7.3.

Two details in the table repay attention. First, 7.4.A.4.i is worth reading twice: the minimum kinetic energy is zero. Not the minimum potential energy, which depends on where you put your zero. Second, both equations in this topic are labelled Relevant equation by the framework rather than Derived equation, and neither one is printed on the equation sheet. The two labels do not track what is printed, so read each one as a label and check the sheet separately.

Where the spring potential energy comes from in this course

AP Physics 1 hands you $U_s = \frac{1}{2}k(\Delta x)^2$ and asks you to use it. This course prints the same line, and also prints the line it comes from:

\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}

where the subscript marks a conservative force. That relation appears nowhere on the AP Physics 1 sheet, and it turns the spring energy from a fact into a two-line derivation. Take the spring force $F_x = -kx$ from 7.1.A.2, put the zero of potential energy at the equilibrium position, and integrate from there out to a displacement $x$ :

U_s(x) = -\int_0^x (-k x')\, dx' = k\int_0^x x'\, dx' = \tfrac{1}{2}kx^2

The integral rule that step needs, $\int x^n dx = \frac{1}{n+1}x^{n+1}$ , is printed in the booklet's Calculus table.

The reverse direction is printed too, as $F_x = -dU(x)/dx$ , and it is the more useful of the two on this topic. It says the force is the negative slope of the energy curve, so:

Wherever the potential energy graph has zero slope, the net force is zero. That is the equilibrium position of statement 7.1.A.2.ii.
A minimum of $U$ gives a restoring force, since the slope is positive on the right and negative on the left. That is the condition for oscillation.
The motion is simple harmonic exactly when $U$ is quadratic in the displacement from that minimum, because only a quadratic differentiates to something proportional to displacement. The curvature is the effective force constant.

So in this course "the energy is $\frac{1}{2}kA^2$ " and "the motion is simple harmonic" are two readings of the same statement about the shape of a curve, which is a connection the algebra-based course has no equipment to make. Topic 7.1 works a numerical example of identifying SHM from a potential energy curve.

One consequence worth carrying for vertical springs. The sheet prints $\Delta U_g = mg\Delta y$ separately, and it is tempting to think a hanging block has two potential energies to track. Add them and measure the displacement from the hanging equilibrium, and the cross terms cancel against each other exactly the way the forces did, leaving $\frac{1}{2}ky^2$ plus a constant. So a vertical spring oscillator obeys $E_{\text{total}} = \frac{1}{2}kA^2$ unchanged, provided $A$ is measured from where the block hangs at rest.

The total energy is fixed by the amplitude, and the identity that keeps it constant

Statement 7.4.A.4.ii is the quantitative core of the topic: changing the amplitude changes the maximum potential energy and therefore the total energy of the system, and for a spring and object system

E_{\text{total}} = \tfrac{1}{2}kA^2

The argument behind it is one sentence. At maximum displacement the object is momentarily at rest, so by 7.4.A.4.i the kinetic energy is zero and by 7.4.A.1 the total is entirely potential. Evaluate $U_s = \frac{1}{2}k(\Delta x)^2$ at $\Delta x = A$ and you have the total energy of the whole cycle, because 7.4.A.2 says that total is constant.

Notice the exponent. The energy goes as the square of the amplitude, so doubling the amplitude quadruples the energy while leaving the period untouched, per 7.3.A.5. That pair of facts, energy quadrupling and period unmoved, is the single most tested consequence of Unit 7's energy statements, because the two statements sit in different topics and a question can set them against each other.

Why is the total constant instant by instant, and not just at the two ends? Substitute the printed solution into the two energy expressions. With $x = A\cos(\omega t + \phi)$ and $v = -A\omega\sin(\omega t + \phi)$ :

U = \tfrac{1}{2}kA^2\cos^2(\omega t + \phi) \qquad K = \tfrac{1}{2}mA^2\omega^2\sin^2(\omega t + \phi)

and since $\omega^2 = k/m$ for a spring, $mA^2\omega^2 = kA^2$ , so both prefactors are the same $\frac{1}{2}kA^2$ . Adding them:

E = \tfrac{1}{2}kA^2\left[\cos^2(\omega t + \phi) + \sin^2(\omega t + \phi)\right] = \tfrac{1}{2}kA^2

The bracket is 1 by the Pythagorean identity, and that identity is printed: the booklet's Identities table gives $\sin^2\theta + \cos^2\theta = 1$ . So conservation of energy for a simple harmonic oscillator is, algebraically, that one identity. Statement 7.4.A.2 is a physical claim and this is its arithmetic.

What 7.4.A.2 assumes is worth naming, since it is the assumption a real oscillator breaks. Constant total energy means nothing is removing energy from the system: no friction, no drag, no inelastic losses in the spring. The exam conventions box states that air resistance is assumed to be negligible unless otherwise stated and that springs and strings are assumed to be ideal unless otherwise stated, which is where that permission comes from. A real oscillator that visibly dies away has an amplitude falling with time, and therefore, by 7.4.A.4.ii, a total energy falling with it. Its period, by 7.3.A.5, is essentially unchanged, which is why a dying pendulum still keeps time.

Energy against position: two parabolas that add to a horizontal line

Plot energy against position rather than against time and the whole topic becomes one picture.

U(x) = \tfrac{1}{2}kx^2 \qquad K(x) = E - U = \tfrac{1}{2}k\left(A^2 - x^2\right)

$U$ is an upward parabola with its minimum at the equilibrium position. $K$ is a downward parabola with its maximum there, reaching zero at $x = \pm A$ . Their sum is a horizontal line at $\frac{1}{2}kA^2$ . The motion is confined to the region where $K \geq 0$ , which is $|x| \leq A$ , so the turning points are where the horizontal line meets the $U$ parabola.

Four readings that answer most Topic 7.4 questions:

At $x = 0$ : $U = 0$ , $K = E$ , speed maximum, acceleration zero. This is 7.4.A.3.
At $x = \pm A$ : $U = E$ , $K = 0$ , speed zero, acceleration maximum. This is 7.4.A.4 with 7.4.A.4.i.
At $x = \pm A/\sqrt{2}$ : $U = K = E/2$ , since $U/E = (x/A)^2 = 1/2$ . That is at about 0.707 of the amplitude, not at half of it, and "halfway out" is the standard wrong answer.
At $x = \pm A/2$ : $U = E/4$ and $K = 3E/4$ , so the object still has $\sqrt{3}/2 = 0.866$ of its maximum speed at half amplitude. Speed falls off slowly near the middle and fast near the ends, which is why an oscillator spends more of its time near the turning points than near the centre.

Energy bar charts are the representation the CED asks for here, and the unit's Preparing for the AP Exam note names them: its worked Unit 7 example asks a student 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. The consistency is the point. At maximum displacement the free-body diagram shows the largest net force and the bar chart shows all the energy as potential. At equilibrium the free-body diagram shows zero net force and the bar chart shows all of it as kinetic.

Energy against time: both forms cycle twice per period

The unit's fourth sample instructional activity asks for energy against time, not just against position: give students a graph of position, velocity, or acceleration for SHM and have them produce the other two on the same time scale, along with force, momentum, kinetic energy, potential energy and total energy against time. The energy graphs behave in a way the position graph does not prepare you for, and it is worth seeing why.

From the previous section, with $\phi = 0$ for simplicity:

U(t) = \tfrac{1}{2}kA^2\cos^2(\omega t) \qquad K(t) = \tfrac{1}{2}kA^2\sin^2(\omega t)

Both are squares of sinusoids, and a squared sinusoid is never negative. That single observation gives every feature of the graphs:

Both curves sit entirely on or above the time axis, oscillating between 0 and $\frac{1}{2}kA^2$ . Energy graphs never go negative here, unlike position, velocity and acceleration, which all spend half their time below the axis.
Each completes a full cycle in half a period. A cosine returns to $+1$ after $2\pi$ , but a cosine squared returns to 1 after $\pi$ , because it takes the same value at $\theta$ and at $\theta + \pi$ . So the energy curves repeat at $T/2$ , twice as often as the motion.
$U$ and $K$ are exact mirror images in the horizontal line at $E/2$ , since they add to a constant.
They cross four times per period, at every point where $|x| = A/\sqrt{2}$ .

That doubling is a favourite question, and reasoning it out beats memorising it: the oscillator passes through the equilibrium position twice per cycle and reaches a turning point twice per cycle, so the kinetic energy peaks twice and the potential energy peaks twice. Energy does not care which direction the object is moving, and it does not care which side of the equilibrium it is on.

One small honesty. The trigonometric identity that rewrites $\cos^2\theta$ as a single cosine at twice the angle is not printed on this sheet, and neither is the corresponding form for $\sin^2\theta$ . The Identities table prints $\sin^2\theta + \cos^2\theta = 1$ and $\sin(2\theta) = 2\sin\theta\cos\theta$ , and that is all. You do not need the missing one: the period-halving argument above is complete without it.

The other two graphs the activity asks for are simpler and often forgotten. The force graph is $F = -kx$ , which is the position graph flipped and rescaled, exactly like the acceleration graph of Topic 7.3. The momentum graph is $p = mv$ , which is the velocity graph rescaled with no change of shape at all.

Speed at any position, and the relation that is not printed

The most-used result of this topic is not on the equation sheet in any form, and producing it is two lines. Set the total energy at an arbitrary position equal to the total energy at the turning point:

\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2

v = \pm\sqrt{\frac{k}{m}\left(A^2 - x^2\right)} = \pm\,\omega\sqrt{A^2 - x^2}

The second form is the one to carry, because $\omega$ is already the quantity you extracted from the differential equation and it needs neither $k$ nor $m$ separately. Setting $x = 0$ recovers $v_{\max} = A\omega$ , which is statement 7.3.A.3.ii, so the two routes agree as they must.

The $\pm$ is not decoration. Energy is a scalar and cannot tell you which way the object is going, so an energy calculation always returns a speed and leaves the direction to be argued separately. If a question asks for a velocity rather than a speed, that argument is part of the answer.

A useful second identity falls out of the same equation. Divide through by $\frac{1}{2}kA^2$ :

\left(\frac{v}{v_{\max}}\right)^2 + \left(\frac{x}{A}\right)^2 = 1

So a graph of velocity against position, which is called a phase plot, is an ellipse, and the fractional speed and fractional displacement behave like the sine and cosine of the same angle. That is why the numbers in the previous section came out as they did: at half amplitude the speed fraction is $\sqrt{1 - 1/4} = 0.866$ , and at $1/\sqrt{2}$ of the amplitude both fractions are $1/\sqrt{2}$ and the energies are equal. It is also the same Pythagorean structure that determined the amplitude from initial conditions in Topic 7.3.

For a rotational oscillator the same construction runs with rotational quantities. Replace $\frac{1}{2}mv^2$ with the printed $K_{\text{rot}} = \frac{1}{2}I\omega_{\text{rot}}^2$ and the maximum angular speed is $\theta_{\max}$ times the angular frequency, giving a total energy of $\frac{1}{2}I\omega^2\theta_{\max}^2$ . Worked example 3 evaluates that for a physical pendulum and checks it against the gravitational energy directly.

The half k A squared result belongs to a spring, and not to every oscillator

Statement 7.4.A.4.ii is careful in a way that is easy to skip. Its equation is introduced as the "relevant equation for a spring and object system", not as a general result for simple harmonic motion. Everything above it in the statement is general; the formula is not.

What is general:

The total energy is the sum of kinetic and potential, and it is constant (7.4.A.1 and 7.4.A.2).
Kinetic energy peaks where potential energy is least, and the other way round (7.4.A.3 and 7.4.A.4).
The minimum kinetic energy is zero (7.4.A.4.i).
Changing the amplitude changes the maximum potential energy and therefore the total energy (7.4.A.4.ii's own first sentence).
The total energy goes as the square of the amplitude, for any simple harmonic oscillator, because the potential energy is quadratic in the displacement by definition.

What is not general is the identification of $k$ with a spring constant and $A$ with a linear displacement. For a pendulum there is no spring and no spring constant, and the stored energy is gravitational. Its total energy is still proportional to the square of its amplitude, but written in the right variables.

Oscillator	Stored energy	Total energy
Object and ideal spring	elastic, $\frac{1}{2}kx^2$	$\frac{1}{2}kA^2$
Simple or physical pendulum	gravitational, from the rise of the centre of mass	$\frac{1}{2}mgd\,\theta_{\max}^2$ for small angles
Torsion pendulum	elastic, in the twisted wire	$\frac{1}{2}k\,\theta_{\max}^2$ , with $k$ a torque per unit angle

Only the first row's formula is in the CED. The other two rows are worth knowing because they show the pattern rather than a list: in every case the total energy is one half the restoring coefficient times the square of the amplitude, where the restoring coefficient is whatever multiplies the displacement in the restoring force or torque. For a spring it is $k$ ; for a pendulum, from statement 7.5.A.2.ii's $\tau = -mgd\theta$ , it is $mgd$ ; for a torsion pendulum, from 7.5.A.4's $I\alpha = -k\Delta\theta$ , it is that $k$ . Worked example 3 derives the pendulum row and checks it numerically against the height the centre of mass actually rises.

A warning on notation, because the last row is where it bites. The symbol $k$ means a spring constant in newtons per metre in the first row and a torsion constant in newton metres per radian in the third. The equation sheet's variable list defines $k$ as "spring constant" and the CED reuses the symbol in 7.5.A.4 for the torsion case. Check the units before substituting.

What the sheet prints, and how Topic 7.4 is tested

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

Relation	On the C: Mechanics sheet
$K = \frac{1}{2}mv^2$	yes
$U_s = \frac{1}{2}k(\Delta x)^2$	yes
$\Delta U_g = mg\Delta y$	yes
$K_{\text{rot}} = \frac{1}{2}I\omega^2$	yes
$\Delta U = -\int_a^b \vec{F}_{\text{cf}}(r) \cdot d\vec{r}$	yes, and absent from the AP Physics 1 sheet
$F_x = -\dfrac{dU(x)}{dx}$	yes, and absent from the AP Physics 1 sheet
$E_{\text{total}} = U + K$	no, it is statement 7.4.A.1 only
$E_{\text{total}} = \frac{1}{2}kA^2$	no, it is statement 7.4.A.4.ii only
$v = \omega\sqrt{A^2 - x^2}$	no
$v_{\max} = A\omega$	no, it is statement 7.3.A.3.ii

So the two equations this topic is built on are both unprinted, and both are one line from things that are printed. Also printed and useful here: $W = \int_a^b \vec{F} \cdot d\vec{r}$ , $\Delta K = \sum W_i$ , and, in the booklet's Calculus and Identities tables, the power rule for integration and $\sin^2\theta + \cos^2\theta = 1$ .

On assessment: Unit 7 is weighted 10 to 15% of the multiple-choice section, at about 12 to 17 class periods. The exam is 3 hours, with 42 multiple-choice questions in 85 minutes and 4 free-response questions in 95 minutes, each worth half the score, and a calculator is allowed throughout. Skill 2.B carries 20 to 25% of the multiple-choice section and 2.D carries 10 to 15%; both are suggested here. Skill 3.C carries 5 to 10% there and Science Practice 3 carries 30 to 35% of the free-response section. Skill 1.C points at the free-response section only, since Practice 1 is not assessed on the multiple-choice section. Required course content can be assessed with any skill.

Neither of the CED's two Unit 7 sample multiple-choice questions aligns to 7.4.A: question 5 aligns to 7.2.A and question 12 to 7.3.A. None of the four sample free-response questions aligns to Unit 7 at all.

Skill 3.C is the one to prepare for in words rather than numbers. The unit's Preparing for the AP Exam note, in its guidance on justifications elsewhere in the framework, is blunt that simply referencing an equation, law, or physical principle is not sufficient. On an energy question that means naming which energy grew, which shrank, and by what mechanism, not writing "conservation of energy" and stopping. The conservation of energy guide owns the general accounting routine and the simple harmonic motion guide owns the oscillator routines.

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

AP Physics 1 has a Topic 7.4 with the same title, and the four qualitative statements about where each energy peaks are close to identical in the two frameworks. Say the honest part plainly: on the bare statements, these two topics are nearly the same topic.

The difference is what surrounds them. The [AP Physics 1 Topic 7.4 page](/ap-physics-1/unit-7-oscillations/7-4-energy-of-simple-harmonic-oscillators) is for students in the algebra-based course, who are handed the spring potential energy as a formula and work with energy at a small number of named positions. This page is for students in AP Physics C: Mechanics, whose sheet also prints the potential energy as an integral of a conservative force and the force as the negative derivative of the potential energy, so the energy can be produced from the force law, read off the curvature of a graph, or written as a function of time and differentiated. If you are in Physics 1, neither of those two printed lines is on your sheet and no integral 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 7.1 identifies SHM from the shape of the potential energy curve, Topic 7.2 supplies the $\omega$ that appears in every energy expression here, Topic 7.3 is the graph work, and Topic 7.5 supplies the rotational oscillators in worked example 3 and has no algebra-based counterpart. The conservation of energy guide is the general accounting routine this topic specialises, and Topic 6.1 is where the rotational kinetic energy in worked example 3 comes from. The Unit 7 hub lists all five topics.

One oscillator, four positions, two independent routes to the speed

A 0.40 kg block on a frictionless surface oscillates on an ideal spring of force constant $k = 90\ \mathrm{N/m}$ with amplitude $A = 0.060$ m. Find (a) the total energy, the angular frequency and the period, (b) the maximum speed, two ways, (c) the kinetic energy, potential energy and speed at $x = 0.036$ m, two ways, and (d) the position where the kinetic and potential energies are equal.

Declare the convention: $x$ is the displacement from the equilibrium position, and the zero of potential energy is at $x = 0$ , which is what makes $U_s = \frac{1}{2}k(\Delta x)^2$ the right expression.
(a) Statement 7.4.A.4.ii gives the total energy from the amplitude: $E = \frac{1}{2}kA^2 = \frac{1}{2}(90)(0.060)^2 = \frac{1}{2}(90)(3.600 \times 10^{-3}) = 0.1620\ \mathrm{J}$ .
The equation of motion is $d^2x/dt^2 = -(k/m)x$ , so $\omega^2 = 90/0.40 = 225\ \mathrm{rad^2/s^2}$ and $\omega = 15.0\ \mathrm{rad/s}$ . The printed $T = 2\pi/\omega$ gives $T = 2\pi/15.0 = 0.4189 \approx 0.419\ \mathrm{s}$ .
(b) Energy route: at $x = 0$ all of $E$ is kinetic, so $\frac{1}{2}mv_{\max}^2 = 0.1620$ J and $v_{\max} = \sqrt{2(0.1620)/0.40} = \sqrt{0.8100} = 0.900\ \mathrm{m/s}$ .
Kinematic route: statement 7.3.A.3.ii gives $v_{\max} = A\omega = (0.060)(15.0) = 0.900\ \mathrm{m/s}$ . The two agree exactly, which they must, since $\frac{1}{2}m(A\omega)^2 = \frac{1}{2}mA^2(k/m) = \frac{1}{2}kA^2$ .
For completeness, $a_{\max} = A\omega^2 = (0.060)(225) = 13.5\ \mathrm{m/s^2}$ , at the turning points where the speed is zero.
(c) Energy route at $x = 0.036$ m, which is $0.600A$ : $U = \frac{1}{2}(90)(0.036)^2 = \frac{1}{2}(90)(1.296 \times 10^{-3}) = 0.05832\ \mathrm{J}$ .
Then $K = E - U = 0.1620 - 0.05832 = 0.10368\ \mathrm{J}$ by 7.4.A.1, and $v = \sqrt{2K/m} = \sqrt{2(0.10368)/0.40} = \sqrt{0.5184} = 0.720\ \mathrm{m/s}$ .
Direct route with the relation derived above: $v = \omega\sqrt{A^2 - x^2} = 15.0\sqrt{(0.060)^2 - (0.036)^2} = 15.0\sqrt{3.600 \times 10^{-3} - 1.296 \times 10^{-3}} = 15.0\sqrt{2.304 \times 10^{-3}} = 15.0(0.04800) = 0.720\ \mathrm{m/s}$ . Identical.
Sanity check the fractions rather than the joules: at $0.600A$ the potential energy fraction is $(0.600)^2 = 0.360$ and the speed fraction is $\sqrt{1 - 0.360} = 0.800$ , and indeed $0.05832/0.1620 = 0.360$ and $0.720/0.900 = 0.800$ . The 3, 4, 5 triangle again.
(d) Setting $U = K$ means each is $E/2$ , so $\frac{1}{2}kx^2 = \frac{1}{2}\left(\frac{1}{2}kA^2\right)$ , giving $x = \pm A/\sqrt{2} = \pm 0.060/1.4142 = \pm 0.04243 \approx \pm 0.0424\ \mathrm{m}$ .
Each energy there is $0.1620/2 = 0.0810$ J, and the speed is $v_{\max}/\sqrt{2} = 0.900/1.4142 = 0.6364 \approx 0.636\ \mathrm{m/s}$ . Note that this is at 70.7% of the amplitude, not at 50%, which is the standard error on this question.

(a) $E = 0.162$ J, $\omega = 15.0\ \mathrm{rad/s}$ , $T = 0.419$ s. (b) $v_{\max} = 0.900$ m/s from energy and from $A\omega$ , identically. (c) At $x = 0.036$ m: $U = 0.0583$ J, $K = 0.1037$ J and $v = 0.720$ m/s, confirmed by $\omega\sqrt{A^2 - x^2}$ . (d) The energies are equal at $x = \pm 0.0424$ m, that is $A/\sqrt{2}$ , where each is 0.0810 J and the speed is 0.636 m/s.

The same block against time, and why the energy graphs repeat twice as often

The block of worked example 1 is released from rest at $x = +A$ at $t = 0$ . Write the kinetic and potential energies as functions of time, evaluate both at $t = 0$ , $T/8$ , $T/4$ , $3T/8$ and $T/2$ , and state the period of each energy graph.

Released from rest at $+A$ makes the phase angle zero, so the printed position function is $x = (0.060\ \mathrm{m})\cos\left[(15.0\ \mathrm{rad/s})t\right]$ and its derivative is $v = -(0.900\ \mathrm{m/s})\sin(15.0t)$ .
Potential energy: $U = \frac{1}{2}kx^2 = \frac{1}{2}(90)(0.060)^2\cos^2(15.0t) = (0.1620\ \mathrm{J})\cos^2(15.0t)$ .
Kinetic energy: $K = \frac{1}{2}mv^2 = \frac{1}{2}(0.40)(0.900)^2\sin^2(15.0t) = (0.1620\ \mathrm{J})\sin^2(15.0t)$ . The two prefactors came out equal, and that is not a coincidence: $\frac{1}{2}m(A\omega)^2$ and $\frac{1}{2}kA^2$ are the same number whenever $\omega^2 = k/m$ .
The sum is $(0.1620)\left[\cos^2 + \sin^2\right] = 0.1620$ J at every instant, by the identity printed in the booklet's Identities table. That is statement 7.4.A.2 in algebra.
With $T = 0.4189$ s, the sampling times are $T/8 = 0.05236$ s, $T/4 = 0.1047$ s, $3T/8 = 0.1571$ s and $T/2 = 0.2094$ s. The corresponding arguments $15.0t$ are 0, $\pi/4$ , $\pi/2$ , $3\pi/4$ and $\pi$ radians, that is 0, 45, 90, 135 and 180 degrees, all of which appear in the sheet's table of trigonometric values.

$t$	$15.0t$	$x$ (m)	$U$ (J)	$K$ (J)	$U + K$ (J)
0	0	$+0.060$	0.1620	0	0.1620
$T/8$	45 $^\circ$	$+0.0424$	0.0810	0.0810	0.1620
$T/4$	90 $^\circ$	0	0	0.1620	0.1620
$3T/8$	135 $^\circ$	$-0.0424$	0.0810	0.0810	0.1620
$T/2$	180 $^\circ$	$-0.060$	0.1620	0	0.1620

Read the first and last rows together. At $t = 0$ the block is at $+A$ and at $t = T/2$ it is at $-A$ , which are opposite ends of the motion, and yet the energies are identical. Energy is a scalar, so it cannot distinguish the two ends.
That is the whole reason for the doubling. The potential energy has returned to its maximum after only half a period, so the potential energy graph has period $T/2$ , and so does the kinetic energy graph. Algebraically, $\cos^2\theta$ takes the same value at $\theta$ and at $\theta + \pi$ .
The energy graphs therefore repeat at $0.4189/2 = 0.2094$ s while the position graph repeats at 0.4189 s. Both energy curves stay between 0 and 0.1620 J and never go negative, unlike position, velocity and acceleration, which are all negative half the time.
The two curves cross wherever $U = K = 0.0810$ J, which happens at $T/8$ , $3T/8$ , $5T/8$ and $7T/8$ : four times per period of the motion, at the four instants when $|x| = A/\sqrt{2} = 0.0424$ m, matching part (d) of worked example 1.
Two more graphs the CED's fourth sample activity asks for. Force: $F = -kx = -(5.40\ \mathrm{N})\cos(15.0t)$ , the position graph flipped and rescaled, peaking at 5.40 N at the turning points. Momentum: $p = mv = -(0.360\ \mathrm{kg \cdot m/s})\sin(15.0t)$ , the velocity graph rescaled with the same shape. Both of those keep the period $T$ ; only the energy graphs halve it.

$U = (0.162\ \mathrm{J})\cos^2(15.0t)$ and $K = (0.162\ \mathrm{J})\sin^2(15.0t)$ , summing to 0.162 J at every instant. The sampled values are 0.162 and 0, then 0.081 and 0.081, then 0 and 0.162, then 0.081 and 0.081, then 0.162 and 0. Each energy graph has period $T/2 = 0.209$ s, half the period of the motion, because a squared sinusoid repeats twice as often and energy cannot tell the two ends of the swing apart.

A physical pendulum, where the stored energy is gravitational

A uniform rod of mass $M = 0.45$ kg and length $L = 0.55$ m is pivoted at one end and swings in a vertical plane with angular amplitude $\theta_{\max} = 0.12$ rad. Its rotational inertia about the pivot is $\frac{1}{3}ML^2$ . Take $g = 9.8\ \mathrm{m/s^2}$ . Find (a) the rotational inertia, the distance to the centre of mass and the period, (b) the maximum angular speed, (c) the total energy, and (d) check that total against the height the centre of mass actually rises.

Declare the convention: $\theta$ is the angular displacement from the vertical, positive on one side, and the zero of gravitational potential energy is at the lowest point of the swing.
(a) $I = \frac{1}{3}ML^2 = \frac{1}{3}(0.45)(0.55)^2 = \frac{1}{3}(0.45)(0.3025) = 0.045375\ \mathrm{kg \cdot m^2}$ , and for a uniform rod the centre of mass is at $d = L/2 = 0.275$ m.
$Mgd = (0.45)(9.8)(0.275) = 1.21275\ \mathrm{N \cdot m}$ , so the printed $T_{\text{phys}} = 2\pi\sqrt{I/(Mgd)} = 2\pi\sqrt{0.045375/1.21275} = 2\pi\sqrt{0.0374150} = 2\pi(0.193429) = 1.2154 \approx 1.22\ \mathrm{s}$ .
The angular frequency of the oscillation is $\omega = 2\pi/T = \sqrt{Mgd/I} = \sqrt{1.21275/0.045375} = \sqrt{26.7273} = 5.1698\ \mathrm{rad/s}$ . Keep the two angular quantities apart: $\omega$ is the angular frequency of the swing, and $d\theta/dt$ is the rod's instantaneous angular speed.
(b) The angular position is $\theta = \theta_{\max}\cos(\omega t)$ , so differentiating gives $d\theta/dt = -\theta_{\max}\omega\sin(\omega t)$ , whose magnitude peaks at $\theta_{\max}\omega = (0.12)(5.1698) = 0.62038 \approx 0.620\ \mathrm{rad/s}$ . This is the rotational twin of 7.3.A.3.ii's $v_{\max} = A\omega$ , not a separate printed result.
(c) All the energy is kinetic at the lowest point, where the printed $K_{\text{rot}} = \frac{1}{2}I\omega_{\text{rot}}^2$ applies with the maximum angular speed: $E = \frac{1}{2}(0.045375)(0.62038)^2 = \frac{1}{2}(0.045375)(0.384873) = 8.7318 \times 10^{-3}\ \mathrm{J}$ , about 8.73 mJ.
Notice what that equals symbolically: $E = \frac{1}{2}I\omega^2\theta_{\max}^2$ , and since $\omega^2 = Mgd/I$ , the rotational inertia cancels to leave $E = \frac{1}{2}Mgd\,\theta_{\max}^2$ . Numerically $\frac{1}{2}(1.21275)(0.12)^2 = \frac{1}{2}(1.21275)(0.0144) = 8.7318 \times 10^{-3}$ J, the same.
That is the pattern from the section above: one half the restoring coefficient times the square of the amplitude, with $Mgd$ playing the part $k$ plays for a spring. It follows from statement 7.5.A.2.ii, whose small-angle restoring torque is $\tau = -Mgd\theta$ .
(d) Check it against the actual gravitational energy. The centre of mass rises by $\Delta y = d\left(1 - \cos\theta_{\max}\right) = 0.275(1 - \cos 0.12) = 0.275(1 - 0.9928086) = 0.275(0.0071914) = 1.9776 \times 10^{-3}\ \mathrm{m}$ .
The printed $\Delta U_g = Mg\Delta y$ gives $(0.45)(9.8)(1.9776 \times 10^{-3}) = 8.7213 \times 10^{-3}\ \mathrm{J}$ , about 8.72 mJ.
The two totals differ by $(8.7318 - 8.7213)/8.7213 = 0.12\%$ . They are not identical because the simple harmonic model itself is an approximation for a pendulum: statement 7.5.A.2.ii replaces $\sin\theta$ with $\theta$ , and part (c) inherits that. At 0.12 rad, which is 6.9 degrees, the approximation costs about one part in a thousand of the energy.
The direction of the discrepancy is the useful part. The linear model slightly overstates the restoring effect, so it overstates the energy stored at a given amplitude, which is why 8.73 mJ came out above 8.72 mJ. The same approximation makes the model's period slightly short compared with a real pendulum's.

(a) $I = 0.0454\ \mathrm{kg \cdot m^2}$ , $d = 0.275$ m, $T = 1.22$ s. (b) The maximum angular speed is $\theta_{\max}\omega = 0.620$ rad/s. (c) $E = 8.73$ mJ, which equals $\frac{1}{2}Mgd\,\theta_{\max}^2$ with the rotational inertia cancelling out. (d) The centre of mass rises 1.98 mm, giving 8.72 mJ directly, 0.12% below the harmonic value, which is the cost of the small-angle approximation.

Frequently asked questions

What is the total energy of a simple harmonic oscillator in AP Physics C?

Essential knowledge 7.4.A.1 says the total energy of a system exhibiting simple harmonic motion is the sum of its kinetic and potential energies, and 7.4.A.2 says conservation of energy makes that total constant. For a spring and object system, statement 7.4.A.4.ii gives it as one half the force constant times the amplitude squared. Neither of those equations is printed on the AP Physics C: Mechanics equation sheet, though the pieces are: the sheet prints one half m v squared for kinetic energy and one half k times the displacement squared for spring potential energy. Because the energy goes as the square of the amplitude, doubling the amplitude quadruples the energy while leaving the period unchanged.

Where is the kinetic energy maximum in simple harmonic motion?

At the equilibrium position, where the potential energy is at its minimum and the speed is greatest. Essential knowledge 7.4.A.3 states that pairing directly, and 7.4.A.4 states the reverse: the potential energy is at a maximum when the kinetic energy is at a minimum, which is at the two turning points. Statement 7.4.A.4.i adds that the minimum kinetic energy of a system exhibiting simple harmonic motion is zero, so at maximum displacement the object is momentarily at rest and every joule in the system is stored. That is the fact that lets you compute the total energy from the amplitude alone.

How do you find the speed of an oscillator at a given position?

Set the total energy at that position equal to the total energy at the turning point: one half m v squared plus one half k x squared equals one half k A squared. Solving gives the speed as the angular frequency times the square root of the amplitude squared minus the position squared. That relation is not printed on any AP equation sheet, and it is two lines from things that are. Setting the position to zero recovers the maximum speed as the amplitude times the angular frequency, which is essential knowledge 7.3.A.3.ii. The result is a speed rather than a velocity, because energy is a scalar and cannot say which way the object is moving.

Why do the kinetic and potential energy graphs oscillate twice as fast as the position graph?

Because both energies depend on the square of a sinusoid, and a squared sinusoid repeats twice as often as the sinusoid itself. Physically, the oscillator passes through the equilibrium position twice in every cycle, so the kinetic energy peaks twice, and it reaches a turning point twice, so the potential energy peaks twice. Energy is a scalar, so it cannot tell the left turning point from the right one, or a rightward pass through equilibrium from a leftward one. Both energy curves also stay on or above the axis, unlike position, velocity and acceleration, which are negative half the time, and they cross each other four times per period.

If you double the amplitude of an oscillator, what happens to its energy and its period?

The energy quadruples and the period does not change. Essential knowledge 7.4.A.4.ii says changing the amplitude changes the maximum potential energy and therefore the total energy, and for a spring and object system that energy is one half the force constant times the amplitude squared, so it scales as the square. Essential knowledge 7.3.A.5, in the previous topic, says changing the amplitude will not change the period. The maximum speed and maximum acceleration also double, since both are proportional to the amplitude. So everything about the strength of the motion scales up and the timing is untouched.

Does the one half k A squared formula work for a pendulum?

Not as written. Statement 7.4.A.4.ii introduces that equation as the relevant equation for a spring and object system, and a pendulum has no spring and no spring constant. Its stored energy is gravitational and comes from the rise of the centre of mass. The pattern still holds in the right variables: for any simple harmonic oscillator the total energy is one half the restoring coefficient times the square of the amplitude, where the restoring coefficient is whatever multiplies the displacement in the restoring force or torque. For a small-angle pendulum that coefficient is the mass times the gravitational field strength times the distance from the pivot to the centre of mass, and the amplitude is an angle in radians.

Where does the spring potential energy formula come from in AP Physics C?

From integrating the spring force. The AP Physics C: Mechanics equation sheet prints the change in potential energy as minus the integral of a conservative force over displacement, a line that appears nowhere on the AP Physics 1 sheet. Integrating the spring force from the equilibrium position out to a displacement gives one half the force constant times that displacement squared, using the power rule printed in the booklet's calculus table. The sheet also prints the reverse relation, the force as minus the derivative of the potential energy with respect to position, which is how you go from an energy curve back to a force and how you identify simple harmonic motion from the curvature of a potential energy graph.