9 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 9
Two loudspeakers side by side emit sound continuously, one at 300 Hz and the other at 500 Hz. Which statement about the two sound waves, in the region where both are present, is correct?
Answer and reasoning
AThey do not interfere, because waves of different frequencies pass through without interacting. A student who thinks interference requires equal frequencies picks this. Steady interference patterns do need matched frequencies, but any overlapping waves interact: their displacements add. Beats are interference between waves of different frequencies.
BThey interfere just at the points where a crest of one wave meets a crest of the other. A student who thinks interference happens only at special points picks this. Superposition applies at every point where the waves overlap; crest-on-crest points are just where the result is largest.
CThey interfere: at each point the air's displacement is the sum of both displacements.Correct Interference is the interaction of any waves that overlap; they need not share a frequency. Wherever both waves are present, the displacement of the air is the sum of the displacements each would produce alone. Because the frequencies differ, the waves move in and out of step and the combined disturbance keeps changing, but they still interfere.
DThey do not interfere, because the louder wave takes over wherever the two overlap. A student who thinks the stronger wave overrides the weaker one picks this. Neither wave takes over: both displacements are present and add, so the quieter wave still changes the result.
The diagram shows a long rope at one instant, with pulse P moving right and pulse Q moving left, as the arrows show. The horizontal grid lines are equally spaced. Which statement describes the rope after the pulses have completely passed each other?
Answer and reasoning
AP, still upright, is to the left of Q and moves left; Q, still inverted, moves right. A student who pictures the pulses as objects that collide and bounce back picks this. Pulses are disturbances of the rope, not objects: they overlap and pass through each other, so P keeps moving right and ends up to the right of Q.
BA single upright pulse, half as tall as P, is left on the rope and keeps moving along it. A student who thinks the combined shape formed during the overlap stays on the rope picks this. The sum of two grid spaces up and one grid space down exists only while the pulses overlap; once they separate, each pulse reappears with its original shape.
CP moves right with its shape unchanged, and Q has been absorbed into P. A student who thinks the larger pulse overwhelms the smaller one picks this. Neither pulse absorbs the other: Q emerges unchanged, still inverted and one grid space deep, moving left.
DP, still upright, is to the right of Q and moves right; Q, still inverted, moves left.Correct Pulses pass through each other. While they overlap the displacements add, but afterward each pulse continues in its original direction with its original shape: P, upright and two grid spaces tall, is now on the right moving right, and Q, inverted and one grid space deep, is on the left moving left.
Working Pulses pass through each other unchanged. After passing: P (upright, height 2 grid spaces) is on the right moving right; Q (inverted, depth 1 grid space) is on the left moving left.
The graph shows, for an instant when two pulses on a rope overlap, the displacement y that pulse 1 and pulse 2 would each give the rope on its own. Upward is positive. What is the displacement of the rope at point P at this instant?
Answer and reasoning
A2.0 cm A student who adds the sizes of the displacements and ignores their directions gets 1.5 cm + 0.5 cm and picks this. Pulse 2 displaces the rope downward, so its displacement is negative and reduces the total.
B1.0 cmCorrect Superposition: add the individual displacements with their signs. At P, pulse 1 gives +1.5 cm and pulse 2 gives −0.5 cm, so the rope's displacement is +1.5 cm − 0.5 cm = +1.0 cm.
C1.5 cm A student who takes the displacement of the larger pulse at P picks this. Both pulses act at P; pulse 2's −0.5 cm must be added to pulse 1's +1.5 cm.
D0.0 cm A student who thinks an upward and a downward displacement always cancel completely picks this. They cancel completely only when they are equal in size; here +1.5 cm and −0.5 cm leave +1.0 cm.
Working From the graph at P: pulse 1 rises from 0 to +3.0 cm over two grid columns, and P is one column from its start, so pulse 1 alone gives +1.5 cm; pulse 2 alone gives −0.5 cm. y = +1.5 cm + (−0.5 cm) = +1.0 cm.
Two pulses on a rope overlap at point X. At that instant, pulse 1 on its own would displace the rope at X by 2.0 mm downward, and pulse 2 on its own would displace it by 3.0 mm downward. Taking upward as positive, what is the displacement of the rope at X?
Answer and reasoning
A−5.0 mmCorrect Both displacements are downward, in the same direction, so the interference is constructive: y = (−2.0 mm) + (−3.0 mm) = −5.0 mm, a larger downward displacement than either pulse gives alone.
B+1.0 mm A student who thinks two downward displacements interfere destructively subtracts one from the other, (−2.0 mm) − (−3.0 mm) = +1.0 mm, and picks this. Interference is constructive whenever the displacements are in the same direction, downward as well as upward, so the rope is displaced 5.0 mm downward.
C+5.0 mm A student who applies 'two negatives make a positive' picks this. That rule is for multiplication; adding two negative displacements gives a more negative total, so the rope is displaced farther downward.
D−3.0 mm A student who takes the displacement of the larger pulse alone picks this. Both pulses displace the rope at X, and their displacements add.
Working Upward positive: y₁ = −2.0 mm, y₂ = −3.0 mm. Same direction, so constructive: y = y₁ + y₂ = −5.0 mm. (Treating two downward displacements as destructive and subtracting them, y₁ − y₂ = (−2.0 mm) − (−3.0 mm) = +1.0 mm.)
The graph shows the displacement y of a long rope against position x at time t = 0. Pulse A moves right and pulse B moves left, each at the speed shown. What is the displacement of the rope at x = 5.0 cm at time t = 1.25 s?
Answer and reasoning
A3.0 cm A student who adds the sizes of the two displacements, 2.0 cm and 1.0 cm, picks this. Pulse B displaces the rope downward, so its −1.0 cm reduces the total.
B1.0 cmCorrect Each pulse moves (2.0 cm/s)(1.25 s) = 2.5 cm. Pulse A then covers x = 3.5 cm to 5.5 cm and pulse B covers x = 4.5 cm to 6.5 cm, so both are at x = 5.0 cm: y = +2.0 cm + (−1.0 cm) = +1.0 cm.
C2.0 cm A student who takes the displacement of the larger pulse, A, picks this. Both pulses cover x = 5.0 cm at t = 1.25 s, so both displacements must be added.
D0.0 cm A student who thinks an upward and a downward displacement always cancel completely picks this. The displacements are +2.0 cm and −1.0 cm, which leave +1.0 cm.
Working Distance moved by each pulse: (2.0 cm/s)(1.25 s) = 2.5 cm. A (height +2.0 cm, from x = 1.0 cm to 3.0 cm at t = 0) now spans 3.5 cm to 5.5 cm; B (height −1.0 cm, from 7.0 cm to 9.0 cm) now spans 4.5 cm to 6.5 cm. At x = 5.0 cm both are present: y = +2.0 cm − 1.0 cm = +1.0 cm.
Two tuning forks, of frequencies 440 Hz and 444 Hz, are struck together. A listener hears the loudness rise and fall four times each second. Which statement correctly explains this?
Answer and reasoning
AThe sound given out by each fork on its own grows louder and quieter four times each second. A student who thinks the pulsing is in the sources picks this. Each fork alone gives a steady tone that fades only slowly; the rise and fall appears only when both sound together, because it comes from their interference.
BThe two waves combine into a new sound wave of frequency 4 Hz, which is heard as the pulsing. A student who thinks the beat is a separate low-frequency wave picks this. The combined wave still oscillates at about 442 Hz; what changes four times a second is its amplitude, heard as the loudness rising and falling.
CWaves of different frequencies cannot interfere, so the ear switches back and forth between the two sounds. A student who thinks only waves of equal frequency interfere picks this. Waves of any frequencies superpose where they overlap; beats are the result of that superposition for two slightly different frequencies.
DThe waves drift in and out of step, so their displacements alternately add together and oppose.Correct Because the frequencies differ slightly, the waves are in phase at some instants (displacements in the same direction: constructive interference, loud) and out of phase a little later (opposite displacements: destructive interference, quiet). The cycle repeats |444 Hz − 440 Hz| = 4 times each second.
A string fixed at both ends vibrates in a standing wave. Which statement correctly describes how the standing wave is formed?
Answer and reasoning
AA single wave travels along the string until it fills it, and then the wave stops moving. A student who takes 'standing' to mean that the wave has stopped picks this. The pattern stays in place, but it is made of two waves traveling in opposite directions; one traveling wave cannot produce fixed nodes.
BTwo waves of slightly different frequencies travel in the same direction along the string and overlap. A student who confuses standing waves with beats picks this. Two waves of slightly different frequency give an amplitude that rises and falls in time, not a fixed pattern of nodes and antinodes along the string.
CWaves traveling in opposite directions collide at the nodes and bounce back off each other there. A student who pictures waves as objects that collide picks this. The two waves pass through each other everywhere; nodes are points where their displacements are always equal and opposite, not places where the waves rebound.
DReflections from the fixed ends produce two oppositely moving waves whose displacements add at every point.Correct A standing wave on a string is the superposition of two waves of equal frequency traveling in opposite directions, produced here by reflection at the fixed ends. Their interference gives points that never move (nodes) and points of largest oscillation (antinodes).
The figure shows the displacement envelope of a standing sound wave in a pipe of length L that is closed at one end and open at the other. Which expression gives the wavelength of this standing wave?
Answer and reasoning
Aλ = (2/5)L A student who takes each loop to be a whole wavelength counts two and a half wavelengths in L and picks this. Each node-to-node loop is half a wavelength.
Bλ = (4/5)LCorrect The pattern runs from a node at the closed end to an antinode at the open end. It contains two node-to-node loops (each λ/2) and one node-to-antinode section (λ/4): L = 2(λ/2) + λ/4 = 5λ/4, so λ = 4L/5. This is the fifth harmonic, an odd harmonic, as a node–antinode pipe requires.
Cλ = (5/4)L A student who applies the fraction the wrong way round, writing λ = 5L/4 instead of L = 5λ/4, picks this. More than one wavelength fits in the pipe, so λ must be less than L.
Dλ = (4/3)L A student who takes the harmonic number to be the number of antinodes (three) and uses λ = 4L/3 picks this. For a pipe closed at one end the harmonic number is the number of quarter-wavelengths in L, which here is five.
Working Closed end: node; open end: antinode. Pattern: node, antinode, node, antinode, node, antinode = five quarter-wavelengths. L = 5(λ/4), so λ = 4L/5.
A string fixed at both ends is made to vibrate at a frequency of 150 Hz, forming the standing wave shown in the figure. What is the speed of waves on the string?
Answer and reasoning
A6.0 × 10¹ m/s A student who takes each loop to be a whole wavelength uses λ = 1.2 m/3 = 0.40 m and picks this. Each loop is half a wavelength, so λ = 0.80 m.
B1.2 × 10² m/sCorrect The pattern has three loops between the fixed ends, and each loop is half a wavelength: 1.2 m = 3(λ/2), so λ = 0.80 m. Then v = fλ = (150 Hz)(0.80 m) = 120 m/s.
C3.6 × 10² m/s A student who uses the fundamental wavelength, λ = 2L = 2.4 m, whatever pattern is shown picks this. The string is vibrating in three loops, so λ = 2L/3 = 0.80 m.
D1.9 × 10² m/s A student who writes λ = f/v, so that v = f/λ = (150 Hz)/(0.80 m), picks this. The wave moves one wavelength each period, so v = fλ.
Working Three loops: L = 3(λ/2), λ = 2(1.2 m)/3 = 0.80 m. v = fλ = (150 Hz)(0.80 m) = 120 m/s = 1.2 × 10² m/s.
In preparation: 0 of 9 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
14.6.A.1 Wave interference Fix
Wave interference
The interaction of two or more wave pulses or waves that are in the same region of a medium at the same time. It happens whenever waves overlap, whatever their frequencies or amplitudes, and the result at each point is found by superposition.
Students often think Only waves of the same frequency interfere; waves of different frequencies pass through the same region without interacting. In fact Yes. Interference is the interaction of any waves that overlap: at every point their displacements add, whatever their frequencies. Waves of equal frequency can give a steady pattern; waves of slightly different frequency give beats.
Students often think Waves interfere just at particular points, such as where a crest meets a crest or a crest meets a trough; elsewhere they pass without interacting. In fact Everywhere they overlap. Superposition applies at every point; the points where crest meets crest, or crest meets trough, are simply where the result is largest or smallest.
14.6.A.2 Pulses passing through each other Fix
Pulses passing through each other
Wave pulses or waves that meet do not collide or bounce off each other. Each travels through the region of overlap and emerges with its original shape, speed and direction; the combined shape of the medium exists only while they overlap.
Students often think Pulses or waves that meet collide like objects and bounce off each other, reversing direction. In fact They pass through each other. While they overlap the displacements add; afterward each pulse continues in its original direction with its original shape.
Students often think When two pulses overlap they merge into one new pulse with the combined shape, and that pulse stays on the rope afterward. In fact No. The combined shape exists only while the pulses overlap. Afterward each pulse emerges with its original shape, size and direction.
14.6.A.3 Principle of superposition Fix
Principle of superposition
Where waves overlap, the displacement of the medium at each point is the sum of the displacements each wave would produce there on its own, added with their signs: for example, +1.5 cm + (−1.0 cm) = +0.5 cm. SI unit of displacement: meter (m).
Students often think When two waves or pulses overlap, the larger (or louder) one takes over, and the displacement is that of the larger wave alone. In fact No. Both waves contribute at every point where they overlap: the displacement is the sum of the two, so a smaller wave still changes the result.
Students often think In superposition the sizes of the displacements are added, whatever their directions, so the resultant is larger than either wave alone. In fact No. Displacements are added with their signs: an upward displacement and a downward displacement partly or completely cancel.
14.6.A.4 Constructive interference Fix
Constructive interference
Interference at a point where the superposed displacements are in the same direction (both upward or both downward). The resultant displacement is larger in magnitude than either individual displacement.
Destructive interference
Interference at a point where the superposed displacements are in opposite directions. The resultant displacement is smaller in magnitude than the larger individual displacement, and it is zero where the two are equal in size.
Classifying interference
At each point and instant, interference is classified by comparing the directions of the individual displacements, not the directions in which the waves travel: displacements in the same direction give constructive interference, displacements in opposite directions give destructive interference.
Amplitude variation from interference
Overlapping waves can produce a resultant whose amplitude varies from place to place (for example, loud and quiet places in front of two loudspeakers emitting the same frequency) or from time to time (beats), according to whether their displacements reinforce or oppose each other.
Students often think Two downward displacements (troughs) interfere destructively, because both are negative or 'below', so the result is smaller. In fact No. Displacements in the same direction interfere constructively, whether both are upward or both are downward; two troughs make a deeper trough.
Students often think Adding two negative displacements gives a positive result, because two negatives make a positive. In fact A larger negative displacement: (−3.0 mm) + (−2.0 mm) = −5.0 mm.
14.6.A.5 Pulse diagrams at successive instants Fix
Pulse diagrams at successive instants
A graph of each pulse's displacement against position. To find the shape of the medium at a later time, move each pulse a distance (speed × time) in its direction of travel and then add the displacements point by point.
14.6.A.6 Beats Fix
Beats
The periodic rise and fall in loudness (amplitude) heard when two sound waves of slightly different frequency overlap. The combined wave oscillates at about the average of the two frequencies; its amplitude varies at the beat frequency.
In phase and out of phase
Two oscillations at a point are in phase at an instant when they are at the same stage of their cycles (for example, both at a crest) and out of phase when they are at opposite stages (one at a crest, the other at a trough). Waves of different frequencies drift from in phase to out of phase and back.
Beat frequency
The number of loudness maxima per second: |fbeat| = |f1 − f2|. The beat period, the time from one loudness maximum to the next, is 1/|fbeat|. SI unit: hertz (Hz).
Tuning fork
A two-pronged metal fork that vibrates at a single, stable frequency when struck. Sounding a fork of known frequency with another source gives beats at the difference of the frequencies; adding a small mass (such as clay) to a prong slightly lowers the fork's frequency.
Students often think Beats occur because the loudness of each source rises and falls on its own. In fact No. Each source on its own gives a steady sound. The loudness varies only when both sound together, because their waves drift in and out of phase.
Students often think Two sounds of slightly different frequency combine into a new sound wave whose frequency is the difference between them, and that wave is what is heard as the beats. In fact No. The combined wave oscillates at about the average of the two frequencies; what varies at the beat frequency is its amplitude, heard as loudness rising and falling.
14.6.B.1 Standing wave Fix
Standing wave
A vibration pattern that does not travel, formed by the superposition of two waves of equal frequency and amplitude traveling in opposite directions in a confined region, such as a wave and its reflection. Different points oscillate with different amplitudes, all at the same frequency.
Node
A point on a standing wave where the amplitude is always zero, so the medium there does not move. Adjacent nodes are half a wavelength apart.
Antinode
A point on a standing wave where the amplitude is greatest. The medium there still passes through equilibrium twice each cycle. Adjacent antinodes are half a wavelength apart, and a node and the nearest antinode are a quarter-wavelength apart.
Standing-wave envelope
The outline traced by the extreme positions of the medium in a standing wave. Its width at each point shows that point's amplitude: zero at nodes, greatest at antinodes. One loop, from node to node, is half a wavelength.
Boundary conditions
Requirements a standing wave must meet at the ends of its region: a fixed string end or a closed pipe end must be a displacement node; a loose string end or an open pipe end must be a displacement antinode. Only wavelengths that satisfy both ends can form a standing wave.
Possible wavelengths
For a region of length L with the same kind of end at both ends (two nodes or two antinodes), L = n(λ/2) with n = 1, 2, 3, …, so λ = 2L/n. With a node at one end and an antinode at the other, L = n(λ/4) with n = 1, 3, 5, …, so λ = 4L/n.
Standing-wave (resonant) frequencies
The frequencies at which a standing wave can be established in a region: f = v/λ for each possible wavelength. They depend on the length and end conditions (through λ) and on the wave speed in the medium.
Strings with fixed or loose ends
A fixed (clamped) end of a string is a node. A loose end, such as one tied to a light ring that slides freely on a smooth vertical rod, is an antinode.
Pipes with open or closed ends
A standing sound wave in a pipe is described here by the displacement of the air along the pipe. At a closed end the air cannot move, so the closed end is a displacement node; at an open end the air moves freely, so the open end is a displacement antinode.
Students often think A standing wave is a single wave that has stopped traveling and stands still in the medium. In fact No. A standing wave is the superposition of two waves of equal frequency traveling in opposite directions. The pattern stays in place, but the medium keeps oscillating.
Students often think Standing waves are formed by two waves of slightly different frequency traveling in the same direction, the same way beats are formed. In fact No. Beats come from two waves of slightly different frequency and give an amplitude that varies in time. A standing wave comes from two waves of equal frequency traveling in opposite directions and gives an amplitude that varies with position (nodes and antinodes).
14.6.B.2 Fundamental (first harmonic) Fix
Fundamental (first harmonic)
The standing wave with the longest possible wavelength, and so the lowest frequency, in a region: λ = 2L for a string fixed at both ends or a pipe open at both ends; λ = 4L for a region with a node at one end and an antinode at the other.
Harmonics
The standing waves a region can support, numbered by frequency as whole-number multiples of the fundamental, fn = n f1. For a region with a node at one end and an antinode at the other, only odd harmonics (n = 1, 3, 5, …) can be established.
Students often think A pipe closed at one end has every harmonic, so its second-lowest frequency is twice its fundamental. In fact No. With a node at one end and an antinode at the other, only odd harmonics can be established: f1, 3f1, 5f1, …
Students often think Closing one end of an open pipe keeps its fundamental and removes the even harmonics, leaving the odd multiples of the open pipe's fundamental. In fact No. Closing an end changes the fundamental wavelength from 2L to 4L, halving the fundamental frequency. The closed pipe's standing-wave frequencies are odd multiples of this new, lower fundamental.
14.6.B.3 Wave speed on a string Fix
Wave speed on a string
vstring = √(FT/(m/ℓ)), where FT is the tension and m/ℓ the mass per length. The speed is set by the string and its tension, not by the frequency of the source. SI unit: m/s.
Reading a standing-wave diagram
Count the half-wavelengths (node to node) or quarter-wavelengths (node to antinode) that fit in the length L to find λ, then use λ = v/f. For a given region and wave speed, a pattern with more loops has a shorter wavelength and a higher frequency.
Students often think Each loop of a standing wave, from one node to the next, is one whole wavelength. In fact No. One loop, from node to node, is half a wavelength. A string fixed at both ends vibrating in three loops has L = 3(λ/2).
Students often think The speed of waves on a string is the tension divided by the mass per length, with no square root, so it is proportional to the tension. In fact No. v = √(FT/(m/ℓ)), so the speed is proportional to the square root of the tension: four times the tension gives twice the speed.
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
Two pulses with the same shape and size, one upright and one inverted, travel toward each other along a long rope. At one instant they overlap exactly, and the whole rope is straight. Which statement about the rope at that instant is correct?
Answer and reasoning
APart of the rope is moving, and both pulses will re-form and travel on.Correct At that instant the displacements cancel, so the rope is straight, but the section where the pulses overlap is moving: there the rope's energy is all kinetic. That motion carries the rope out of line again, and the two pulses re-form and continue in their original directions.
BThe pulses have cancelled for good, and their energy has disappeared. A student who thinks cancelling pulses destroy each other picks this. The displacements cancel only for an instant; the energy is still in the rope as kinetic energy of the moving section, and both pulses reappear.
CEvery point on the rope is at rest, since no point is displaced. A student who thinks zero displacement means zero velocity picks this. Being at equilibrium says nothing about speed; in the overlap region the rope is passing through equilibrium quickly.
DThe pulses are turning around and will travel back toward where they started. A student who pictures pulses as objects that collide and rebound picks this. The pulses pass through each other: each continues in its original direction.
Working Equal and opposite displacements cancel, so y = 0 everywhere at this instant, but the velocities of the rope in the overlap region add (both pulses move the rope the same way there), so that section is moving and carries all the energy as kinetic energy. The pulses then separate unchanged.
Noise-cancelling headphones use a microphone to detect incoming noise and a small speaker inside each ear cup to produce a second sound wave. Near the ear, the noise is greatly reduced. Which explanation is correct?
Answer and reasoning
AThe speaker's sound is louder than the noise, so it overpowers the noise and is heard in its place. A student who thinks a stronger wave overrides a weaker one picks this. A louder sound added to the noise would make the total louder, not quieter; the reduction comes from displacements in opposite directions adding to nearly zero.
BThe speaker's wave travels in the direction opposite to the noise, and oppositely traveling waves cancel. A student who thinks the type of interference is set by the directions in which the waves travel picks this. What matters is the direction of the displacements at each point: opposite displacements interfere destructively whichever way the waves move.
CThe speaker's wave collides with the incoming noise and reflects it back out of the ear cup. A student who pictures waves as objects that bounce off each other picks this. Two sound waves pass through each other; neither reflects the other. Where they overlap, their displacements add.
DThe speaker's wave displaces the air opposite to the noise, so the sum is nearly zero.Correct This is destructive interference. The speaker produces a wave whose displacement at each instant is opposite to that of the noise and nearly equal in size, so near the ear the sum of the two displacements is close to zero.
Two loudspeakers a short distance apart are connected to the same signal generator and emit sound of a single frequency. A student walks slowly along a line several meters in front of the speakers, parallel to the line joining them, and hears the sound become loud and quiet several times. Which explanation is correct?
Answer and reasoning
AThe sound is loud where the student is close to one of the speakers and quiet where she is farther from both. A student who puts every change in loudness down to distance from the source picks this. Several meters away, the distances to the speakers change only slowly along the line, which cannot give several loud and quiet places; the alternation comes from interference.
BIn the quiet places the two waves destroy each other, and the energy they carry there is lost. A student who thinks waves that cancel are destroyed picks this. At a quiet place the displacements cancel, but the waves travel on unchanged and no energy is lost: more energy arrives at the loud places.
CAt some places the displacements point the same way and at others opposite, so the amplitude varies.Correct The waves from the two speakers overlap everywhere in front of them. Where they arrive with displacements in the same direction they interfere constructively and the resultant amplitude is large (loud); where the displacements are opposite they interfere destructively and it is small (quiet). Moving along the line changes which case applies.
DThe waves interfere in the loud places, but in the quiet places they pass through without interacting. A student who thinks waves interfere only at special points picks this. The waves overlap and add at every point; the quiet places are where their displacements are opposite, which is interference too.
Two sound sources of slightly different frequency play together. The graph shows the displacement y of the air at a microphone against time t; the dashed curves outline the oscillations. What is the beat frequency?
Answer and reasoning
A8.0 Hz A student who takes one beat to be the change from loud to quiet uses the 0.125 s from a maximum to the next minimum and picks this. A beat runs from one loudness maximum to the next, 0.25 s.
B4.0 HzCorrect One beat is one cycle of loud, quiet, loud. The oscillations are largest at t = 0, 0.25 s and 0.50 s, so a beat lasts 0.25 s and fbeat = 1/(0.25 s) = 4.0 Hz.
C2.0 Hz A student who takes the period of one dashed curve, 0.50 s, as the beat period picks this. The loudness is greatest both where a dashed curve is at its top and where it is at its bottom, so there are two loudness maxima in each cycle of that curve.
D3.0 Hz A student who counts the three loudness maxima on the graph and reports the count as the beat frequency picks this. A frequency is a number per second: the maxima are 0.25 s apart, which is 4.0 per second.
Working Loudness maxima (largest swings) at t = 0, 0.25 s and 0.50 s, so Tbeat = 0.25 s and fbeat = 1/Tbeat = 1/(0.25 s) = 4.0 Hz.
A 440.0 Hz tuning fork and a second tuning fork are struck together, and a student counts 12 beats in 4.00 s. Which of the following could be the frequency of the second fork?
Answer and reasoning
A443.0 HzCorrect The beat frequency is 12 beats/4.00 s = 3.00 Hz, and it equals the difference between the frequencies. The second fork is therefore at 440.0 Hz + 3.00 Hz = 443.0 Hz or 440.0 Hz − 3.00 Hz = 437.0 Hz; 443.0 Hz is the option offered.
B452.0 Hz A student who treats the 12 beats counted as the beat frequency, without dividing by 4.00 s, picks this. The beat frequency is beats per second: 12/4.00 s = 3.00 Hz.
C446.0 Hz A student who thinks the beat frequency is half the difference between the frequencies doubles 3.00 Hz and picks this. The beat frequency equals the whole difference, |f1 − f2|.
D440.3 Hz A student who uses the time per beat, 4.00 s/12 = 0.333 s, as if it were the beat frequency gets 440.0 Hz + 0.3 Hz and picks this. The time per beat is the beat period; the beat frequency is its reciprocal, 3.00 Hz.
Working fbeat = 12/(4.00 s) = 3.00 Hz; f2 = 440.0 Hz ± 3.00 Hz = 443.0 Hz or 437.0 Hz.
A student strikes a 440 Hz reference tuning fork together with fork U and hears 4 beats each second. The student claims that U's frequency is 444 Hz rather than 436 Hz. Sticking a small piece of clay to a fork's prongs slightly lowers its frequency. Which observation supports the student's claim?
Answer and reasoning
AWith clay stuck to U, the beats heard with the reference fork speed up. A student who thinks beats get faster as two frequencies get closer picks this. The beat frequency equals the difference between the frequencies, so it falls as they approach; beats that speed up after adding clay would show that U started below 440 Hz, at 436 Hz.
BWhen both forks are struck equally hard, fork U sounds louder than the reference fork. A student who links loudness with pitch picks this. Loudness depends on amplitude, not frequency, so how loud U sounds says nothing about whether its frequency is above or below 440 Hz.
CWith clay stuck to U, the beats heard with the reference fork slow down.Correct If U is at 444 Hz, clay lowers it toward 440 Hz, so the difference in frequencies, and with it the beat frequency, decreases: the beats slow down. If U were at 436 Hz, clay would take it farther from 440 Hz and the beats would speed up. Slower beats therefore support 444 Hz.
DThe 4 beats heard each second exactly match the difference 444 Hz − 440 Hz. A student who thinks a beat frequency shows which fork is higher picks this. 4 beats per second fits 440 Hz − 436 Hz just as well; the beat frequency gives the size of the difference, not which fork is higher.
A string fixed at both ends vibrates in a standing wave. In the figure, the dashed curves show the extreme positions of the string, and the solid line shows the string at an instant when it is straight. Points J, K, and M are on the string. Which ranking of the speeds vJ, vK, and vM of these points at that instant is correct?
Answer and reasoning
AvJ = vM = vK = 0 A student who thinks a point with zero displacement must be at rest picks this. The string is straight because every point is passing through equilibrium, which is where each moving point is fastest.
BvJ > vM > vK = 0Correct J is at an antinode, K at a node, and M between them, so their amplitudes rank J > M > K = 0. All points oscillate with the same period, and a point with a larger amplitude travels farther each cycle, so as the straight string passes through equilibrium J moves fastest, M more slowly, and K not at all.
CvK > vM > vJ = 0 A student who has the meanings of node and antinode reversed picks this. A node, like K, is where the amplitude is always zero; an antinode, like J, is where it is greatest.
DvJ = vM = vK > 0 A student who thinks every point on a standing wave oscillates with the same amplitude, as on a traveling wave, picks this. The amplitude varies along a standing wave, from zero at a node to a maximum at an antinode.
Working From the envelope: J is at an antinode (largest amplitude), K is at a node (amplitude zero), M is between a node and an antinode. At the straight instant every point is at equilibrium, moving at its greatest speed; with a common period, greatest speed increases with amplitude. So vJ > vM > vK = 0.
A student claims that point X on a string vibrating in a standing wave is a node. Which observation supports this claim?
Answer and reasoning
AA slow-motion video shows the string at X never moving away from equilibrium.Correct A node is a point where the amplitude is always zero. Video over whole cycles showing that X never leaves equilibrium is exactly that evidence.
BA single photograph of the vibrating string shows the string at X at equilibrium. A student who thinks any point at equilibrium at one instant is a node picks this. Every point passes through equilibrium twice each cycle, and at some instants the whole string is straight; one photograph cannot show that X stays there.
CA slow-motion video shows the string at X moving farther from equilibrium than anywhere else. A student who has node and antinode reversed picks this. A point that moves farther than any other is an antinode, the opposite of a node.
DX is the end of the string tied to a light ring that slides freely on a smooth rod. A student who thinks every end of a string is a node picks this. A loose end, free to slide on the rod, is an antinode: the string there moves more than anywhere else, so this observation counts against the claim.
A string of length L is fixed at both ends. The tension in the string is FT and its mass per length is m/ℓ. Which expression gives the frequency of the fundamental standing wave on the string?
Answer and reasoning
A(1/L)√(FT/(m/ℓ)) A student who takes one loop, from node to node, to be a whole wavelength uses λ = L and picks this. Node to node is half a wavelength, so λ = 2L.
B(1/(2L))(FT/(m/ℓ)) A student who takes the wave speed to be FT/(m/ℓ), leaving out the square root, picks this. The speed is v = √(FT/(m/ℓ)); without the root the expression does not have units of frequency.
C(1/(2L))√(FT/(m/ℓ))Correct Both fixed ends are nodes, and the fundamental has one loop between them, which is half a wavelength: λ = 2L. With v = √(FT/(m/ℓ)) and f = v/λ, f = (1/(2L))√(FT/(m/ℓ)).
D(2L)√(FT/(m/ℓ)) A student who writes f = λv instead of f = v/λ multiplies 2L by the speed and picks this. Frequency is speed divided by wavelength; the product λv has units m²/s, not Hz.
Working Nodes at both ends; fundamental: L = λ/2, so λ = 2L. v = √(FT/(m/ℓ)). f1 = v/λ = (1/(2L))√(FT/(m/ℓ)). Units: (1/m)(m/s) = 1/s = Hz.
A string fixed at both ends vibrates in its fundamental mode. The tension in the string is then increased to four times its original value, with the length and the string unchanged. By what factor does the frequency of the fundamental change?
Answer and reasoning
A×2Correct The fundamental wavelength, λ = 2L, is set by the length and the fixed ends and does not change. The wave speed v = √(FT/(m/ℓ)) grows by √4 = 2, so f = v/λ doubles.
B×4 A student who takes the wave speed to be proportional to the tension itself, rather than to its square root, picks this. Four times the tension gives √4 = 2 times the speed.
C×1 A student who thinks a string's standing-wave frequencies are set by its length alone picks this. The length fixes the wavelength, but the frequency f = v/λ also depends on the wave speed, which the tension changes.
D×½ A student who inverts the ratio in the wave-speed equation, taking v = √((m/ℓ)/FT), finds that the speed halves and picks this. Greater tension makes waves travel faster, not slower.
Working λ1 = 2L is unchanged. v ∝ √FT, so vnew = √4 v = 2v. f = v/λ, so fnew = 2f.
Three strings are each fixed at both ends and under the same tension. String 1 has length L and mass per length m/ℓ. String 2 has length L and mass per length 4(m/ℓ). String 3 has length L/2 and mass per length 4(m/ℓ). Which ranking of the frequencies f1, f2, and f3 of their fundamental standing waves is correct?
Answer and reasoning
Af3 > f1 = f2 A student who thinks the frequency depends only on the length, so that equal lengths give equal frequencies and the shorter string is higher, picks this. String 2's greater mass per length makes waves on it slower, which lowers its frequency.
Bf1 > f3 > f2 A student who takes the wave speed to be inversely proportional to the mass per length, leaving out the square root, gets speeds v, v/4 and v/4, frequencies v/(2L), v/(8L) and v/(4L), and picks this. The speed depends on the square root: √4 = 2, not 4.
Cf1 = f3 > f2Correct The fundamental wavelength is twice the length: 2L, 2L and L. The wave speed v = √(FT/(m/ℓ)) is halved by four times the mass per length: v, v/2, v/2. So f = v/λ gives f1 = v/(2L), f2 = v/(4L) and f3 = (v/2)/L = v/(2L).
Df1 > f2 > f3 A student who writes f = λv, making frequency grow with wavelength, gets 2Lv, Lv and Lv/2 and picks this. Frequency is v/λ: a shorter wavelength at the same speed means a higher frequency.
Working λ: 2L, 2L, L. v = √(FT/(m/ℓ)): v, v/√4 = v/2, v/2. f = v/λ: f1 = v/(2L); f2 = (v/2)/(2L) = v/(4L); f3 = (v/2)/L = v/(2L). So f1 = f3 > f2.
Each numbered diagram shows the displacement envelope (dashed) of a proposed standing wave: sound waves in two pipes and waves on two strings. The end conditions are labeled under each diagram. Which of patterns 1 to 4 can actually be established?
Answer and reasoning
APattern 2 A student who swaps the end conditions of a pipe picks this. The closed end must be a displacement node, because the air there cannot move along the pipe, and the open end must be an antinode; pattern 2 has them the other way round.
BPattern 3 A student who thinks every end is a node picks this. The loose end, tied to a ring that slides freely on the rod, must be an antinode, not a node.
CPattern 4 A student who thinks a standing wave of any wavelength can form picks this. The right-hand end is fixed, so it must be a node; pattern 4 is displaced there because its wavelength does not fit a whole number of half-wavelengths into the length.
DPattern 1Correct An open pipe end is a displacement antinode, so a pipe open at both ends can have antinodes at both ends with a node between them: this is its fundamental.
Working Displacement end conditions: fixed string end or closed pipe end = node; loose string end or open pipe end = antinode. 1: antinodes at both open ends, one node between (open-pipe fundamental), allowed. 2: antinode at the closed end, node at the open end, reversed. 3: node at the loose end, not allowed. 4: nonzero displacement at a fixed end, not allowed.
Pipe X is open at both ends. The frequencies at which standing sound waves form in it are 400 Hz, 800 Hz, 1200 Hz, and so on. One end of pipe X is then closed, with its length and the air inside unchanged. What is now the second-lowest frequency at which a standing wave can form in the pipe?
Answer and reasoning
A6.0 × 10² HzCorrect Open at both ends, f1 = v/(2L) = 400 Hz. With one end closed, the fundamental wavelength becomes 4L, so f1 = v/(4L) = 200 Hz, and only odd harmonics can form: 200 Hz, 600 Hz, 1000 Hz, … The second-lowest is 600 Hz.
B4.0 × 10² Hz A student who thinks a pipe closed at one end has every harmonic, so that its second-lowest frequency is 2 × 200 Hz, picks this. With a node at one end and an antinode at the other, only odd harmonics form.
C1.2 × 10³ Hz A student who thinks closing an end just removes the open pipe's even harmonics, leaving 400 Hz, 1200 Hz, …, picks this. Closing the end also doubles the fundamental wavelength to 4L, halving the fundamental to 200 Hz.
D8.0 × 10² Hz A student who thinks a pipe's standing-wave frequencies depend only on its length, so nothing changes, picks this. The end conditions matter too: a closed end must be a displacement node.
Working Open–open: f1 = v/(2L) = 400 Hz, so v/L = 800 Hz. Closed–open: λ1 = 4L, f1 = v/(4L) = 200 Hz; odd harmonics only: 200, 600, 1000 Hz, … Second-lowest = 3 × 200 Hz = 600 Hz = 6.0 × 10² Hz.
The figure shows the displacement envelope of a standing sound wave in a pipe of length L that is open at both ends. The speed of sound in the air is v. Which expression gives the frequency of this standing wave?
Answer and reasoning
Af = 2v/L A student who takes the harmonic number to be the number of antinodes, four, uses f = 4v/(2L) and picks this. For a pipe open at both ends the harmonic number is the number of half-wavelengths in L, which here is three.
Bf = v/(2L) A student who uses the fundamental wavelength, λ = 2L, whatever pattern is shown picks this. The pattern holds three half-wavelengths, so λ = 2L/3.
Cf = 2Lv/3 A student who writes f = λv instead of f = v/λ picks this. Its units are m²/s, not Hz.
Df = 3v/(2L)Correct Both open ends are antinodes and there are three nodes between them. Antinode to antinode is half a wavelength, and the pipe holds three such sections: L = 3(λ/2), so λ = 2L/3 and f = v/λ = 3v/(2L), the third harmonic.
Working Antinodes at both ends, three nodes inside: three half-wavelengths in L. λ = 2L/3, f = v/λ = 3v/(2L).
An oscillator vibrating at 200 Hz drives one end of a string whose other end is fixed, forming the standing wave shown in the figure. The tension is kept constant. At what frequency must the oscillator be set for the string to form a standing wave with one more loop?
Answer and reasoning
A400 Hz A student who thinks each next harmonic doubles the frequency, like an octave, picks this. The harmonics of a string are whole-number multiples of the fundamental: 100 Hz, 200 Hz, 300 Hz, 400 Hz, …
B133 Hz A student who takes frequency to grow with wavelength (writing f = λv) scales 200 Hz by the new wavelength ratio, (2L/3)/L = 2/3, and picks this. At a fixed wave speed, a shorter wavelength needs a higher frequency: f = v/λ.
C300 HzCorrect The figure shows two loops, so the string holds one full wavelength: this is the second harmonic, and the fundamental is 200 Hz/2 = 100 Hz. One more loop is the third harmonic, at 3 × 100 Hz = 300 Hz.
D200 Hz A student who thinks more loops come from driving the string harder at the same frequency picks this. A larger amplitude makes the loops bigger but does not change the wavelength; only a different frequency gives another loop.
Working Two loops: L = 2(λ/2) = λ, second harmonic, so f1 = 200 Hz/2 = 100 Hz (v unchanged because the tension is constant). Three loops: third harmonic, f3 = 3f1 = 300 Hz. Equivalently λ goes from L to 2L/3, so f = v/λ rises by 3/2.
One end of a horizontal string is attached to an oscillator. The string passes over a frictionless pulley a distance L from the oscillator, and a block of mass M hangs at rest from its other end. The string has mass per length m/ℓ. At one frequency of the oscillator, the string between the oscillator and the pulley forms a standing wave with three loops, with nodes at the oscillator and at the pulley. The acceleration due to gravity is g. Which expression gives this frequency?
Answer and reasoning
A(3/(2L))√(M/(m/ℓ)) A student who puts the block's mass M in place of the tension picks this. The tension balances the block's weight, so FT = Mg; with M alone, √(M/(m/ℓ)) has units of √m, not m/s.
B(3/(2L))√(Mg/(m/ℓ))Correct The tension is the block's weight, Mg, so v = √(Mg/(m/ℓ)). Three loops fill L and each loop is half a wavelength, so λ = 2L/3. Then f = v/λ = (3/(2L))√(Mg/(m/ℓ)).
C(3/(2L))√((m/ℓ)/(Mg)) A student who puts the tension in the denominator of the wave-speed equation picks this. Waves travel faster on a tighter string: v = √(FT/(m/ℓ)). The inverted ratio has units of s/m, so this expression is not a frequency.
D(1/(L/3))√(Mg/(m/ℓ)) A student who takes each loop, from node to node, to be a whole wavelength sets L = 3λ, so λ = L/3, and picks this. Each loop is half a wavelength, so three loops give L = 3λ/2 and λ = 2L/3.
Working Block at rest: FT = Mg. Three loops between the nodes, each loop λ/2: L = 3λ/2, so λ = 2L/3. Wave speed v = √(FT/(m/ℓ)) = √(Mg/(m/ℓ)). f = v/λ = (3/(2L))√(Mg/(m/ℓ)). Units: (1/m)·√(N/(kg/m)) = (1/m)(m/s) = Hz.
Two strings, of lengths L₁ and L₂ with L₁ slightly shorter than L₂, are each fixed at both ends, and waves travel at the same speed v on both strings. The strings are plucked at the same time, each vibrates in its fundamental mode, and a listener hears beats. Which expression gives the beat frequency?
Answer and reasoning
A0.25v(L₂ − L₁)/(L₁L₂) A student who thinks the beat frequency is half the difference between the two frequencies halves the correct difference and picks this. The beat frequency equals the whole difference, |f₁ − f₂|.
B0.50v(L₂ − L₁)/(L₁L₂)Correct The ends of each string are nodes and the fundamental is one loop, so λ = 2L and f = v/(2L). The shorter string has the higher frequency, and the beat frequency is the difference of the two frequencies: v/(2L₁) − v/(2L₂) = (v/2)(L₂ − L₁)/(L₁L₂).
C1.00v(L₂ − L₁)/(L₁L₂) A student who takes the single loop of the fundamental as one whole wavelength uses λ = L and f = v/L for each string, and picks this. A loop runs from one node to the next, which is half a wavelength, so λ = 2L and f = v/(2L).
D2.00v(L₂ − L₁)/(L₁L₂) A student who turns L = λ/2 round the wrong way to λ = L/2 uses f = 2v/L for each string and picks this. Solving L = λ/2 for the wavelength gives λ = 2L, so f = v/(2L).
Working Both ends of each string are nodes and the fundamental is a single loop, so L = λ/2, λ = 2L and f = v/λ = v/(2L). f₁ = v/(2L₁) and f₂ = v/(2L₂), with f₁ > f₂ because L₁ < L₂. |fbeat| = f₁ − f₂ = (v/2)(1/L₁ − 1/L₂) = (v/2)(L₂ − L₁)/(L₁L₂) = 0.50v(L₂ − L₁)/(L₁L₂). Units: (m/s)(m)/(m²) = 1/s = Hz.
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
