3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A floating buoy moves up and down as water waves pass it. Which of the following is the period of the waves?
Answer and reasoning
AThe time the buoy takes to go from its highest point back to its highest pointCorrect The period is the time for one complete oscillation. Starting at its highest point, the buoy must go down to its lowest point and come back up to its highest point before its motion starts to repeat.
BThe time the buoy takes to go from its highest point down to its very lowest point A student who counts one swing from top to bottom as a whole oscillation picks this. From highest to lowest is only half an oscillation: the buoy has not yet returned to where it started, so this time is half the period.
CThe time a crest takes to travel from the buoy to the shore close by A student who thinks the period is the time the wave takes to get somewhere picks this. That travel time depends on how far away the shore is and on the wave speed; the period describes how often the motion at one place repeats.
DThe number of times the buoy reaches its highest point each second A student who treats period and frequency as the same quantity picks this. The number of repeats each second is the frequency, f; the period is its reciprocal, T = 1/f, the time taken by one repeat.
As a wave passes, the displacement of one point on a string is described by x(t) = (0.020 m) cos[(10π rad/s)t]. Which statement about the motion of this point is correct?
Answer and reasoning
AIt completes about 31 full oscillations in each second. A student who takes the number multiplying t as the frequency picks this: 10π ≈ 31. That number is the angular frequency ω = 2πf, in rad/s; dividing it by 2π gives f = 5.0 Hz.
BIt completes 5 full oscillations in each second.Correct The equation has the form x(t) = A cos(2πft), so 2πf = 10π rad/s and f = 5.0 Hz. The point completes 5 oscillations each second, one every 0.20 s.
CIn each second it completes 10 full oscillations. A student who reads the number in front of π as the frequency picks this. The coefficient of t is 2πf, so 10π = 2πf and f = 5.0 Hz, not 10 Hz.
DIt completes one full oscillation in every 5 seconds. A student who uses the frequency as the time for one oscillation picks this. The frequency is 5.0 Hz, so the period is T = 1/f = 0.20 s, not 5 s.
Working Compare with x(t) = A cos(2πft): 2πf = 10π rad/s, so f = 5.0 Hz, that is, 5 complete oscillations each second (T = 1/f = 0.20 s).
A wave travels along a stretched string of length L. The distance between a crest and the adjacent trough is D, and a crest takes a time t₀ to travel the full length of the string. Which expression gives the frequency of the wave?
Answer and reasoning
Af = L/(Dt₀) A student who takes the crest-to-trough distance D as the wavelength picks this. Crest to adjacent trough is only half a wavelength, so λ = 2D, which halves the frequency.
Bf = L/(2Dt₀)Correct The wave speed is v = L/t₀. A crest and the adjacent trough are half a wavelength apart, so λ = 2D. From λ = v/f, f = v/λ = (L/t₀)/(2D) = L/(2Dt₀).
Cf = 2DL/t₀ A student who uses λ = f/v, and so f = λv, picks this: (2D)(L/t₀). The correct relation is f = v/λ; 2DL/t₀ has units m²/s, not Hz.
Df = 1/t₀ A student who takes the travel time along the string as the period picks this. The time t₀ for a crest to cross the string depends on the string's length; the period is the time for one oscillation, T = λ/v = 2Dt₀/L.
Working v = L/t₀. Crest to adjacent trough = λ/2, so λ = 2D. f = v/λ = (L/t₀)/(2D) = L/(2Dt₀).
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
14.2.A.1 Periodic wave Fix
Periodic wave
A wave in which the same pattern of disturbance repeats at regular intervals, both in time at any one location and in space along the medium at any one instant. Its regular repetition is described by its period and frequency (and, in space, by its wavelength).
Period, T
The time for one complete oscillation of the wave: for example, the time for a point of the medium to go from its highest position down to its lowest and back up to its highest, or the time between one crest and the next passing a fixed point. SI unit: second (s).
Frequency, f
The rate at which the wave repeats: the number of complete oscillations per unit time. It is the reciprocal of the period, f = 1/T (T = 1/f). For a wave produced by a source, the frequency is that of the source. SI unit: hertz (Hz), where 1 Hz = 1 s⁻¹.
Amplitude and its independence of frequency
The amplitude is the maximum displacement of the medium from equilibrium. It is independent of the period and frequency: a source can make a larger-amplitude wave at the same frequency by moving through a larger distance in the same time, that is, by moving faster. SI unit: meter (m) for a displacement.
Energy and frequency
A wave's energy increases as its frequency increases. For two waves of the same amplitude in the same medium, each part of the medium moves through the same distance in each cycle, but in the higher-frequency wave it completes more cycles each second, so it moves faster and the wave carries more energy.
Pitch
The perceived highness or lowness of a sound, which is related to its frequency: the greater the frequency, the higher the pitch. Pitch is distinct from loudness, which increases with the amplitude of the sound wave.
Wavelength, λ
The distance between successive corresponding positions on a wave, such as from one crest to the next crest or from one trough to the next trough, measured along the direction the wave travels. The distance from a crest to the adjacent trough is half a wavelength. SI unit: meter (m).
Students often think One complete oscillation is a single swing from one extreme position to the other (for example, highest to lowest), so the period is the time between a crest and the next trough, or between successive zero crossings. In fact No. One complete oscillation takes a point back to the same position moving in the same direction: for example, from its highest point down to its lowest point and back up to its highest point. Highest to lowest is half an oscillation and takes half the period.
Students often think The period of a wave is the time the wave takes to travel from its source to a given place, or along the whole medium. In fact No. The period is the time for one complete oscillation at a location. The travel time depends on the distance and the wave speed, and may be many periods or a small fraction of one.
14.2.A.2 Displacement–time description, x(t) Fix
Displacement–time description, x(t)
For a sinusoidal wave, the displacement from equilibrium at one location as a function of time can be written x(t) = A cos(ωt) = A cos(2πft). Its graph repeats every period T; the horizontal axis is time, so the repeat distance on the graph is a period, not a wavelength.
Displacement–position description, y(x)
For a sinusoidal wave, the displacement from equilibrium at one instant as a function of position along the medium can be written y(x) = A cos(2πx/λ). Its graph (a snapshot of the wave) repeats every wavelength λ.
Angular frequency, ω
The quantity multiplying t in x(t) = A cos(ωt): ω = 2πf. It is not the frequency itself; the frequency is f = ω/(2π). SI unit: radian per second (rad/s).
Students often think A displacement–time graph is a picture of the wave's shape, so the horizontal length of one cycle on it is the wavelength. In fact No. A displacement–time graph shows how one point moves as time passes, so the horizontal length of one cycle is the period, a time. The wavelength is a distance: it is read from a displacement–position (snapshot) graph, or found from λ = vT.
Students often think In x(t) = A cos(ωt), the number multiplying t is the frequency f, in hertz. In fact No. The number multiplying t is the angular frequency ω = 2πf, in rad/s. The frequency is f = ω/(2π).
14.2.A.3 Wave speed and λ = v/f Fix
Wave speed and λ = v/f
For a periodic wave, the wavelength is proportional to the wave's speed and inversely proportional to its frequency, λ = v/f (equivalently v = fλ). In one period the wave travels one wavelength. The speed is set by the medium (for a string, vstring = √(FT/(m/ℓ))), the frequency by the source, and the wavelength follows from both.
Students often think The wavelength is given by λ = f/v: it is proportional to the frequency and inversely proportional to the wave speed. In fact No. λ = v/f: the wavelength is proportional to the wave speed and inversely proportional to the frequency. In a given medium, a higher frequency gives a shorter wavelength, and at a given frequency a faster wave has a longer wavelength.
Students often think In the same medium, a higher-frequency (shorter-wavelength) wave travels faster, since v = fλ increases with f. In fact No. The wave speed is set by the medium. In the same medium, waves of all frequencies travel at the same speed; a higher frequency gives a shorter wavelength, λ = v/f.
11 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 11
A wave travels along a string at 1.5 m/s. The graph shows the displacement y of one point on the string as a function of time t. What is the wavelength of the wave?
Answer and reasoning
A0.40 m A student who reads the graph as a picture of the string picks this, taking the horizontal length of one cycle as the wavelength. The horizontal axis is time, so 0.40 s is the period. The wavelength is the distance the wave travels in one period: (1.5 m/s)(0.40 s) = 0.60 m.
B0.30 m A student who takes the period as the time between successive zero crossings, 0.20 s, picks this: (1.5 m/s)(0.20 s) = 0.30 m. Between those crossings the point makes only half an oscillation; a full oscillation takes 0.40 s.
C0.60 mCorrect The motion repeats every 0.40 s (for example, y = 0 moving upward at t = 0 and again at t = 0.40 s), so T = 0.40 s and f = 1/T = 2.5 Hz. The wavelength is λ = v/f = (1.5 m/s)/(2.5 Hz) = 0.60 m: in one period the wave travels 0.60 m.
D0.50 m A student who counts the three cycles on the graph and calls the frequency 3 Hz picks this: (1.5 m/s)/(3 Hz) = 0.50 m. The three cycles take 1.20 s, so the frequency is 3/(1.20 s) = 2.5 Hz.
Working The graph repeats every 0.40 s (zero crossings moving upward at t = 0, 0.40 s, 0.80 s, 1.20 s), so T = 0.40 s and f = 1/T = 2.5 Hz. λ = v/f = vT = (1.5 m/s)(0.40 s) = 0.60 m.
A student makes a wave on a long rope by moving one end up and down. She claims: "If I make the amplitude of the wave twice as large, its frequency has to decrease, because my hand has farther to travel in each cycle." Which response to her claim is correct?
Answer and reasoning
AIt is right: each cycle takes longer when her hand travels farther, so the frequency of the wave will decrease. A student who thinks a larger amplitude means a longer period picks this. The time for each cycle depends on how fast her hand moves as well as how far; she can cover twice the distance in the same time, so the frequency need not change.
BIt is wrong: the frequency of a wave on a rope is fixed by the rope's tension and mass per length. A student who thinks the medium fixes the frequency picks this. The rope's tension and mass per length set the wave speed. The frequency is set by the source: every point of the rope oscillates as often as her hand does, so she can change it.
CIt is wrong: a larger-amplitude wave travels faster, so more crests pass a point on the rope per second. A student who thinks a bigger wave travels faster picks this. The wave speed is set by the rope's tension and mass per length, not by the amplitude, and the number of crests passing a point each second equals the number of cycles her hand makes each second.
DIt is wrong: by moving her hand faster, she can keep the frequency the same with the larger amplitude too.Correct Amplitude and frequency are independent. The frequency is set by how often her hand repeats its motion, and the amplitude by how far it moves. Moving her hand twice as far in the same time, at twice the speed, doubles the amplitude and leaves the frequency, and the period, unchanged.
Two sinusoidal waves travel along two identical strings under the same tension. The waves have the same amplitude. Wave X has a frequency of 20 Hz, and wave Y has a frequency of 60 Hz. Which claim about the energy carried by one meter of each wave, with its reasoning, is correct?
Answer and reasoning
AThey carry equal amounts of energy, because the two waves have the same amplitude. A student who thinks a wave's energy depends only on its amplitude picks this. Energy increases with amplitude, but it also increases with frequency: at equal amplitudes, the parts of string Y oscillate faster and so have more energy.
BY carries more energy, because each part of its string moves faster as it oscillates.Correct The energy of a wave increases with its frequency. The amplitudes are equal, so a point on either string moves through the same distance in each cycle, but a point on Y completes three times as many cycles each second. It therefore moves three times as fast on average and has more kinetic energy, so each meter of wave Y carries more energy.
CX carries more energy, because its wavelength is longer, which makes it a bigger wave. A student who links a longer wavelength with a bigger, more energetic wave picks this. X's wavelength is three times as long because its frequency is one third as great; at the same amplitude, the lower-frequency wave carries less energy, not more.
DY carries more energy, because it travels along its string at the greater speed. A student who thinks higher-frequency waves travel faster picks this. The strings are identical and under the same tension, so the two waves travel at the same speed and Y simply has the shorter wavelength. Y does carry more energy, but because its string oscillates faster, not because the wave moves faster.
The two graphs shown give the change in air pressure ΔP at a microphone as a function of time t for two sounds, A and B, drawn to the same scales. Which claim about the pitches of the two sounds, with its reasoning, is supported by the graphs?
Answer and reasoning
AB has the higher pitch, as its pressure variations are the larger ones. A student who links pitch with loudness picks this. B's larger pressure variations (larger amplitude) make B the louder sound; pitch depends on how often the variations repeat, and A's repeat twice as often.
BB has the higher pitch, as each of its cycles lasts a longer time. A student who treats a longer period as a higher frequency picks this. B's cycles last 4 ms and A's last 2 ms, so B has the LOWER frequency: f = 1/T gives 250 Hz for B and 500 Hz for A.
CA has the higher pitch, as its pressure variations repeat more often.Correct Pitch is related to frequency. Sound A completes 4 cycles in 8 ms (one every 2 ms, 500 Hz), and sound B completes 2 cycles in 8 ms (one every 4 ms, 250 Hz). A has the greater frequency, so it has the higher pitch, even though B is the louder sound.
DThey have the same pitch, as both travel through the air at one speed. A student who thinks pitch is set by the speed of sound picks this. The two sounds do travel through the air at the same speed, but pitch depends on frequency, and the graphs show that A's frequency is twice B's.
Two sounds travel through the air in a room. Sound 1 has a wavelength of 0.50 m, and sound 2 has a wavelength of 1.0 m. Which claim about the pitches of the two sounds, with its reasoning, is correct?
Answer and reasoning
ASound 1 has the higher pitch, as its frequency is the greater one.Correct Both sounds travel through the same air at the same speed, so by f = v/λ the sound with the shorter wavelength has the greater frequency: sound 1's frequency is twice sound 2's. Pitch is related to frequency, so sound 1 has the higher pitch.
BSound 2 has the higher pitch, as its longer wavelength means a greater frequency. A student who thinks wavelength increases with frequency (λ = f/v) picks this. At the same speed, λ = v/f, so the longer wavelength goes with the LOWER frequency and sound 2 has the lower pitch.
CTheir pitches are equal, as both travel through the air at the same speed. A student who thinks pitch is set by the speed of sound picks this. The sounds do travel at the same speed, but their wavelengths differ, so their frequencies, and hence their pitches, differ.
DSound 1 has the higher pitch, as its shorter waves travel faster through air. A student who thinks shorter-wavelength waves travel faster picks this. All sounds travel through the same air at the same speed; sound 1's shorter wavelength shows that its frequency is greater, and that is why its pitch is higher.
Working Same medium, so the same speed v for both sounds. f = v/λ: f₁/f₂ = λ₂/λ₁ = (1.0 m)/(0.50 m) = 2, so sound 1 has twice the frequency of sound 2 and the higher pitch.
The graph shows the shape of a string at one instant as a wave of frequency 4.0 Hz travels along it. What is the speed of the wave?
Answer and reasoning
A1.6 m/s A student who takes the wavelength as the distance from a crest to the next trough, 0.40 m, picks this: (4.0 Hz)(0.40 m) = 1.6 m/s. Crest to trough is half a wavelength; crest to next crest is 0.80 m.
B6.4 m/s A student who takes the wavelength as the length of the whole wave shown, 1.60 m, picks this: (4.0 Hz)(1.60 m) = 6.4 m/s. The graph shows two repeats of the pattern; one repeat, 0.80 m, is the wavelength.
C5.0 m/s A student who uses λ = f/v, and so v = f/λ, picks this: (4.0 Hz)/(0.80 m) = 5.0. The correct relation is λ = v/f, so v = fλ = 3.2 m/s; f/λ does not even have the units of a speed.
D3.2 m/sCorrect Successive crests are at x = 0.20 m and x = 1.00 m, so the wavelength is 0.80 m. v = fλ = (4.0 Hz)(0.80 m) = 3.2 m/s.
Working From the graph, successive crests are at x = 0.20 m and x = 1.00 m, so λ = 0.80 m. v = fλ = (4.0 Hz)(0.80 m) = 3.2 m/s.
The graph shows the displacement x of one point on a string as a function of time t while a sinusoidal wave passes. Which equation describes the motion of the point?
Answer and reasoning
Ax(t) = A₀ cos(2πt/t₁) A student who takes the time from maximum to minimum as a full oscillation picks this, using T = t₁. That is half a cycle: the point is not back at A₀ until 2t₁, so the period is 2t₁.
Bx(t) = A₀ cos(t/(2t₁)) A student who puts the frequency, f = 1/(2t₁), where 2πf belongs picks this. The quantity multiplying t in the cosine is 2πf; without the 2π the cosine would take 4πt₁, not 2t₁, to repeat.
Cx(t) = A₀ cos(2πt/(2t₁))Correct The point is at its maximum displacement A₀ at t = 0 and at its minimum −A₀ at t₁, halfway through a cycle, so one full cycle takes T = 2t₁ and f = 1/(2t₁). Substituting into x(t) = A₀ cos(2πft) gives x(t) = A₀ cos(2πt/(2t₁)), which can also be written A₀ cos(πt/t₁).
Dx(t) = A₀ cos(2π(2t₁)t) A student who uses the period, 2t₁, in place of the frequency picks this. The frequency is the reciprocal of the period, f = 1/(2t₁); with 2t₁ in its place, the argument of the cosine is not even dimensionless.
Working From the graph: x = A₀ at t = 0 and x = −A₀ at t = t₁ (half a cycle), so T = 2t₁ and f = 1/T = 1/(2t₁). x(t) = A₀ cos(2πft) = A₀ cos(2πt/(2t₁)) = A₀ cos(πt/t₁).
A wave generator of fixed frequency makes waves on a stretched string, and the wavelength of the waves is 2.0 m. The tension in the string is then increased to four times its original value, with the frequency of the generator unchanged. What is the new wavelength?
Answer and reasoning
A4.0 mCorrect The wave speed on a string is vstring = √(FT/(m/ℓ)), so four times the tension gives √4 = 2 times the speed. The frequency is set by the generator and does not change, so λ = v/f doubles: 2 × 2.0 m = 4.0 m.
B8.0 m A student who takes the wave speed as proportional to the tension picks this, making the speed, and so the wavelength, four times as great. The speed depends on the square root of the tension, so it only doubles.
C2.0 m A student who thinks the generator alone sets the wavelength picks this. The generator sets the frequency; the tension changes the wave speed, and λ = v/f changes with it.
D1.0 m A student who thinks the wavelength is inversely proportional to the wave speed (λ = f/v) picks this: the speed doubles, so the wavelength is halved. In fact λ = v/f, so at a fixed frequency a faster wave has a longer wavelength.
Working vstring = √(FT/(m/ℓ)): FT → 4FT gives v → √4 v = 2v. f unchanged, so λ = v/f → 2λ = 2 × 2.0 m = 4.0 m.
A student measures the wavelength λ of waves on a stretched string for several frequencies f of the wave generator, keeping the tension constant. The graph shows λ plotted against 1/f. What is the speed of the waves on the string?
Answer and reasoning
A2.4 m/s A student who reads the answer off the vertical axis picks this, taking the height of the last data point, 2.4 m, as the speed. That value is a wavelength; the speed is the slope, 2.4 m ÷ 0.20 s.
B0.24 m/s A student who uses the area under the line, (1/2)(0.20 s)(2.4 m) = 0.24 m·s, picks this. The area has units m·s, not m/s; the speed is the slope of the line.
C0.083 m/s A student who divides the horizontal change by the vertical change, 0.20 s ÷ 2.4 m, picks this. That ratio has units s/m; the slope is rise over run, 2.4 m ÷ 0.20 s = 12 m/s.
D12 m/sCorrect From λ = v/f = v × (1/f), a graph of λ against 1/f is a straight line through the origin whose slope is the wave speed. Slope = Δλ/Δ(1/f) = (2.4 m − 0)/(0.20 s − 0) = 12 m/s.
Working λ = v(1/f), so the slope of λ against 1/f is v. Using the origin and the point (0.20 s, 2.4 m): v = (2.4 m)/(0.20 s) = 12 m/s. (Every data point gives the same ratio, e.g. 0.6 m/0.05 s = 12 m/s.)
A student makes waves on a long rope by moving one end up and down. Which of the following changes would produce waves with a longer wavelength?
Answer and reasoning
AMoving the end up and down fewer times in a secondCorrect The wave speed is set by the rope. With λ = v/f, a lower frequency at the same speed gives a longer wavelength: each cycle of the source lasts longer, so the wave travels farther during one cycle.
BMoving the end through a larger distance each cycle A student who thinks a larger-amplitude wave travels faster, and so has a longer wavelength, picks this. The amplitude changes neither the wave speed nor the frequency, so the wavelength is unchanged.
CMoving the end up and down more times each second A student who thinks wavelength increases with frequency picks this. At the same speed, λ = v/f: more cycles each second give a shorter wavelength.
DUsing a heavier rope that is kept at the same tension A student who thinks waves travel faster in a heavier medium picks this. At the same tension, waves travel more slowly on a heavier rope, vstring = √(FT/(m/ℓ)), so the wavelength becomes shorter.
A sinusoidal wave of frequency f travels along a string. At one instant, the displacement of the string from equilibrium varies with position x along the string as A cos(Kx), where A and K are positive constants. Which expression gives the speed of the wave?
Answer and reasoning
Aπf/K A student who takes the wavelength as the distance from a crest to the adjacent trough picks this: the crest is at x = 0 and the next trough at Kx = π, so this student uses λ = π/K and gets v = πf/K. A wavelength is the distance from one crest to the next, Kx = 2π, so λ = 2π/K.
B2πf/KCorrect Matching A cos(Kx) to A cos(2πx/λ) gives 2π/λ = K, so λ = 2π/K. From λ = v/f, v = fλ = 2πf/K.
CfK/(2π) A student who uses λ = f/v, so v = f/λ, picks this: f ÷ (2π/K) = fK/(2π). The relationship is λ = v/f, so v = fλ.
Df/K A student who takes the constant multiplying x as 1/λ, so λ = 1/K, picks this: v = f/K. K equals 2π/λ, so λ = 2π/K and v = 2πf/K.
Working Match A cos(Kx) to A cos(2πx/λ): K = 2π/λ ⇒ λ = 2π/K. λ = v/f ⇒ v = fλ = 2πf/K. Errors: crest-to-trough as λ → λ = π/K, v = πf/K; v = f/λ → fK/(2π); λ = 1/K → f/K.
Compiled from the AP Physics 2 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
