AP Physics C: Mechanics · Unit 7 of 7

Unit 7: Oscillations

10-15% of the multiple-choice section5 topics

Topics in this unit

7.1Defining Simple Harmonic Motion (SHM)
7.2Frequency and Period of SHM
7.3Representing and Analyzing SHM
7.4Energy of Simple Harmonic Oscillators
7.5Simple and Physical Pendulums

Oscillations is Unit 7 of AP Physics C: Mechanics, worth 10 to 15 percent of the multiple-choice section over about 12 to 17 class periods. Five topics, five learning objectives, one per topic. Simple harmonic motion here is the solution of a second-order differential equation.

AP Physics: Unit 7 (topics 7.1 Defining Simple Harmonic Motion (SHM), 7.2 Frequency and Period of SHM, 7.3 Representing and Analyzing SHM, 7.4 Energy of Simple Harmonic Oscillators, 7.5 Simple and Physical Pendulums). Unit 7 of the current AP Physics C: Mechanics course and exam description, weighted 10 to 15% of the multiple-choice section at about 12 to 17 class periods, against 5 to 8% and four topics for the identically titled AP Physics 1 Unit 7. Five topics and five learning objectives, one per topic, all using the task verb describe. The unit prints exactly one boundary statement, under Topic 7.3: AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes SHM, as well as be able to identify SHM, and does not expect students to mathematically prove that the solution is correct. Topics 7.1, 7.2, 7.4 and 7.5 print none. Equation-sheet facts verified against the AP Physics C: Mechanics Table of Information: the only position function printed is x = x_max cos(omega t + phi), and neither of the two 2*pi*f*t forms given in essential knowledge 7.3.A.1 appears, which is the reverse of the AP Physics 1 sheet, where both 2*pi*f*t forms are printed and no phase angle exists in the variable list. Both pendulum periods are printed on the C: Mechanics sheet, the simple pendulum 2*pi*sqrt(l/g) and the physical pendulum 2*pi*sqrt(I/(mgd)); the AP Physics 1 sheet prints the simple pendulum only. Not printed despite appearing in required content: the second-order differential equation itself, a = -omega^2 x, v_max = A*omega, a_max = A*omega^2, E_total = U + K, E_total = kA^2/2, the physical pendulum restoring torque, the small-angle approximation, and the torsion pendulum relation I*alpha = -k*delta-theta. Suggested skills by topic: 7.1 uses 1.A, 2.C, 3.B, 3.C; 7.2 uses 1.C, 2.A, 2.D, 3.B; 7.3 uses 1.C, 2.A, 2.C, 3.B; 7.4 uses 1.C, 2.B, 2.D, 3.C; 7.5 uses 1.B, 2.A, 2.B, 3.A, 3.B.

What the CED requires across Unit 7

Unit 7 of AP Physics C: Mechanics is Oscillations. The course and exam description weights it at 10 to 15% of the multiple-choice section and suggests about 12 to 17 class periods. That is the same band as Units 1, 5 and 6, and it is roughly double the 5 to 8% that AP Physics 1 gives its own Unit 7.

Five topics, five learning objectives, one per topic. No other structure in this unit; the depth is inside the essential knowledge statements.

Topic	Learning objective	Suggested skills
7.1 Defining Simple Harmonic Motion (SHM)	7.1.A	1.A, 2.C, 3.B, 3.C
7.2 Frequency and Period of SHM	7.2.A	1.C, 2.A, 2.D, 3.B
7.3 Representing and Analyzing SHM	7.3.A	1.C, 2.A, 2.C, 3.B
7.4 Energy of Simple Harmonic Oscillators	7.4.A	1.C, 2.B, 2.D, 3.C
7.5 Simple and Physical Pendulums	7.5.A	1.B, 2.A, 2.B, 3.A, 3.B

All five objectives open with the task verb "describe". The CED says that verb, used in nearly all learning objectives, "encompasses the range of possible graphical, mathematical, or verbal skill applications", and that students should be able to describe a physical concept graphically, mathematically, and verbally.

The CED's framing is that in Unit 7 students apply previously encountered models and methods of analysis to simple harmonic motion, and are reminded that even in new situations the fundamental laws of physics remain the same. It adds that because this unit is the first in which students possess all the tools of force, energy, and momentum analysis, including energy bar charts, free-body diagrams, and momentum diagrams, scaffolding lessons will enhance student understanding of fundamental physics principles and their limitation as they relate to oscillating systems.

The essential questions on the unit opener are how oscillations can be used to make our lives easier and more comfortable, how an astronaut can be weighed in space, how you could measure the length of a long string with a stopwatch, and what a child on a swing, a beating heart and a metronome have in common. Topic 7.5's torsion pendulum answers the second of those, because its period contains no $g$ at all.

SHM is the solution of a differential equation, not an assumed sinusoid

This is the difference between the two courses, and the CED states it in one statement and one boundary statement.

Statement 7.3.A.2 says the position as a function of time for an object exhibiting SHM "is a solution of the second-order differential equation derived from the application of Newton's second law", and gives that equation, labelled a Derived equation:

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

So the logic runs forward, not backward. You write Newton's second law for the restoring force, rearrange it into that shape, read $\omega$ off the coefficient, and only then write down a cosine. For a spring, $ma = -k\Delta x$ becomes $d^2x/dt^2 = -(k/m)x$ , so $\omega^2 = k/m$ , so $T_s = 2\pi\sqrt{m/k}$ . The period is not memorised; it falls out of a coefficient.

Unit 7 prints exactly one boundary statement across all five topics, and it sits here, under Topic 7.3. Quoted whole, both sentences:

"AP Physics C: Mechanics only expects students to know the solution to the second-order differential equation that describes SHM, as well as be able to identify SHM. AP Physics C: Mechanics does not expect students to mathematically prove that the solution is correct."

Read that carefully in both directions. You are expected to know the solution and to identify SHM when you see it, which means recognising the differential equation's shape in a new situation. You are not expected to substitute the cosine back in and verify it. Topics 7.1, 7.2, 7.4 and 7.5 print no boundary statement at all.

The second-order form returns in Topic 7.5. Statement 7.5.A.2.iii says the small-angle approximation and Newton's second law in rotational form yield a second-order differential equation that describes SHM, and prints its angular twin:

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

That is the same equation with a different variable, which is why one method covers springs, pendulums, physical pendulums and torsion pendulums instead of four remembered period formulas.

The rest of the unit hangs off that coefficient. Statement 7.3.A.3.i labels $a = -\omega^2 x$ a Derived equation, and 7.3.A.3.ii labels $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$ Derived equations, saying "it can be shown that" the maximum velocity and acceleration are related to the angular frequency of the object's motion. All three follow from differentiating the solution.

The equation-sheet trap: which position function is printed

This is the row most worth checking before an exam, because the two courses print different things and the framework prints a third.

Statement 7.3.A.1 of the AP Physics C: Mechanics framework says that for an object exhibiting SHM the displacement measured from equilibrium "can be represented by the equations" $x = A\cos(2\pi ft)$ or $x = A\sin(2\pi ft)$ . Statement 7.3.A.3 then says characteristics of SHM can be determined by or derived from $x = A\cos(\omega t + \phi)$ .

The AP Physics C: Mechanics equation sheet prints one position function:

x = x_{\max}\cos(\omega t + \phi)

It does not print either $2\pi ft$ form. It uses $x_{\max}$ where the framework uses $A$ , and its variable list defines $\phi$ as the phase angle. There is no $A$ in that list at all.

The AP Physics 1 sheet prints the opposite pair: $x = A\cos(2\pi ft)$ and $x = A\sin(2\pi ft)$ , and no $\omega t + \phi$ form anywhere. Its variable list defines $A$ as amplitude or area and contains no phase angle.

So a student who learned SHM in the algebra-based course and walks into this exam expecting the sine option will not find it, and a student who only ever wrote $\cos(\omega t + \phi)$ will find that the AP Physics 1 sheet has no place to put a phase constant. Both are the same physics; only one line is printed per course, and it is a different line.

The practical consequence is that on this exam the phase constant is the tool for handling any starting condition other than release from maximum displacement. Released from rest at $x = +A$ , $\phi = 0$ . Released from rest at $x = -A$ , $\phi = \pi$ . Passing through equilibrium moving in the $+x$ direction, $\phi = -\pi/2$ , which is the sine option written as a cosine. That is why the sine form is not needed.

Four more Unit 7 results are absent from the sheet, and all four are labelled Derived equations or their close relatives: the second-order differential equation itself, $a = -\omega^2 x$ , $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$ . Statement 7.4.A.1's $E_{\text{total}} = U + K$ is not printed either, nor is 7.4.A.4.ii's $E_{\text{total}} = \frac{1}{2}kA^2$ for a spring-object system. What the sheet does print, and what you build all of them from, is $\vec{F}_s = -k\Delta\vec{x}$ , $U_s = \frac{1}{2}k(\Delta x)^2$ , $K = \frac{1}{2}mv^2$ , $T = \frac{2\pi}{\omega} = \frac{1}{f}$ and the cosine solution.

The pendulums, and what Topic 7.5 adds

Topic 7.5, Simple and Physical Pendulums, has no AP Physics 1 counterpart. The algebra-based Unit 7 stops at four topics; this one has five, and the fifth is where the second-order differential equation earns its keep.

Statement 7.5.A.1 defines a physical pendulum as a rigid body that undergoes oscillation about a fixed axis. Statement 7.5.A.2 says that for small amplitudes its period is derived from the application of Newton's second law in rotational form, and gives

T_{\text{phys}} = 2\pi\sqrt{\frac{I}{mgd}}

where $d$ is the distance from the axis to the centre of mass. The derivation is printed in pieces. Statement 7.5.A.2.i gives the restoring torque as a Derived equation, $\tau = -mgd\sin\theta$ , and says that when displaced from equilibrium the gravitational force exerted on the pendulum's centre of mass provides a restoring torque. Statement 7.5.A.2.ii gives the small-angle approximation $\sin\theta \approx \theta$ and the resulting $\tau = -mgd\theta = I\alpha$ , also as Derived equations. Statement 7.5.A.2.iii is the differential equation.

Statement 7.5.A.3 then makes the simple pendulum a special case rather than a separate fact: a simple pendulum is a physical pendulum in which the hanging object can be modelled as a point mass at a distance $\ell$ from the pivot, with period $T_p = 2\pi\sqrt{\ell/g}$ .

Both period formulas are printed on the sheet. $T_p = 2\pi\sqrt{\ell/g}$ and $T_{\text{phys}} = 2\pi\sqrt{I/(mgd)}$ both appear in the rotational column, alongside $T_s = 2\pi\sqrt{m/k}$ . The AP Physics 1 sheet prints $T_s$ and $T_p$ and has no physical-pendulum entry, which follows from its course not having the topic.

What is not printed is everything between them: the restoring torque, the small-angle approximation, and the differential equation. Those are yours, and they are what a Mathematical Routines free-response question asks for.

Statement 7.5.A.4 closes the unit with a case that belongs to no other AP physics course. A torsion pendulum is SHM where the restoring torque is proportional to the angular displacement of a rotating system, and the CED gives the example of a horizontal disk suspended from a wire attached to its centre of mass undergoing rotational oscillations about the wire in the horizontal plane. Its Derived equation is $I\alpha = -k\Delta\theta$ . Note what is missing from it: there is no $g$ , no length, and no mass except through $I$ .

How the five topics build

7.1 Defining Simple Harmonic Motion is a definition and two conditions. Statement 7.1.A.1 says SHM is a special case of periodic motion. Statement 7.1.A.2 says SHM results when the magnitude of the restoring force exerted on an object is proportional to that object's displacement from its equilibrium position, with a Derived equation for the spring restoring force that the sheet prints in vector form as $\vec{F}_s = -k\Delta\vec{x}$ . Then 7.1.A.2.i defines a restoring force as one exerted in a direction opposite to the object's displacement from equilibrium, and 7.1.A.2.ii defines an equilibrium position as a location where the net force on the object or system is zero. Proportional and opposite: those two words are the test for whether a given motion is SHM at all.

7.2 Frequency and [Period](/glossary/period) of SHM gives $T = \frac{2\pi}{\omega} = \frac{1}{f}$ , then the spring period in 7.2.A.1.i and the small-angle simple-pendulum period in 7.2.A.1.ii. Note that the pendulum result arrives here as a fact and is derived later, in Topic 7.5.

7.3 Representing and Analyzing SHM is the largest topic in the unit by essential-knowledge count, with six statements under a single objective. It carries the position functions, the differential equation, the derived maximum values, and then three statements that get asked constantly: 7.3.A.4 on resonance, which says a system may exhibit resonance in the presence of a sinusoidal external force, that resonance occurs when an external force is exerted at the natural frequency of an oscillating system, that resonance increases the amplitude of the motion, and that the natural frequency is the frequency at which the system oscillates when displaced from equilibrium; 7.3.A.5, which says changing the amplitude of a system exhibiting SHM will not change its period; and 7.3.A.6, that properties of SHM can be determined and analyzed using graphical representations. Statements 7.3.A.1.i and 7.3.A.1.ii add that the minima, maxima and zeros of displacement, velocity and acceleration are features of harmonic motion, and that recognising where those extrema and zeros occur helps in qualitatively describing the behaviour of the motion.

7.4 Energy of Simple Harmonic Oscillators is conservation of energy applied to an oscillator: $E_{\text{total}} = U + K$ , constant by 7.4.A.2, with the kinetic energy maximum where the potential energy is minimum and the other way round. Statement 7.4.A.4.i says the minimum kinetic energy of a system exhibiting SHM is zero. Statement 7.4.A.4.ii then says that changing the amplitude changes the maximum potential energy and therefore the total energy of the system, and gives $E_{\text{total}} = \frac{1}{2}kA^2$ for a spring-object system. Read 7.3.A.5 and 7.4.A.4.ii together: amplitude changes the energy and does not change the period.

7.5 Simple and Physical Pendulums is the section above.

Traps that span more than one topic

Amplitude changes the energy, not the period. Statement 7.3.A.5 says changing the amplitude will not change the period; 7.4.A.4.ii says it will change the maximum potential energy and therefore the total energy. Both are true at once, and a question that changes the amplitude is testing whether you know which quantity moves.

Mass changes a spring's period and not a pendulum's. The CED's own sample multiple-choice question 5, aligned to objective 7.2.A and essential knowledge 7.2.A.1 with skill 2.C, is exactly this test: a 1 kg sphere on a string and a 1 kg sphere on a spring, equal lengths at equilibrium, oscillating with the same period. Replace both with 2 kg spheres at unchanged amplitudes, and the keyed answer is that the pendulum's period remains the same while the spring's period increases. The reason is visible in the two printed formulas: $T_s = 2\pi\sqrt{m/k}$ contains the mass and $T_p = 2\pi\sqrt{\ell/g}$ does not.

A physical pendulum's period is not a simple pendulum's with the same length. For a uniform rod pivoted at one end, the period is $\sqrt{2/3}$ of the simple-pendulum value for a bob at the rod's far end, because $I/(md)$ is $2\ell/3$ , not $\ell$ . Using $T_p = 2\pi\sqrt{\ell/g}$ for an extended body is the most available error in Topic 7.5.

The acceleration graph is the position graph flipped, not shifted. Statement 7.3.A.3.i gives $a = -\omega^2 x$ , so the acceleration is a negative constant times the position at every instant. The CED's sample multiple-choice question 12, aligned to 7.3.A and 7.3.A.3 with skill 3.B, gives a position-against-time graph for a block on a spring and asks for the matching acceleration graph. The minus sign is the entire question.

Maximum speed and maximum acceleration never happen together. Speed peaks at the equilibrium position, where the acceleration is zero; acceleration peaks at the extremes, where the speed is zero. They are a quarter period apart, and $v_{\max} = A\omega$ against $a_{\max} = A\omega^2$ is why doubling the frequency does different things to them.

A large restoring force is not enough for SHM. Statement 7.1.A.2 requires the magnitude of the restoring force to be proportional to the displacement. A ball rolling in a bowl is only approximately SHM, which is why one of the unit's own sample activities asks students to predict whether it is and then take data to check, testing whether the period is independent of amplitude and whether the force is proportional to displacement.

Resonance increases amplitude, not natural frequency. Statement 7.3.A.4.ii says resonance increases the amplitude of oscillating motion, and 7.3.A.4.iii defines the natural frequency as a property of the system. The driving frequency is what you tune; the natural frequency is what you tune it to.

Angles in the small-angle approximation are radians. $\sin\theta \approx \theta$ from 7.5.A.2.ii is false in degrees. Every $\theta$ in the pendulum derivations is a radian measure.

How Unit 7 is assessed

The AP Physics C: Mechanics exam is 3 hours long: 42 multiple-choice questions in 85 minutes for half the score, then 4 free-response questions in 95 minutes for the other half, always in the order Mathematical Routines (10 points, suggested 20 to 25 minutes), Translation Between Representations (12 points, 25 to 30 minutes), Experimental Design and Analysis (10 points), and Qualitative/Quantitative Translation (8 points). A four-function, scientific, or graphing calculator is allowed on both sections.

On the multiple-choice section, skill 2.A carries 25 to 30%, skill 2.B 20 to 25%, and skills 2.C and 2.D 10 to 15% each; skill 3.B carries 15 to 25% and 3.C carries 5 to 10%. Practice 1 is not assessed there. On the free-response section, Practice 1 carries 20 to 35%, Practice 2 carries 40 to 45%, and Practice 3 carries 30 to 35%.

The unit's Building the Science Practices page flags 1.A, 1.C, 2.A and 3.C, and says there are many opportunities in this unit for students to create graphs that may include force, energy, or momentum as either a function of position or time for a single scenario and to make connections between physical concepts based on those graphs.

The Preparing for the AP Exam note ties Unit 7 to the second free-response question, the Translation Between Representations question, which requires students to create graphical and verbal models of scenarios and compare them to mathematical representations of the same situation. It 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-spring system at those same two positions, then make connections between the two representations by explaining how they are consistent with each other. It closes with a caution worth reading: 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.

Two of the CED's fifteen sample multiple-choice questions align to Unit 7: question 5 on the mass dependence of the two periods, and question 12 on translating a position graph into an acceleration graph. Both appear in the traps section above. Neither of the four sample free-response questions aligns to Unit 7, which is a reminder that with four questions and seven units a Unit 7 free-response question is not guaranteed in a given year. The reliable figure is the multiple-choice weighting of 10 to 15%.

Skill 3.A, creating experimental procedures, is listed for Topic 7.5 alone, and this is the one AP Physics C: Mechanics unit whose lab work has an obvious right answer to check against. One of the unit's sample activities has student groups use a pendulum to determine the acceleration of gravity in the classroom, with the winning group being the one whose procedure includes the most components for reducing error: timing multiple periods, linearizing the data, and very precisely finding the centre of mass of the bob. Another clamps a steel ruler to a table, hangs various masses from the hole in its end, and uses the measured periods to determine the ruler's spring constant. Because the accepted value of $g$ is itself worth care on this exam, the is g 9.8 or 10 guide is worth reading before you compare a lab result to it. The Progress Check for Unit 7 runs about 18 multiple-choice questions and 4 free-response questions, one of each type.

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

AP Physics C: Mechanics is a calculus-based introductory college-level course, equivalent to the first course in an introductory college sequence in calculus-based physics. Its only stated prerequisite is that students should have taken, or be concurrently taking, calculus.

AP Physics 1 also has a Unit 7 called Oscillations, and the two differ in three measurable ways. It has four topics rather than five, stopping before Simple and Physical Pendulums. It is weighted 5 to 8% of that exam's multiple-choice section against 10 to 15% here. And its equation sheet prints the two $2\pi ft$ position functions with no phase constant, where this one prints the cosine with a phase constant and neither $2\pi ft$ form.

If you are in the algebra-based course, the page you want is the AP Physics 1 Unit 7 hub. It is written to the AP Physics 1 framework and stops where that framework stops. That page is for AP Physics 1 students; this one is for AP Physics C: Mechanics students. The shared title is a College Board naming decision, not a claim that the units cover the same ground.

Unit 7 also leans on Unit 5 and Unit 6 harder than the algebra-based version can, because the physical pendulum needs Newton's second law in rotational form and a rotational inertia. For the step-by-step routines rather than the framework, the simple harmonic motion guide owns the procedures. The AP Physics C: Mechanics course hub lists all seven units.

From Newton's second law to the period, without memorising it

A 0.45 kg block on a frictionless surface is attached to an ideal spring of force constant $k = 72\ \mathrm{N/m}$ and pulled 0.080 m from equilibrium, then released from rest. Find (a) the angular frequency from the differential equation, (b) the period and frequency, (c) the maximum speed and maximum acceleration, (d) the total energy, and (e) the position function with the correct phase constant.

Declare the convention: let $+x$ point away from equilibrium in the direction of the initial pull, so the block starts at $x = +0.080$ m.
(a) Statement 7.1.A.2 says the restoring force is proportional to the displacement and opposite to it, and the sheet prints $\vec{F}_s = -k\Delta\vec{x}$ . Newton's second law gives $m\dfrac{d^2x}{dt^2} = -kx$ , so $\dfrac{d^2x}{dt^2} = -\dfrac{k}{m}x$ .
Compare that with statement 7.3.A.2's $\dfrac{d^2x}{dt^2} = -\omega^2 x$ and read the coefficient: $\omega^2 = \dfrac{k}{m} = \dfrac{72}{0.45} = 160\ \mathrm{rad^2/s^2}$ , so $\omega = \sqrt{160} = 12.649\ \mathrm{rad/s}$ .
That is the whole method, and it is why $T_s = 2\pi\sqrt{m/k}$ does not have to be recalled. The boundary statement under Topic 7.3 confirms the scope: you need the solution, not a proof that it is the solution.
(b) The sheet prints $T = \dfrac{2\pi}{\omega} = \dfrac{1}{f}$ , so $T = \dfrac{2\pi}{12.649} = 0.4967 \approx 0.497$ s and $f = 1/T = 2.01$ Hz. Check against the printed spring period: $2\pi\sqrt{0.45/72} = 2\pi\sqrt{6.25 \times 10^{-3}} = 2\pi(0.0790569) = 0.4967$ s, identical.
(c) Statement 7.3.A.3.ii gives $v_{\max} = A\omega = (0.080)(12.649) = 1.012\ \mathrm{m/s}$ and $a_{\max} = A\omega^2 = (0.080)(160) = 12.8\ \mathrm{m/s^2}$ . Neither is on the equation sheet; both are Derived equations.
(d) Statement 7.4.A.4.ii gives $E_{\text{total}} = \frac{1}{2}kA^2 = \frac{1}{2}(72)(0.080)^2 = \frac{1}{2}(72)(6.4 \times 10^{-3}) = 0.2304\ \mathrm{J}$ , about 0.230 J.
Check it the other way, through the printed $K = \frac{1}{2}mv^2$ at the equilibrium position where all the energy is kinetic: $\frac{1}{2}(0.45)(1.012)^2 = \frac{1}{2}(0.45)(1.0240) = 0.2304\ \mathrm{J}$ . The two agree to every digit carried, which is 7.4.A.2 stated numerically.
(e) The sheet's position function is $x = x_{\max}\cos(\omega t + \phi)$ . Released from rest at $+A$ means $x = +A$ at $t = 0$ , so $\cos\phi = 1$ and $\phi = 0$ : $x = (0.080\ \mathrm{m})\cos[(12.6\ \mathrm{rad/s})t]$ .
Had the block been released from rest at $x = -A$ instead, the same function with $\phi = \pi$ handles it. Had it been pushed through equilibrium in the $+x$ direction at $t = 0$ , $\phi = -\pi/2$ , which reproduces the sine form that this course's equation sheet does not print.

(a) $\omega = 12.6\ \mathrm{rad/s}$ , read off the coefficient of the differential equation. (b) $T = 0.497$ s and $f = 2.01$ Hz. (c) $v_{\max} = 1.01\ \mathrm{m/s}$ and $a_{\max} = 12.8\ \mathrm{m/s^2}$ . (d) $E_{\text{total}} = 0.230$ J, confirmed twice. (e) $x = (0.080\ \mathrm{m})\cos[(12.6\ \mathrm{rad/s})t]$ , with $\phi = 0$ .

A physical pendulum against the simple-pendulum answer

A uniform rod of mass $M = 0.60$ kg and length $L = 0.80$ m is pivoted at one end and swings in a vertical plane through a small angle. Its rotational inertia about the pivot is $\frac{1}{3}ML^2$ . Using $g = 9.8\ \mathrm{m/s^2}$ , find (a) the rotational inertia and the distance to the centre of mass, (b) the period, (c) the period of a simple pendulum of the same 0.80 m length, and (d) the length of the simple pendulum that would match the rod.

Set up the derivation the way 7.5.A.2 says it is done, from Newton's second law in rotational form. The gravitational force acts at the centre of mass a distance $d$ from the pivot, so 7.5.A.2.i gives the restoring torque $\tau = -Mgd\sin\theta$ .
Apply the small-angle approximation of 7.5.A.2.ii, $\sin\theta \approx \theta$ with $\theta$ in radians, so $-Mgd\theta = I\alpha$ , and therefore $\dfrac{d^2\theta}{dt^2} = -\dfrac{Mgd}{I}\theta$ .
Compare with 7.5.A.2.iii's $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$ and read $\omega^2 = \dfrac{Mgd}{I}$ , which gives the printed $T_{\text{phys}} = 2\pi\sqrt{I/(Mgd)}$ . The formula is on the sheet; the three steps above are not.
(a) $I = \frac{1}{3}ML^2 = \frac{1}{3}(0.60)(0.80)^2 = \frac{1}{3}(0.60)(0.64) = 0.128\ \mathrm{kg \cdot m^2}$ , and for a uniform rod $d = L/2 = 0.40$ m.
(b) $Mgd = (0.60)(9.8)(0.40) = 2.352\ \mathrm{N \cdot m}$ , so $\dfrac{I}{Mgd} = \dfrac{0.128}{2.352} = 0.054422\ \mathrm{s^2}$ and $T_{\text{phys}} = 2\pi\sqrt{0.054422} = 2\pi(0.233285) = 1.4658 \approx 1.47$ s.
(c) The printed simple-pendulum result with $\ell = 0.80$ m gives $T_p = 2\pi\sqrt{0.80/9.8} = 2\pi\sqrt{0.081633} = 2\pi(0.285714) = 1.7952 \approx 1.80$ s. The rod swings faster than a bob at its tip, by a factor $1.4658/1.7952 = 0.8165$ .
That factor is exactly $\sqrt{2/3} = 0.81650$ , because $\dfrac{I}{Md} = \dfrac{ML^2/3}{ML/2} = \dfrac{2L}{3}$ , and the period depends on the square root of that length.
(d) So the matching simple pendulum has $\ell_{\text{eq}} = \dfrac{I}{Md} = \dfrac{0.128}{(0.60)(0.40)} = \dfrac{0.128}{0.24} = 0.5333$ m, two thirds of the rod's length. Check: $2\pi\sqrt{0.5333/9.8} = 1.4658$ s, matching part (b) to every digit.
Note that the mass cancelled out of $\ell_{\text{eq}}$ entirely, as it must, since $I$ and $Md$ are both proportional to $M$ . A heavier rod of the same shape has the same period.

(a) $I = 0.128\ \mathrm{kg \cdot m^2}$ and $d = 0.40$ m. (b) $T_{\text{phys}} = 1.47$ s. (c) A simple pendulum of the same 0.80 m length gives 1.80 s, longer by a factor $\sqrt{3/2}$ . (d) The equivalent simple-pendulum length is $I/(Md) = 0.533$ m, two thirds of the rod, and the rod's mass cancels out of it.

A torsion pendulum, where g never appears

A uniform disk of mass 0.35 kg and radius 0.12 m hangs horizontally from a vertical wire attached at its centre, and oscillates in the horizontal plane. The wire's restoring torque per unit angle is $k = 0.040\ \mathrm{N \cdot m/rad}$ . The disk's rotational inertia about the wire is $\frac{1}{2}MR^2$ . Find (a) the rotational inertia, (b) the angular frequency and period, (c) the maximum angular speed if the disk is released from rest at $\Delta\theta = 0.35$ rad, and (d) what would change if the whole apparatus were taken to the Moon.

Declare the convention: positive $\Delta\theta$ is the direction of the initial twist, so the restoring torque is negative there.
(a) $I = \frac{1}{2}MR^2 = \frac{1}{2}(0.35)(0.12)^2 = \frac{1}{2}(0.35)(0.0144) = 2.52 \times 10^{-3}\ \mathrm{kg \cdot m^2}$ .
(b) Statement 7.5.A.4 gives the Derived equation $I\alpha = -k\Delta\theta$ , so $\dfrac{d^2\theta}{dt^2} = -\dfrac{k}{I}\theta$ . Match it to $\dfrac{d^2\theta}{dt^2} = -\omega^2\theta$ and read $\omega^2 = \dfrac{k}{I} = \dfrac{0.040}{2.52 \times 10^{-3}} = 15.873\ \mathrm{rad^2/s^2}$ .
$\omega = \sqrt{15.873} = 3.9841\ \mathrm{rad/s}$ , and the printed $T = 2\pi/\omega$ gives $T = \dfrac{2\pi}{3.9841} = 1.5771 \approx 1.58$ s.
(c) Released from rest at maximum twist, the solution is $\theta = \Delta\theta_{\max}\cos(\omega t)$ with $\phi = 0$ , the same form as the printed position function. Differentiating gives an angular speed of magnitude $\Delta\theta_{\max}\,\omega\lvert\sin(\omega t)\rvert$ , which peaks at $\Delta\theta_{\max}\omega$ .
That is $(0.35)(3.9841) = 1.394 \approx 1.39\ \mathrm{rad/s}$ , reached each time the disk passes through the untwisted position. This is the rotational analogue of the Derived equation $v_{\max} = A\omega$ in 7.3.A.3.ii, not a separate printed result.
(d) Nothing. Neither $k$ nor $I$ contains $g$ , so the period is unchanged on the Moon, in orbit, or anywhere else. Compare a physical pendulum, whose period goes as $1/\sqrt{g}$ and so lengthens wherever the gravitational field is weaker.
That contrast is one answer to the unit's own essential question about weighing an astronaut in space: a torsion or spring oscillator measures inertia, and a pendulum measures inertia and gravity together.
Sanity check the units: $\mathrm{N \cdot m/rad}$ divided by $\mathrm{kg \cdot m^2}$ is $\mathrm{s^{-2}}$ , since a newton is $\mathrm{kg \cdot m/s^2}$ and the radian is dimensionless. So $\sqrt{k/I}$ is a frequency, as required.

(a) $I = 2.52 \times 10^{-3}\ \mathrm{kg \cdot m^2}$ . (b) $\omega = 3.98\ \mathrm{rad/s}$ and $T = 1.58$ s. (c) The maximum angular speed is $1.39\ \mathrm{rad/s}$ , at the untwisted position. (d) Nothing changes, because no $g$ appears in $\omega = \sqrt{k/I}$ , unlike a pendulum's period.

Frequently asked questions

How much of the AP Physics C Mechanics exam is Unit 7?

Unit 7, Oscillations, is weighted at 10 to 15% of the multiple-choice section of the AP Physics C: Mechanics exam, and the course description suggests about 12 to 17 class periods for it. Units 1, 5 and 6 carry the same band; Unit 2 is heaviest at 20 to 25%, Unit 3 is 15 to 25%, and Unit 4 is 10 to 20%. AP Physics 1 weights its own Unit 7, which is also called Oscillations, at only 5 to 8% and gives it four topics rather than five, so this unit is worth roughly twice as much and covers more ground.

Which SHM position function is on the AP Physics C Mechanics equation sheet?

Only one: the position as a maximum value times the cosine of the angular frequency times time plus a phase angle. The sheet does not print either of the two forms written in terms of two pi f t, even though essential knowledge 7.3.A.1 of the framework gives both a cosine and a sine version in that form. It also uses a maximum-position symbol rather than an amplitude symbol, and its variable list defines the phase angle. The AP Physics 1 sheet prints exactly the opposite pair, both two pi f t forms and no phase angle at all. On this exam the phase constant is how you handle a start anywhere other than maximum displacement, which is why no separate sine form is needed.

Does AP Physics C Unit 7 have any boundary statements?

One, under Topic 7.3. It 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 adds that AP Physics C: Mechanics does not expect students to mathematically prove that the solution is correct. Topics 7.1, 7.2, 7.4 and 7.5 print no boundary statement at all. So you are expected to recognise the differential equation's shape in a new situation and to know that a cosine solves it, but not to verify the solution by substitution.

How do you get the period of an oscillator from the differential equation?

Write Newton's second law for the restoring force, rearrange it until the second derivative of the position sits alone on one side, and compare the coefficient with the framework's derived equation, which sets the second derivative of position equal to minus the angular frequency squared times the position. Whatever multiplies the position is the angular frequency squared. For a mass on a spring, the force constant over the mass gives the spring period. For a physical pendulum, the weight times the distance to the centre of mass over the rotational inertia gives the physical-pendulum period. For a torsion pendulum, the torsion constant over the rotational inertia. One method, four results, and the equation sheet supplies the step from angular frequency to period.

What is a physical pendulum in AP Physics C Mechanics?

Essential knowledge 7.5.A.1 defines it as a rigid body that undergoes oscillation about a fixed axis, and 7.5.A.2 gives its small-amplitude period as two pi times the square root of the rotational inertia divided by the product of mass, gravitational field strength, and the distance from the axis to the centre of mass. That formula is printed on the AP Physics C: Mechanics equation sheet. A simple pendulum is a special case of it, per 7.5.A.3, where the hanging object can be modelled as a point mass. The topic has no AP Physics 1 counterpart: the algebra-based Unit 7 has only four topics and stops before pendulums as rigid bodies, and its equation sheet has no physical-pendulum entry.

Does changing the amplitude change the period in simple harmonic motion?

No. Essential knowledge 7.3.A.5 says plainly that changing the amplitude of a system exhibiting simple harmonic motion will not change its period. What amplitude does change is the energy: statement 7.4.A.4.ii says changing the amplitude changes the maximum potential energy and therefore the total energy of the system, and gives the total energy of a spring-object system as one half the force constant times the amplitude squared. The maximum speed and maximum acceleration also scale with the amplitude, through the derived relations that set them equal to the amplitude times the angular frequency and the amplitude times the angular frequency squared. The period is the one thing that does not move.

Does the mass affect a pendulum's period?

Not for a simple pendulum, and not for a physical pendulum of fixed shape either. The printed simple-pendulum period contains only the length and the gravitational field strength, so the mass of the bob cancels out. For a physical pendulum, the rotational inertia and the product of mass and centre-of-mass distance are both proportional to the mass, so it cancels there too. A mass on a spring is the opposite case: its printed period contains the mass explicitly, so doubling the mass lengthens the period. The AP Physics C: Mechanics course description uses that exact contrast as one of its sample multiple-choice questions, keyed so that the pendulum's period stays the same while the spring's period increases.