Summary to follow. 6 syllabus statements · 22 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Standing wave
Standing wave
A wave pattern formed by the superposition of two identical travelling waves (same frequency, wavelength, amplitude and speed) moving in opposite directions through the same medium, usually an incident wave and its reflection. The pattern does not travel: the positions of the nodes and antinodes are fixed. Every point between the nodes oscillates, with an amplitude that depends on its position.
Principle of superposition
When two waves meet at a point, the resultant displacement at that point is the vector sum of the displacements that each wave would produce on its own. In a standing wave the two travelling waves pass through each other unchanged; their displacements simply add at every point and every instant.
Energy in a standing wave
The two travelling waves carry energy at equal rates in opposite directions, so a standing wave produces no net transfer of energy along the medium. The energy stays within each section between nodes, changing between kinetic energy (greatest when the medium passes through its undisplaced position) and elastic potential energy (greatest at maximum displacement). Nodes never move, so no energy passes through them.
Students often think A standing wave is one wave that has stopped moving, so its crests and troughs stay at fixed places on the string. In fact No. A standing wave is the superposition of two identical travelling waves moving in opposite directions. Each wave still travels; it is the pattern of nodes and antinodes that stays in place.
Students often think A standing wave carries energy along the medium from the source to the far end, as every wave does. In fact No. The two travelling waves carry energy at equal rates in opposite directions, so there is no net transfer of energy along the medium. The energy stays in each section between nodes.
Node (displacement node)
Node (displacement node)
A point on a standing wave at which the displacement is zero at all times, because the two travelling waves always arrive there in antiphase and cancel. Adjacent nodes are half a wavelength (λ/2) apart.
Antinode (displacement antinode)
A point on a standing wave at which the amplitude of oscillation is greatest. If each travelling wave has amplitude A, the amplitude at an antinode is 2A. Adjacent antinodes are λ/2 apart, and each antinode lies midway between two nodes, λ/4 from each.
Relative amplitude along a standing wave
In a standing wave, unlike a travelling wave, the amplitude varies with position: it is zero at the nodes, greatest (2A) at the antinodes and takes intermediate values in between. Points at the same distance from a node have equal amplitudes.
Phase difference in a standing wave
All points between two adjacent nodes oscillate in phase: they reach their maximum displacements at the same instant. Points in adjacent sections (on opposite sides of a node) oscillate in antiphase, a phase difference of π rad. The phase does not change steadily with distance as it does along a travelling wave.
Students often think The phase difference between two points on a standing wave is 2πx/λ for points a distance x apart, as for a travelling wave. In fact No. All points between two adjacent nodes oscillate in phase, and points in adjacent sections oscillate in antiphase (π rad). The phase changes abruptly by π at each node, not steadily with distance.
Students often think Every point on a standing wave oscillates with the same amplitude, as the points on a travelling wave do. In fact No. The amplitude varies with position: zero at the nodes, greatest at the antinodes, and intermediate between them.
Boundary conditions for strings
Boundary conditions for strings
A fixed end of a string cannot move, so it is always a displacement node. A free end (for example, a light ring sliding on a smooth rod) moves with the greatest amplitude, so it is a displacement antinode. The three cases are two fixed boundaries (node at each end), one fixed and one free boundary (node at one end, antinode at the other), and two free boundaries (antinode at each end).
Boundary conditions for air in pipes
Sound in air is longitudinal, so the air oscillates along the length of the pipe. At a closed end the air cannot move along the pipe, so a closed end is a displacement node. At an open end the air moves with the greatest amplitude, so an open end is a displacement antinode. Sound reflects at both kinds of end, so standing waves form in pipes with two closed ends, one closed and one open end, or two open ends.
First harmonic and harmonics
The first harmonic is the lowest-frequency mode of a standing wave in a given string or pipe. A higher mode whose frequency is n times that of the first harmonic is the nth harmonic. Which values of n exist depends on the boundary conditions.
Harmonics with the same kind of boundary at both ends
For a string with two fixed or two free ends, or air in a pipe with two closed or two open ends, the length L holds a whole number of half-wavelengths. The nth harmonic has λ_n = 2L/n and f_n = v/λ_n = nv/(2L), where v is the wave speed, and every whole number n = 1, 2, 3, … occurs.
Harmonics with different boundaries at the two ends
For a string with one fixed and one free end, or air in a pipe closed at one end and open at the other, there is a node at one end and an antinode at the other, so L holds an odd number of quarter-wavelengths. The nth harmonic has λ_n = 4L/n and f_n = nv/(4L) with n odd only (1, 3, 5, …). The mode above the first harmonic is therefore the third harmonic, at three times the first-harmonic frequency.
Students often think Every end of a vibrating string or air column is a node, as the ends of a guitar string are. In fact No. Only a boundary where the medium cannot move is a node: a fixed end of a string, or a closed end of a pipe. A free end of a string and an open end of a pipe are displacement antinodes.
Students often think The closed end of a pipe is a displacement antinode, because the air hits the closed end hardest. In fact No. The air at a closed end cannot move along the pipe, so a closed end is always a displacement node.
Natural frequency
Natural frequency
The frequency at which a system oscillates when it is displaced and released and left to oscillate freely, with no periodic driving force. SI unit: Hz. A string or air column has a whole series of natural frequencies, one for each harmonic.
Driving frequency and forced oscillation
When a periodic external force acts on a system, the system is forced to oscillate. Once the initial transient motion has died away, it oscillates at the driving frequency (the frequency of the external force), not at its own natural frequency. The amplitude of this steady oscillation depends on how close the driving frequency is to the natural frequency.
Resonance
The large increase in the amplitude of a forced oscillation that occurs when the driving frequency equals (for a lightly damped system, very nearly equals) a natural frequency of the system. The driving force then does positive work on the system throughout each cycle, so energy builds up until the energy supplied per cycle equals the energy dissipated per cycle by damping.
Useful and destructive effects of resonance
Useful: the tuned circuit of a radio receiver is adjusted so its natural frequency matches one station's frequency; a quartz crystal in a clock or watch is kept oscillating at its natural frequency to keep time; the air columns and strings of musical instruments resonate to produce loud notes. Destructive: a glass shattered by a sustained note at its natural frequency; buildings shaken by earthquake waves at a natural frequency of the building; machine parts, vehicle panels or washing machines that vibrate violently at particular engine or drum speeds; bridges and walkways set swaying by periodic forces from wind or walkers.
Students often think A system that is driven by a periodic force still oscillates at its own natural frequency; the driving force only feeds it energy. In fact No. After the transient motion dies away, a driven system oscillates at the driving frequency. Its natural frequency determines how large the amplitude is, not the frequency of the steady oscillation.
Students often think The faster a system is pushed (the higher the driving frequency), the more energy it receives and the larger its amplitude becomes. In fact No. The amplitude is greatest when the driving frequency is close to the natural frequency and becomes smaller when the driving frequency is increased further.
Damping
Damping
The removal of energy from an oscillating system by resistive forces (such as friction or air resistance), which transfer the energy to the internal energy of the system and its surroundings. For a driven oscillator, damping limits the amplitude at resonance.
Effect of damping on the frequency response
A graph of steady amplitude against driving frequency (the frequency response) has a peak near the natural frequency. As the damping increases, the maximum amplitude decreases, the peak becomes wider, and the frequency at which it occurs moves slightly lower. Far from the natural frequency the amplitude is almost unaffected by damping.
Students often think Damping makes a system stiffer or more resistant, so the frequency at which it resonates increases. In fact No. Increasing the damping moves the frequency of maximum amplitude slightly lower, as well as lowering and widening the peak.
Students often think Damping only affects how quickly a free oscillation dies away; when a driving force keeps supplying energy, the damping makes no difference to the steady amplitude. In fact No. Damping sets the maximum amplitude: the amplitude grows until the energy dissipated per cycle equals the energy supplied per cycle, and with more damping this balance is reached at a smaller amplitude.
Light damping
Light damping
Damping small enough that the displaced system oscillates about equilibrium with an amplitude that decreases gradually over many cycles. The time period stays almost constant as the amplitude decreases (it is very slightly longer than with no damping).
Critical damping
The smallest degree of damping at which a displaced system returns to equilibrium without oscillating. It brings the system to equilibrium in the shortest possible time without passing through the equilibrium position; it is used in car suspensions, door closers and meter needles.
Heavy damping
Damping greater than critical. The displaced system does not oscillate but returns to equilibrium slowly, taking longer than with critical damping; the heavier the damping, the slower the return.
Students often think The more damping there is, the faster a displaced system comes back to rest, so heavy damping is quicker than critical damping. In fact No. Critical damping gives the quickest return to equilibrium without oscillation. Heavy damping (more than critical) makes the return slower, and the heavier the damping, the slower it is.
Students often think With critical damping the system passes through equilibrium once, overshooting, and then comes to rest there. In fact No. A critically damped system does not pass through the equilibrium position at all. It returns to equilibrium in the shortest possible time without oscillating.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Which statement about a standing wave on a string is correct?
Answer and reasoning
It forms where two identical waves travelling in opposite directions superpose. — A standing wave is the superposition of two waves with the same frequency, wavelength, amplitude and speed moving in opposite directions, on a string usually the incident wave and its reflection. Their displacements add at every point, producing fixed nodes and antinodes.
It is a single wave that has stopped travelling, so its crests stay in fixed places. — A student who takes "standing" literally picks this. No crest stays in place: each antinode moves from a crest to a trough and back every cycle. The pattern is fixed because two travelling waves pass through each other.
It carries energy along the string from the source towards the far end. — A student who applies "waves transfer energy" to every wave picks this. The two component waves carry energy at equal rates in opposite directions, so a standing wave has no net energy transfer along the string.
Each point oscillates slightly later than its neighbour nearer the source. — A student who carries the travelling-wave idea of a steadily changing phase over to standing waves picks this. All points between two adjacent nodes oscillate in phase, and points on opposite sides of a node are in antiphase.
2 A vibrator of frequency 320 Hz sets up a standing wave on a stretched string. The distance between adjacent nodes is 0.45 m. What is the speed of the travelling waves on the string?
Answer and reasoning
144 m s⁻¹ — A student who takes one loop (node to node) as a whole wavelength uses λ = 0.45 m and gets 320 × 0.45 = 144 m s⁻¹. Each loop is half a wavelength, so λ = 0.90 m.
288 m s⁻¹ — Adjacent nodes are half a wavelength apart, so λ = 2 × 0.45 = 0.90 m. v = fλ = 320 × 0.90 = 288 m s⁻¹.
576 m s⁻¹ — A student who thinks adjacent nodes are λ/4 apart uses λ = 4 × 0.45 = 1.80 m and gets 576 m s⁻¹. A quarter-wavelength is the node-to-antinode distance; adjacent nodes are λ/2 apart.
356 m s⁻¹ — A student who writes v = f/λ gets 320/0.90 = 356. The units of f/λ are s⁻¹ m⁻¹, not m s⁻¹. The wave equation is v = fλ = 288 m s⁻¹.
Working Adjacent nodes are λ/2 apart: λ = 2 × 0.45 m = 0.90 m. v = fλ = 320 Hz × 0.90 m = 288 m s⁻¹.
3 A pipe is closed at one end and open at the other. The air in the pipe vibrates in its first harmonic. Where along the pipe is the displacement of the air zero at all times?
Answer and reasoning
At the closed end of the pipe — A point where the displacement is zero at all times is a displacement node. The air at the closed end cannot move along the pipe, so it is a node; the air at the open end moves with the greatest amplitude, so it is an antinode. The first harmonic is the simplest pattern with a node at one end and an antinode at the other (L = λ/4), so the closed end is the only node.
Midway between the two ends — A student who pictures the air hitting the closed end hardest treats that end as an antinode, so has antinodes at both ends and puts the node between them. The solid end stops the air next to it moving, so the closed end is a displacement node, and the open end is the only antinode.
At both ends of the pipe — A student who thinks every end is a node, as for a guitar string, picks this. Only the closed end is a node. The air at the open end is free to move and oscillates with the greatest amplitude: it is a displacement antinode.
At no point along the pipe — A student who thinks every point of a standing wave oscillates with the same amplitude expects no point to be still. The amplitude varies along the pipe: zero at the closed end, where the air cannot move (a node), and greatest at the open end (an antinode).
4 A radio receiver contains a circuit whose natural frequency can be adjusted. Signals from many stations, each broadcasting at a different frequency, reach its aerial at the same time. How does the receiver pick out one station?
Answer and reasoning
It responds most strongly to the strongest signal, whatever its frequency. — A student who thinks the size of the response depends only on the size of the driving force picks this. A tuned circuit responds strongly to a signal at its natural frequency, even if that signal is weaker than others.
It responds most strongly to the signal with the highest frequency reaching it. — A student who thinks a higher driving frequency always produces a larger amplitude picks this. The response is greatest when the signal frequency matches the circuit's natural frequency, not when it is highest.
It oscillates at its own natural frequency, whatever signals reach the aerial. — A student who thinks a driven system always oscillates at its natural frequency picks this. The circuit is driven by the signals and oscillates at their frequencies; it responds strongly only to the one that matches its natural frequency.
Its natural frequency is set equal to that station's frequency, so it resonates with it. — This is a useful application of resonance. The circuit responds with a large amplitude only to a signal at (or very near) its natural frequency, so adjusting that frequency to match one station selects it while the others produce only small responses.
5 A lightly damped oscillator is driven by a periodic force of constant amplitude, and its steady amplitude is measured over a range of driving frequencies. The damping is then increased slightly, and the measurements are repeated. Which describes the new response curve compared with the first?
Answer and reasoning
The maximum amplitude decreases, and it occurs at a slightly higher frequency. — A student who thinks damping makes a system stiffer picks this. Damping opposes the velocity; it does not act like a stiffer spring. The frequency of maximum amplitude moves slightly lower, not higher.
The maximum amplitude decreases, and it occurs at a slightly lower frequency. — With more damping, more energy is dissipated per cycle, so the energy balance at resonance is reached at a smaller amplitude. The peak also moves slightly to a lower frequency and becomes wider. This holds provided the damping is still light, as it is after a slight increase.
The maximum amplitude is unchanged, and it occurs at the same frequency. — A student who thinks the driver simply makes up any losses picks this. The maximum amplitude is reached when the energy supplied per cycle equals the energy dissipated per cycle; with more damping this happens at a smaller amplitude.
The amplitude falls by the same factor at every frequency; the peak frequency is unchanged. — A student who thinks damping scales the whole response curve down picks this. Damping acts mainly near resonance: far from the natural frequency the amplitude is almost unchanged, while the peak becomes lower and wider and moves to a slightly lower frequency.
6 A mass–spring system is displaced from equilibrium and released. Which statement about the effect of damping on its motion is correct?
Answer and reasoning
Heavy damping: no oscillation, but a slower return to equilibrium than with critical damping. — Heavy damping is greater than critical damping. The system does not oscillate, but the large resistive force slows its return, so it takes longer to reach equilibrium than a critically damped system does.
Heavy damping: no oscillation, and a quicker return to equilibrium than with critical damping. — A student who thinks more damping always stops the system sooner picks this. Critical damping gives the fastest return without oscillation; heavier damping slows the return.
Critical damping: it passes through equilibrium once, then comes to rest at equilibrium. — A student who thinks critical damping allows a single swing picks this. A critically damped system does not pass through equilibrium at all; any overshoot means the damping is light.
Light damping: the time period decreases steadily as the amplitude of oscillation decreases. — A student who thinks a smaller oscillation takes less time picks this. With light damping the amplitude decreases while the time period stays almost constant.
7 A string fixed at both ends vibrates in its first harmonic. At one instant every point on the string has zero displacement, so the string is straight. Which statement about the string at this instant is correct?
Answer and reasoning
Every point is momentarily at rest; its energy is all stored as elastic potential energy. — A student who thinks zero displacement means zero velocity picks this. Rest occurs at maximum displacement. A point passing through its undisplaced position is moving at its greatest speed, and a straight string has no extra stretch to store energy.
The two waves have cancelled each other out, so the string holds no energy at this instant. — A student who thinks waves that cancel have destroyed each other picks this. Superposition adds displacements, not energies: the velocities do not cancel, so the energy is all kinetic, and the string is displaced again a quarter of a period later.
Energy is being carried along the string from one fixed end to the other. — A student who thinks every wave transfers energy along the medium picks this. The two component waves carry energy equally in opposite directions, so the standing wave has no net energy transfer; the energy stays in the vibrating string.
Points between the ends are moving at their greatest speeds; its energy is all kinetic. — Each point on the string oscillates about its undisplaced position and passes through it at its greatest speed. When the whole string is straight, every point except the fixed ends is at maximum speed, so the energy of the standing wave is all kinetic energy at this instant.
8 A string fixed at both ends vibrates in a standing wave. Its nodes are at x = 0, 0.40 m, 0.80 m and 1.20 m, measured from one end. Points P, Q and R on the string are at x = 0.10 m, 0.20 m and 0.70 m. Which statement about the oscillations of P, Q and R is correct?
Answer and reasoning
P and Q oscillate π/4 rad out of phase, being λ/8 apart. — A student who uses the travelling-wave rule Δφ = 2πx/λ picks this: λ = 0.80 m and x = 0.10 m give π/4 rad. In a standing wave, P and Q lie between the same pair of nodes (0 and 0.40 m), so they oscillate in phase.
P and Q oscillate in phase, with amplitudes equal to each other. — A student who thinks every point of a standing wave has the same amplitude picks this. P and Q are in phase, but Q is an antinode (maximum amplitude) while P, halfway between a node and the antinode, has a smaller amplitude.
P and R oscillate in antiphase, with equal amplitudes. — P (0.10 m) is 0.10 m from the node at 0, and R (0.70 m) is 0.10 m from the node at 0.80 m, so they have equal amplitudes. P lies between the nodes at 0 and 0.40 m, and R between the nodes at 0.40 m and 0.80 m: they are in adjacent sections, on opposite sides of the node at 0.40 m, so they oscillate in antiphase (phase difference π rad).
P and R oscillate in phase, with equal amplitudes. — A student who thinks the whole standing wave moves up and down together picks this. P and R do have equal amplitudes (each is 0.10 m from a node), but they are on opposite sides of the node at 0.40 m, so when P is displaced upwards R is displaced downwards: they are in antiphase.
9 A string of length 0.80 m is fixed at both ends. The speed of waves on the string is 240 m s⁻¹. What is the frequency of the third harmonic?
Answer and reasoning
450 Hz — With a node at each fixed end, the third harmonic has three loops, each λ/2 long: λ₃ = 2L/3 = 2 × 0.80/3 = 0.533 m. f₃ = v/λ₃ = 240/0.533 = 450 Hz (equivalently 3 × 150 Hz).
900 Hz — A student who takes each loop as a whole wavelength uses λ = 0.80/3 = 0.267 m and gets 900 Hz. Each loop is half a wavelength, so λ₃ = 2L/3 = 0.533 m and f₃ = 450 Hz.
300 Hz — A student who thinks the third harmonic has three nodes, counting the ends, draws two loops and gets λ = 0.80 m, f = 300 Hz. That pattern is the second harmonic. The third harmonic has three loops and four nodes.
150 Hz — A student who thinks a string has only one natural frequency gives the first-harmonic value, 240/1.60 = 150 Hz. The third harmonic has one-third of the wavelength and, at the same wave speed, three times the frequency: 450 Hz.
Working Two fixed ends: λ_n = 2L/n. λ₃ = 2 × 0.80 m/3 = 0.533 m. f₃ = v/λ₃ = 240 m s⁻¹/0.533 m = 450 Hz. (Check: f₁ = v/2L = 240/1.60 = 150 Hz, and f₃ = 3f₁ = 450 Hz.)
10 A lightly damped mass–spring system has a natural frequency of 2.0 Hz. It is driven by a periodic force of constant amplitude with a frequency of 3.5 Hz. Which describes its oscillation once the amplitude has become steady?
Answer and reasoning
At 3.5 Hz, with a smaller amplitude than when driven at 2.0 Hz — In the steady state a forced oscillator moves at the driving frequency, 3.5 Hz. The amplitude is greatest when the driving frequency is close to the natural frequency (2.0 Hz, resonance) and smaller away from it.
At 2.0 Hz, its natural frequency, with a small steady amplitude — A student who thinks a driven system keeps oscillating at its natural frequency picks this. Once the transient dies away, the system follows the driving force and oscillates at 3.5 Hz.
At 3.5 Hz, with a larger amplitude than when it is driven at 2.0 Hz — A student who thinks faster pushing always gives a larger amplitude picks this. Above the natural frequency the force is increasingly out of step with the motion; the amplitude is largest at resonance, near 2.0 Hz.
At 3.5 Hz, with the same amplitude as when it is driven at 2.0 Hz — A student who thinks the amplitude depends only on the size of the driving force picks this. For a force of fixed size, the amplitude depends strongly on the driving frequency and is greatest at resonance.
Working Steady forced oscillation is at the driving frequency, 3.5 Hz. Driving at 2.0 Hz equals the natural frequency (resonance), so the amplitude at 3.5 Hz is smaller than at 2.0 Hz.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
12 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A string of length 0.60 m is fixed at one end. Its other end is free: it is tied to a light ring that slides without friction on a vertical rod. The speed of waves on the string is 240 m s⁻¹. Which gives the harmonic number n, the wavelength λ and the frequency f of the standing wave with the next frequency above that of the first harmonic?
Answer and reasoning
n = 2; λ = 1.20 m; f = 200 Hz — A student who assumes every system has harmonics at all whole-number multiples of f₁ gives 2 × 100 = 200 Hz. With a node at one end and an antinode at the other, L must hold an odd number of quarter-wavelengths, so there is no mode at 2f₁.
n = 3; λ = 0.40 m; f = 600 Hz — A student who takes one loop of the pattern as a whole wavelength identifies the mode correctly but reads L = 3λ/4 (one and a half loops) as 1.5λ, so uses λ = 0.60/1.5 = 0.40 m and gets f = 240/0.40 = 600 Hz. Each loop is half a wavelength, so λ = 4L/3 = 0.80 m and f = 300 Hz.
n = 2; λ = 0.80 m; f = 300 Hz — A student who numbers harmonics in order finds the correct wavelength and frequency but calls the mode the second harmonic. The harmonic number is the ratio f/f₁ = 300/100 = 3, so this is the third harmonic; there is no second harmonic.
n = 3; λ = 0.80 m; f = 300 Hz — There is a node at the fixed end and an antinode at the free end, so L holds an odd number of quarter-wavelengths. The first harmonic has λ = 4L = 2.40 m and f₁ = 100 Hz. The next mode has L = 3λ/4, so λ = 0.80 m and f = 240/0.80 = 300 Hz = 3f₁: the third harmonic.
Working One fixed end (node) and one free end (antinode): λ_n = 4L/n with n odd. First harmonic: λ₁ = 4 × 0.60 m = 2.40 m, f₁ = 240/2.40 = 100 Hz. Next mode: n = 3, λ₃ = 4 × 0.60/3 = 0.80 m, f₃ = 240/0.80 = 300 Hz.
2 A pipe of length 0.68 m is closed at one end and open at the other. The speed of sound in the air in the pipe is 340 m s⁻¹. Which gives, in increasing order, the lowest frequencies at which the air in the pipe resonates?
Answer and reasoning
f = 125, 250 and 375 Hz — A student who assumes all whole-number harmonics exist, as for a string fixed at both ends, gives 125, 250, 375 Hz. With a node at one end and an antinode at the other, only odd multiples of 125 Hz can occur.
f = 125, 375 and 625 Hz — A closed end is a displacement node and an open end an antinode, so L = λ/4, 3λ/4, 5λ/4, … The first harmonic has λ = 4 × 0.68 = 2.72 m and f₁ = 340/2.72 = 125 Hz; only odd harmonics exist, giving 125, 375 and 625 Hz.
f = 250, 500 and 750 Hz — A student who treats the closed end as a displacement antinode has an antinode at each end, so uses λ₁ = 2L = 1.36 m and gets 250, 500, 750 Hz. The closed end is a node, so λ₁ = 4L and f₁ = 125 Hz.
f = 125 Hz and no other — A student who thinks a pipe has a single natural frequency gives only 125 Hz. The air column has a series of natural frequencies, one for each odd harmonic: 125, 375, 625 Hz, …
Working Closed end: node; open end: antinode. λ_n = 4L/n, n odd. λ₁ = 4 × 0.68 m = 2.72 m; f₁ = 340 m s⁻¹/2.72 m = 125 Hz. f₃ = 3 × 125 = 375 Hz; f₅ = 5 × 125 = 625 Hz.
3 A singer shatters a wine glass by singing one steady note close to it. Which condition is essential for this to happen?
Answer and reasoning
The note is very loud, whatever frequency the singer chooses. — A student who thinks a large vibration needs only a large force picks this. A loud note at the wrong frequency drives the glass out of step with its natural vibration, so the amplitude stays small; the frequency must match.
The note is the highest frequency that the singer is able to produce. — A student who thinks a higher driving frequency always gives a larger amplitude picks this. The amplitude is greatest when the driving frequency equals a natural frequency of the glass, not when it is as high as possible.
The note's frequency is equal to a natural frequency of the glass. — This is resonance. At a natural frequency of the glass, the sound pushes the glass in step with its own vibration, so energy builds up cycle after cycle until the amplitude is large enough to break the glass.
The sound echoes inside the bowl of the glass and builds up. — A student who uses "resonance" in its everyday sense of echoing picks this. Reflection of sound is not what breaks the glass; the glass itself is driven at one of its natural frequencies.
4 The same oscillator is driven by a periodic force of the same constant amplitude in two experiments, X and Y, with different damping. In X the maximum amplitude is 25 mm, at 5.00 Hz, and the amplitude falls to half this value at 4.82 Hz and 5.17 Hz. In Y the maximum amplitude is 3.4 mm, at 4.89 Hz, and the amplitude falls to half this value at 3.3 Hz and 6.1 Hz. Which conclusion do the data support?
Answer and reasoning
X is more heavily damped, because its peak is at the higher frequency. — A student who thinks damping raises the resonant frequency picks this. Damping moves the peak slightly lower, and X's tall, narrow peak shows light damping; Y is the more heavily damped.
Only X shows resonance; Y's damping is too heavy for resonance to occur. — A student who thinks resonance needs an unlimited amplitude picks this. Y's amplitude has a clear maximum of 3.4 mm at 4.89 Hz and falls to half of it on either side: Y resonates too, with a lower, wider peak.
Y is more heavily damped, so at 1.0 Hz its amplitude is also only about a seventh of X's. — A student who thinks damping scales the whole response curve down by the peak ratio (25/3.4 ≈ 7) picks this. Damping acts mainly near resonance, where it limits the amplitude; far below resonance the amplitude is set mainly by the spring and the size of the force, so at 1.0 Hz the two amplitudes would be almost the same.
Y is more heavily damped: its peak is lower, wider and at a slightly lower frequency. — All three features of Y point to heavier damping: a lower maximum amplitude (3.4 mm against 25 mm), a much wider peak (2.8 Hz between the half-maximum points against 0.35 Hz), and a peak at a slightly lower frequency (4.89 Hz against 5.00 Hz).
5 A door closer damps the motion of a heavy door. With the damping first set low, the door, released from wide open, swings past the closed position and back several times before settling. The damping is then increased in small steps. Which describes how the door's motion changes?
Answer and reasoning
The swinging past stops at critical damping; beyond it, the door closes even more quickly. — A student who thinks more damping always brings a system to rest sooner picks this. Beyond critical damping the resistive force slows the door's return, so it takes longer to close.
The swinging past stops at critical damping; beyond it, the door closes more slowly. — As the damping increases, the oscillation dies away faster until, at critical damping, the door no longer swings past the closed position and reaches it in the shortest time. Heavier damping than this makes the door creep shut more slowly.
At critical damping the door swings past just once, then stops at the closed position. — A student who thinks critical damping allows one swing picks this. At critical damping the door does not pass the closed position at all; a door that swings past even once is still lightly damped.
The door keeps swinging past, by less each time, however much damping is added to it. — A student who has only seen lightly damped oscillations picks this. At and beyond critical damping the door returns to the closed position without swinging past it.
6 A pipe is open at both ends. The air in the pipe vibrates in its first harmonic. Which statement about the motion of the air is correct?
Answer and reasoning
The air at the midpoint of the pipe does not oscillate. — Each open end is a displacement antinode and, in the first harmonic of a pipe open at both ends (L = λ/2), there is a single displacement node midway along the pipe. At the node the displacement of the air is zero at all times, so the air there does not oscillate.
The air at each open end oscillates side to side across the pipe. — A student who reads the standing-wave diagram as a picture of the air's motion picks this. The open ends are indeed antinodes, but the curves drawn across the pipe only show how large the displacement is at each position. Sound is longitudinal, so the air oscillates along the pipe's axis, not across it.
The air at each open end of the pipe does not oscillate. — A student who thinks every end is a node, as the ends of a guitar string are, picks this. The air at an open end is free to move, so each open end is a displacement antinode, where the air oscillates with the greatest amplitude. The node of the first harmonic is midway along the pipe.
The air oscillates with the same amplitude all along the pipe. — A student who thinks every point on a standing wave has the same amplitude picks this. The amplitude varies along the pipe: greatest at the open ends (antinodes) and zero at the midpoint (the node), where the air does not oscillate.
7 The diagram shows the envelope of a standing wave on a string of length L = 1.20 m that is fixed at both ends. What is the wavelength of the waves on the string?
Answer and reasoning
0.40 m — A student who takes one loop, from a node to the next node, as a whole wavelength picks this: 1.20 m / 3 = 0.40 m. One loop is only half a wavelength, because the string bulges the other way in the next loop; a whole wavelength spans two loops, so λ = 0.80 m.
0.80 m — Three loops fit between the fixed ends. Adjacent nodes are half a wavelength apart, so each loop is λ/2: 1.20 m = 3 × λ/2, giving λ = 0.80 m. This is the third harmonic of the string.
0.60 m — A student who counts the four nodes (including the ends) and calls the pattern the fourth harmonic, with λ = 2L/4, picks this. The harmonic number is the number of loops, three, not the number of nodes: λ = 2L/3 = 0.80 m.
1.60 m — A student who thinks adjacent nodes are a quarter of a wavelength apart picks this: 4 × 0.40 m = 1.60 m. It is a node and the next antinode that are λ/4 apart; adjacent nodes are λ/2 apart, so λ = 2 × 0.40 m = 0.80 m.
Working The envelope has three loops between the fixed ends. Each loop (node to node) is half a wavelength, so L = 3λ/2 and λ = 2L/3 = 2 × 1.20 m / 3 = 0.80 m.
8 The diagram shows the displacement envelope of the air in a pipe of length L = 0.85 m that is closed at one end and open at the other. The speed of sound in the air is 340 m s⁻¹. What is the frequency of the sound?
Answer and reasoning
600 Hz — A student who takes the full loop between the two nodes as a whole wavelength (and the last quarter of the pattern as half a wavelength) gets L = 3λ/2, λ = 0.567 m and f = 600 Hz. A loop from node to node is half a wavelength, so L = 3λ/4 and f = 300 Hz.
300 Hz — The envelope shows a node at the closed end, an antinode a quarter-wavelength in, a node at the half-wavelength point and an antinode at the open end: the pipe holds three quarter-wavelengths, so λ = 4L/3 = 1.13 m and f = 340/1.13 = 300 Hz.
200 Hz — A student who counts the two nodes and calls the pattern the second harmonic, f = 2 × v/(4L), gets 200 Hz. A pipe closed at one end has only odd harmonics; with three quarter-wavelengths in the pipe this is the third harmonic, f = 3 × v/(4L) = 300 Hz.
385 Hz — A student who multiplies the speed by the wavelength, f = vλ = 340 × 1.13, gets 385 Hz. Frequency is speed divided by wavelength: f = v/λ = 340/1.13 = 300 Hz.
Working The closed end is a displacement node and the open end an antinode. The pattern has a node at the closed end, one more node inside the pipe and an antinode at the open end, so L = 3λ/4: λ = 4L/3 = 4 × 0.85 m / 3 = 1.13 m. f = v/λ = 340 m s⁻¹ / 1.133 m = 300 Hz (the third harmonic of the pipe).
9 The diagram shows the envelope of a standing wave on a string of length L, driven by a vibrator at its left end and fixed at its right end. The amplitude of the vibrator is so small that both ends of the string can be treated as nodes. P and Q are points on the string. At one instant, P has its maximum displacement, upward. Which describes Q at that instant?
Answer and reasoning
Q is displaced upward by exactly the same amount as P. — A student who thinks the whole string moves up and down together picks this. Points in adjacent loops are in antiphase: while the left-hand loop bulges upward the right-hand loop bulges downward, so Q is displaced downward when P is displaced upward.
Q is moving through its equilibrium position. — A student who applies the travelling-wave rule Δφ = 2πx/λ picks this: P and Q are λ/4 apart (λ = L here), giving a phase difference of π/2, so Q would be at zero displacement when P is at its maximum. In a standing wave all points within one loop are in phase and points in adjacent loops are in antiphase, so Q is at its maximum displacement, downward, at that instant.
Q is displaced upward but by a smaller amount than P. — A student who pictures the wave weakening as it travels away from the vibrator, with the string rising and falling as one, picks this. The amplitude of a point on a standing wave depends only on its distance from the nearest node: P and Q are the same distance from the midpoint node, so their amplitudes are equal, and because they lie in adjacent loops they oscillate in antiphase, so Q is displaced downward by the same amount as P is displaced upward.
Q is displaced downward by the same amount as P. — P and Q are the same distance from the node at the midpoint, so they have equal amplitudes, and they lie in adjacent loops, on opposite sides of that node, so they oscillate in antiphase. When P is at its maximum displacement upward, Q is at its maximum displacement downward, and the two displacements are equal in size.
10 The graph shows how the steady amplitude of a lightly damped mass–spring system depends on the frequency of the periodic force that drives it. The amplitude of the driving force is kept constant while its frequency is reduced slowly from 8.0 Hz to 2.0 Hz. Which describes the amplitude of the mass–spring system during this change?
Answer and reasoning
It falls throughout, because a slower driving force transfers less energy to the system. — A student who thinks a faster driver always gives a bigger response picks this. The amplitude depends on how close the driving frequency is to the natural frequency, not on how high it is: the graph rises as the frequency comes down towards 5 Hz and only falls once the frequency is below that.
It stays the same throughout, because the amplitude of the driving force does not change. — A student who thinks the size of the response is set only by the size of the force picks this. With the same force amplitude the response varies strongly with frequency, as the graph shows: small far from 5 Hz and largest at resonance.
It rises to a maximum as the driving frequency passes about 5 Hz, then falls again. — The graph peaks at about 5 Hz, the natural frequency of the system. As the driving frequency falls from 8.0 Hz it approaches the natural frequency and the amplitude grows, is largest when the driving frequency passes about 5 Hz (resonance), and then decreases again as the driving frequency moves further below the natural frequency towards 2.0 Hz.
It rises to a maximum near 5 Hz then stays there, as the system now oscillates at its natural frequency. — A student who thinks a driven system settles at its own natural frequency picks this: once the driver has passed 5 Hz the system is imagined to keep resonating on its own. After the transient dies away a driven system oscillates at the driving frequency, whatever that is; the graph shows that when the driving frequency has fallen to 2.0 Hz the steady amplitude is small again.
Working The peak of the response curve, read from the graph, is at about 5 Hz: this is the natural frequency of the system. Reducing the driving frequency from 8.0 Hz brings it towards 5 Hz, so the amplitude rises; it is greatest as the driving frequency passes about 5 Hz (resonance) and then falls as the driving frequency continues down to 2.0 Hz.
11 Curve X on the graph shows how the steady amplitude of a driven oscillator varies with driving frequency. One change is then made to the apparatus, and the measurements are repeated to give curve Y. Which change was made?
Answer and reasoning
The damping acting on the oscillator was made smaller. — A student who thinks damping makes a system stiffer, raising its resonant frequency, reads the lower peak frequency of Y as a sign of less damping. Damping lowers the peak frequency slightly, and Y's lower, broader peak confirms that the damping was increased.
The amplitude of the driving force was reduced. — A student who thinks damping cannot alter the steady amplitude of a driven system looks for another cause and blames the driver. A smaller driving force would scale the whole curve down, including the parts far from resonance, and would not shift or broaden the peak. Only Y's peak region differs from X, which is the effect of increased damping.
The natural frequency of the oscillator was lowered. — A student who thinks damping scales the whole response curve down rules damping out, because the curves coincide away from the peak, and attributes the small shift to a change in the natural frequency. A lower natural frequency would move the whole peak, not make it three times lower and much broader; a lower, broader peak at a slightly lower frequency is the signature of increased damping.
The damping acting on the oscillator was increased. — Curve Y has the three signs of heavier damping: a lower maximum amplitude, a wider peak, and a peak at a slightly lower frequency. Away from resonance the curves coincide, which is also expected, because damping matters most near resonance, where the amplitude is limited by the energy dissipated per cycle.
12 The graph shows displacement–time curves X, Y and Z for the same mass–spring system, released from rest at the same displacement with three different amounts of damping. Which statement about the curves is correct?
Answer and reasoning
Y shows heavy damping: the heaviest damping gives the quickest return to equilibrium. — A student who thinks more damping always brings a system to rest sooner picks this. Beyond critical damping the resistive force slows the return, which is why Z creeps back more slowly than Y. Y, the quickest non-oscillating return, shows critical damping and Z shows heavy damping.
X shows critical damping: it crosses equilibrium once and then settles at it. — A student who thinks a critically damped system overshoots once picks this. A critically damped system does not cross equilibrium at all. Any crossing, even a single small one, means the damping is less than critical, so X shows light damping.
Y shows critical damping: it reaches equilibrium in the least time without crossing it. — Critical damping is the smallest damping for which the system returns to equilibrium without oscillating, and it gives the quickest return. Y is on the axis sooner than Z and, unlike X, does not cross it, so Y is critically damped; X is lightly damped and Z heavily damped.
X shows heavy damping: even heavy damping leaves a small oscillation before rest. — A student who thinks every damped system oscillates a little before it stops picks this. With critical or heavy damping the system returns to equilibrium without crossing it, as Y and Z do; X, which crosses the axis and dips below it, is the lightly damped case.
That was your twenty minutes. Real practice on C.4 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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