Wave Interference and Standing Waves
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AP Physics 2 › Wave Interference and Standing Waves
Two identical loudspeakers emit coherent sound waves in phase. A listener stands where the path-length difference from the speakers is $\Delta L=1.5\lambda$. Assuming equal amplitudes, which interference is produced at the listener’s location?
Constructive interference because the waves always add when $\Delta L$ is a multiple of $\lambda/2$.
Destructive interference that permanently cancels the sound at that point even if the listener moves slightly.
Destructive interference because the waves arrive $180^\circ$ out of phase.
No interference because sound waves pass through each other without superposition.
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.
A string of length $1.20,\text{m}$ is fixed at both ends and driven at a frequency that produces the standing-wave pattern shown. Points $N_1$, $N_2$, and $N_3$ are labeled as nodes, and points $A_1$ and $A_2$ are labeled as antinodes. The string’s shape oscillates in time, but the node locations remain fixed. Which statement correctly describes the motion at a node compared with an antinode?
A node oscillates with maximum amplitude, while an antinode remains at zero displacement.
A node remains at zero displacement at all times, while an antinode oscillates with maximum amplitude.
A node slowly moves along the string, while an antinode remains fixed in position.
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.
Two coherent water waves of the same frequency overlap at a point. The phase difference at that point is $\pi$ radians and the amplitudes are equal. Which result occurs at that point at that instant?
Destructive interference that removes energy from the system permanently at that point.
No interference because phase difference prevents superposition.
Maximum constructive interference because a phase difference of $\pi$ makes crests align.
Destructive interference producing zero displacement at that instant.
Explanation
This question tests understanding of wave interference and standing waves. When two coherent waves of equal amplitude meet with a phase difference of π radians (180°), one wave's crest aligns with the other's trough, producing destructive interference. At that instant, the waves cancel completely, resulting in zero displacement at that point. Choice A incorrectly claims that π phase difference causes constructive interference, but constructive interference requires phase differences of 0, 2π, 4π, etc. (even multiples of π), while odd multiples of π cause destructive interference. The strategy is to remember that phase differences of nπ produce constructive interference when n is even and destructive interference when n is odd.
A string fixed at both ends shows a standing wave with nodes at $x=0$ and $x=L$ and antinodes at $x=\frac{L}{4}$ and $x=\frac{3L}{4}$. This pattern remains stationary while the string oscillates. Which statement about the nodes is correct?
Nodes are points where the string’s displacement is always maximum.
Nodes are points of zero displacement at all times due to destructive interference.
Energy disappears at nodes, so the wave loses power each cycle.
Nodes move back and forth along the string as the wave oscillates.
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.
Two coherent water-wave sources in a ripple tank are in phase and produce circular wavefronts. At point $Q$, the distance to source 1 is $0.30,\text{m}$ and to source 2 is $0.45,\text{m}$. The wavelength is $0.15,\text{m}$. Which interference occurs at $Q$?
No interference, because the waves cancel and then stop propagating.
Destructive interference, because the path-length difference is $1\lambda$.
Constructive interference, because the path-length difference is $1\lambda$.
Constructive interference, because the path-length difference is $\tfrac{1}{2}\lambda$.
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.
A string fixed at both ends vibrates at $f=120,\text{Hz}$ with 2 antinodes (one loop). Which statement best identifies the nodes and antinodes?
Antinodes at both ends and a node at the midpoint.
Nodes at both ends and an antinode at the midpoint.
Nodes at both ends and at the midpoint, with antinodes between them.
Nodes drift along the string, so their locations cannot be identified.
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.
In a two-slit interference experiment, point $R$ on a screen is at a path difference of $1.5\lambda$ from the slits. Which best describes the intensity at $R$?
A permanent dark spot, because once destructive interference occurs the light is eliminated.
Neither, because interference only occurs for sound waves, not light.
A minimum, because a half-integer multiple of $\lambda$ produces destructive interference.
A maximum, because any multiple of $\lambda$ produces constructive interference.
Explanation
This question tests understanding of wave interference and standing waves. A path difference of 1.5λ = (3/2)λ represents an odd multiple of λ/2, producing destructive interference and a minimum intensity at point R. This occurs because one wave travels an extra 1.5 wavelengths, arriving 180° out of phase with the other. Choice B incorrectly claims any multiple of λ produces constructive interference, missing that 1.5 is not an integer. Choice D shows the misconception that destructive interference permanently eliminates light, when actually the energy appears at bright fringes elsewhere. For two-slit interference, remember: integer multiples of λ give maxima, half-integer multiples give minima.
Two identical sinusoidal waves travel in opposite directions on a string, producing a standing wave with an antinode at $x=0.30\ \text{m}$. Which statement about that point is correct?
The point is motionless because energy is destroyed at antinodes.
The point moves along the string because antinodes travel at the wave speed.
The point oscillates with maximum amplitude because it is always constructive interference there.
The point always has zero displacement because antinodes are locations of destructive interference.
Explanation
This question tests understanding of wave interference and standing waves. An antinode is a point of permanent constructive interference where the two oppositely traveling waves always add constructively, resulting in maximum amplitude oscillation. The point oscillates up and down with twice the amplitude of either individual wave. Choice A incorrectly claims antinodes have zero displacement, confusing them with nodes. Choice D wrongly suggests antinodes move along the string, but standing wave patterns are stationary with fixed node and antinode positions. Remember that in standing waves, energy oscillates between kinetic and potential forms at antinodes rather than propagating.
A string of length $1.20\ \text{m}$ is fixed at both ends and driven at $60\ \text{Hz}$ until a standing wave forms with three antinodes and nodes at both ends. Which statement correctly describes the standing-wave structure on the string?
The string has $3$ loops, so $L=\tfrac{3\lambda}{2}$ and adjacent nodes are separated by $\tfrac{\lambda}{2}$.
Because of destructive interference at nodes, the string cancels permanently and no antinodes can exist.
The nodes move back and forth along the string as time passes, carrying energy to the ends.
At each node the energy of the wave disappears, so the string is motionless everywhere after a short time.
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.
A standing wave on a string has adjacent nodes separated by $0.30,\text{m}$. Which wavelength $\lambda$ corresponds to this standing wave?
$\lambda = 0.30,\text{m}$ because node-to-node spacing is one wavelength.
$\lambda = 0.60,\text{m}$ because node-to-node spacing is $\lambda/2$.
$\lambda = 1.20,\text{m}$ because nodes are where energy accumulates.
$\lambda = 0.15,\text{m}$ because nodes move with time so spacing halves.
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.