AP Physics C: Mechanics · Topic 7.3

Topic 7.3: Representing and Analyzing SHM

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

Only one: the maximum position 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 writes the amplitude as a maximum position rather than as A, and there is no A in either of its variable lists, while there is an entry defining 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.

Q: How do you find the phase angle in simple harmonic motion?

Evaluate the position function and its derivative at time zero. That gives the initial position as the amplitude times the cosine of the phase angle, and the initial velocity as minus the amplitude times the angular frequency times the sine of the phase angle. Solve the pair: the amplitude is the square root of the initial position squared plus the initial velocity divided by the angular frequency, all squared. Then get the phase angle from the signs of both its cosine and its sine, not from an inverse tangent, because an angle and that angle plus pi have the same tangent and describe opposite motions. Work in radians throughout.

Q: How do you get velocity and acceleration from the SHM position equation?

Differentiate. The velocity is minus the amplitude times the angular frequency times the sine of the same argument, and the acceleration is minus the amplitude times the angular frequency squared times the cosine of that argument. Three results follow immediately: the acceleration equals minus the angular frequency squared times the position at every instant, the maximum speed is the amplitude times the angular frequency, and the maximum acceleration is the amplitude times the angular frequency squared. None of those three is printed on the AP Physics C: Mechanics equation sheet; all three are labelled Derived equations in the framework. The derivative rules for sine and cosine are printed, in the booklet's calculus table.

Q: Why is the acceleration graph in SHM upside down compared to the position graph?

Because the acceleration is a negative constant times the position at every instant, which is essential knowledge 7.3.A.3.i. The constant is the angular frequency squared, so the acceleration graph is the position graph reflected in the time axis and rescaled vertically by that factor. It is reflected, not shifted, which is the distinction the AP Physics C: Mechanics sample multiple-choice question 12 is built on. The velocity graph is the one that is shifted, by a quarter period, since it is the slope of the position graph and a slope vanishes at a turning point.

Q: Where are the velocity and acceleration maximum in simple harmonic motion?

The speed is maximum at the equilibrium position and zero at both turning points. The acceleration is maximum in magnitude at both turning points and zero at the equilibrium position. So they never peak at the same moment: they are a quarter period apart. The maximum speed is the amplitude times the angular frequency and the maximum acceleration is the amplitude times the angular frequency squared, which is why doubling the angular frequency at fixed amplitude doubles one and quadruples the other. Both relations are essential knowledge 7.3.A.3.ii and neither is printed on the equation sheet.

Q: What is resonance in AP Physics C Mechanics?

Essential knowledge 7.3.A.4 says that in the presence of a sinusoidal external force, a system may exhibit resonance, and 7.3.A.4.i says resonance occurs when an external force is exerted at the natural frequency of an oscillating system. Statement 7.3.A.4.ii says what resonance does: it increases the amplitude of oscillating motion. Statement 7.3.A.4.iii defines the natural frequency as the frequency at which the system will oscillate when it is displaced from its equilibrium position, so it is a property of the system, set by its inertia and its restoring coefficient. The driving frequency is what you tune; the natural frequency is what you tune it to, and resonance does not change it.

Q: Does AP Physics C ask you to prove the cosine solves the SHM differential equation?

No. The only boundary statement in Unit 7 of AP Physics C: Mechanics sits under Topic 7.3 and says the course 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 it does not expect students to mathematically prove that the solution is correct. So you must be able to recognise the equation's shape in an unfamiliar system and to write down and use the cosine solution, but you will not be asked to substitute it back in and verify it, and no technique for solving the equation from scratch is examinable.

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

The AP Physics C Mechanics sheet prints one position function for SHM: the maximum position times the cosine of the angular frequency times time plus a phase angle. Differentiate it once for velocity, twice for acceleration. The phase angle lets you start the motion anywhere.

AP Physics: Unit 7 (topics 7.3 Representing and Analyzing SHM). Topic 7.3 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. The largest topic in the unit by essential-knowledge count: one learning objective, 7.3.A, and six numbered statements. 7.3.A.1 gives x = A cos(2*pi*f*t) or A sin(2*pi*f*t), with 7.3.A.1.i on minima, maxima and zeros being features of harmonic motion and 7.3.A.1.ii on recognising them helping to describe the motion qualitatively. 7.3.A.2 gives the Derived equation d2x/dt2 = -omega^2 x. 7.3.A.3 gives x = A cos(omega t + phi), with Derived equations a = -omega^2 x (7.3.A.3.i) and v_max = A*omega and a_max = A*omega^2 (7.3.A.3.ii). 7.3.A.4 covers resonance, with 7.3.A.4.i that it occurs when an external force is exerted at the natural frequency, 7.3.A.4.ii that it increases the amplitude, and 7.3.A.4.iii defining the natural frequency. 7.3.A.5 says changing the amplitude will not change the period, and 7.3.A.6 says properties of SHM can be determined and analyzed using graphical representations. Suggested skills 1.C, 2.A, 2.C, 3.B. This topic carries the only boundary statement in Unit 7, quoted whole on the page: 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. Equation-sheet facts verified against both Tables of Information rendered as images: the C: Mechanics sheet prints x = x_max cos(omega t + phi) as its only position function, with phi defined as phase angle and no A in either variable list; it prints neither 2*pi*f*t form. The AP Physics 1 sheet prints both 2*pi*f*t forms, no phase angle, and A as amplitude or area. Not printed on either sheet: the differential equation, a = -omega^2 x, v_max = A*omega, a_max = A*omega^2. The C: Mechanics booklet does print a Calculus table containing the derivative rules for sine and cosine, a table of trigonometric values at 0, 30, 37, 45, 53, 60 and 90 degrees, and an Identities table containing sin^2 + cos^2 = 1; the AP Physics 1 booklet has no calculus table. Sample multiple-choice question 12 aligns to 7.3.A and 7.3.A.3 with skill 3.B and is keyed A: a block hung vertically from an ideal spring, pulled down and released, with a position against time graph, asking which graph best represents the acceleration against time.

What Topic 7.3 requires

Topic 7.3 is the largest topic in Unit 7 by essential-knowledge count: one learning objective and six numbered statements, three of which carry sub-statements. It also carries the only boundary statement in the whole unit.

Learning objective 7.3.A: describe the displacement, velocity, and acceleration of an object exhibiting SHM.

Statement	What it says
7.3.A.1	For an object exhibiting SHM, the displacement of that object measured from its equilibrium position can be represented by the equations $x = A\cos(2\pi ft)$ or $x = A\sin(2\pi ft)$
7.3.A.1.i	Minima, maxima, and zeros of displacement, velocity, and acceleration are features of harmonic motion
7.3.A.1.ii	Recognizing the positions or times at which the displacement, velocity, and acceleration for SHM have extrema or zeros can help in qualitatively describing the behavior of the motion
7.3.A.2	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. Derived equation: $\dfrac{d^2x}{dt^2} = -\omega^2 x$
7.3.A.3	Characteristics of SHM, such as velocity and acceleration, can be determined by or derived from the equation $x = A\cos(\omega t + \phi)$
7.3.A.3.i	The acceleration of an object exhibiting SHM is related to the object's angular frequency and position. Derived equation: $a = -\omega^2 x$
7.3.A.3.ii	It can be shown that the maximum velocity and acceleration of an object exhibiting SHM are related to the angular frequency of the object's motion. Derived equations: $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$
7.3.A.4	In the presence of a sinusoidal external force, a system may exhibit resonance
7.3.A.4.i	Resonance occurs when an external force is exerted at the natural frequency of an oscillating system
7.3.A.4.ii	Resonance increases the amplitude of oscillating motion
7.3.A.4.iii	The natural frequency of a system is the frequency at which the system will oscillate when it is displaced from its equilibrium position
7.3.A.5	Changing the amplitude of a system exhibiting SHM will not change its period
7.3.A.6	Properties of SHM can be determined and analyzed using graphical representations

The suggested skills are 1.C (create qualitative sketches of graphs that represent features of a model or the behavior of the physical system), 2.A (derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway), 2.C (compare physical quantities between two or more scenarios or at different times and locations within a single scenario) and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Notice that this topic prints two position functions in 7.3.A.1 and a third, different one in 7.3.A.3. That is not redundancy, and which of the three is on the equation sheet is the single most checkable fact on this page.

The one boundary statement in Unit 7, read in both directions

Topics 7.1, 7.2, 7.4 and 7.5 print no boundary statement at all. This one does, and here it is 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 it forwards first. You are expected to know the solution, which means being able to write $x = A\cos(\omega t + \phi)$ and use it. You are also expected to identify SHM, which means recognising the shape of $d^2x/dt^2 = -\omega^2 x$ in a situation you have not seen before, and reading $\omega$ off the coefficient. Both halves are required.

Now read it backwards. You are not expected to prove the solution is correct. So you will not be asked to substitute the cosine into the differential equation and verify that both sides match, and you will not be asked to solve the differential equation from scratch by any technique. If a method in a textbook involves a characteristic equation, complex exponentials or separation of variables, it is past the line this statement draws.

What this leaves is an unusually clean division of labour, and it is worth naming because it tells you what to practise:

Getting to the differential equation is required, and it is Newton's second law plus a force law. That is Topics 7.1 and 7.5.
Solving it is not required. You quote the solution.
Working with the solution is required, and it is calculus. Differentiate it, evaluate it, impose initial conditions on it, graph it. That is this topic.

Statement 7.3.A.2 is what sets up the first bullet, and the label on the equation is a signal too. The CED marks $d^2x/dt^2 = -\omega^2 x$ a Derived equation, which is its wording for something you produce rather than look up, and consistent with that, the equation is not printed on the equation sheet.

The equation-sheet reversal, line by line

This is the row worth checking before the exam, because the framework prints one thing, this course's sheet prints another, and the algebra-based course's sheet prints the opposite of both. All three checked against the source rather than recalled: the Table of Information appendices of both course and exam descriptions, rendered as images and read at magnification.

What the framework says. Statement 7.3.A.1 gives $x = A\cos(2\pi ft)$ or $x = A\sin(2\pi ft)$ . Statement 7.3.A.3 gives $x = A\cos(\omega t + \phi)$ . Three expressions across two statements.

What the AP Physics C: Mechanics sheet prints. One position function:

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

It prints neither $2\pi ft$ form. It writes $x_{\max}$ where the framework writes $A$ , and there is no entry for $A$ in either of its two variable lists. There is an entry for $\phi$ , defined as the phase angle, and $\omega$ is defined as "angular frequency or angular speed".

What the AP Physics 1 sheet prints. Exactly the reverse pair:

x = A\cos(2\pi ft) \qquad 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 at all, and it defines $\omega$ as "angular speed" only.

	C: Mechanics sheet	AP Physics 1 sheet
$x = x_{\max}\cos(\omega t + \phi)$	printed	absent
$x = A\cos(2\pi ft)$	absent	printed
$x = A\sin(2\pi ft)$	absent	printed
$A$ in the variable list	absent	amplitude or area
$\phi$ in the variable list	phase angle	absent
$T = 2\pi/\omega$	printed	absent

So a student who learned SHM in the algebra-based course and sits this exam looking for the sine option will not find it, and a student who only ever wrote $\cos(\omega t + \phi)$ would find no place to put a phase constant on the Physics 1 sheet. It is the same physics. Only one line is printed per course, and it is a different line.

Four more Unit 7 results are absent from the C: Mechanics sheet, and all four are labelled Derived equations in the framework: the differential equation of 7.3.A.2, the acceleration relation $a = -\omega^2 x$ of 7.3.A.3.i, and both of 7.3.A.3.ii's maxima, $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$ . Neither sheet prints any of them.

The phase angle is how an arbitrary start gets posed

Why does this course's sheet carry $\phi$ and the algebra-based one not? Because with a phase angle you never need a second function.

The two $2\pi ft$ forms of 7.3.A.1 handle exactly two starting conditions. The cosine starts at maximum displacement with zero velocity; the sine starts at equilibrium moving in the positive direction. Anything in between has no printed form in that course, which is why AP Physics 1 questions release oscillators from rest at maximum displacement almost every time.

Add $\phi$ and the whole family collapses into one function. The special cases become choices of one number:

Starting condition at $t = 0$	$\phi$
At rest at $x = +A$	$0$
At rest at $x = -A$	$\pi$
At equilibrium moving toward $+x$	$-\pi/2$
At equilibrium moving toward $-x$	$+\pi/2$
Anywhere else	whatever the initial conditions give

The third row is the sine form written as a cosine, which is why this course's sheet does not need to print a sine option.

The last row is the one that matters, and it is the question type the algebra-based course cannot pose. Given any starting position $x_0$ and any starting velocity $v_0$ , differentiate the solution to get $v = -A\omega\sin(\omega t + \phi)$ , evaluate both at $t = 0$ , and solve the pair:

x_0 = A\cos\phi \qquad v_0 = -A\omega\sin\phi

A = \sqrt{x_0^2 + \left(\frac{v_0}{\omega}\right)^2} \qquad \tan\phi = -\frac{v_0}{\omega x_0}

Two cautions on that second relation, because both cost points. First, $\tan\phi$ alone does not fix the quadrant: $\phi$ and $\phi + \pi$ have the same tangent and describe opposite motions. Get the quadrant from the signs of $\cos\phi = x_0/A$ and $\sin\phi = -v_0/(A\omega)$ together, not from an inverse tangent on a calculator. Second, $\omega t$ is in radians, so $\phi$ is in radians too, and a calculator left in degree mode will produce a plausible wrong answer.

The expression for $A$ is worth a second look. It is a Pythagorean sum of the starting position and the starting velocity divided by $\omega$ , so those two quantities behave like perpendicular legs. That is the same structure as the energy statement in Topic 7.4: squaring it and multiplying by $\frac{1}{2}k$ turns it into $\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ , which is the total energy. Worked example 1 checks that identity numerically.

Differentiate the solution, and note which rules are printed

Statement 7.3.A.3 says characteristics of SHM such as velocity and acceleration "can be determined by or derived from" the cosine solution. Determined by or derived from means differentiated, and this is where the calculus prerequisite earns its place in the course.

Start from the printed function and differentiate twice:

x(t) = A\cos(\omega t + \phi)

v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)

a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2} = -A\omega^2\cos(\omega t + \phi)

Three results fall out immediately, and all three are the framework's Derived equations rather than printed formulas:

The last line is $-\omega^2$ times the first line, so $a = -\omega^2 x$ , which is 7.3.A.3.i.
The sine and cosine are bounded by 1, so $v_{\max} = A\omega$ and $a_{\max} = A\omega^2$ , which is 7.3.A.3.ii. The CED's own wording is "it can be shown that", and this is the showing.
Differentiating twice returned the original function times a negative constant, which is the differential equation of 7.3.A.2 satisfied. Note that observing this is not the same as proving the solution is correct, which the boundary statement excludes.

Here is the detail worth carrying into the exam, verified against the booklet. The derivative rules you need are printed, and the results you get from them are not. The C: Mechanics Table of Information includes a Calculus table, which the physics-equation tables on this site do not transcribe, and among its entries are

\frac{d}{dx}\left[\sin(ax)\right] = a\cos(ax) \qquad \frac{d}{dx}\left[\cos(ax)\right] = -a\sin(ax)

along with the power rule, the exponential and logarithm rules, and the matching integrals. The AP Physics 1 booklet has no Calculus table at all.

So the exam's structure on this topic is deliberate: it hands you the machinery and withholds the output. That is the practical meaning of the Derived equation label. Two other printed tables help on Topic 7.3 questions. A table of trigonometric values covers 0, 30, 37, 45, 53, 60 and 90 degrees, which is why phase angles on this exam so often come out at those values, and worked example 1 lands on one. An Identities table prints $\sin^2\theta + \cos^2\theta = 1$ , which is the identity that makes the total energy constant.

The three graphs, and where the extrema and zeros sit

Statements 7.3.A.1.i and 7.3.A.1.ii are about reading a graph rather than computing from one, and 7.3.A.6 says properties of SHM can be determined and analyzed using graphical representations. The whole of it fits in one table, for the case $\phi = 0$ .

Position	$x$	$v$	$a$	Net force
$x = +A$	maximum	zero	most negative	maximum, toward equilibrium
Moving toward equilibrium	decreasing	negative, growing	negative, shrinking	shrinking
$x = 0$	zero	extreme, $\mp A\omega$	zero	zero
$x = -A$	minimum	zero	most positive	maximum, toward equilibrium

Four facts to carry out of it:

Velocity is a quarter period out of step with position, and acceleration is a half period out of step. The velocity graph is the position graph shifted; the acceleration graph is the position graph flipped, not shifted.
Speed and acceleration never peak together. Speed peaks where the acceleration is zero and acceleration peaks where the speed is zero. They are a quarter period apart, which is $T/4$ in time and the full amplitude in position.
The velocity graph's zeros are the position graph's extrema, which is the calculus statement that the derivative vanishes at a turning point.
The acceleration is zero exactly where the net force is zero, which is the equilibrium position of statement 7.1.A.2.ii, not the spring's natural length. On a vertical spring those are different places.

The amplitudes stack in powers of $\omega$ : the position graph peaks at $A$ , the velocity graph at $A\omega$ , the acceleration graph at $A\omega^2$ . So doubling the frequency at fixed amplitude doubles the peak speed and quadruples the peak acceleration. That is the fastest way to compare two graph sets, and it is also the trap in a question that changes $\omega$ and asks which graph changed more.

One reading habit worth building. To go from a position graph to an acceleration graph, do not try to differentiate twice by eye. Use $a = -\omega^2 x$ from 7.3.A.3.i, which says the acceleration at any instant is a fixed negative multiple of the position at that same instant. Reflect the position graph in the time axis and rescale the vertical axis by $\omega^2$ . No shifting is involved, and the minus sign is the entire content of the CED's sample question on this topic.

Resonance, natural frequency, and the amplitude that does not move the period

Statement 7.3.A.4 and its three sub-statements are the only place in Unit 7 that a driven oscillator appears, and they are short enough to hold exactly.

7.3.A.4: in the presence of a sinusoidal external force, a system may exhibit resonance. 7.3.A.4.i: resonance occurs when an external force is exerted at the natural frequency of an oscillating system. 7.3.A.4.ii: resonance increases the amplitude of oscillating motion. 7.3.A.4.iii: the natural frequency of a system is the frequency at which the system will oscillate when it is displaced from its equilibrium position.

Read 7.3.A.4.iii as a definition of where the natural frequency comes from. It is the frequency the system produces on its own, so for a block on a spring it is $\frac{1}{2\pi}\sqrt{k/m}$ and for a small-angle pendulum it is $\frac{1}{2\pi}\sqrt{g/\ell}$ . It is a property of the system, fixed by its inertia and its restoring coefficient, and nothing you do with the driving force changes it.

Read 7.3.A.4.ii for what it does and does not say. Resonance increases the amplitude. It does not change the natural frequency, it does not change the period of the resulting oscillation, and the CED does not describe damping, quality factors or resonance curves anywhere in this unit. A question that asks what resonance does has one answer in this framework's vocabulary: the amplitude grows.

What you tune is the driving frequency. What you tune it to is the natural frequency. Getting those two the wrong way round is the standard error here, and resonance is worth a glance if the vocabulary is new.

Statement 7.3.A.5 belongs beside this, because the two are easy to run together: changing the amplitude of a system exhibiting SHM will not change its period. Amplitude and period are independent, in both directions. Resonance grows the amplitude and the period stays put. Pulling an oscillator twice as far back changes its speeds, its accelerations, its forces and its energy, and leaves its period exactly where it was. Read it with 7.4.A.4.ii, which says changing the amplitude changes the maximum potential energy and therefore the total energy: amplitude moves the energy, and it does not move the period.

One honest limit on 7.3.A.5, the same one that qualifies Topic 7.2. For a pendulum, period independence from amplitude holds only inside the small-angle approximation. It is a statement about SHM, and a pendulum is only SHM to that approximation.

How Topic 7.3 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.A is weighted 25 to 30%, skill 2.C 10 to 15% and skill 3.B 15 to 25%, and all three are suggested skills for this topic. Skill 1.C points at the free-response section, where Science Practice 1 carries 20 to 35%, since Practice 1 is not assessed on the multiple-choice section at all. The exam-weighting page adds that required course content can be assessed with any skill.

One of the CED's fifteen sample multiple-choice questions aligns to this topic. Question 12 is aligned to learning objective 7.3.A and essential knowledge 7.3.A.3 with skill 3.B, and it is keyed A. Its stem: a block is hung vertically from an ideal spring, pulled down, released, and allowed to oscillate; the vertical position of the block as a function of time is shown in a graph, and the question asks which of four graphs best represents the corresponding acceleration of the block as a function of time. Every option is a sinusoid of the same period, so the question is entirely about sign and phase, and $a = -\omega^2 x$ settles it in one step.

That stem is worth noticing for a second reason: the oscillator is a vertical spring. The position graph oscillates about the hanging equilibrium, and the acceleration is zero there even though the gravitational force is not.

The unit's Preparing for the AP Exam note ties Unit 7 to the second free-response question, Translation Between Representations, worth 12 points at a suggested 25 to 30 minutes, which requires students to create graphical and verbal models of scenarios and compare them to mathematical representations of the same situation. Its Unit 7 example is a block oscillating on a spring: sketch free-body diagrams at maximum displacement and at equilibrium, 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. It 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.

The unit's fourth sample instructional activity is a pure 7.3 exercise: give students a graph of position, velocity, or acceleration for SHM and have them make the other two on the same time scale, along with force, momentum, kinetic energy, potential energy and total energy against time, then draw energy bar charts for various instants.

For the procedures rather than the framework, the simple harmonic motion guide owns the routines 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 7.3 with the same title, and on this topic the two courses diverge more than anywhere else in the unit, because their equation sheets print opposite things.

The [AP Physics 1 Topic 7.3 page](/ap-physics-1/unit-7-oscillations/7-3-representing-and-analyzing-shm) is for students in the algebra-based course, whose sheet prints a cosine and a sine written in terms of two pi f t, with no phase angle anywhere, so choosing between the two functions is the whole of the initial-condition question. This page is for students in AP Physics C: Mechanics, whose sheet prints one cosine with a phase angle and neither of the other two, and who are expected to differentiate that function to obtain velocity and acceleration rather than quote them. Neither page is a reading-level variant of the other. If you are in Physics 1, there is no phase angle on your sheet and no calculus table in your booklet.

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 is where the differential equation comes from, Topic 7.2 turns its coefficient into a period, Topic 7.4 graphs the same motion in energy, and Topic 7.5 applies all of it to rigid bodies and has no algebra-based counterpart. Topic 2.8 is where the force law behind all of these graphs comes from, including how a force against position graph yields the force constant as a slope. The Unit 7 hub lists all five topics.

A start that is neither at maximum displacement nor at equilibrium

A 0.50 kg block on a frictionless surface is attached to an ideal spring of force constant $k = 32\ \mathrm{N/m}$ . At $t = 0$ the block is at $x = +0.030$ m and moving in the $+x$ direction at $0.32\ \mathrm{m/s}$ . Find (a) the angular frequency and period, (b) the amplitude, (c) the phase angle, (d) the maximum speed and maximum acceleration, and (e) the first time the block reaches $x = +A$ . Confirm the amplitude through the energy.

Declare the convention: $+x$ is the direction the block is moving at $t = 0$ , and $x$ is measured from the equilibrium position.
(a) The restoring force is $-kx$ , so Newton's second law gives $\dfrac{d^2x}{dt^2} = -\dfrac{k}{m}x = -\dfrac{32}{0.50}x = -(64\ \mathrm{s^{-2}})x$ . Matching 7.3.A.2 gives $\omega^2 = 64\ \mathrm{rad^2/s^2}$ , so $\omega = 8.0\ \mathrm{rad/s}$ and the printed $T = 2\pi/\omega$ gives $T = 0.7854 \approx 0.785$ s.
(b) Use the printed position function and its derivative at $t = 0$ : $x_0 = A\cos\phi = 0.030$ m and $v_0 = -A\omega\sin\phi = +0.32$ m/s.
The convenient combination is $v_0/\omega = 0.32/8.0 = 0.040$ m, which has units of length. Then $A = \sqrt{x_0^2 + (v_0/\omega)^2} = \sqrt{(0.030)^2 + (0.040)^2} = \sqrt{0.000900 + 0.001600} = \sqrt{0.002500} = 0.050\ \mathrm{m}$ .
(c) Get both trigonometric functions rather than a tangent, so the quadrant is not in doubt: $\cos\phi = \dfrac{x_0}{A} = \dfrac{0.030}{0.050} = 0.600$ and $\sin\phi = -\dfrac{v_0}{A\omega} = -\dfrac{0.32}{(0.050)(8.0)} = -\dfrac{0.32}{0.40} = -0.800$ .
Positive cosine with negative sine puts $\phi$ in the fourth quadrant, so $\phi = -0.9273\ \mathrm{rad}$ , which is $-53.13$ degrees. Those are the 3, 4, 5 values, and the sheet's table of trigonometric functions prints $\sin 53^\circ = 4/5$ and $\cos 53^\circ = 3/5$ , which is why exam phase angles land here so often.
A negative $\phi$ is the right sign, and it is worth a sanity check: the block is on its way out toward $+A$ rather than on its way back, so it reaches maximum displacement after $t = 0$ , and a cosine peaks when its argument is zero, which needs $\omega t = -\phi > 0$ .
The full position function is $x = (0.050\ \mathrm{m})\cos\left[(8.0\ \mathrm{rad/s})t - 0.927\right]$ .
(d) From 7.3.A.3.ii, $v_{\max} = A\omega = (0.050)(8.0) = 0.40\ \mathrm{m/s}$ and $a_{\max} = A\omega^2 = (0.050)(64) = 3.2\ \mathrm{m/s^2}$ . Neither is printed on the sheet.
(e) The block is at $+A$ when the cosine's argument is zero: $8.0t - 0.9273 = 0$ , so $t = 0.9273/8.0 = 0.1159 \approx 0.116\ \mathrm{s}$ . That is $0.1159/0.7854 = 0.148$ of a period, less than the quarter period it would take from equilibrium, which is right because the block started part of the way out already.
Confirm the amplitude independently through the energy, which is statement 7.4.A.4.ii. Total energy from the initial state: $\frac{1}{2}mv_0^2 + \frac{1}{2}kx_0^2 = \frac{1}{2}(0.50)(0.32)^2 + \frac{1}{2}(32)(0.030)^2 = 0.02560 + 0.01440 = 0.04000\ \mathrm{J}$ .
Total energy from the amplitude: $\frac{1}{2}kA^2 = \frac{1}{2}(32)(0.050)^2 = \frac{1}{2}(32)(0.002500) = 0.04000\ \mathrm{J}$ . Identical to every digit carried, which is the same Pythagorean statement as part (b) multiplied through by $\frac{1}{2}k$ .

(a) $\omega = 8.0\ \mathrm{rad/s}$ and $T = 0.785$ s. (b) $A = 0.050$ m, confirmed by energy at 0.0400 J both ways. (c) $\phi = -0.927$ rad, that is $-53.1$ degrees, fixed by taking the signs of the sine and cosine together. (d) $v_{\max} = 0.40\ \mathrm{m/s}$ and $a_{\max} = 3.2\ \mathrm{m/s^2}$ . (e) 0.116 s, which is 0.148 of a period.

Differentiating the printed solution, and checking the acceleration relation

An object exhibiting SHM has $x = (0.12\ \mathrm{m})\cos\left[(5.0\ \mathrm{rad/s})t + \pi/3\right]$ . Find (a) the velocity and acceleration as functions of time, (b) the period and the maximum speed and acceleration, (c) the position, velocity and acceleration at $t = 0$ , verifying $a = -\omega^2 x$ there, and (d) the first time the object passes through equilibrium and its speed then.

Read off the parameters by comparing with the printed $x = x_{\max}\cos(\omega t + \phi)$ : $A = 0.12$ m, $\omega = 5.0$ rad/s, $\phi = \pi/3$ rad, which is 60 degrees.
(a) Differentiate using the rule printed in the booklet's Calculus table, $\dfrac{d}{dx}\left[\cos(ax)\right] = -a\sin(ax)$ , applied with the chain rule for the constant inside: $v = \dfrac{dx}{dt} = -(0.12)(5.0)\sin(5.0t + \pi/3) = -(0.60\ \mathrm{m/s})\sin(5.0t + \pi/3)$ .
Differentiate again, now with $\dfrac{d}{dx}\left[\sin(ax)\right] = a\cos(ax)$ : $a = \dfrac{dv}{dt} = -(0.60)(5.0)\cos(5.0t + \pi/3) = -(3.0\ \mathrm{m/s^2})\cos(5.0t + \pi/3)$ .
(b) $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{5.0} = 1.2566 \approx 1.26\ \mathrm{s}$ . The coefficients found above are the maxima, since sine and cosine are bounded by 1: $v_{\max} = A\omega = 0.60\ \mathrm{m/s}$ and $a_{\max} = A\omega^2 = (0.12)(25) = 3.0\ \mathrm{m/s^2}$ , matching 7.3.A.3.ii.
(c) At $t = 0$ the argument is $\pi/3$ , that is 60 degrees, and the sheet's trigonometric table prints $\cos 60^\circ = 1/2$ and $\sin 60^\circ = \sqrt{3}/2$ .
$x(0) = (0.12)(0.500) = 0.060\ \mathrm{m}$ ; $v(0) = -(0.60)(0.8660) = -0.5196 \approx -0.520\ \mathrm{m/s}$ ; $a(0) = -(3.0)(0.500) = -1.50\ \mathrm{m/s^2}$ .
Check 7.3.A.3.i: $-\omega^2 x = -(25)(0.060) = -1.50\ \mathrm{m/s^2}$ , which is exactly $a(0)$ . It has to be, because the acceleration function is $-\omega^2$ times the position function at every instant, not just at this one.
Read the signs physically before moving on. The object is at positive $x$ , moving in the $-x$ direction, and accelerating in the $-x$ direction. It is past the maximum and heading back toward equilibrium, gaining speed. All three signs are consistent.
(d) Equilibrium is $x = 0$ , so the cosine must vanish: $5.0t + \pi/3 = \pi/2$ , giving $t = \dfrac{\pi/2 - \pi/3}{5.0} = \dfrac{\pi/6}{5.0} = 0.10472 \approx 0.105\ \mathrm{s}$ .
The speed there is $|v| = (0.60)\left|\sin(\pi/2)\right| = 0.60\ \mathrm{m/s}$ , which is the maximum, as it must be at equilibrium. Check the timing against the phase: the object was 60 degrees past its peak at $t = 0$ and needed the remaining 30 degrees to reach equilibrium, which is one twelfth of a cycle, $1.2566/12 = 0.1047$ s. Consistent.

(a) $v = -(0.60\ \mathrm{m/s})\sin(5.0t + \pi/3)$ and $a = -(3.0\ \mathrm{m/s^2})\cos(5.0t + \pi/3)$ . (b) $T = 1.26$ s, $v_{\max} = 0.60\ \mathrm{m/s}$ , $a_{\max} = 3.0\ \mathrm{m/s^2}$ . (c) $x = 0.060$ m, $v = -0.520\ \mathrm{m/s}$ , $a = -1.50\ \mathrm{m/s^2}$ , and $-\omega^2 x$ reproduces the acceleration exactly. (d) At $t = 0.105$ s, moving at the maximum speed of 0.60 m/s.

From a position graph to an acceleration graph, on a vertical spring

A block hangs from an ideal vertical spring. It is pulled down 0.045 m from its hanging rest position, released from rest, and oscillates. A position against time graph shows the motion repeating every 0.60 s. Take $g = 9.8\ \mathrm{m/s^2}$ and take upward as positive. Find (a) the angular frequency, (b) the position and acceleration functions, (c) the maximum acceleration and where it occurs, and (d) whether the spring ever goes slack.

Declare the convention: $y$ is measured upward from the hanging equilibrium position, which is the position of zero net force, not the spring's natural length. Pulled down means the motion starts at $y = -A$ with $A = 0.045$ m.
(a) Read the period straight off the graph, $T = 0.60$ s, then use the printed $T = 2\pi/\omega$ backwards: $\omega = \dfrac{2\pi}{0.60} = 10.472\ \mathrm{rad/s}$ , so $\omega^2 = 109.66\ \mathrm{rad^2/s^2}$ .
(b) Released from rest at $y = -A$ means $\phi = \pi$ , since $\cos\pi = -1$ gives $y(0) = -A$ and the derivative $-A\omega\sin\pi$ is zero. Writing it with the sign pulled out is tidier: $y = -(0.045\ \mathrm{m})\cos\left[(10.47\ \mathrm{rad/s})t\right]$ .
Apply 7.3.A.3.i rather than differentiating twice by eye: $a = -\omega^2 y$ , so $a = +(0.045)(109.66)\cos(10.47t) = +(4.93\ \mathrm{m/s^2})\cos(10.47t)$ .
The acceleration function is the position function reflected in the time axis and rescaled by $\omega^2$ . It is not shifted. That reflection is the whole content of the CED's sample multiple-choice question 12, which uses this exact scenario and is keyed A.
(c) $a_{\max} = A\omega^2 = (0.045)(109.66) = 4.935 \approx 4.93\ \mathrm{m/s^2}$ , reached at both turning points, where the speed is zero. At the lowest point the block is at $y = -A$ and the acceleration is $+4.93\ \mathrm{m/s^2}$ , that is upward, back toward equilibrium.
For contrast, $v_{\max} = A\omega = (0.045)(10.472) = 0.471\ \mathrm{m/s}$ , reached at $y = 0$ , where the acceleration is zero. The two maxima are a quarter period apart, that is 0.15 s.
(d) The spring stays stretched as long as the amplitude is smaller than the static stretch $y_0 = mg/k$ . Since $\omega^2 = k/m$ , that stretch is $y_0 = g/\omega^2 = 9.8/109.66 = 0.0894\ \mathrm{m}$ , and $A = 0.045$ m is smaller, so the spring never goes slack.
The same test written the other way is quicker on an exam: the condition $A < g/\omega^2$ is identical to $a_{\max} < g$ . Here $a_{\max}/g = 4.93/9.8 = 0.50$ , so the block's acceleration never exceeds half of $g$ and the spring is doing nothing exotic.
Note what did not enter any of this: the mass and the force constant were never needed separately, because the graph supplied $\omega$ directly and $\omega$ is the only combination of them the motion depends on.

(a) $\omega = 10.5\ \mathrm{rad/s}$ . (b) $y = -(0.045\ \mathrm{m})\cos(10.47t)$ and $a = +(4.93\ \mathrm{m/s^2})\cos(10.47t)$ , the position graph reflected in the time axis and scaled by $\omega^2$ . (c) $a_{\max} = 4.93\ \mathrm{m/s^2}$ at both turning points, a quarter period away from the maximum speed of 0.471 m/s at equilibrium. (d) No: the static stretch is 0.0894 m, larger than the 0.045 m amplitude, which is the same condition as $a_{\max} < g$ .

Frequently asked questions

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

Only one: the maximum position 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 writes the amplitude as a maximum position rather than as A, and there is no A in either of its variable lists, while there is an entry defining 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.

How do you find the phase angle in simple harmonic motion?

Evaluate the position function and its derivative at time zero. That gives the initial position as the amplitude times the cosine of the phase angle, and the initial velocity as minus the amplitude times the angular frequency times the sine of the phase angle. Solve the pair: the amplitude is the square root of the initial position squared plus the initial velocity divided by the angular frequency, all squared. Then get the phase angle from the signs of both its cosine and its sine, not from an inverse tangent, because an angle and that angle plus pi have the same tangent and describe opposite motions. Work in radians throughout.

How do you get velocity and acceleration from the SHM position equation?

Differentiate. The velocity is minus the amplitude times the angular frequency times the sine of the same argument, and the acceleration is minus the amplitude times the angular frequency squared times the cosine of that argument. Three results follow immediately: the acceleration equals minus the angular frequency squared times the position at every instant, the maximum speed is the amplitude times the angular frequency, and the maximum acceleration is the amplitude times the angular frequency squared. None of those three is printed on the AP Physics C: Mechanics equation sheet; all three are labelled Derived equations in the framework. The derivative rules for sine and cosine are printed, in the booklet's calculus table.

Why is the acceleration graph in SHM upside down compared to the position graph?

Because the acceleration is a negative constant times the position at every instant, which is essential knowledge 7.3.A.3.i. The constant is the angular frequency squared, so the acceleration graph is the position graph reflected in the time axis and rescaled vertically by that factor. It is reflected, not shifted, which is the distinction the AP Physics C: Mechanics sample multiple-choice question 12 is built on. The velocity graph is the one that is shifted, by a quarter period, since it is the slope of the position graph and a slope vanishes at a turning point.

Where are the velocity and acceleration maximum in simple harmonic motion?

The speed is maximum at the equilibrium position and zero at both turning points. The acceleration is maximum in magnitude at both turning points and zero at the equilibrium position. So they never peak at the same moment: they are a quarter period apart. The maximum speed is the amplitude times the angular frequency and the maximum acceleration is the amplitude times the angular frequency squared, which is why doubling the angular frequency at fixed amplitude doubles one and quadruples the other. Both relations are essential knowledge 7.3.A.3.ii and neither is printed on the equation sheet.

What is resonance in AP Physics C Mechanics?

Essential knowledge 7.3.A.4 says that in the presence of a sinusoidal external force, a system may exhibit resonance, and 7.3.A.4.i says resonance occurs when an external force is exerted at the natural frequency of an oscillating system. Statement 7.3.A.4.ii says what resonance does: it increases the amplitude of oscillating motion. Statement 7.3.A.4.iii defines the natural frequency as the frequency at which the system will oscillate when it is displaced from its equilibrium position, so it is a property of the system, set by its inertia and its restoring coefficient. The driving frequency is what you tune; the natural frequency is what you tune it to, and resonance does not change it.

Does AP Physics C ask you to prove the cosine solves the SHM differential equation?

No. The only boundary statement in Unit 7 of AP Physics C: Mechanics sits under Topic 7.3 and says the course 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 it does not expect students to mathematically prove that the solution is correct. So you must be able to recognise the equation's shape in an unfamiliar system and to write down and use the cosine solution, but you will not be asked to substitute it back in and verify it, and no technique for solving the equation from scratch is examinable.