C.4 Standing waves and resonance flashcards: IB Physics SL and HL

All 72 C.4 Standing waves and resonance cards

1. DefinitionSL & HL

   Define a standing (stationary) wave.

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   A wave formed by the superposition of two waves of the same frequency and wavelength, and similar amplitude, travelling in opposite directions along the same line; the resulting pattern has fixed points of permanently zero amplitude (nodes) and points of maximum amplitude (antinodes), and no net energy is transferred along the medium. Exam tip: the two marking points are 'superposition of two identical/same-frequency waves' and 'travelling in opposite directions'; answers such as 'a wave that does not move' score zero. Not a quantity, so no unit; the displacement of each particle is a vector.

2. DefinitionSL & HL

   Define a node on a standing wave.

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   A point on a standing wave where the displacement is always zero (permanently zero amplitude), because the two superposing waves always arrive there in antiphase (path difference (n + ½)λ) and interfere destructively. Exam tip: say 'always/permanently zero displacement' — 'zero displacement' alone is not enough because every particle passes through zero twice per cycle. Adjacent nodes are λ/2 apart. Unit of the position: m; displacement is a vector, amplitude a scalar.

3. DefinitionSL & HL

   Define an antinode on a standing wave.

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   A point on a standing wave where the amplitude of oscillation is a maximum (equal to twice the amplitude of each component wave for equal-amplitude waves), because the two superposing waves always arrive in phase (path difference nλ) and interfere constructively. Exam tip: 'maximum amplitude' earns the mark; 'maximum displacement' is loosely accepted but weaker. Adjacent antinodes are λ/2 apart; a node and the adjacent antinode are λ/4 apart. Amplitude is a scalar measured in m.

4. DefinitionSL & HL

   State the principle of superposition.

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   When two or more waves meet at a point, the resultant displacement at that point is the vector sum of the individual displacements each wave would produce alone. Exam tip: the mark is for 'sum of the displacements' — writing 'sum of the amplitudes' or 'sum of the intensities' loses it; say vector sum (or 'taking sign/direction into account') because displacements can cancel. After overlapping, the waves continue unchanged. Displacement is a vector, unit m; superposition underlies interference, beats and standing waves.

5. DefinitionSL & HL

   Define the fundamental (first harmonic) of a vibrating system.

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   The lowest frequency standing wave (mode of vibration) that can be set up in the system consistent with its boundary conditions; it has the longest wavelength and the smallest number of nodes and antinodes. Exam tip: the 2025 syllabus wording is 'first harmonic', so prefer that in answers ('fundamental' is still accepted), and never call the second harmonic the 'first overtone' in an IB answer unless the question uses that term. For a string of length L fixed at both ends, λ₁ = 2L and f₁ = v/2L. Unit: Hz (s⁻¹), scalar.

6. DefinitionSL & HL

   Define a harmonic (the nth harmonic) of a standing wave system.

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   A standing wave mode whose frequency is an integer multiple n of the first-harmonic frequency: f_n = n f₁. Exam tip: for a string fixed at both ends and for a pipe open at both ends all integers n = 1, 2, 3 … are allowed; for a pipe closed at one end only odd n = 1, 3, 5 … exist, so the 'third harmonic' of a closed pipe is 3f₁ and there is no second harmonic. Losing marks here almost always comes from numbering the modes 1, 2, 3 for a closed pipe. Unit: Hz, scalar.

7. DefinitionSL & HL

   Define natural frequency.

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   The frequency at which a system oscillates when it is displaced and then allowed to oscillate freely, with no external periodic driving force and negligible damping. Exam tip: the mark requires 'freely / no external driving force'; a system may have several natural frequencies (the harmonics). Symbol f₀, unit Hz (s⁻¹), scalar. Examples: a mass–spring system, a pendulum, a stretched string, an air column, an LC circuit.

8. DefinitionSL & HL

   Define forced (driven) oscillation and driving frequency.

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   A forced oscillation is one in which an external periodic force continuously supplies energy to a system, making it oscillate at the frequency of that force rather than at its own natural frequency. The driving frequency f is the frequency of that external periodic force. Exam tip: state that in the steady state the system oscillates AT the driving frequency — this is the commonly missed mark; the amplitude, not the frequency, depends on how close f is to f₀. Unit: Hz, scalar.

9. DefinitionSL & HL

   Define resonance.

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   Resonance occurs when a system is forced to oscillate at (or very close to) its natural frequency, so that energy is transferred from the driver to the system most efficiently and the amplitude of oscillation reaches a maximum. Exam tip: full marks need BOTH 'driving frequency equals natural frequency' AND 'amplitude is a maximum'; add 'maximum energy transfer' for the third mark in a 3-mark question. With damping present the peak occurs at a frequency slightly below f₀. Amplitude unit m; frequency Hz.

10. DefinitionSL & HL

    Define damping.

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    Damping is the dissipation of the energy of an oscillating system to the surroundings by a resistive (dissipative) force such as friction or air resistance, so the amplitude of the oscillation decreases with time. Exam tip: the mark is for energy loss to the surroundings AND decreasing amplitude; note that for light damping the period/frequency is essentially unchanged — students wrongly say the oscillation 'slows down'. Amplitude decays exponentially for light damping. Energy in J, amplitude in m.

11. DefinitionSL & HL

    Define light (under) damping.

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    Damping in which the resistive force is small, so the system oscillates many times with an amplitude that decreases exponentially with time; the period is only slightly greater than the undamped natural period. Exam tip: sketch answers must show a decaying sinusoid with an exponential envelope and unchanged spacing of the zeros — many students draw the peaks getting closer together, which is wrong. Amplitude in m; time constant in s.

12. DefinitionSL & HL

    Define critical damping.

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    The minimum amount of damping that returns a displaced system to its equilibrium position in the shortest possible time without it oscillating (without overshooting). Exam tip: the mark scheme wants 'shortest time' AND 'no oscillation' — saying only 'returns quickly' or only 'does not oscillate' is a half answer that scores 1 of 2. Applications: car suspension (shock absorbers), analogue meter needles, door closers. Time in s.

13. DefinitionSL & HL

    Define heavy (over) damping.

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    Damping in which the resistive force is so large that the displaced system returns to equilibrium slowly and without oscillating, taking longer than the critically damped case. Exam tip: distinguish from critical damping by the TIME taken — both are non-oscillatory, so an answer that only says 'no oscillation' cannot separate them. Sketching all three curves on the same axes (light, critical, heavy) is a common 3-mark question. Displacement in m, time in s.

14. DefinitionSL & HL

    State the boundary conditions for standing waves at a fixed end and at a free/open end.

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    A fixed (closed) boundary — a string tied to a post, or the closed end of an air column — must be a displacement node, because the medium there cannot move; the reflected wave undergoes a π (180°) phase change. A free or open boundary — the open end of a pipe, or a string end free to slide — is a displacement antinode, and there is no phase change on reflection. Exam tip: for pipes always label the diagram 'displacement' — a displacement node at the closed end is a PRESSURE antinode, and confusing the two is the classic trap.

15. DefinitionSL & HL

    Define the end correction for a resonance tube.

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    The small distance e beyond the open end of a pipe at which the displacement antinode actually forms, because the air just outside the tube also oscillates; the effective length of the air column is therefore L + e rather than L. Exam tip: e ≈ 0.6r for a cylindrical tube of radius r, and it is a systematic error that makes a single-length measurement of v too small — the mark is usually for identifying it as systematic and for saying it is eliminated by using the difference of two resonance lengths. Unit: m, scalar.

16. EquationSL & HLData booklet: No – memorise

    Give the relationship between node spacing, antinode spacing and wavelength in a standing wave.

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    Distance between adjacent nodes = λ/2; distance between adjacent antinodes = λ/2; distance from a node to the nearest antinode = λ/4. Symbols: λ = wavelength of the component travelling waves (m). Valid for any standing wave in a uniform medium. Data booklet status: not printed — memorise/derive from a diagram. Common misuse: reading the distance between two adjacent nodes as one whole wavelength, which halves the calculated speed. Sanity check: a string with 3 loops over L = 1.2 m has λ/2 = 0.40 m so λ = 0.80 m.

17. EquationSL & HLData booklet: Yes

    State the wave equation linking speed, frequency and wavelength, as used for standing waves.

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    c = fλ. c = wave speed of the travelling waves in the medium (m s⁻¹), f = frequency (Hz = s⁻¹), λ = wavelength (m). Rearranged: f = c/λ and λ = c/f. Valid for the SPEED OF THE WAVE IN THE MEDIUM, which is fixed by the medium (tension and mass per unit length for a string, temperature for air), not by the harmonic. Data booklet: yes (C.2, written c = fλ). Common misuse: thinking a higher harmonic travels faster — it is λ that changes, not c. Check: f = 340/0.80 = 425 Hz.

18. EquationSL & HLData booklet: Yes

    State the relationship between period and frequency for a standing wave.

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    T = 1/f. T = period, the time for one complete oscillation of a particle at an antinode (s); f = frequency (Hz = s⁻¹). Valid for any periodic oscillation; every particle of a standing wave has the same T. Data booklet: yes (T = 1/f). Common misuse: using the time for the string to go from one extreme to the other as T — that is only T/2. Check: f = 425 Hz gives T = 2.35 × 10⁻³ s.

19. EquationSL & HLData booklet: No – derive

    Give the wavelengths of the harmonics on a string fixed at both ends.

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    λ_n = 2L/n for n = 1, 2, 3, … . λ_n = wavelength of the nth harmonic (m), L = length of the string between the fixed ends (m), n = harmonic number (dimensionless). Valid when both ends are nodes and the string is uniform. Data booklet: not printed — derive by drawing the mode and counting half-wavelengths (n loops in length L). Common misuse: using L = 2λ instead of L = nλ/2. Check: L = 0.60 m, third harmonic gives λ₃ = 1.20/3 = 0.40 m.

20. EquationSL & HLData booklet: No – derive

    Give the frequencies of the harmonics on a string fixed at both ends.

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    f_n = nc/(2L) = n f₁, with f₁ = c/(2L), n = 1, 2, 3, … . f_n = frequency of nth harmonic (Hz), c = wave speed on the string (m s⁻¹), L = length (m). Valid for a uniform string with a node at each end. Data booklet: not printed — derive from c = fλ with λ_n = 2L/n. Common misuse: assuming the harmonics of every system are all integer multiples — true here and for an open pipe, false for a closed pipe. Check: c = 200 m s⁻¹, L = 0.50 m gives f₁ = 200 Hz, f₃ = 600 Hz.

21. EquationSL & HLData booklet: No – derive

    Give the wavelengths and frequencies of the harmonics in a pipe open at both ends.

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    λ_n = 2L/n and f_n = nc/(2L) = n f₁, n = 1, 2, 3, … . L = length of the air column (m), c = speed of sound in air (m s⁻¹, ≈ 340 m s⁻¹ at room temperature), λ_n in m, f_n in Hz. Valid with a displacement antinode at each open end (ignoring end corrections). Data booklet: not printed — derive from the diagram. Common misuse: applying 4L/n to an open–open pipe. Check: L = 0.34 m gives λ₁ = 0.68 m and f₁ = 340/0.68 = 500 Hz.

22. EquationSL & HLData booklet: No – derive

    Give the wavelengths and frequencies of the harmonics in a pipe closed at one end.

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    λ_n = 4L/n and f_n = nc/(4L) = n f₁ with n ODD only (n = 1, 3, 5, …). L = length of air column (m), c = speed of sound (m s⁻¹), f_n in Hz. Valid with a displacement node at the closed end and antinode at the open end. Data booklet: not printed — derive. Common misuse: writing n = 1, 2, 3 and predicting a frequency (2f₁) that does not exist; also forgetting that a closed pipe of the same length sounds an octave LOWER than an open one. Check: L = 0.17 m gives λ₁ = 0.68 m, f₁ = 500 Hz, next f₃ = 1500 Hz.

23. EquationSL & HLData booklet: No – memorise

    State the equation for the speed of a transverse wave on a stretched string.

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    c = √(T/μ). c = wave speed (m s⁻¹), T = tension in the string (N), μ = mass per unit length (linear density) of the string (kg m⁻¹), where μ = m/L. Valid for a uniform flexible string with small-amplitude transverse waves. Data booklet: NOT printed in the 2025 booklet — if it is needed it will be given in the question stem, so quote the stem. Common misuse: using total mass instead of mass per unit length. Note c ∝ √T, so f₁ ∝ √T: quadrupling the tension doubles the pitch. Check: T = 50 N, μ = 2.0 × 10⁻³ kg m⁻¹ gives c = 158 m s⁻¹.

24. EquationSL & HLData booklet: No – derive

    Give the resonance condition for the first and second resonance positions of a resonance tube (closed pipe).

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    First resonance: L₁ + e = λ/4. Second resonance: L₂ + e = 3λ/4. L₁, L₂ = measured air-column lengths (m), e = end correction (m), λ = wavelength of the sound (m). Valid for a tube closed at one end driven by a tuning fork of fixed frequency f. Data booklet: not printed — derive. Common misuse: omitting e and then using λ = 4L₁, which gives a speed of sound several per cent too low — 6.6% for the numbers below (a systematic error, always low). Check: f = 512 Hz, L₁ = 0.155 m, e = 0.011 m gives λ = 0.664 m and c = 340 m s⁻¹.

25. EquationSL & HLData booklet: No – derive

    Give the two-resonance-length equation used to find the speed of sound without knowing the end correction.

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    L₂ − L₁ = λ/2, so λ = 2(L₂ − L₁) and c = 2f(L₂ − L₁). L₁, L₂ = first and second resonance lengths (m), f = tuning-fork frequency (Hz), c = speed of sound (m s⁻¹). Valid for a closed resonance tube where the end correction e is the same at both lengths, so it cancels. Data booklet: not printed — derive by subtracting L₁ + e = λ/4 from L₂ + e = 3λ/4. Common misuse: still subtracting an end correction after using this method (double counting). Check: f = 512 Hz, L₁ = 0.155 m, L₂ = 0.487 m gives c = 2 × 512 × 0.332 = 340 m s⁻¹.

26. EquationSL & HLData booklet: No – derive

    Give the expression for the end correction in terms of the first two resonance lengths.

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    e = (L₂ − 3L₁)/2, obtained from L₁ + e = λ/4 and L₂ + e = 3λ/4. e = end correction (m), L₁, L₂ = first and second resonance lengths (m). Also e ≈ 0.6r for a cylindrical tube of internal radius r (m). Data booklet: not printed — derive. Common misuse: quoting e = 0.6d using the diameter instead of the radius, which doubles the value. Check: L₁ = 0.155 m, L₂ = 0.487 m gives e = (0.487 − 0.465)/2 = 0.011 m, consistent with a tube of radius ≈ 1.8 cm.

27. EquationSL & HLData booklet: No – derive

    Give the proportionality relationships between first-harmonic frequency, length and tension for a string.

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    f₁ = c/(2L) = (1/2L)√(T/μ), so at constant c, f₁ ∝ 1/L; at constant L and μ, f₁ ∝ √T; at constant L and T, f₁ ∝ 1/√μ. f₁ in Hz, L in m, T in N, μ in kg m⁻¹. Data booklet: not printed — derive. Common misuse: assuming f ∝ T rather than √T, so predicting double the pitch for double the tension when the true factor is √2 ≈ 1.41. Check: halving the vibrating length of a guitar string (fretting at the 12th fret) doubles the frequency — one octave.

28. EquationSL & HLData booklet: No – derive

    Give the relationship between the amplitude of a standing wave and the amplitudes of the two component waves.

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    For two identical waves of amplitude a superposing, the amplitude at an antinode is 2a and at a node it is 0; at a general point a distance x from a node the amplitude is A(x) = 2a sin(2πx/λ). a = amplitude of each travelling wave (m), A = amplitude of the standing wave at that point (m), x = distance from a node (m), λ = wavelength (m). Data booklet: not printed. Common misuse: assuming every particle has amplitude 2a. Since I ∝ A², the intensity at an antinode is 4× that of one component wave, and energy is redistributed, not created.

29. EquationSL & HLData booklet: No – memorise

    State how the frequency of a driven system relates to its natural frequency at resonance.

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    Resonance occurs when f_driving = f₀ (the natural frequency); for a lightly damped system the amplitude peaks at f ≈ f₀, and with heavier damping the peak occurs at a frequency slightly LESS than f₀. f_driving and f₀ in Hz. In the steady state the system always oscillates at f_driving, never at f₀. Data booklet: not printed. Common misuse: writing that the driven system oscillates at its natural frequency — it oscillates at the driving frequency with maximum amplitude when the two match. Check: a 512 Hz fork resonates an air column whose f₁ = c/4(L + e) = 512 Hz.

30. Graph/diagramSL & HL

    Describe the snapshot (displacement–position) graph of a standing wave on a string.

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    Axes: displacement y / mm (or cm) on the vertical, distance x along the string / m on the horizontal. Shape: a sinusoid whose zeros (nodes) are fixed and equally spaced λ/2 apart; draw the extreme position as a solid curve and the mirror-image extreme as a dashed curve, with the envelope between them showing the amplitude at each point. Reading off: node spacing gives λ/2 → λ; the maximum of the envelope gives 2a. There is no gradient/area quantity to extract here. Changing to the next harmonic adds one more loop in the same length L, so λ decreases to 2L/n and f rises to nf₁; increasing the tension does not change the shape but increases f.

31. Graph/diagramSL & HL

    Describe the diagrams of the first three harmonics on a string fixed at both ends.

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    Axes: displacement (vertical) against position along the string of length L (horizontal). Both ends are nodes. n = 1: one loop, L = λ/2, λ₁ = 2L, f₁ = c/2L. n = 2: two loops with one central node, L = λ, λ₂ = L, f₂ = 2f₁. n = 3: three loops with two internal nodes, L = 3λ/2, λ₃ = 2L/3, f₃ = 3f₁. Extract λ by counting loops: number of loops = n and each loop is λ/2. Changing L: shortening the string raises every f_n as f ∝ 1/L; increasing tension raises c and so all f_n by √T, without changing the shapes.

32. Graph/diagramSL & HL

    Describe the displacement diagrams for the harmonics of a pipe open at both ends.

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    Axes: displacement of the air particles (vertical) against distance along the pipe of length L (horizontal). Both open ends are displacement ANTINODES. n = 1: a single node at the centre, L = λ/2, λ₁ = 2L. n = 2: two nodes, L = λ. n = 3: three nodes, L = 3λ/2. All integer harmonics are present, f_n = nc/2L. Extract λ from the node spacing (λ/2). Note the corresponding PRESSURE graph is the inverse: pressure nodes at the open ends, pressure antinodes where displacement nodes are. Lengthening the pipe lowers all frequencies (f ∝ 1/L); warming the air raises c and so raises f.

33. Graph/diagramSL & HL

    Describe the displacement diagrams for the harmonics of a pipe closed at one end.

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    Axes: air-particle displacement (vertical) against distance along the pipe of length L (horizontal). The closed end is a displacement NODE, the open end a displacement ANTINODE. n = 1: a quarter-wave shape, L = λ/4, λ₁ = 4L. Next mode: L = 3λ/4 (third harmonic, 3f₁); then L = 5λ/4 (fifth harmonic). Only ODD harmonics exist, so the frequency spectrum is f₁, 3f₁, 5f₁ … Extract λ using L = nλ/4 with n odd, remembering the end correction shifts the antinode a distance e outside the open end. Halving the length doubles all frequencies; a closed pipe sounds an octave lower than an open pipe of the same length.

34. Graph/diagramSL & HL

    Describe the resonance curve (amplitude of a driven oscillator against driving frequency) and the effect of damping.

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    Axes: amplitude A / m (vertical) against driving frequency f / Hz (horizontal). Shape: rises from a small non-zero value at f = 0 to a sharp peak near the natural frequency f₀, then falls towards zero at high f. The peak position gives f₀ and the peak height gives the maximum amplitude; the width of the peak (at, say, half the maximum) indicates how heavily damped the system is. Increasing the damping lowers the peak, broadens it, and shifts the maximum to a slightly LOWER frequency; all curves converge at very high f. Exam trap: students draw the damped curves peaking at the same height or at a higher frequency.

35. Graph/diagramSL & HL

    Describe the linearised graph used to find the speed of sound from a resonance tube (first-harmonic length against 1/f).

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    Axes: resonance length L / m (vertical) against 1/f / s (horizontal), using several tuning forks at the FIRST resonance. Since L + e = c/(4f), L = (c/4)(1/f) − e, so the graph is a straight line. Gradient = c/4, so c = 4 × gradient (m s⁻¹); the vertical intercept is −e, giving the end correction directly. Shape: straight line of positive gradient, intercepting the L-axis below the origin. Extract c from the max/min gradients of lines through the error bars to obtain the uncertainty. Using warmer air increases c and so steepens the gradient; a wider tube increases e and shifts the whole line down.

36. Graph/diagramSL & HL

    Describe the linearised graph of first-harmonic frequency against 1/length for a stretched string (Melde/sonometer).

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    Axes: f₁ / Hz (vertical) against 1/L / m⁻¹ (horizontal) at constant tension. Because f₁ = (1/2L)√(T/μ) = (c/2)(1/L), the graph is a straight line through the origin. Gradient = c/2, so the wave speed c = 2 × gradient, and μ can then be found from c = √(T/μ) → μ = T/c². Extract the uncertainty in c from max/min gradient lines drawn through the error bars. A non-zero intercept indicates a systematic error, e.g. the bridge position or the vibrator end not being an exact node. Increasing the tension increases c and so steepens the line.

37. Graph/diagramSL & HL

    Describe the displacement–time graphs of particles at a node, at an antinode and at points either side of a node on a standing wave.

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    Axes: displacement y / mm (vertical) against time t / s (horizontal). At a node: a flat line along y = 0 at all times. At an antinode: a sinusoid of amplitude 2a and period T = 1/f. At an intermediate point: a sinusoid of smaller amplitude 2a sin(2πx/λ) but the SAME period, in phase with the antinode in its own loop. For a point in the adjacent loop (across a node): the same period but exactly π (180°) out of phase — an inverted sinusoid. Extract T (hence f) from the spacing of the peaks and 2a from the antinode amplitude. Increasing the driving amplitude scales all curves vertically without changing T.

38. Concept/explainSL & HL

    Explain how a standing wave is formed on a string fixed at both ends.

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    • A wave travels along the string and is reflected at the fixed end
    • Reflection at a fixed boundary inverts the wave, giving a π phase change
    • The reflected wave is identical in frequency, wavelength, speed and amplitude but travels in the opposite direction
    • The two waves superpose, so displacements add algebraically at every point
    • Where the waves are permanently in antiphase, destructive interference gives zero amplitude — a node
    • Where they are permanently in phase, constructive interference gives maximum amplitude 2a — an antinode
    • The pattern is stationary because the nodes do not move, so there is no net energy transfer along the string
    • Exam tip: incomplete answers just say the waves cancel; marks require two identical waves travelling in opposite directions plus superposition.
39. Concept/explainSL & HL

    Explain the spacing of nodes and antinodes in a standing wave in terms of wavelength.

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    • Adjacent nodes are separated by λ/2, and adjacent antinodes are also separated by λ/2
    • A node and the neighbouring antinode are λ/4 apart
    • Reason: the path difference between the two superposing waves changes by a full λ over a distance λ/2, so the interference condition alternates every half wavelength
    • The λ of a standing wave is the wavelength of the original travelling waves, not the length of one loop
    • One loop, node to node, is therefore half a wavelength
    • Measuring across several loops and dividing gives λ with a much smaller fractional uncertainty
    • Exam tip: the commonest error is calling the distance between adjacent nodes one wavelength, which halves the calculated speed.
40. Concept/explainSL & HL

    Compare and contrast a travelling wave and a standing wave.

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    • Energy: a travelling wave transfers energy through the medium; a standing wave stores energy with zero net transfer along it
    • Amplitude: every particle of a travelling wave oscillates with the same amplitude; in a standing wave the amplitude depends on position, from zero at a node to 2a at an antinode
    • Phase: in a travelling wave phase varies continuously with distance; in a standing wave all particles within one loop are in phase and adjacent loops are exactly π out of phase
    • Profile: a travelling waveform moves along; a standing waveform stays fixed in space
    • Both obey v = fλ and both can be transverse or longitudinal
    • Exam tip: answer in paired statements covering energy, amplitude and phase; saying the wave does not move earns almost nothing.
41. Concept/explainSL & HL

    Explain the phase relationship between particles of a medium carrying a standing wave.

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    • All particles in one loop, between two adjacent nodes, oscillate in phase: they reach maximum displacement at the same instant
    • Particles in adjacent loops are exactly π rad out of phase, moving in opposite directions at any instant
    • Phase difference between any two points is therefore only 0 or π, never an intermediate value
    • Nodes are permanently at rest, so phase has no meaning there
    • The whole medium passes through the zero-displacement position simultaneously, twice per cycle
    • Amplitude varies with position but frequency is the same everywhere
    • Exam tip: students wrongly apply the travelling-wave rule phase difference = 2πΔx/λ; in a standing wave the phase does not vary smoothly with distance.
42. Concept/explainSL & HLData booklet: No – derive

    Explain the boundary conditions that determine which standing waves can exist on a string or in a pipe.

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    • A fixed end of a string cannot move, so a displacement node must form there
    • A free or open end of an air column can oscillate freely, so a displacement antinode forms there
    • At a closed pipe end the air cannot move along the pipe, so a displacement node forms (this is a pressure antinode)
    • Only patterns fitting a whole number of half or quarter wavelengths between the boundaries survive, so only discrete wavelengths and frequencies resonate
    • String or open–open pipe: λ_n = 2L/n with n = 1, 2, 3…
    • Open–closed pipe: λ_n = 4L/n with n odd only
    • Exam tip: always state whether your diagram shows displacement or pressure; IB diagrams normally show displacement, so mark a node at a closed end.
43. Concept/explainSL & HLData booklet: No – derive

    Explain why a pipe closed at one end can only resonate at odd harmonics of its fundamental.

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    • The closed end must be a displacement node and the open end a displacement antinode
    • The shortest pattern satisfying this is a quarter wavelength, so λ₁ = 4L and f₁ = v/4L
    • Each successive resonance inserts one more half wavelength, giving L = λ/4, 3λ/4, 5λ/4…
    • Hence λ_n = 4L/n and f_n = nv/4L with n = 1, 3, 5, 7 only
    • An even harmonic would need a node and an antinode to exchange ends, which the boundary conditions forbid
    • Consequence: a closed pipe of length L sounds an octave below an open pipe of the same length, and its timbre lacks even harmonics
    • Exam tip: the second harmonic of a closed pipe does not exist — 3f₁ is its second resonance.
44. Concept/explainSL & HLData booklet: No – derive

    Show how the relationship λ_n = 2L/n arises for a string fixed at both ends, and outline its consequences.

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    • Both ends are fixed, so both must be nodes
    • The simplest pattern has a single loop with one antinode at the centre: L = λ₁/2, so λ₁ = 2L and f₁ = v/2L, the fundamental or first harmonic
    • Adding one loop at a time gives L = nλ_n/2, hence λ_n = 2L/n
    • Since v = fλ and v depends only on the string (tension and mass per unit length), f_n = nv/2L = nf₁
    • The resonant frequencies therefore form the integer series f₁, 2f₁, 3f₁…
    • This relationship is not printed in the data booklet and must be derived from a labelled diagram
    • Exam tip: count loops (antinodes) to find n, not nodes; n loops means n half wavelengths fit the length.
45. Concept/explainSL & HL

    Explain what is meant by natural frequency, forced oscillation and resonance.

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    • Every oscillating system has one or more natural frequencies f₀ at which it oscillates freely after being displaced and released
    • In a forced oscillation an external periodic driver supplies energy at a driving frequency f_d, and the system settles at f_d, not at f₀
    • Resonance occurs when f_d equals f₀
    • At resonance the transfer of energy from driver to system is a maximum, because the driving force stays in phase with the velocity and so does positive work throughout the cycle
    • The amplitude therefore rises to a maximum, limited only by damping
    • Well away from resonance energy is alternately given and returned, so the amplitude stays small
    • Exam tip: resonance means maximum energy transfer and maximum amplitude, not the highest frequency.
46. Concept/explainSL & HL

    Explain how increasing the degree of damping changes the resonance curve of a driven system.

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    • A resonance curve plots the amplitude of the driven system against driving frequency for a fixed driving amplitude
    • Increasing damping lowers the peak amplitude, because more energy is dissipated as thermal energy per cycle
    • The peak also becomes broader, so the system responds appreciably over a wider range of frequencies and is less sharply tuned
    • The peak shifts slightly to a frequency below the undamped natural frequency
    • With very heavy damping no distinct peak appears at all
    • All curves converge at very low and very high driving frequencies
    • Applications: heavy damping protects bridges and buildings; light damping gives sharp selectivity in radio tuning
    • Exam tip: sketched curves must share the same low-frequency amplitude and show the damped peak lower, broader and slightly left-shifted.
47. Concept/explainSL & HL

    Distinguish between light, critical and heavy damping and give a real example of each.

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    • Damping is the dissipation of oscillation energy to the surroundings, usually as thermal energy, through resistive forces
    • Light damping: the amplitude decays exponentially over many oscillations while the period is almost unchanged — a pendulum swinging in air, a plucked guitar string
    • Critical damping: the system returns to equilibrium in the shortest possible time without oscillating — car suspension, analogue meter needles, self-closing doors
    • Heavy (over) damping: the system returns to equilibrium slowly without oscillating — a pointer moving through thick oil
    • The greater the damping, the faster the energy loss and the lower and broader the resonance peak
    • Exam tip: critical damping is not the fastest stop of all; it is the fastest return without overshoot or oscillation.
48. Concept/explainSL & HLData booklet: No – derive

    Explain how a resonance tube is used to determine the speed of sound and why an end correction is needed.

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    • A tuning fork of known frequency is sounded over a tube whose air column length is varied by changing the water level
    • Resonance is heard as a sharp increase in loudness when a displacement node sits at the water surface and an antinode near the open end
    • The first resonance is at l₁ ≈ λ/4 and the second at l₂ ≈ 3λ/4
    • The antinode actually lies a short distance outside the tube, the end correction c, because the air just beyond the mouth also oscillates: l₁ + c = λ/4 and l₂ + c = 3λ/4
    • Subtracting eliminates c, giving l₂ − l₁ = λ/2 and so v = 2f(l₂ − l₁)
    • Exam tip: using l₁ alone gives a systematically low speed; the two-length method removes this systematic error.
49. Concept/explainSL & HL

    Explain why a standing wave does not transfer energy along the medium, and where the energy resides.

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    • A travelling wave transfers energy in its direction of propagation
    • A standing wave is the superposition of two identical waves carrying equal energy in opposite directions, so the net energy flow is zero
    • Energy is therefore trapped between the boundaries and interchanges continuously between kinetic and potential form
    • When the string is momentarily straight all the energy is kinetic, concentrated near the antinodes where the speed is greatest
    • At maximum displacement all the energy is elastic potential, greatest near the nodes where the curvature is largest
    • Nodes never move, so no energy can pass through them
    • Exam tip: say that the net energy transfer along the wave is zero, not that the wave has no energy — a resonating instrument clearly radiates sound.
50. Concept/explainSL & HL

    Outline real-world examples of resonance and identify the driver and the natural frequency in each.

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    • Musical instruments: a bow, reed or plucking drives the string or air column, which responds only at its harmonics; the mixture of harmonics present determines the timbre
    • Radio and television tuning: the incoming electromagnetic wave drives an LC circuit whose natural frequency is tuned to one station, which is then amplified far above the others
    • Microwave ovens: 2.45 GHz radiation drives rotational resonance of water molecules, heating food, while reflections in the metal cavity form standing waves and hot spots, so the food is rotated
    • MRI: radio-frequency photons resonate with proton spin transitions in a strong magnetic field
    • Structures: wind, traffic or earthquakes can drive bridges and buildings near a natural frequency, so dampers are fitted
    • Exam tip: always name both the driver and the natural frequency.
51. Concept/explainSL & HLData booklet: No – memorise

    Explain how the speed of a transverse wave on a stretched string depends on its properties, and the effect on the harmonics.

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    • The wave speed depends only on the medium: v = √(T/μ), where T is the tension in N and μ is the mass per unit length in kg m⁻¹
    • Greater tension gives a larger restoring force, so waves travel faster; a heavier string has more inertia per unit length, so waves travel more slowly
    • This equation is not printed in the 2025 data booklet, so it is supplied when required
    • Combining with f₁ = v/2L gives f₁ = (1/2L)√(T/μ), so at fixed L, f₁ ∝ √T and f₁ ∝ 1/√μ
    • This is why a guitar is tuned by adjusting tension and why bass strings are thicker
    • Exam tip: v is a property of the string, not of the source; changing T changes v and hence all the resonant frequencies.
52. Worked problemSL & HLData booklet: Yes

    A string of length 0.80 m is fixed at both ends and vibrates in its fundamental mode at 256 Hz. Determine the wavelength of the fundamental and the speed of transverse waves on the string.

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    Principle: both ends are nodes, so the fundamental is one loop and L = λ₁/2, giving λ₁ = 2L. Substitution: λ₁ = 2 × 0.80 m = 1.60 m. Then v = fλ (data booklet): v = 256 Hz × 1.60 m. Intermediate: 256 × 1.60 = 409.6 m s⁻¹. Answer: λ₁ = 1.6 m and v = 4.1 × 10² m s⁻¹ (2 s.f.). Check/Trap: do not set λ = L — the string length is only half a wavelength in the fundamental, and that error halves the speed. The speed is a property of the string, set by its tension and mass per unit length, so it is the same for every harmonic; use this as a check on later parts of the question.

53. Worked problemSL & HLData booklet: No – derive

    The 0.80 m string of the previous item is driven so that it vibrates in its third harmonic. Calculate the frequency and wavelength of this mode and state the number of nodes and antinodes.

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    Principle: for a string fixed at both ends λ_n = 2L/n and f_n = nf₁, because v is unchanged. Substitution: λ₃ = 2 × 0.80/3 = 0.533 m; f₃ = 3 × 256 Hz = 768 Hz. Verification: v = f₃λ₃ = 768 × 0.533 = 4.1 × 10² m s⁻¹, the same speed as the fundamental, confirming the working. Answer: λ₃ = 0.53 m, f₃ = 768 Hz, with 3 loops, 3 antinodes and 4 nodes counting both fixed ends. Check/Trap: the third harmonic has n = 3 loops, so students who count nodes instead of antinodes get n wrong. Higher harmonics have shorter wavelengths but the same wave speed — only f and λ change.

54. Worked problemSL & HLData booklet: No – derive

    A pipe open at both ends has a length of 0.50 m. Taking the speed of sound in air as 340 m s⁻¹, determine the fundamental frequency and the frequency of the second harmonic.

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    Principle: an open–open pipe has a displacement antinode at each end, so λ_n = 2L/n, exactly like a string. Substitution: λ₁ = 2 × 0.50 m = 1.00 m, so f₁ = v/λ₁ = 340/1.00. Intermediate: f₁ = 340 Hz. Second harmonic: λ₂ = L = 0.50 m, so f₂ = 340/0.50 = 680 Hz = 2f₁. Answer: f₁ = 3.4 × 10² Hz and f₂ = 6.8 × 10² Hz (2 s.f.). Check/Trap: an open pipe supports every integer harmonic, so successive resonances are separated by exactly f₁ = 340 Hz — a quick way to check an answer. Remember the centre of an open–open pipe is a displacement node in the fundamental, not an antinode.

55. Worked problemSL & HLData booklet: No – derive

    A pipe of length 0.25 m is closed at one end. Taking the speed of sound as 340 m s⁻¹, determine the two lowest resonant frequencies of the air column.

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    Principle: a closed pipe has a displacement node at the closed end and an antinode at the open end, so λ_n = 4L/n with n odd only. Substitution: λ₁ = 4 × 0.25 m = 1.00 m, so f₁ = 340/1.00 = 340 Hz. Next allowed mode is n = 3: λ₃ = 4 × 0.25/3 = 0.333 m, so f₃ = 340/0.333 = 1020 Hz = 3f₁. Answer: 3.4 × 10² Hz and 1.0 × 10³ Hz (2 s.f.). Check/Trap: there is no resonance at 680 Hz because even harmonics are forbidden — this is the standard multiple-choice distractor. A closed pipe of length L has the same fundamental as an open pipe of length 2L, which is why closed pipes sound an octave lower.

56. Worked problemSL & HLData booklet: Yes

    A tuning fork of frequency 512 Hz is held over a resonance tube. Resonance is first heard when the air column is 0.155 m long and next when it is 0.487 m long. Determine the speed of sound in the tube and the end correction.

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    Principle: successive resonances of a closed air column differ by λ/2, and taking the difference removes the end correction c. Substitution: λ/2 = l₂ − l₁ = 0.487 − 0.155 = 0.332 m, so λ = 0.664 m. Then v = fλ = 512 Hz × 0.664 m. Intermediate: v = 340.0 m s⁻¹. End correction: l₁ + c = λ/4 = 0.166 m, so c = 0.166 − 0.155 = 0.011 m. Answer: v = 3.40 × 10² m s⁻¹ and c = 1.1 × 10⁻² m (2 s.f.). Check/Trap: using v = 4fl₁ alone gives 317 m s⁻¹, about 7 % low — a systematic error from ignoring the end correction. c is typically about 0.6 of the tube radius (equivalently 0.3 of the diameter).

57. Worked problemSL & HLData booklet: Yes

    In a microwave oven operating at 2.45 GHz the turntable is removed and a tray of marshmallows melts in evenly spaced lines 6.1 cm apart. Determine the speed of the microwaves.

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    Principle: the melted lines lie at the antinodes of a standing wave formed by reflection from the metal walls, and adjacent antinodes are λ/2 apart. Substitution: λ = 2 × 6.1 × 10⁻² m = 0.122 m. Then v = fλ = 2.45 × 10⁹ Hz × 0.122 m. Intermediate: 2.45 × 0.122 = 0.2989, so v = 2.99 × 10⁸ m s⁻¹. Answer: v ≈ 3.0 × 10⁸ m s⁻¹ (2 s.f.), the speed of light, as expected for an electromagnetic wave. Check/Trap: treating 6.1 cm as a full wavelength halves the answer to 1.5 × 10⁸ m s⁻¹. Ovens rotate the food precisely because the nodes are fixed in space and would otherwise leave cold regions.

58. Worked problemSL & HLData booklet: No – memorise

    A string of mass per unit length 5.0 × 10⁻⁴ kg m⁻¹ is stretched between fixed points 1.2 m apart under a tension of 20 N. Using v = √(T/μ), calculate the fundamental frequency.

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    Principle: find the wave speed from the string properties, then use λ₁ = 2L and f₁ = v/λ₁. Substitution: v = √(20 N / 5.0 × 10⁻⁴ kg m⁻¹) = √(4.0 × 10⁴ m² s⁻²). Intermediate: v = 200 m s⁻¹; λ₁ = 2 × 1.2 m = 2.4 m. Then f₁ = 200/2.4 = 83.3 Hz. Answer: f₁ = 83 Hz (2 s.f.). Check/Trap: v = √(T/μ) is not in the 2025 data booklet, so it will be given in the question. Verify the units: N/(kg m⁻¹) = kg m s⁻²/(kg m⁻¹) = m² s⁻², whose square root is m s⁻¹. Forgetting the factor 2 in λ₁ = 2L doubles the frequency to 167 Hz.

59. Worked problemSL & HLData booklet: No – memorise

    A string sounds a fundamental of 120 Hz at a tension of 40 N. Determine the tension required to raise the fundamental to 180 Hz, with the length and mass per unit length unchanged.

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    Principle: f₁ = (1/2L)√(T/μ), so at fixed L and μ, f₁ ∝ √T and therefore T ∝ f₁². Substitution: T₂/T₁ = (f₂/f₁)² = (180/120)² = (1.50)². Intermediate: ratio = 2.25, so T₂ = 40 N × 2.25. Answer: T₂ = 90 N (2 s.f.). Check/Trap: assuming T ∝ f gives 60 N, the standard wrong option — a 50 % rise in frequency needs a 125 % rise in tension. In practice tightening a real string also stretches it slightly, reducing μ and altering L, so the tuned value differs a little. Note the length is unchanged here, so λ₁ = 2L is the same and only v changes.

60. Worked problemSL & HLData booklet: No – derive

    The vibrating length of a guitar string is 0.65 m and it sounds a fundamental of 330 Hz. The player presses the string onto a fret so the vibrating length becomes 0.58 m. Calculate the new fundamental frequency.

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    Principle: fretting does not alter the tension or mass per unit length, so v is unchanged; since f₁ = v/2L, f₁ ∝ 1/L. Substitution: v = 2Lf₁ = 2 × 0.65 × 330 = 429 m s⁻¹. New value: f₁′ = v/(2L′) = 429/(2 × 0.58) = 429/1.16. Intermediate: 369.8 Hz. Answer: f₁′ = 3.7 × 10² Hz (2 s.f.). Check/Trap: a shorter string must give a higher frequency, so check the direction of change before trusting the arithmetic. A one-line route is f₁′ = 330 × 0.65/0.58, avoiding the need to find v at all. Do not assume the wavelength stays constant — it is L that is fixed by the fret, and λ₁ = 2L changes with it.

61. Worked problemSL & HLData booklet: No – derive

    A pipe resonates at 350 Hz, 490 Hz and 630 Hz and at no frequency in between. Deduce whether the pipe is open at both ends or closed at one end, and determine its fundamental frequency and length. Take v = 340 m s⁻¹.

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    Principle: an open pipe resonates at all integer multiples of f₁; a closed pipe only at odd multiples. Analysis: the spacing is a constant 140 Hz, and 350 : 490 : 630 = 5 : 7 : 9, which are consecutive odd integers. Deduction: the pipe is closed at one end; the spacing between successive odd harmonics is 2f₁, so f₁ = 140/2 = 70 Hz. Length: f₁ = v/4L, so L = 340/(4 × 70) = 340/280 = 1.214 m. Answer: closed pipe, f₁ = 70 Hz, L = 1.2 m (2 s.f.). Check/Trap: if the ratios had been consecutive integers the spacing would equal f₁ itself and the pipe would be open. Never assume the lowest quoted frequency is the fundamental.

62. Worked problemSL & HLData booklet: Yes

    In a standing wave the distance from a node to the nearest antinode is 0.15 m and the frequency is 400 Hz. Calculate the speed of the waves that formed it.

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    Principle: adjacent nodes are λ/2 apart, so the distance from a node to the neighbouring antinode is λ/4. Substitution: λ/4 = 0.15 m, so λ = 4 × 0.15 = 0.60 m. Then v = fλ = 400 Hz × 0.60 m. Intermediate: 400 × 0.60 = 240. Answer: v = 2.4 × 10² m s⁻¹ (2 s.f.). Check/Trap: the two favourite Paper 1 distractors come from treating 0.15 m as λ/2, giving 120 m s⁻¹, or as a whole λ, giving 60 m s⁻¹. Sketch one loop and mark the quarter wavelength before substituting — it takes a few seconds and secures the mark. The answer is the speed of the two travelling waves, not of the pattern, which does not move.

63. Worked problemSL & HLData booklet: Yes

    An organ pipe open at both ends has a fundamental frequency of 500 Hz when the speed of sound is 340 m s⁻¹. The building warms until the speed of sound is 350 m s⁻¹. Determine the new fundamental frequency.

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    Principle: the boundary conditions fix the wavelength through λ₁ = 2L, and L is unchanged, so f₁ = v/λ₁ gives f₁ ∝ v. Substitution: λ₁ = v/f₁ = 340/500 = 0.680 m, so L = 0.340 m. New frequency: f₁′ = 350/0.680. Intermediate: 514.7 Hz. Answer: f₁′ = 5.1 × 10² Hz (2 s.f.), about 3 % sharp. Check/Trap: it is the wavelength, not the frequency, that is fixed by the pipe — the reverse of a fixed-source situation where f is fixed and λ changes with medium. This is why wind instruments must be warmed up before they play in tune. Thermal expansion of the pipe changes L by far less and is normally neglected.

64. Worked problemSL & HLData booklet: Yes

    Two identical waves of amplitude 3.0 mm travel in opposite directions along a string of length 2.4 m fixed at both ends, forming a standing wave of 4 loops at 50 Hz. Determine the wavelength, the wave speed and the amplitude at an antinode.

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    Principle: n loops means n half wavelengths fit the length, so λ_n = 2L/n; superposition gives an antinode amplitude of 2a. Substitution: λ = 2 × 2.4/4 = 1.2 m. Then v = fλ = 50 Hz × 1.2 m = 60 m s⁻¹. Antinode amplitude = 2 × 3.0 mm = 6.0 mm. Answer: λ = 1.2 m, v = 60 m s⁻¹, amplitude 6.0 mm (2 s.f.). Check/Trap: the antinode amplitude is twice one wave amplitude, not four times — it is the intensity that quadruples, since I ∝ A². Amplitude varies with position from 0 at a node to 6.0 mm at an antinode, so quoting one amplitude for the whole standing wave is wrong. This is the 4th harmonic, so f₁ = 12.5 Hz.

65. Exam technique/trapSL & HL

    State the distance relationships in a standing wave that are most often confused, and how to avoid the resulting errors.

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    The trap: taking the distance between adjacent nodes as one wavelength. Why students fall for it: a single loop looks like one complete hump, so it is mistaken for a whole wave. Correct approach: adjacent nodes and adjacent antinodes are λ/2 apart, a node and the next antinode are λ/4 apart, and one full wavelength spans two loops. Mark λ on your diagram across two loops before substituting into v = fλ. In experiments, measure across as many loops as possible and divide, because this reduces the fractional uncertainty in λ. A halved wavelength halves the calculated speed, which usually shows up as an absurd speed of sound near 170 m s⁻¹ — always sanity-check the magnitude.

66. Exam technique/trapSL & HL

    Explain the naming of resonances in a pipe closed at one end and the marks lost through careless numbering.

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    The trap: calling the second resonance of a closed pipe the second harmonic. Why: for strings and open pipes the resonances run f₁, 2f₁, 3f₁, so students assume the pattern continues. Correct approach: a closed pipe supports only odd harmonics f₁, 3f₁, 5f₁, so its second resonance is the third harmonic, also called the first overtone. Overtone and harmonic numbering differ, so use harmonic language unless the question says overtone. When given successive resonant frequencies, take differences: constant spacing equal to f₁ with integer ratios means an open pipe or string, while spacing equal to 2f₁ with odd ratios means a closed pipe. State clearly which end is closed in your answer.

67. Exam technique/trapSL & HL

    Outline what a sketch question on standing waves must show to gain full marks.

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    Command term: sketch means a labelled qualitative drawing with the correct shape and key features, not a scale plot. For a harmonic on a string or in a pipe, draw the correct number of loops, place nodes at fixed and closed ends and antinodes at open ends, label at least one node N and one antinode A, and mark λ spanning two loops or mark L. Show both extremes of displacement, drawing the mirror image as a dashed curve, because a single curve does not show the oscillation. For a pipe, state that the curves represent displacement, not the physical shape of the air. For a resonance curve, label the axes amplitude and driving frequency and mark f₀. Labels earn the marks, not neatness.

68. Exam technique/trapSL & HL

    Identify the errors students make when describing or sketching resonance and damping curves.

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    Trap 1: labelling the horizontal axis time or driver amplitude instead of driving frequency. Trap 2: drawing all damping curves with the same peak height, or with peaks at exactly the same frequency. Correct approach: for greater damping draw a peak that is lower, broader and shifted slightly to lower frequency, with all curves converging at very low and very high driving frequencies. Trap 3: stating only that the frequencies are equal, without saying that amplitude and energy transfer are then a maximum. Trap 4: confusing damping with resonance — damping removes energy from the system while the driver supplies it. For explain, add that at resonance the driving force stays in phase with the velocity, so the work done on the system per cycle is greatest.

69. Exam technique/trapSL & HLData booklet: No – derive

    Describe a resonance tube experiment to measure the speed of sound, including variables, analysis, limitations and improvements.

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    Apparatus: tuning forks of known frequency, a tube standing in a water reservoir so the air column length can be varied, a metre rule and a thermometer. Method: sound a fork above the tube, lower the water until loudness is a clear maximum, record l₁, continue to the second maximum l₂; repeat three times and average. Independent variable f, dependent variable resonant length, controlled variables tube diameter and air temperature. Analysis: v = 2f(l₂ − l₁) removes the end correction, or plot l₁ against 1/f, where gradient = v/4 and the intercept on the l₁ axis is −c. Limitations: judging maximum loudness by ear, an unsteady water surface, temperature drift. Improvement: use a signal generator with a microphone and oscilloscope to locate resonance objectively.

70. Exam technique/trapSL & HLData booklet: No – memorise

    Describe an experiment with a vibrating string to test how the fundamental frequency depends on tension, and explain how the data are analysed.

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    Apparatus: a string passing over a pulley and loaded with slotted masses, a vibration generator driven by a signal generator at the other end, a metre rule, and a balance to obtain μ = m/L. Method: keep L and μ fixed; for each tension T = mg adjust the driving frequency until a single-loop standing wave of maximum amplitude is seen, and record f₁; repeat for at least six tensions with repeats at each. Analysis: theory gives f₁ = (1/2L)√(T/μ), so plot f₁² against T — a straight line through the origin confirms f₁ ∝ √T, and gradient = 1/(4L²μ) yields μ for comparison with the measured value. Limitations: the driver end is not a perfect node, and pulley friction makes T slightly less than mg.

71. Exam technique/trapSL & HLData booklet: Yes

    Explain how uncertainties are handled when a wave speed is determined from standing-wave measurements.

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    Absolute vs fractional: a metre rule gives about ±1 mm at each end, so ±2 mm absolute on l₂ − l₁; if that difference is 0.332 m the fractional uncertainty is only 0.6 %, whereas measuring a single 0.02 m loop would give 10 %. Rule: measure across many loops and divide. Combining: for v = fλ add the fractional uncertainties in f and λ, convert back to an absolute uncertainty, round it to 1 s.f. and quote v to the matching decimal place. Graphs: plot error bars, then draw maximum and minimum gradient lines through the extremes of the bars, with uncertainty in gradient = (max − min)/2. Repeats reduce random error; the end correction is a systematic error that repeats can never remove.

72. Exam technique/trapSL & HL

    State the wording the mark scheme expects for compare, explain and deduce questions on standing waves, and the phrases that lose marks.

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    Compare and contrast: give paired statements about both cases in the same sentence, for example a travelling wave transfers energy along the medium whereas a standing wave does not. Explain: give a causal chain — two identical waves in opposite directions, superposition, constructive and destructive interference, hence antinodes and nodes. Deduce: reach a conclusion with the reasoning shown, for example odd frequency ratios therefore a pipe closed at one end. Suggest: apply physics to an unfamiliar context, naming the driver and the natural frequency. Phrases that lose marks: the waves cancel out, the wave stands still, the string has no energy, resonance is when the frequency is highest. Instead say net energy transfer is zero and amplitude is a maximum at resonance.

Practise this topic with exam-style questions: C.4 Standing waves and resonance questions (SL) · C.4 Standing waves and resonance questions (HL) · all flashcards

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