6 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 6
A loudspeaker plays a steady note to a listener 5 m away in a room with still air. Which of the following travels from the loudspeaker to the listener?
Answer and reasoning
AEnergy, passed along by air molecules that vibrate about fixed positionsCorrect A sound wave transfers energy from the loudspeaker to the listener without transferring matter. Each air molecule vibrates back and forth about its own position and passes the disturbance on to its neighbors; the molecules near the loudspeaker never reach the listener.
BAir molecules, pushed across the room from the loudspeaker to the ear A student who thinks a wave carries the material of the medium with it picks this. The loudspeaker cone pushes the nearby air back and forth, but those molecules only vibrate about their positions. The air in a room with still air does not flow to the listener; only the disturbance and its energy travel.
CSound particles, which the loudspeaker gives off and the air carries A student who pictures sound as a substance of its own picks this. There are no sound particles: sound is the disturbance of the air molecules themselves, a pattern of pressure changes passed from molecule to molecule.
DAn electromagnetic signal, which needs no air to carry it across A student who groups sound with light and radio waves picks this. Sound is a mechanical wave: it is carried by the air (or another medium) and would not reach the listener through a vacuum.
Light from the Sun reaches Earth, but the loud sounds produced by the churning gas at the Sun's surface do not. Most of the space between the Sun and Earth is a near-perfect vacuum. Which reasoning accounts for this observation?
Answer and reasoning
ASound spreads out and fades over the distance, but light does not. A student who thinks sound could cross empty space but would arrive too weak picks this. Sound cannot travel through a vacuum at all, however loud it is at the source. (Light also spreads out as it travels; it reaches Earth because it needs no medium.)
BLight is carried across space by its thin gas, but sound is not. A student who thinks every wave needs a medium picks this. Light is an electromagnetic wave and needs no medium; it would cross a perfect vacuum. The reason sound does not arrive is that sound, unlike light, needs a medium.
CSound travels only through air, and space contains no air. A student who links sound with air in particular picks this. Sound travels through any medium, solid, liquid or gas, not only air. It fails to cross the space between the Sun and Earth because there is almost no medium of any kind there.
DSound needs a medium to travel through, but light does not.Correct Sound is a mechanical wave: it is a disturbance of the particles of a medium, so it cannot cross a near-vacuum. Light is an electromagnetic wave, which does not require a medium, so it crosses the space between the Sun and Earth.
A distant star flares, emitting radio waves, visible light and X-rays at the same instant. All three travel to Earth through the vacuum of space. How do their arrival times at Earth compare?
Answer and reasoning
AThe X-rays arrive first, as higher-energy radiation travels faster through space. A student who thinks more energetic radiation moves faster picks this. X-rays differ from radio waves and visible light in wavelength and frequency, not in their speed in a vacuum, which is c for all three.
BThe radio waves arrive first, as waves with longer wavelengths travel faster. A student who thinks a longer wave is a faster wave picks this. Radio waves have the longest wavelengths of the three but also the lowest frequencies; in a vacuum all three travel at c.
CAll three arrive together, as all electromagnetic waves travel at c in a vacuum.Correct Radio waves, visible light and X-rays are all electromagnetic waves. In a vacuum every electromagnetic wave travels at the same speed, c = 3.00 × 10⁸ m/s, whatever its wavelength or frequency, so radiation emitted together over the same distance arrives together.
DThe visible light arrives first, as c is the speed of visible light itself. A student who thinks c belongs to visible light alone picks this. c is the speed in a vacuum of every electromagnetic wave, including radio waves and X-rays, which are the same kind of wave as light with different wavelengths.
Which of the following is an example of a transverse wave or wave pulse?
Answer and reasoning
AA sound traveling through the air from a singer on a stage to a listener in a hall A student who pictures sound as the wavy line used to graph it picks this. In a sound wave the air molecules vibrate back and forth along the direction the sound travels, so sound is longitudinal.
BA pulse on a slinky made by pushing one end along its length and back again A student who thinks 'transverse' means motion along the direction of travel picks this. Pushing the end along the slinky displaces the coils parallel to the direction the pulse travels: that pulse is longitudinal.
CA marble rolling along a wavy track, rising and falling as it moves along A student who thinks a wave is something that moves along a wavy path picks this. The marble itself is carried along the track; no disturbance is passed through a medium, so this is not a wave at all.
DA stadium wave, in which fans stand up and sit down as it travels around the standsCorrect Each fan moves up and down, while the disturbance travels horizontally around the stadium. The disturbance is perpendicular to the direction of propagation, so the pulse is transverse.
A long slinky is stretched along a table. A student repeatedly pushes one end forward along the slinky's length and pulls it back, sending a wave along the slinky. How does one coil in the middle of the slinky move as the wave passes?
Answer and reasoning
ASide to side across the table, perpendicular to the slinky's length A student who pictures every wave as a sideways (transverse) motion picks this. The student's hand moves along the slinky, so the coils are displaced along the slinky too: the disturbance is parallel to the direction of travel.
BAlong the slinky to the far end, carried there by the passing wave A student who thinks a wave carries the medium with it picks this. The wave transfers energy to the far end, but each coil only moves back and forth about its own position and stays in the middle of the slinky.
CBack and forth along the slinky's length, about a fixed positionCorrect Pushing and pulling the end along the slinky makes a longitudinal wave: each coil is displaced parallel to the direction the wave travels, moving back and forth about its equilibrium position. The wave shows as regions of bunched-up and spread-out coils.
DAlong a wavy path, following the shape of the wave as it passes A student who thinks each part of the medium follows the shape of the wave picks this. The coil moves back and forth along one straight line, the length of the slinky; no part of the slinky follows a wavy path.
The graph shows the air pressure P as a function of position x along a sound wave at one instant. P₀ is the pressure of the undisturbed air, and the vertical axis is broken between 0 and P₀. Which labeled arrow represents the amplitude of the sound wave?
Answer and reasoning
AArrow X A student who measures from a trough to a crest picks this. Arrow X spans the maximum decrease and the maximum increase together, twice the amplitude; the amplitude is measured from P₀.
BArrow Y A student who takes the amplitude as the highest pressure reached picks this. Arrow Y runs from zero pressure to a crest, P₀ plus the amplitude. The undisturbed air already has pressure P₀; the amplitude is only the change from P₀.
CArrow Z A student who measures along the wave from one crest to the next picks this. Arrow Z is horizontal, a distance along x between neighboring crests (compressions), which is the wavelength, not the amplitude.
DArrow WCorrect The amplitude of a longitudinal pressure wave is the maximum increase (or decrease) in pressure from the equilibrium pressure P₀. Arrow W runs from the P₀ line up to a crest, the maximum increase above P₀.
Working Amplitude = maximum departure from equilibrium pressure P₀ = arrow W (P₀ to crest). X = crest-to-trough (2 × amplitude); Y = from P = 0 to crest (P₀ + amplitude); Z = crest to crest along x (wavelength).
In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
14.1.A.1 Wave Fix
Wave
A disturbance that travels from one location to another, transferring energy between the two locations without transferring the matter of the medium between them. Each part of the medium moves about its own equilibrium position and returns to it.
Wave pulse
A single disturbance that travels through a medium, transferring energy without transferring matter between two locations; for example, the bump that travels along a rope when one end is moved up and back down once.
Periodic wave
A continuous disturbance in which the same motion repeats at regular intervals of time at each location and at regular intervals of distance along the medium, so that it has a well-defined wavelength and frequency.
Wavelength and frequency
Wavelength is the distance between successive corresponding points of a periodic wave, such as two neighboring crests (SI unit: m). Frequency is the number of repetitions of the wave per unit time at a location (SI unit: Hz = 1/s). A single pulse does not repeat, so it has neither.
Students often think A wave carries the material of the medium along with it: the rope, air or water that is disturbed travels from the source to the far end. In fact No. A wave transfers energy between two locations without transferring the matter between them. Each part of the medium is disturbed and moves about its own equilibrium position; it is the disturbance, not the material, that travels.
Students often think Sound is a thing in its own right (sound particles or a sound substance) that is given off by the source and carried through the air to the ear. In fact No. Sound is a disturbance of the medium: air molecules vibrate about their positions and pass the disturbance on to their neighbors. No sound particles or sound substance travel from the source to the ear.
14.1.A.2 Medium Fix
Medium
The material (solid, liquid or gas, such as a rope, a spring, water or air) through which a mechanical wave travels. The medium is disturbed as the wave passes, but it is not carried along with the wave.
Mechanical wave
A wave that is a disturbance of a material medium, such as a wave on a string or a sound wave. It requires a medium and cannot travel through a vacuum.
Electromagnetic wave
A wave, such as light, radio waves or X-rays, that does not require a medium, so it can travel through a vacuum as well as through some materials.
Students often think Sound can travel through a vacuum; across a large empty space it only becomes fainter, or takes longer, than it would in air. In fact No. Sound is a mechanical wave and needs a medium. In a vacuum there are no particles to be disturbed, so sound cannot travel at all, however loud it is at the source.
Students often think Light, like sound, needs a medium to travel through; in space it is carried by the thin gas there. In fact No. Light is an electromagnetic wave and does not need a medium. It travels through the near-vacuum of space from the Sun and stars.
14.1.A.3 Wave speed, v Fix
Wave speed, v
The speed at which a disturbance travels through a medium. It is set by the type of wave and the properties of the medium (for a string, its tension and mass per length; for sound, the material and its temperature), not by the amplitude of the disturbance. SI unit: m/s.
Speed of light, c
The speed of every electromagnetic wave in a vacuum, whatever its wavelength or frequency: c = 3.00 × 10⁸ m/s, a universal physical constant.
Tension, FT
The magnitude of the pulling force that each part of a taut string exerts on the neighboring part. For a string held taut by a hanging object at rest, FT equals the object's weight, Mg. SI unit: newton (N).
Mass per length, m/ℓ
The mass of a string divided by its length (also called linear mass density, μ). It is a property of the string itself, not of the amount of string used. SI unit: kg/m.
Speed of a wave on a string
vstring = √(FT/(m/ℓ)). The speed increases with the tension and decreases with the mass per length: quadrupling the tension doubles the speed, and quadrupling the mass per length halves it.
Speed of sound and temperature
In a given medium, sound travels faster when the medium is at a higher temperature. In air near room temperature the speed is about 343 m/s at 20°C, and it is greater on a hot day than on a cold one.
Students often think The thunder is produced some time after the flash, so the delay between seeing the flash and hearing the thunder is a delay at the source. In fact No. The flash and the thunder are produced together: the lightning heats the air along its path suddenly, and the air expands violently, producing the sound. The thunder is heard later only because sound travels much more slowly through air than light does.
Students often think A wave or pulse of larger amplitude, such as a bigger flick of a rope or a louder sound, travels faster than a smaller one. In fact No. The speed of a wave depends on the type of wave and the properties of the medium, not on the amplitude. A large pulse and a small pulse on the same rope, or a loud and a quiet sound in the same air, travel at the same speed.
14.1.A.4 Transverse wave Fix
Transverse wave
A wave in which each part of the medium is displaced perpendicular to the direction in which the wave travels, as in a wave on a taut rope.
Students often think In a transverse wave the medium moves back and forth along the direction in which the wave travels. In fact No. In a transverse wave each part of the medium moves perpendicular to the direction the wave travels. Motion parallel to the direction of travel is the defining feature of a longitudinal wave.
Students often think Each part of the medium follows the wavy shape of the wave, traveling along the curve over each crest and down into each trough. In fact No. Each part of the medium only moves back and forth about its own equilibrium position (up and down for a wave on a rope). The curved shape is a picture of many different parts of the medium at one instant, not the path of any one part.
14.1.A.5 Longitudinal wave Fix
Longitudinal wave
A wave in which each part of the medium is displaced parallel to the direction in which the wave travels, back and forth about its equilibrium position, as in a wave made by pushing and pulling the end of a slinky along its length.
Sound wave
A mechanical, longitudinal wave: the molecules of the medium vibrate back and forth along the direction in which the sound travels, producing regions of higher and lower pressure that travel through the medium.
Compression
A region of a longitudinal (sound) wave where the particles of the medium are closer together than at equilibrium, so the pressure is higher than the equilibrium pressure.
Rarefaction
A region of a longitudinal (sound) wave where the particles of the medium are farther apart than at equilibrium, so the pressure is lower than the equilibrium pressure. The region still contains the medium.
Students often think In a longitudinal wave, like any wave, the medium moves up and down, perpendicular to the direction of travel. In fact No. In a longitudinal wave each part of the medium moves back and forth parallel to the direction the wave travels, producing compressions and rarefactions.
Students often think Sound is a transverse wave, with air molecules vibrating up and down across the direction of travel, as the wavy line used to draw it suggests. In fact No. Sound is a longitudinal wave: air molecules vibrate along the direction the sound travels. The wavy line on a graph of a sound wave plots pressure (or displacement) against time or position; it is not a picture of up-and-down motion.
14.1.A.6 Amplitude Fix
Amplitude
The maximum displacement of the medium from its equilibrium position as a wave passes, measured from equilibrium to a crest (or to a trough), not from crest to trough. For a displacement, SI unit: m.
Pressure amplitude
For a longitudinal pressure (sound) wave, the maximum increase or decrease in pressure from the equilibrium pressure P₀, not the maximum pressure itself. SI unit: pascal (Pa).
Loudness
How loud a sound is perceived to be. The loudness of a sound increases with its amplitude; it is a different property from pitch, which is related to frequency.
Energy carried by a wave
The energy that a wave transfers through a medium. For waves of the same kind in the same medium, a wave or pulse of larger amplitude carries more energy.
Students often think The amplitude of a wave is the distance from the top of a crest to the bottom of a trough, the full range of the motion. In fact No. The amplitude is the maximum displacement from the equilibrium position: from equilibrium to a crest, or from equilibrium to a trough. The crest-to-trough distance is twice the amplitude.
Students often think The amplitude at a point, and so the loudness there, is the displacement or pressure change of the medium at that point at the instant shown. In fact No. The amplitude is the maximum displacement (or maximum pressure change), the same for every point of a wave that is not dying away. At any one instant most points have a displacement smaller than the amplitude, and some have zero displacement.
17 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 17
A student moves one end of a long, taut horizontal rope up and then back down once, sending a single pulse along the rope. A small piece of tape is stuck to the middle of the rope. Which statement describes the motion of the tape as the whole pulse passes it?
Answer and reasoning
AIt moves along the rope with the pulse, riding on its peak. A student who thinks the pulse carries the rope along with it picks this. The peak of the pulse is made of different pieces of rope at different instants; each piece only rises and falls as the peak passes, so the tape is left behind at its starting position.
BIt moves along the rope toward the far end, then back again. A student who thinks the rope moves along the direction of travel picks this. On a taut rope the pulse is transverse: the tape is displaced up and down, perpendicular to the rope, not back and forth along it.
CIt moves up and then back down, ending where it started.Correct A pulse transfers energy along the rope without transferring the rope itself. As the pulse reaches the tape, that piece of rope is pulled up by its neighbor, and as the pulse moves on it returns to its equilibrium position. When the pulse has passed, the tape is where it started.
DIt travels over the pulse's curved shape, up one side and down. A student who reads the shape of the pulse as the path of the rope picks this. The curve is the shape of many pieces of rope at one instant. The tape moves only up and then down at its own position along the rope; it does not travel along the curve.
A student moves the far end of a long rope, and a disturbance travels along it. The graph shows the vertical displacement y of one point on the rope as a function of time t. Which claim is supported by the graph?
Answer and reasoning
AThe rope has crests 0.25 m apart along its length, as the curve shows. A student who reads the graph as a picture of the rope picks this. The horizontal axis is time, t (s), so the graph follows one point through time: 0.25 is a time interval in seconds, not a distance between crests in meters. The graph cannot show where the crests are along the rope.
BThe same motion repeats every 0.25 s, so a periodic wave is passing the point.Correct The displacement goes through the same cycle, up to 2.0 cm, down to −2.0 cm and back, every 0.25 s, four times in 1.00 s. A disturbance that repeats the same motion at regular intervals is a periodic wave, with a well-defined frequency, rather than a single pulse.
CThe amplitude of the point's motion is 4.0 cm, from its lowest to highest position. A student who measures the amplitude from the lowest position to the highest picks this. The point does move from −2.0 cm to +2.0 cm, a range of 4.0 cm, but the amplitude is the maximum displacement from equilibrium, 2.0 cm.
DThe point travels along the rope on a wavy path, rising and falling as it goes. A student who reads the wavy curve as the path of the point picks this. The graph plots the point's vertical displacement against time; the point moves only up and down about its equilibrium position and does not travel along the rope.
Working Read the graph: y goes 0 → +2.0 cm → 0 → −2.0 cm → 0 in each 0.25 s, four identical cycles from t = 0 to t = 1.00 s. Identical cycles at equal time intervals → periodic wave. The amplitude is 2.0 cm (m27 error: taking the 4.0 cm range as the amplitude); the horizontal axis is time, not position.
During a storm, a student sees a lightning flash and hears the thunder from the same strike several seconds later. Which explanation of the delay is correct?
Answer and reasoning
AThe thunder is produced a few seconds after the flash, once the air has heated up. A student who takes the order in which the events are perceived as the order in which they happen picks this. The lightning heats the air suddenly and the thunder is produced at once; the delay is travel time, and it is longer for a more distant strike.
BLight and sound are different types of wave that travel at different speeds in air.Correct The flash and the thunder are produced together. The speed of a wave depends on the type of wave and the medium: in air, light travels at almost 3.00 × 10⁸ m/s and sound at about 343 m/s, so over a few kilometers the sound arrives several seconds after the light.
CThe sound slows down as it gets fainter, while the light keeps a steady speed. A student who links the speed of a wave to its amplitude picks this. The sound does fade as it spreads, but its speed is set by the air, not by its loudness; it travels at the same speed all the way.
DLight has a much higher frequency, and higher-frequency waves travel faster. A student who thinks a wave's frequency sets its speed picks this. Light does have a far higher frequency, but that is not why it is faster: sounds of high and low frequency travel through air at the same speed. Light is faster because it is a different type of wave.
A radio pulse is sent from Earth toward the Moon. It reflects from the Moon's surface and is detected back on Earth 2.56 s after it was sent. The speed of light in a vacuum is c = 3.00 × 10⁸ m/s, and the speed of sound in air is 343 m/s. What is the distance from Earth to the Moon?
Answer and reasoning
A3.84 × 10⁸ mCorrect The radio pulse is an electromagnetic wave, so it travels at c. In 2.56 s it goes to the Moon and back, a distance 2d: 2d = ct, so d = ct/2 = (3.00 × 10⁸ m/s)(2.56 s)/2 = 3.84 × 10⁸ m.
B7.68 × 10⁸ m A student who uses the whole 2.56 s as the time to reach the Moon picks this: ct = 7.68 × 10⁸ m. That is the distance there and back; the distance to the Moon is half of it.
C1.17 × 10⁸ m A student who divides the speed by the time picks this: (3.00 × 10⁸ m/s)/(2.56 s) = 1.17 × 10⁸, which has units of m/s², not m. Distance is speed multiplied by time: d = c(2.56 s)/2 = 3.84 × 10⁸ m.
D4.39 × 10² m A student who treats the radio pulse as a sound wave picks this, using 343 m/s: (343 m/s)(2.56 s)/2 = 439 m. Radio waves are electromagnetic waves, which travel at c, and sound could not cross the vacuum between Earth and the Moon at all.
Working Round trip: 2d = ct. d = (3.00 × 10⁸ m/s)(2.56 s)/2 = 3.84 × 10⁸ m. Errors: no halving → 7.68 × 10⁸ m; c/t → 1.17 × 10⁸; speed of sound (343 m/s)(2.56 s)/2 = 439 m.
A pulse travels at 5.00 m/s along a taut rope. The rope is replaced by a second rope that has the same total mass but twice the length, and the tension is doubled. What is the speed of a pulse on the second rope?
Answer and reasoning
A20.0 m/s A student who drops the square root, taking v proportional to FT/(m/ℓ), picks this: 4 × 5.00 m/s = 20.0 m/s. The ratio FT/(m/ℓ) is multiplied by 4, but the speed goes as its square root, so it only doubles.
B2.50 m/s A student who puts the mass per length on top, v = √((m/ℓ)/FT), picks this: √((1/2)/2) = 1/2, giving 2.50 m/s. A larger tension and a smaller mass per length both make pulses faster, not slower.
C7.07 m/s A student who uses the total mass instead of the mass per length picks this: the total mass is unchanged, so only the doubled tension counts, √2 × 5.00 m/s = 7.07 m/s. The speed depends on the mass per length, which is halved.
D10.0 m/sCorrect The second rope has the same mass spread over twice the length, so its mass per length is half that of the first. v = √(FT/(m/ℓ)): doubling FT and halving m/ℓ multiplies FT/(m/ℓ) by 4, so the speed is multiplied by √4 = 2: v = 2(5.00 m/s) = 10.0 m/s.
Working m/ℓ: same m, 2ℓ → (m/ℓ)/2. FT → 2FT. v' = √(2FT/((m/ℓ)/2)) = √4 · v = 2(5.00 m/s) = 10.0 m/s. Errors: no square root → 4v = 20.0 m/s; inverted ratio → v/2 = 2.50 m/s; total mass → √2 v = 7.07 m/s.
A string of length 1.50 m has a mass of 3.0 g. One end is tied to a wall, and the string runs horizontally over a pulley to a block of mass 2.0 kg that hangs at rest. Use g = 9.8 m/s². What is the speed of a pulse on the horizontal part of the string?
Answer and reasoning
A3.2 × 10¹ m/s A student who uses the block's mass, 2.0, in place of the tension picks this: √(2.0/0.0020) ≈ 32 m/s. The tension is a force, the block's weight Mg = 19.6 N.
B9.9 × 10¹ m/sCorrect The tension equals the block's weight: FT = Mg = (2.0 kg)(9.8 m/s²) = 19.6 N. The mass per length is m/ℓ = 0.0030 kg/1.50 m = 0.0020 kg/m. v = √(FT/(m/ℓ)) = √(19.6 N/0.0020 kg/m) = √9800 m²/s² ≈ 99 m/s.
C8.1 × 10¹ m/s A student who divides the tension by the string's whole mass instead of its mass per length picks this: √(19.6 N/0.0030 kg) ≈ 81, whose units are not even m/s. The equation needs m/ℓ = 0.0020 kg/m.
D9.8 × 10³ m/s A student who forgets the square root picks this: FT/(m/ℓ) = 19.6/0.0020 = 9800. That is v², in m²/s²; its square root, about 99 m/s, is the speed.
Working FT = Mg = (2.0 kg)(9.8 m/s²) = 19.6 N. m/ℓ = 3.0 × 10⁻³ kg / 1.50 m = 2.0 × 10⁻³ kg/m. v = √(19.6/2.0 × 10⁻³) = √9800 = 99 m/s = 9.9 × 10¹ m/s. Errors: M for FT → √(2.0/0.0020) = 32 m/s; total mass → √(19.6/0.0030) = 81; no square root → 9800.
Two strings, 1 and 2, are under equal tension. At the same instant, a pulse is started at the left end (x = 0) of each string. The diagram shows the two strings at a later instant. How does the mass per length μ₁ of string 1 compare with the mass per length μ₂ of string 2?
Answer and reasoning
Aμ₁ = 4μ₂, since speed is proportional to 1/√μCorrect In the same time the pulse on string 2 has traveled 2.0 m and the pulse on string 1 only 1.0 m, so v₂ = 2v₁. With equal tensions, v = √(FT/μ) gives μ = FT/v², so μ₁/μ₂ = (v₂/v₁)² = 4.
Bμ₁ = 2μ₂, since speed is proportional to 1/μ A student who drops the square root, taking v ∝ 1/μ, picks this. Because v = √(FT/μ), the mass per length goes as 1/v²: halving the speed means four times the mass per length.
Cμ₂ = 4μ₁, since speed is proportional to √μ A student who thinks a heavier string carries pulses faster picks this. The mass per length is in the denominator of v = √(FT/μ): the slower string, string 1, has the larger mass per length.
Dμ₁ = μ₂, since the taller pulse travels faster A student who links the speed of a pulse to its amplitude picks this. The pulse on string 2 is taller, but amplitude does not affect the speed; with equal tensions, the only reason for the different speeds is a different mass per length.
Working Same time interval: Δx₁ = 1.0 m, Δx₂ = 2.0 m → v₂/v₁ = 2. v = √(FT/μ), equal FT → μ ∝ 1/v² → μ₁/μ₂ = (v₂/v₁)² = 4. Errors: v ∝ 1/μ → μ₁ = 2μ₂; v ∝ √μ → μ₂ = 4μ₁; amplitude explains speed → μ₁ = μ₂.
A student times a sound pulse from a starting pistol as it travels 100 m across a field, once on a winter morning when the air is at −10°C and once on a summer afternoon when the air is at 30°C. How do the two travel times compare?
Answer and reasoning
AThe time is shorter in the morning, when the air is denser. A student who thinks sound is faster wherever the particles are closer together picks this. The cold morning air is denser, but for air the temperature decides: sound travels more slowly in colder air.
BThe times are equal, as sound travels at 343 m/s in air. A student who treats 343 m/s as a constant picks this. That is the speed of sound in air at about 20°C; the speed increases with temperature, so it is lower at −10°C than at 30°C.
CThe time is shorter in the afternoon, when the air is warmer.Correct In a given medium, the speed of sound increases with the temperature of the medium. The air is warmer in the afternoon, so the sound travels faster and covers the 100 m in less time.
DThe time is shorter on whichever occasion the pistol is louder. A student who links the speed of sound to its loudness picks this. A louder sound has a larger amplitude, but amplitude does not change the speed; the temperature of the air does.
Which of the following correctly describes a sound wave traveling through air?
Answer and reasoning
AAir molecules vibrate up and down, perpendicular to the direction of travel. A student who takes the wavy line of a sound graph as a picture of the motion picks this. The graph plots pressure against time or position; the molecules themselves move along the direction of travel, so sound is longitudinal, not transverse.
BElectromagnetic waves from the vibrating source travel through the air to the ear. A student who groups sound with light and radio picks this. Sound is a mechanical wave, a disturbance of the air itself; unlike an electromagnetic wave, it cannot travel through a vacuum.
CSound particles given off by the source travel through the air to the listener. A student who pictures sound as a substance of its own picks this. Nothing called a sound particle is emitted: sound is the vibration of the air molecules, passed from each molecule to its neighbors.
DAir molecules vibrate back and forth along the direction the sound travels.Correct Sound is modeled as a mechanical longitudinal wave. The air molecules are the medium; each one vibrates back and forth parallel to the direction the sound travels, producing the compressions and rarefactions that make up the wave.
The diagram shows thin layers of air in front of a loudspeaker at one instant while it plays a steady note. Each line marks the position of one layer of air. Which statement about the air in region X is correct?
Answer and reasoning
AIts pressure is greater than the pressure of the undisturbed air.Correct In region X the layers of air are closer together than elsewhere, so X is a compression: a region where the air is squeezed and its pressure is above the equilibrium pressure. (Region Y, where the layers are spread out, is a rarefaction, at lower pressure.)
BIts molecules came from the loudspeaker and will travel on to a listener. A student who thinks the wave carries the air with it picks this. The compression moves away from the loudspeaker, but it is made of different molecules at different instants; each layer only vibrates back and forth about its own position.
CThe sound there is louder than the sound in region Y at this instant. A student who takes the pressure at one instant as the loudness picks this. X is at higher pressure than Y only at this instant; a moment later X will be a rarefaction. The amplitude, and so the loudness, of the wave is the same at X and Y.
DIts layers of air vibrate up and down, perpendicular to the direction of travel. A student who pictures every wave as an up-and-down motion picks this. Sound is longitudinal: the layers move back and forth along the direction of travel, which is why they bunch up into compressions and spread out into rarefactions.
The graph shows the shape of a rope at one instant as a wave travels along it. P is a point on the rope. What is the amplitude of the wave?
Answer and reasoning
A4.0 cm A student who measures from a trough to a crest picks this: 2.0 cm − (−2.0 cm) = 4.0 cm. That is twice the amplitude; the amplitude is measured from equilibrium to a crest.
B2.0 cmCorrect The amplitude is the maximum displacement from the equilibrium position, y = 0. The crests reach y = 2.0 cm and the troughs y = −2.0 cm, so the amplitude is 2.0 cm.
C1.0 cm A student who reads the displacement of point P at this instant picks this. P happens to be at y = 1.0 cm at the instant shown; the amplitude is the maximum displacement, which every point on the rope reaches at some instant.
D8.0 cm A student who measures along the rope from one crest to the next picks this: the crests are at x = 2 cm and x = 10 cm, 8.0 cm apart. That horizontal distance is the wavelength; the amplitude is measured vertically from equilibrium.
Working Equilibrium: y = 0. Crest: y = +2.0 cm → A = 2.0 cm. Errors: crest to trough 4.0 cm; displacement of P at this instant 1.0 cm; crest-to-crest spacing (x = 2 → 10 cm) 8.0 cm.
Each graph shows the air pressure P at a microphone as a function of time t for one of two sounds, X and Y. The two graphs are drawn to the same scale. P₀ is the pressure of the undisturbed air. Which statement about the two sounds is correct?
Answer and reasoning
AY is louder, since its pressure changes more rapidly. A student who links loudness to frequency picks this. Y's pressure does repeat more often, so Y has the higher frequency and the higher pitch, but pitch and loudness are different properties; loudness goes with amplitude, which is smaller for Y.
BX travels through the air faster, as its amplitude is larger. A student who thinks a sound of larger amplitude travels faster picks this. X does have the larger amplitude, but both sounds travel through the same air at the same speed; amplitude affects loudness, not speed.
CX is louder, since its pressure departs further from P₀.Correct The loudness of a sound increases with its amplitude, the maximum change in pressure from P₀. Sound X's pressure rises and falls about two gridlines from P₀, while Y's changes by less than one, so X has the larger amplitude and is louder.
DEach is louder at the instants when its pressure is higher. A student who takes the pressure at one instant as the loudness picks this. The loudness of a steady sound depends on its amplitude, the maximum pressure change over a cycle, which is constant for each sound and larger for X.
A student sends two pulses, one after the other, along the same taut rope. For the first pulse she moves her hand up 10 cm and back down; for the second she moves it up 20 cm and back down in the same time. Which statement comparing the two pulses is correct?
Answer and reasoning
AThe second pulse travels along the rope at a higher speed. A student who links the speed of a pulse to its size picks this. Both pulses are on the same rope under the same tension, so they travel at the same speed; the larger amplitude changes the energy carried, not the speed.
BThe second pulse transfers more energy to the far end of the rope.Correct The energy carried by a wave or pulse increases with its amplitude. The second pulse has the larger amplitude (the hand moved farther in the same time and did more work on the rope), so it carries more energy along the rope.
CThe second pulse carries more rope material to the far end. A student who thinks a pulse carries the rope along with it picks this. Neither pulse transfers rope material: each piece of rope rises and falls and returns to its position. What the larger pulse carries is more energy.
DBoth pulses transfer equal energy, as each is one up-and-down movement. A student who thinks a wave's energy is set only by how often the source moves picks this. The two movements take the same time, but the second has twice the amplitude, and the energy carried increases with amplitude.
A string of total length ℓ and mass m passes over two pulleys, and a block of mass M hangs at rest from each end of the string, where m is much less than M. The section of string between the pulleys is horizontal and has length D. A pulse is started at one end of the horizontal section. Which expression gives the time the pulse takes to travel to the other end of that section?
Answer and reasoning
AD√(m/(2Mgℓ)) A student who adds the pulls of the two blocks, taking FT = 2Mg, picks this. Each block hangs at rest under its weight Mg and the string's upward pull, so FT = Mg; the two ends are pulled by one tension, not two to be added.
BD√(m/(Mg)) A student who uses the string's whole mass m in place of its mass per length picks this: v = √(Mg/m). The speed depends on m/ℓ; Mg/m has units of m/s², so its square root is not a speed.
CD√(m/(Mgℓ))Correct Each block is in equilibrium, so the string pulls up on it with FT = Mg; that is the tension throughout the string. The mass per length is m/ℓ, so v = √(Mg/(m/ℓ)) = √(Mgℓ/m), and the time to cross length D is t = D/v = D√(m/(Mgℓ)).
DDm/(Mgℓ) A student who drops the square root picks this: v = Mg/(m/ℓ) = Mgℓ/m, so t = Dm/(Mgℓ). The speed is the square root of FT/(m/ℓ); without it the units are m²/s², not m/s.
Working Each block is at rest: FT − Mg = 0 ⇒ FT = Mg (not 2Mg). Mass per length = m/ℓ. v = √(FT/(m/ℓ)) = √(Mgℓ/m). t = D/v = D√(m/(Mgℓ)). Errors: FT = 2Mg → D√(m/(2Mgℓ)); whole mass m for m/ℓ → v = √(Mg/m), t = D√(m/(Mg)); square root dropped → v = Mgℓ/m, t = Dm/(Mgℓ).
A uniform rope of mass m and length L hangs vertically from a ceiling, and a block of mass M hangs from its lower end. The rope's mass is not negligible, and g is the acceleration due to gravity. A small transverse pulse is started near the lower end and travels up the rope. Which expression gives the speed of the pulse just before it reaches the ceiling?
Answer and reasoning
A√(MgL/m) A student who takes the tension to be Mg all along the rope, as for an ideal rope, picks this. Here the rope's own weight also hangs from the top point, so the tension there is (M + m)g and v = √((M + m)gL/m).
B√((M+m)g/m) A student who uses the rope's whole mass m in place of its mass per length picks this. The speed depends on the mass per length, m/L, so v = √((M + m)g/(m/L)) = √((M + m)gL/m); this option does not even have the units of speed.
C√((M+m)gL/m)Correct Just below the ceiling the rope holds up the block and all of the rope, so FT = (M + m)g. With mass per length m/L, v = √(FT/(m/L)) = √((M + m)gL/m).
D√(ML/m) A student who takes the tension to be the block's mass M, in kilograms, picks this. Tension is a force: at the top it equals the weight of everything hanging below, (M + m)g, so v = √((M + m)gL/m).
Working Just below the ceiling the rope supports the block and the whole rope: equilibrium of everything below that point gives FT = Mg + mg = (M + m)g. Mass per length = m/L. v = √(FT/(m/L)) = √((M + m)gL/m). (Lower down the tension is smaller, so the pulse speeds up as it rises.) Errors: tension Mg all along the rope gives √(MgL/m); whole mass m in place of m/L gives √((M + m)g/m); tension equal to the mass M gives √(ML/m).
A string is stretched horizontally between two fixed supports, so the tension is the same all along it. It consists of two sections of equal length ℓ joined end to end, and section 2 has four times the mass of section 1. A pulse started at the support at the end of section 1 travels along section 1 at speed v. Which expression gives the time the pulse takes to reach the support at the far end of section 2?
Answer and reasoning
A1.50ℓ/v A student who thinks a pulse travels faster on a string with more mass per length picks this: speed proportional to √(m/ℓ) gives 2v on section 2 and ℓ/v + ℓ/(2v) = 1.50ℓ/v. The speed is proportional to 1/√(m/ℓ), so the pulse slows to v/2 on section 2 and the total is 3.00ℓ/v.
B5.00ℓ/v A student who drops the square root, so that speed is inversely proportional to the mass per length, picks this: v/4 on section 2 gives ℓ/v + 4ℓ/v = 5.00ℓ/v. Speed depends on the square root, v = √(FT/(m/ℓ)), so four times the mass per length halves the speed: 3.00ℓ/v.
C2.00ℓ/v A student who thinks the pulse keeps the speed it had on section 1, as a moving object would, picks this: 2ℓ/v = 2.00ℓ/v. The speed is set by the string the pulse is on: on section 2 it is v/2, so the total time is 3.00ℓ/v.
D3.00ℓ/vCorrect The tension is the same in both sections, and section 2 has four times the mass per length, so the pulse's speed there is v/√4 = v/2. The times add: ℓ/v + ℓ/(v/2) = 3.00ℓ/v.
Working Same tension and length; section 2 has four times the mass per length. v = √(FT/(m/ℓ)), so v ∝ 1/√(m/ℓ) at fixed tension: v₂ = v/√4 = v/2. Times: t₁ = ℓ/v; t₂ = ℓ/(v/2) = 2ℓ/v. Total = 3.00ℓ/v. Errors: speed increasing with mass per length (v₂ = √4 v = 2v) gives ℓ/v + ℓ/(2v) = 1.50ℓ/v; square root dropped (v₂ = v/4) gives ℓ/v + 4ℓ/v = 5.00ℓ/v; speed v kept on section 2 gives 2.00ℓ/v.
A small firework explodes at the surface of a calm lake. A swimmer some distance away, with one ear under the water and the other above it, hears the bang through the water and then, 0.200 s later, through the air. The speed of sound is 1.48 × 10³ m/s in the water and 343 m/s in the air. How far is the swimmer from the explosion?
Answer and reasoning
A6.86 × 10¹ m A student who uses the thunder-and-lightning rule, distance = (speed of sound in air) × (delay), picks this: (343 m/s)(0.200 s) = 68.6 m. That rule works for lightning because light's travel time is negligible; here the sound through the water takes about 0.06 s, which is not negligible, so L = 89.3 m.
B5.57 × 10¹ m A student who adds the two travel times to make the delay picks this: L/343 + L/1480 = 0.200 s gives 55.7 m. The delay is the time between the two arrivals, the difference of the travel times, which gives 89.3 m.
C8.93 × 10¹ mCorrect Both sounds travel the same distance L, and the delay is the difference of their travel times: L/(343 m/s) − L/(1.48 × 10³ m/s) = 0.200 s. Solving gives L = 89.3 m = 8.93 × 10¹ m.
D2.27 × 10² m A student who multiplies the difference of the two speeds by the delay picks this: (1480 m/s − 343 m/s)(0.200 s) = 227 m. The distance follows from the difference of the travel times, L/343 − L/1480 = 0.200 s, so L = 89.3 m.
Working Both sounds travel the same distance L, and the delay is the difference of the travel times: L/(343 m/s) − L/(1.48 × 10³ m/s) = 0.200 s. L = (0.200 s)(343 m/s)(1480 m/s)/(1480 m/s − 343 m/s) = 89.3 m = 8.93 × 10¹ m. (Travel times: 0.260 s in air, 0.060 s in water.) Errors: water's travel time neglected, L = (343 m/s)(0.200 s) = 68.6 m; travel times added, L/343 + L/1480 = 0.200 s, L = 55.7 m; difference of speeds × delay, (1480 m/s − 343 m/s)(0.200 s) = 227 m.
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
