AP Physics 2 Quiz: Wave Interference And Standing Waves

What this quiz covers

This quiz focuses on Wave Interference And Standing Waves, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Constructive interference because the waves always add when $\Delta L$ is a multiple of $\lambda/2$.
2. Destructive interference because the waves arrive $180^\circ$ out of phase. (correct answer)
3. No interference because sound waves pass through each other without superposition.
4. Destructive interference that permanently cancels the sound at that point even if the listener moves slightly.

Explanation: This question tests understanding of wave interference and standing waves. When two coherent waves travel different distances to reach a point, their phase relationship depends on the path-length difference Δ L. A path difference of 1.5λ means one wave travels exactly 1.5 wavelengths farther than the other, creating a phase difference of 1.5 × 360° = 540° = 180° (since 540° - 360° = 180°), which results in destructive interference. Choice A incorrectly claims that multiples of λ/2 always produce constructive interference, but odd multiples of λ/2 (like 1.5λ = 3 × λ/2) actually produce destructive interference. The strategy is to convert path differences to phase differences: whole wavelengths (nλ) give constructive interference, while odd half-wavelengths ((2n+1)λ/2) give destructive interference.

Question 2

1. A node oscillates with maximum amplitude, while an antinode remains at zero displacement.
2. A node remains at zero displacement at all times, while an antinode oscillates with maximum amplitude. (correct answer)
3. A node slowly moves along the string, while an antinode remains fixed in position.
4. A node permanently removes energy from the wave, while an antinode stores energy without transfer.

Explanation: This question tests understanding of wave interference and standing waves. In a standing wave on a string, nodes are points where destructive interference between two counter-propagating waves creates zero displacement at all times, while antinodes are points where constructive interference produces maximum amplitude oscillation. The string oscillates up and down between its extreme positions, but nodes remain stationary with zero displacement throughout the motion. Choice A incorrectly reverses the definitions, suggesting nodes oscillate while antinodes stay still—this misconception may arise from confusing the fixed spatial positions with the actual motion. The key strategy is to remember that nodes are 'no-goes' for motion: they mark locations of permanent destructive interference where the string never moves.

Question 3

1. Nodes are points of zero displacement at all times due to destructive interference. (correct answer)
2. Nodes move back and forth along the string as the wave oscillates.
3. Energy disappears at nodes, so the wave loses power each cycle.
4. Nodes are points where the string's displacement is always maximum.

Explanation: This question tests understanding of wave interference and standing waves. In a standing wave, nodes are positions where the two counter-propagating waves always interfere destructively, resulting in zero displacement at all times. The waves have opposite displacements at node locations, causing complete cancellation through superposition. This creates fixed points that never move, unlike the oscillating segments between them. Choice C incorrectly suggests energy disappears at nodes, misunderstanding that interference redistributes energy rather than destroying it - the energy flows past nodes to create larger oscillations at antinodes. Remember that nodes are stationary points of permanent destructive interference where displacement remains zero.

Question 4

1. Constructive interference, because the path-length difference is $1\lambda$. (correct answer)
2. Destructive interference, because the path-length difference is $1\lambda$.
3. Constructive interference, because the path-length difference is $\tfrac{1}{2}\lambda$.
4. No interference, because the waves cancel and then stop propagating.

Explanation: This question tests understanding of wave interference and standing waves. The path-length difference is 0.45 m - 0.30 m = 0.15 m, which equals exactly one wavelength (0.15 m = 1λ). When coherent sources are in phase and the path difference equals an integer multiple of wavelengths, the waves arrive in phase at the observation point, producing constructive interference with maximum amplitude. Choice B incorrectly identifies this as destructive interference, revealing the misconception that any whole-wavelength difference causes cancellation. The reliable strategy is to remember that path differences of nλ (where n = 0, 1, 2, ...) produce constructive interference, while odd multiples of λ/2 produce destructive interference.

Question 5

1. Maximum upward, because all points on the string move together in a standing wave.
2. Maximum downward, because nodes are always $180^\circ$ out of phase with antinodes.
3. Zero, because a node has zero displacement at all times. (correct answer)
4. Nonzero and changing, because the node shifts position as the wave oscillates.

Explanation: This question tests understanding of wave interference and standing waves. In a standing wave, nodes are points of permanent destructive interference where the string displacement remains zero at all times, regardless of what neighboring points are doing. While point R at an antinode oscillates between maximum upward and maximum downward positions, point S at a node maintains zero displacement throughout the entire oscillation cycle. Choice B incorrectly assumes nodes oscillate opposite to antinodes, reflecting the misconception that all points on a standing wave must move. The fundamental principle is that nodes are stationary points created by perfect destructive interference—they never move, serving as fixed pivot points between oscillating segments.

Question 6

1. The string has $3$ loops, so $L=\tfrac{3\lambda}{2}$ and adjacent nodes are separated by $\tfrac{\lambda}{2}$. (correct answer)
2. The nodes move back and forth along the string as time passes, carrying energy to the ends.
3. At each node the energy of the wave disappears, so the string is motionless everywhere after a short time.
4. Because of destructive interference at nodes, the string cancels permanently and no antinodes can exist.

Explanation: This question tests understanding of wave interference and standing waves. A standing wave with three antinodes (loops) and nodes at both ends contains exactly 3/2 wavelengths, so L = 3λ/2, which gives λ = 2L/3 = 0.80 m. Since adjacent nodes are always separated by λ/2, the node spacing is 0.40 m. Choice B incorrectly suggests nodes move, but nodes are fixed points of destructive interference where the string never moves. Choice C misunderstands that while nodes have zero amplitude, the antinodes continue oscillating with energy redistributed there. When analyzing standing waves, remember that nodes are λ/2 apart and count the number of half-wavelengths fitting in the length.

Question 7

1. Nodes remain at fixed positions with zero displacement, while antinodes have maximum displacement amplitude. (correct answer)
2. Nodes move along the string with time, while antinodes stay at fixed positions with zero displacement.
3. Nodes are locations where wave energy disappears permanently, while antinodes store all the energy.
4. Nodes and antinodes both have the same displacement amplitude, but differ only in phase.

Explanation: This question tests understanding of wave interference and standing waves. In a standing wave, nodes are points where destructive interference between the incident and reflected waves creates zero displacement at all times, while antinodes are points where constructive interference produces maximum displacement amplitude. The third harmonic on a string of length 1.20 m has three complete loops, with nodes at x = 0, 0.40, 0.80, and 1.20 m (including the fixed ends), and antinodes at x = 0.20, 0.60, and 1.00 m. Choice C incorrectly suggests that energy disappears at nodes, but energy simply flows through nodes without causing displacement—the wave energy oscillates between kinetic and potential forms as it transfers between adjacent segments. The key insight is that nodes and antinodes are fixed positions in a standing wave pattern, with nodes always having zero displacement and antinodes oscillating with maximum amplitude.

Question 8

1. The pulses permanently cancel and the string remains flat afterward.
2. The string displacement is zero during overlap due to destructive superposition. (correct answer)
3. The string displacement doubles to $6.0\ \text{cm}$ during overlap.
4. The pulses reflect off each other and reverse direction at overlap.

Explanation: This question tests understanding of wave interference and standing waves. When two pulses of equal magnitude but opposite sign (+3.0 cm and -3.0 cm) overlap completely, the principle of superposition states that the net displacement equals the algebraic sum of individual displacements. At complete overlap, the sum is (+3.0) + (-3.0) = 0 cm, resulting in momentary destructive interference where the string appears flat. After overlap, the pulses continue past each other unchanged, preserving their original shapes and energies. Choice A incorrectly assumes the cancellation is permanent, failing to understand that superposition is temporary - waves pass through each other without permanent alteration. When analyzing pulse interactions, apply superposition at each instant: waves add algebraically during overlap then continue independently.

Question 9

1. $\lambda=L$, because the string length equals one full wavelength.
2. $\lambda=\tfrac{L}{2}$, because nodes are separated by half a wavelength.
3. $\lambda=2L$, because the string length is half a wavelength. (correct answer)
4. $\lambda$ cannot be defined for standing waves because energy stops at nodes.

Explanation: This question tests understanding of wave interference and standing waves. The standing wave pattern shows nodes at x = 0 and x = L with a single antinode at x = L/2, indicating that the string length L contains exactly half a wavelength (one node-to-node distance). Since adjacent nodes are separated by λ/2, and the string spans from one node to the next node, we have L = λ/2, which gives λ = 2L. Choice A incorrectly assumes the string length equals one full wavelength, likely from misunderstanding that a complete wave cycle requires two node-to-node segments. The key strategy is to recognize that the distance between adjacent nodes (or adjacent antinodes) always equals λ/2 in any standing wave pattern.

Question 10

1. $n=2$, because there are two ends fixed.
2. $n=3$, because there are three interior nodes.
3. $n=4$, because the number of antinodes equals $n$. (correct answer)
4. $n=5$, because nodes move and create extra loops over time.

Explanation: This question tests understanding of wave interference and standing waves. For a string fixed at both ends, the number of antinodes equals the harmonic number n, while the number of nodes equals n + 1 (including the two fixed ends). With four antinodes visible, this corresponds to the fourth harmonic (n = 4), where the string length contains exactly two full wavelengths (4 half-wavelengths). Choice B incorrectly counts interior nodes instead of antinodes, reflecting the common misconception of focusing on nodes rather than the oscillating regions. The reliable strategy for fixed-end strings is to count antinodes: the number of antinodes directly gives the harmonic number n.

Question 11

1. Third harmonic, because the pattern has two nodes and two antinodes along the pipe. (correct answer)
2. First harmonic, because closed-open pipes always have only one node.
3. Fourth harmonic, because energy is zero at the node so more nodes means higher even harmonics.
4. Second harmonic, because nodes move with time to create two nodes total.

Explanation: This question tests understanding of wave interference and standing waves. In a closed-open pipe, standing waves form with a node at the closed end and an antinode at the open end. The fundamental (first harmonic) has only these two features with no additional nodes inside. With one additional node inside the pipe, there are two nodes total (closed end and interior) and two antinodes (interior and open end), which corresponds to the third harmonic pattern. Choice A incorrectly limits closed-open pipes to one node, while choice C misunderstands that nodes are stationary in standing waves. The misconception is thinking harmonics must be even numbers or that nodes move. For closed-open pipes, only odd harmonics (1st, 3rd, 5th...) are possible, with the nth odd harmonic having n nodes.

Question 12

1. Nodes at both ends and an antinode at the midpoint.
2. Nodes drift along the string, so their locations cannot be identified.
3. Nodes at both ends and at the midpoint, with antinodes between them. (correct answer)
4. Antinodes at both ends and a node at the midpoint.

Explanation: This question tests understanding of wave interference and standing waves. The problem states the string has 2 antinodes, which means it vibrates in the second harmonic (n=2) pattern. For a string fixed at both ends, nodes always occur at the fixed ends, and for the second harmonic, there is exactly one additional node at the midpoint. Between these three nodes are two antinodes where the string oscillates with maximum amplitude. Choice B incorrectly places antinodes at the fixed ends, which is impossible since fixed points cannot move. The misconception is not recognizing that fixed ends must be nodes. For strings fixed at both ends, always start by placing nodes at the boundaries, then distribute the remaining nodes and antinodes according to the harmonic number.

Question 13

1. $\lambda = 0.30\,\text{m}$ because node-to-node spacing is one wavelength.
2. $\lambda = 0.60\,\text{m}$ because node-to-node spacing is $\lambda/2$. (correct answer)
3. $\lambda = 0.15\,\text{m}$ because nodes move with time so spacing halves.
4. $\lambda = 1.20\,\text{m}$ because nodes are where energy accumulates.

Explanation: This question tests understanding of wave interference and standing waves. In any standing wave, the distance between adjacent nodes is exactly half a wavelength (λ/2). Given that adjacent nodes are separated by 0.30 m, we have λ/2 = 0.30 m, which gives λ = 0.60 m. This fundamental relationship holds for all standing waves, whether on strings, in pipes, or other media. Choice B incorrectly claims node spacing equals one full wavelength, while choice D suggests nodes move, which contradicts the definition of standing waves. The misconception is confusing node-to-node distance with wavelength. To find wavelength from standing wave patterns, remember that node-to-node (or antinode-to-antinode) spacing always equals λ/2.

Question 14

1. $0\,\text{mm}$ because equal waves cancel when they overlap.
2. $2.0\,\text{mm}$ because only one wave determines the displacement.
3. $4.0\,\text{mm}$ because displacements add by superposition. (correct answer)
4. $4.0\,\text{mm}$ and the string stays at that displacement permanently after overlap.

Explanation: This question tests understanding of wave interference and standing waves. The principle of superposition states that when waves overlap, the net displacement equals the algebraic sum of individual displacements. Two in-phase waves with displacements of +2.0 mm each produce a total displacement of (+2.0) + (+2.0) = +4.0 mm through constructive interference. Choice D incorrectly suggests the displacement remains permanent, but the waves continue propagating, so the enhanced displacement only occurs while the waves overlap at that location. The strategy is straightforward: for superposition, simply add the signed displacements algebraically, remembering that the result is instantaneous, not permanent.

Question 15

1. $\lambda=2L$, because the tube contains one full wavelength in the fundamental.
2. $\lambda=L$, because the node-to-antinode spacing equals one wavelength.
3. $\lambda=4L$, because the tube contains one-quarter wavelength in the fundamental. (correct answer)
4. $\lambda=\tfrac{L}{4}$, because the tube contains four nodes in the fundamental.

Explanation: This question tests understanding of wave interference and standing waves. In a tube closed at one end, the closed end must be a displacement node (pressure antinode) and the open end must be a displacement antinode (pressure node). The fundamental mode contains exactly one-quarter wavelength, so L = λ/4, which means λ = 4L. Choice A incorrectly assumes the tube contains a full wavelength, confusing this with open-open tubes. Choice B misunderstands that node-to-antinode spacing is λ/4, not λ. For closed-open tubes, remember that odd multiples of quarter wavelengths fit: L = (2n-1)λ/4 where n = 1, 2, 3...

Question 16

1. Permanent cancellation, because once destructive interference occurs the sound energy disappears.
2. Constructive interference, because the path difference equals one wavelength.
3. No interference, because waves from different sources cannot superpose in air.
4. Destructive interference, because the path difference equals half a wavelength. (correct answer)

Explanation: This question tests understanding of wave interference and standing waves. The wavelength is λ = v/f = 340/500 = 0.68 m, and the path difference is 2.34 - 2.00 = 0.34 m, which equals λ/2. When the path difference equals an odd multiple of λ/2, destructive interference occurs because the waves arrive 180° out of phase. Choice A incorrectly calculates the path difference as one wavelength, which would produce constructive interference. Choice D reflects the misconception that destructive interference permanently destroys energy, when actually the energy is redistributed to other locations. To analyze interference, always calculate the path difference as a fraction of wavelength: odd multiples of λ/2 give destructive interference.

Question 17

1. First harmonic
2. Second harmonic
3. Third harmonic (correct answer)
4. Fourth harmonic

Explanation: This question tests understanding of wave interference and standing waves. In a tube closed at one end and open at the other, standing waves form with a displacement node at the closed end (where air cannot move) and a displacement antinode at the open end (where air moves freely). The description states there is one additional node inside the tube, giving a total pattern of: node (closed end) → antinode → node → antinode (open end), which spans 3/4 of a wavelength. For a closed-open tube, only odd harmonics exist: the first harmonic has L = λ/4, the third harmonic has L = 3λ/4, the fifth has L = 5λ/4, and so on. Choice B incorrectly suggests this could be the second harmonic, but even harmonics don't exist in closed-open tubes. The key strategy is to count quarter-wavelengths: from node to antinode is λ/4, so three segments (N→A→N→A) equals 3λ/4, identifying the third harmonic.

Question 18

1. Point $P$ has greater amplitude than point $Q$ because nodes concentrate energy.
2. Point $P$ and point $Q$ have equal amplitude but opposite directions of motion.
3. Point $P$ has zero amplitude while point $Q$ has maximum amplitude. (correct answer)
4. Point $P$ has zero amplitude because the wave stops there and cannot transmit energy past it.

Explanation: This question tests understanding of wave interference and standing waves. In a standing wave, nodes are points of permanent destructive interference where the string never moves (zero amplitude), while antinodes are points of constructive interference with maximum oscillation amplitude. Point P, being a node, has exactly zero transverse displacement at all times, while point Q, the nearest antinode, oscillates with the maximum possible amplitude for that standing wave. Choice D incorrectly suggests that waves cannot transmit energy past nodes, but energy flows continuously through nodes—it simply doesn't cause displacement there because the forward and backward waves cancel. The key principle is that nodes mark positions of perfect destructive interference, not energy barriers.

Question 19

1. The string's displacement is $0\,\text{cm}$ there due to destructive interference at that instant. (correct answer)
2. The pulses permanently cancel and the string remains flat after they pass.
3. The string's displacement is $6.0\,\text{cm}$ there because magnitudes always add.
4. No superposition occurs because pulses reflect before overlapping.

Explanation: This question tests understanding of wave interference and standing waves. When two pulses overlap on a string, the principle of superposition states that the net displacement at any point equals the algebraic sum of the individual displacements. With one pulse at +3.0 cm and another at -3.0 cm at the same location, the total displacement is (+3.0) + (-3.0) = 0 cm, resulting in destructive interference at that instant. Choice B incorrectly suggests permanent cancellation, but after the pulses pass through each other, they continue traveling with their original shapes—superposition is temporary, not permanent. The crucial concept is that interference redistributes wave energy momentarily but doesn't destroy it; the pulses emerge unchanged after overlap.

Question 20

1. Destructive interference occurs there at all times, keeping displacement zero.
2. Constructive interference occurs there, producing maximum displacement magnitude. (correct answer)
3. Energy is destroyed there each cycle, so amplitude decreases along the string.
4. The antinode position drifts along the string as time increases.

Explanation: This question tests understanding of wave interference and standing waves. At an antinode in a standing wave, the two counter-propagating waves interfere constructively, meaning they have the same phase and their amplitudes add together. This produces a location where the string oscillates with maximum amplitude - twice the amplitude of each individual traveling wave. The constructive interference occurs continuously at the antinode position, creating a stable pattern where that point oscillates between maximum positive and negative displacements. Choice A incorrectly describes destructive interference, which actually occurs at nodes, not antinodes - this reverses the fundamental distinction between these two features. Remember: antinodes are points of constructive interference with maximum oscillation amplitude.