AP Physics 2 · Topic 14.2

Topic 14.2: Periodic Waves

Unit 14: Waves, Sound, and Physical Optics12-15% of the multiple-choice section

A periodic wave repeats. Its period is the time for one complete oscillation and its frequency is the rate at which it repeats, linked by T = 1 over f. Wavelength is the distance between successive corresponding positions on the wave, and amplitude is independent of both period and frequency.

AP Physics: Unit 14 (topics 14.2 Periodic Waves). AP Physics 2 Unit 14, Topic 14.2. One learning objective, 14.2.A, describe the physical properties of a periodic wave. Three essential knowledge statements. 14.2.A.1 (periodic waves have regular repetitions that can be described using period and frequency) carries six sub-statements: 14.2.A.1.i (the period is the time for one complete oscillation of the wave), 14.2.A.1.ii (the frequency is the rate at which the wave repeats, relevant equation T = 1/f), 14.2.A.1.iii (the amplitude of a wave is independent of the period and the frequency of that wave), 14.2.A.1.iv (the energy of a wave increases with increasing frequency), 14.2.A.1.v (the frequency of a sound wave is related to its pitch) and 14.2.A.1.vi (wavelength is the distance between successive corresponding positions, such as peaks or troughs, on a wave). 14.2.A.2 states that a sinusoidal wave can be described by equations for the displacement from equilibrium at a specific location as a function of time, and also by an equation for the displacement from equilibrium at a specific time as a function of position; the CED labels x(t) = A cos(omega t) = A cos(2 pi f t) and y(x) = A cos(2 pi x / lambda) as example equations rather than relevant equations. 14.2.A.3 states that for a periodic wave the wavelength is proportional to the wave's speed and inversely proportional to the wave's frequency, relevant equation lambda = v/f. The topic prints no boundary statement; Unit 14's only three sit under Topics 14.4, 14.5 and 14.9. Suggested skills are 1.A, 2.C, 3.B and 3.C, listed identically on the topic page and in the Unit at a Glance table; 2.B is not among them. Four of the 15 equations in the Waves, Sound, and Optics group of the equation sheet belong to this topic: T = 1/f, the two sinusoidal example forms, and lambda = v/f. The sheet prints lambda = v/f rather than v = f lambda. No equation relating wave energy to amplitude or frequency is printed anywhere on the AP Physics 2 sheet. Unit 14 is weighted at 12 to 15 percent of the multiple-choice section across a suggested 14 to 23 class periods.

What Topic 14.2 requires

One learning objective, 14.2.A: describe the physical properties of a periodic wave. Three essential knowledge statements sit under it, and the first carries six sub-statements.

14.2.A.1. Periodic waves have regular repetitions that can be described using period and frequency.

14.2.A.1.i. The period is the time for one complete oscillation of the wave.
14.2.A.1.ii. The frequency is the rate at which the wave repeats. Relevant equation: $T = \dfrac{1}{f}$ .
14.2.A.1.iii. The amplitude of a wave is independent of the period and the frequency of that wave.
14.2.A.1.iv. The energy of a wave increases with increasing frequency.
14.2.A.1.v. The frequency of a sound wave is related to its pitch.
14.2.A.1.vi. Wavelength is the distance between successive corresponding positions (such as peaks or troughs) on a wave.

14.2.A.2. A sinusoidal wave can be described by equations for the displacement from equilibrium at a specific location as a function of time. A wave can also be described by an equation for the displacement from equilibrium at a specific time as a function of position. The CED labels the two forms that follow as example equations rather than as a relevant equation.

14.2.A.3. For a periodic wave, the wavelength is proportional to the wave's speed and inversely proportional to the wave's frequency. Relevant equation: $\lambda = \dfrac{v}{f}$ .

Suggested skills: 1.A create diagrams, tables, charts, or schematics to represent physical situations, 2.C compare physical quantities between two or more scenarios or at different times and locations in a single scenario, 3.B apply an appropriate law, definition, theoretical relationship, or model to make a claim, and 3.C justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Identical on the topic page and in the Unit at a Glance table.

Topic 14.2 prints no boundary statement. The three in Unit 14 sit under Topics 14.4, 14.5 and 14.9.

Note the change from Topic 14.1: 14.1's skill list opens with 1.C, qualitative sketches, and 14.2's opens with 1.A, diagrams and tables. This is the topic where a wave becomes something you tabulate and label rather than sketch.

Period and frequency are one fact counted two ways

14.2.A.1.i defines the period as the time for one complete oscillation of the wave. 14.2.A.1.ii defines the frequency as the rate at which the wave repeats, and gives the relevant equation:

T = \frac{1}{f}

They are the same information stated in opposite units. Period is seconds per repetition; frequency is repetitions per second. If a buoy rises and falls every $4.0$ seconds, its period is $4.0$ s and its frequency is $0.25$ Hz. Neither is more fundamental.

Two details worth pinning down.

"One complete oscillation" means back to the same state, not merely to the same displacement. A point on a wave passes through zero displacement twice per cycle, once going up and once going down. A full period returns it to the same displacement and the same direction of travel.

Frequency is measured in hertz, which is inverse seconds. One hertz is one repetition per second. Nothing in the unit is per minute.

$T = 1/f$ is printed on the AP Physics 2 equation sheet, in the Waves, Sound, and Optics group. So is the angular-frequency-bearing form $x(t) = A\cos(\omega t) = A\cos(2\pi f t)$ , which is where $\omega = 2\pi f$ comes from if you need it.

This same period-and-frequency pair drives oscillations in AP Physics 1, where a mass on a spring has a period set by the mass and the spring constant. The definitions are identical; only the thing oscillating changed. If the oscillation side is shaky, the simple harmonic motion guide covers the period formulas that AP Physics 2 assumes you already have.

Wavelength: read the CED's definition, not the picture

14.2.A.1.vi: wavelength is the distance between successive corresponding positions (such as peaks or troughs) on a wave.

Most textbooks say "crest to crest", and that is a special case of what the CED actually says. The definition is about corresponding positions, and the parenthesis names peaks and troughs as examples rather than as the definition. Any two successive points in the same phase are a wavelength apart: crest to next crest, trough to next trough, and a zero crossing going upward to the next zero crossing going upward.

That matters on a graph-reading question. If the graph gives you a clean pair of zero crossings but the crests fall between gridlines, measure the zero crossings, provided you take two crossings of the same kind. A crossing going up to the very next crossing going down is a half-wavelength, and that is the most common way a graph question is misread.

For a longitudinal wave the same definition holds with the pressure trace substituted: compression to next compression, or rarefaction to next rarefaction. Topic 14.1 names those regions.

Note also which quantity is a distance and which is a time. Wavelength is metres. Period is seconds. The two get confused because both are described as "one full cycle", and they are: one full cycle in space and one full cycle in time. Which one a graph shows you depends entirely on what its horizontal axis is, which is the next section.

The two graphs of one wave, and the two equations that go with them

14.2.A.2 is the statement students skim and then lose marks to. It says a sinusoidal wave can be described by equations for the displacement from equilibrium at a specific location as a function of time, and also by an equation for the displacement from equilibrium at a specific time as a function of position. The CED gives both as example equations:

x(t) = A \cos(\omega t) = A \cos(2\pi f t)

y(x) = A \cos\left(\frac{2\pi x}{\lambda}\right)

Read the phrases the CED attaches to each. The first is displacement at a specific location as a function of time: pick one point of the medium, stand there, and record how it moves. Its horizontal axis is time, and one full cycle across that axis is a period. The second is displacement at a specific time as a function of position: freeze the wave, walk along it, and record its shape. Its horizontal axis is position, and one full cycle across that axis is a wavelength.

Equation	Held fixed	Horizontal axis	One cycle reads as
$x(t) = A \cos(2\pi f t)$	location	time, in seconds	period $T$
$y(x) = A \cos\left(\dfrac{2\pi x}{\lambda}\right)$	time	position, in metres	wavelength $\lambda$

Both curves look like the same cosine. Only the axis label tells them apart, and both are printed on the equation sheet with no note saying which is which. So on any wave graph, the first thing to do is read the horizontal axis, and only then decide whether the repeat distance you can measure is a period or a wavelength.

Two more things the CED's forms tell you, once you look at them as equations.

Both start at maximum displacement. At $t = 0$ the first gives $x = A$ , and at $x = 0$ the second gives $y = A$ . The CED prints cosines, not sines, so its example wave is at a crest at the origin. That is a choice of where to start counting, not a physical requirement.

$A$ appears out front in both, multiplying nothing else. The amplitude scales the whole curve and appears nowhere inside the argument. That is 14.2.A.1.iii written algebraically, which is the next section.

Skill 1.A, create diagrams, tables, charts, or schematics, is on this topic's list, and this pair of graphs is what it most likely means here. A labelled sketch that names its horizontal axis and marks either $T$ or $\lambda$ on it does the job.

Amplitude is independent of period and frequency

14.2.A.1.iii states it flatly: the amplitude of a wave is independent of the period and the frequency of that wave.

This is a short sentence with a lot of consequences, and it is the statement most worth memorizing in Topic 14.2.

Shaking a rope harder without shaking it faster changes the amplitude and leaves the period and frequency alone.
Shaking it faster without shaking it further changes the period and frequency and leaves the amplitude alone.
Turning up the volume on a pure tone raises the amplitude. It does not change the pitch, because pitch is related to frequency (14.2.A.1.v).
Playing a higher note raises the frequency. It does not, by itself, make the note louder, because loudness increases with amplitude (14.1.A.6.ii).

You can see it in the printed equations. In $x(t) = A\cos(2\pi f t)$ , the symbols $A$ and $f$ sit in different places and neither appears in the other's definition. Change one and the other is untouched.

There is one thing amplitude and frequency do share, and the CED keeps the two claims deliberately separate:

14.1.A.6.iii: the energy carried by a wave increases with increasing amplitude.
14.2.A.1.iv: the energy of a wave increases with increasing frequency.

Both raise energy. Neither is given as a proportionality, and no equation relating wave energy to amplitude or to frequency is printed anywhere among the 129 on the AP Physics 2 equation sheet. So the answerable claim on this exam is the direction, not a factor.

There is one exception to that in the course, and it is worth knowing where it lives so you do not import it here. In Unit 15, $E = hf$ gives the energy of a single photon and is printed on the sheet. That is a statement about a photon, not about the energy carried by a classical wave of a given amplitude, and Topic 14.2 is the classical picture.

Pitch, and the octave question the unit opens with

14.2.A.1.v: the frequency of a sound wave is related to its pitch.

Note the wording. Related to, not equal to and not proportional to. Pitch is what a listener hears; frequency is what the wave does. Higher frequency means higher pitch, and that is the extent of what the CED commits to.

The unit opener lists as one of its five essential questions: why do two notes an octave apart sound the same? The CED poses it and does not answer it, so treat the answer as musical rather than as examinable physics: an octave is defined as a doubling of frequency, so the note an octave above $440$ Hz is $880$ Hz, and the two share a great many of their repeats. What is examinable is the physics that surrounds it, that the higher note has the higher frequency, the shorter period, and, in the same air, the shorter wavelength.

Two related traps:

Pitch is not loudness. Loudness tracks amplitude (14.1.A.6.ii), pitch tracks frequency. They are independent, exactly as 14.2.A.1.iii requires.
Frequency belongs to the source, not the air. When a sound crosses into a new medium its speed and wavelength change and its frequency does not, which is 14.3.A.1.iv in Topic 14.3. A note does not change pitch when it passes through a wall.

The other frequency-and-observer effect, where relative motion changes the frequency you measure, is Topic 14.5, and its boundary statement limits it to qualitative treatment.

The relation between speed, frequency and wavelength

14.2.A.3: for a periodic wave, the wavelength is proportional to the wave's speed and inversely proportional to the wave's frequency. The relevant equation:

\lambda = \frac{v}{f}

Look at the form the CED and the equation sheet both use. It is $\lambda = v/f$ , solved for wavelength. The version most students carry, $v = f\lambda$ , is the same relation rearranged, and it is not the form printed on the sheet. Neither is wrong. Just do not spend exam time hunting for a line that says $v = f\lambda$ .

The direction of causation matters more than the algebra. Topic 14.1 established that speed is a property of the medium (14.1.A.3), and frequency is set by whatever is driving the wave. Those two are the inputs. Wavelength is the output: it is whatever it has to be so that a wave moving at $v$ repeats $f$ times a second.

That reading resolves the classic confusion about what happens at a boundary. Frequency does not change when a wave enters a new medium (14.3.A.1.iv), so with $v$ changed and $f$ fixed, $\lambda$ must move. Wavelength is the quantity that absorbs the change, every time.

Three sanity checks the relation gives you for free:

Same medium, higher frequency, shorter wavelength. Speed is fixed, so $\lambda$ and $f$ trade off exactly.
Same frequency, faster medium, longer wavelength.
Units: metres per second divided by inverse seconds gives metres. If a wavelength answer comes out in the wrong units, the rearrangement is upside down.

For the calculation routine itself, working through the algebra case by case and reading values off graphs, the wave speed, frequency and wavelength guide is the dedicated page. This topic's job is the CED framing around it.

How Topic 14.2 shows up on the exam, and where it goes wrong

Topic 14.2's suggested skills are 1.A, 2.C, 3.B and 3.C. Skill 1 is not assessed on the multiple-choice section at all, and 2.B, calculate an unknown quantity, is not on this topic's list, so a Topic 14.2 multiple-choice question tends to look like a comparison or a claim rather than a computation. Typical shapes: two waves on the same string with different frequencies, asked which has the longer wavelength; a graph with an unlabelled axis, asked what you can and cannot determine from it; a claim about loudness and pitch, asked to be justified.

The errors that cost marks:

Reading a period off a graph whose axis is position. Check the horizontal axis label first. 14.2.A.2 defines two different graphs and the CED prints both.
Measuring half a wavelength. 14.2.A.1.vi says successive corresponding positions. Crest to the next trough is half.
Tying amplitude to frequency. 14.2.A.1.iii says they are independent. Turning up the volume does not raise the pitch.
Quantifying an "increases with". 14.2.A.1.iv gives the direction of the energy and frequency relationship and no more, and no equation for it is printed.
Assuming a bigger wave is faster. Speed comes from the medium (14.1.A.3), not from the wave you make.
Expecting frequency to change at a boundary. It does not (14.3.A.1.iv). Speed and wavelength change together and frequency holds.
Hunting for $v = f\lambda$ on the sheet. The printed form is $\lambda = v/f$ .

Where to go next. Topic 14.1 is the prerequisite and owns amplitude, wave speed and the transverse and longitudinal split. Topic 14.3 takes the periodic wave defined here to a boundary and asks what survives. Topic 14.6 overlaps two of them. The unit hub shows how all nine fit together.

One wave, both graphs, both equations

A transverse wave travels along a string. A detector at one fixed point records the string's displacement rising to $3.0 \text{ cm}$ above equilibrium and repeating every $0.25 \text{ s}$ . A photograph of the whole string, taken at one instant, shows crests $1.2 \text{ m}$ apart. Find (a) the frequency, (b) the wave speed, and (c) write the displacement as a function of time at the detector and as a function of position at the instant of the photograph.

Sort the two measurements by which graph they came from. The detector holds location fixed and varies time, so its $0.25 \text{ s}$ repeat is a period: $T = 0.25 \text{ s}$ . The photograph holds time fixed and varies position, so its $1.2 \text{ m}$ repeat is a wavelength: $\lambda = 1.2 \text{ m}$ . This is 14.2.A.2's distinction, and getting it backwards is the whole difficulty of the question.
(a) Frequency from 14.2.A.1.ii: $f = \dfrac{1}{T} = \dfrac{1}{0.25 \text{ s}} = 4.0 \text{ Hz}$ .
(b) Speed from 14.2.A.3, rearranging $\lambda = v/f$ to $v = \lambda f$ : $v = (1.2 \text{ m})(4.0 \text{ Hz}) = 4.8 \text{ m/s}$ . Units check: metres times inverse seconds gives metres per second.
(c) Amplitude is the maximum displacement from equilibrium, so $A = 3.0 \text{ cm} = 0.030 \text{ m}$ (14.1.A.6). Substituting into the first example equation of 14.2.A.2: $x(t) = (0.030 \text{ m})\cos\left(2\pi (4.0 \text{ Hz})\, t\right)$ , which is $x(t) = (0.030 \text{ m})\cos\left(25 \, t\right)$ with $t$ in seconds, since $\omega = 2\pi f = 25.1 \text{ rad/s}$ .
And into the second: $y(x) = (0.030 \text{ m})\cos\left(\dfrac{2\pi x}{1.2 \text{ m}}\right)$ , with $x$ in metres.
Sanity-check the pair. Both have the same $A$ out front, because amplitude does not care which graph you are drawing. The first repeats every $0.25 \text{ s}$ of its argument, the second every $1.2 \text{ m}$ of its argument, and the ratio of those two repeats is $1.2/0.25 = 4.8 \text{ m/s}$ , the speed. The two representations agree.

(a) $f = 4.0 \text{ Hz}$ . (b) $v = 4.8 \text{ m/s}$ . (c) At the detector, $x(t) = (0.030 \text{ m})\cos(2\pi(4.0 \text{ Hz})t)$ . Along the string at that instant, $y(x) = (0.030 \text{ m})\cos\left(\dfrac{2\pi x}{1.2 \text{ m}}\right)$ .

Change one thing at a time (skill 2.C)

The wave of the previous example travels at $4.8 \text{ m/s}$ on its string with $f = 4.0 \text{ Hz}$ , $\lambda = 1.2 \text{ m}$ and $A = 0.030 \text{ m}$ . Two changes are made, each starting from that original wave and each on the same string at the same tension. (a) The source is shaken three times as far but at the same rate. (b) The source is shaken three times as fast but through the same distance. For each, state what happens to speed, frequency, period, wavelength, amplitude and energy.

(a) Three times as far, same rate. The amplitude triples to $0.090 \text{ m}$ . By 14.2.A.1.iii the amplitude of a wave is independent of the period and the frequency, so $f = 4.0 \text{ Hz}$ and $T = 0.25 \text{ s}$ are unchanged.
Speed is unchanged at $4.8 \text{ m/s}$ , because 14.1.A.3 makes it a property of the string and neither the tension nor the mass per length moved. With $v$ and $f$ both fixed, $\lambda = v/f$ leaves the wavelength at $1.2 \text{ m}$ .
Energy increases, by 14.1.A.6.iii, since the amplitude increased. The CED gives the direction only, so stop at "increases".
(b) Three times as fast, same distance. Now $f = 12.0 \text{ Hz}$ and $T = 1/12.0 = 0.0833 \text{ s}$ . The amplitude stays at $0.030 \text{ m}$ , again by 14.2.A.1.iii read the other way.
Speed is still $4.8 \text{ m/s}$ , for the same reason as before: nothing about the string changed. So the wavelength has to absorb the change: $\lambda = \dfrac{v}{f} = \dfrac{4.8 \text{ m/s}}{12.0 \text{ Hz}} = 0.40 \text{ m}$ , one third of the original, exactly as 14.2.A.3's inverse proportionality predicts.
Energy increases here too, this time by 14.2.A.1.iv, the energy of a wave increases with increasing frequency. Two different essential knowledge statements, two different levers, the same one-word conclusion.
The pattern to carry away: the speed never moved, because the string never moved. In (a) only the amplitude and the energy changed. In (b) the frequency, the period, the wavelength and the energy changed, and the amplitude did not. Every "unchanged" in that list traces back to a specific essential knowledge statement rather than to intuition.

(a) Amplitude triples to $0.090 \text{ m}$ and energy increases; speed, frequency, period and wavelength are all unchanged. (b) Frequency triples to $12.0 \text{ Hz}$ , period falls to $0.0833 \text{ s}$ , wavelength falls to $0.40 \text{ m}$ , and energy increases; speed and amplitude are unchanged.

Two notes an octave apart, in the same air

In a room where the speed of sound is $343 \text{ m/s}$ , a tuning fork sounds a note of frequency $512 \text{ Hz}$ . (a) Find its period and wavelength. (b) A second fork sounds the note one octave higher, at twice the frequency. Find its period and wavelength. (c) Which of the two notes is louder, on the information given?

(a) Period from 14.2.A.1.ii: $T = \dfrac{1}{f} = \dfrac{1}{512 \text{ Hz}} = 1.95 \times 10^{-3} \text{ s}$ , about $1.95$ milliseconds.
Wavelength from 14.2.A.3: $\lambda = \dfrac{v}{f} = \dfrac{343 \text{ m/s}}{512 \text{ Hz}} = 0.670 \text{ m}$ .
(b) An octave up is $f = 1024 \text{ Hz}$ . Period: $T = \dfrac{1}{1024 \text{ Hz}} = 9.77 \times 10^{-4} \text{ s}$ , exactly half the first, since $T$ and $f$ are reciprocals.
Wavelength: the speed of sound is a property of the air (14.1.A.3) and the air has not changed, so $v$ stays at $343 \text{ m/s}$ and only $f$ moves. $\lambda = \dfrac{343 \text{ m/s}}{1024 \text{ Hz}} = 0.335 \text{ m}$ , exactly half the first. Doubling the frequency halves the wavelength, which is 14.2.A.3's inverse proportionality.
(c) You cannot tell. Loudness increases with amplitude (14.1.A.6.ii), and no amplitude was given for either fork. Frequency is related to pitch (14.2.A.1.v), not to loudness, and 14.2.A.1.iii states that the amplitude of a wave is independent of the period and the frequency, so the higher note is not louder by virtue of being higher.
One further caution about part (b). The higher note carries more energy per wave by 14.2.A.1.iv, and that is still not a statement about loudness. Energy and loudness are attached to different essential knowledge statements in this course, and the CED never merges them.

(a) $T = 1.95 \times 10^{-3} \text{ s}$ and $\lambda = 0.670 \text{ m}$ . (b) $T = 9.77 \times 10^{-4} \text{ s}$ and $\lambda = 0.335 \text{ m}$ , each exactly half the first, because the air's speed of sound is unchanged. (c) Indeterminate: loudness depends on amplitude (14.1.A.6.ii), which was not given, and amplitude is independent of frequency (14.2.A.1.iii).

Frequently asked questions

What is a periodic wave in AP Physics 2?

A wave with regular repetitions that can be described using period and frequency, per essential knowledge 14.2.A.1. The AP Physics 2 CED models a wave as a continuous, periodic disturbance with well-defined wavelength and frequency (14.1.A.1.ii), which is what separates it from a wave pulse, a single disturbance with no wavelength or frequency defined. Period, frequency, wavelength and amplitude are the four quantities Topic 14.2 uses to describe one.

What does the AP Physics 2 CED say wavelength is?

Essential knowledge 14.2.A.1.vi defines wavelength as the distance between successive corresponding positions, such as peaks or troughs, on a wave. The key phrase is corresponding positions: peaks and troughs are named as examples, not as the definition. Two successive points in the same phase are one wavelength apart, so crest to next crest, trough to next trough, or an upward zero crossing to the next upward zero crossing all work. A crest to the next trough is half a wavelength.

Does amplitude affect the frequency or period of a wave?

No. Essential knowledge 14.2.A.1.iii states that the amplitude of a wave is independent of the period and the frequency of that wave. You can see it in the CED's own example equation x(t) = A cos(2 pi f t): the amplitude A multiplies the whole cosine and appears nowhere inside its argument, so changing A leaves f untouched. In practice, turning up the volume of a pure tone raises its amplitude and its loudness without changing its pitch.

What are the two sinusoidal wave equations in AP Physics 2?

Essential knowledge 14.2.A.2 gives two example equations. x(t) = A cos(omega t) = A cos(2 pi f t) describes the displacement from equilibrium at a specific location as a function of time, so its horizontal axis is time and one cycle across it is a period. y(x) = A cos(2 pi x / lambda) describes the displacement from equilibrium at a specific time as a function of position, so its horizontal axis is position and one cycle across it is a wavelength. Both are printed on the AP Physics 2 equation sheet.

Does the AP Physics 2 equation sheet print v = f lambda?

Not in that form. The Waves, Sound, and Optics group of the AP Physics 2 sheet prints lambda = v/f, which is the same relation solved for wavelength, and the CED cites it as the relevant equation for essential knowledge 14.2.A.3. That statement says the wavelength is proportional to the wave's speed and inversely proportional to the wave's frequency. Rearranging to v = f lambda is fine; just do not expect to find that line printed on the sheet during the exam.

Why does a higher-frequency wave carry more energy?

The AP Physics 2 CED states the relationship without deriving it. Essential knowledge 14.2.A.1.iv says the energy of a wave increases with increasing frequency, and separately 14.1.A.6.iii says the energy carried by a wave increases with increasing amplitude. Neither is given as a proportionality, and no equation relating wave energy to frequency or amplitude appears on the AP Physics 2 equation sheet. E = hf, in the Modern Physics group, is the energy of a photon in Unit 15, not the energy of a classical wave.

Is pitch the same thing as frequency?

Not quite. Essential knowledge 14.2.A.1.v says the frequency of a sound wave is related to its pitch, and the CED stops there. Frequency is a measurable property of the wave, in hertz; pitch is what a listener perceives. Higher frequency means higher pitch, which is all the AP Physics 2 framework commits to. Pitch is also independent of loudness, since loudness tracks amplitude (14.1.A.6.ii) and amplitude is independent of frequency (14.2.A.1.iii).