IB DP Physics: Standing waves, resonance and damping (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover section C.4 (standing waves and resonance) of the IB Physics guide, first assessment 2025. The section is common to SL and HL. Read the study guide first, or use the revision notes for a quick recap. Tick off outcomes on the printable checklist and see the IB Physics hub. Where a question needs the speed of sound in air, it is given as 340 m s⁻¹.

Questions

1. Define the terms node and antinode for a standing wave. [2]

2. A pipe of length 0.45 m is closed at one end and open at the other. The speed of sound in air is 340 m s⁻¹. Determine the frequency of the first harmonic and of the next two harmonics the pipe can sustain. [3]

3. A string of length 0.50 m is fixed at both ends and vibrates in its fourth harmonic at 640 Hz.

(a) Determine the wavelength of the wave. [1] (b) Determine the speed of waves on the string. [1] (c) Determine the frequency of the first harmonic. [2]

4. A standing wave on a string has antinode amplitude 5.0 mm and wavelength 0.48 m. Point P is 0.04 m from a node. Point Q is 0.28 m from the same node, on the same side, and the node adjacent to P on that side lies at 0.24 m.

(a) Calculate the amplitude of P. [2] (b) State the amplitude of Q and the phase difference between P and Q. [3]

5. A string is fixed at both ends and a wave is sent along it. Explain how a standing wave forms on the string. [4]

6. A tuning fork of frequency 512 Hz is held over a tube whose air column length can be varied. Loud resonances are found at air column lengths of 0.165 m and 0.505 m.

(a) State the relationship between the difference in these two lengths and the wavelength. [1] (b) Calculate the wavelength and the speed of sound. [2] (c) Predict the next air column length that gives resonance. [2]

7. An open pipe and a pipe closed at one end both have length 0.85 m.

(a) Calculate the frequency of the first harmonic of each. [2] (b) List every harmonic frequency below 650 Hz for each pipe. [2] (c) For the third harmonic of the closed pipe, state where the displacement nodes lie, measured from the closed end. [2]

8. A mass on a spring is driven by a vibrating support. The table shows the amplitude of its oscillation for two different amounts of damping.

Driving frequency / Hz	2.0	3.0	3.5	4.0	4.5	5.0	6.0
Amplitude, light damping / mm	4	9	18	31	17	8	3
Amplitude, heavier damping / mm	3	6	7.5	7	5.5	4	2.5

(a) Identify the driving frequency of greatest response with light damping. [1] (b) State two differences between the response curves for the two amounts of damping. [2] (c) Calculate the ratio of the peak amplitudes. [1] (d) Explain why damping limits the amplitude at resonance. [2]

9. A door closer is designed so that a door released from an open position swings shut and stops at the frame without bouncing.

(a) State the type of damping the designer aims for. [1] (b) Describe the displacement-time graph for light, critical and heavy damping after a single release. [3]

10. Pedestrians cross a footbridge whose natural frequency is close to the rate at which people step.

(a) Explain why large oscillations of the bridge might develop. [2] (b) Suggest two changes that would reduce the risk. [2]

11. A guitar string of length 0.65 m, fixed at both ends, has a first harmonic of 220 Hz.

(a) Calculate the speed of waves on the string. [2] (b) The string is held near a pipe closed at one end. Calculate the length of the pipe that has a first harmonic of 220 Hz. [2] (c) Show that the third harmonic of the pipe matches the third harmonic of the string. [2] (d) An open pipe of the same length as in (b) is used instead. Explain whether it can resonate with this string. [2]

12. A string of length 0.60 m is fixed at one end and free at the other. Waves on it travel at 90 m s⁻¹.

(a) Determine the frequency of its first harmonic and state which harmonics can occur. [3] (b) A pipe of length 0.50 m is closed at both ends. Determine its first harmonic frequency. [2]

Answers

1. A node is a point on a standing wave where the amplitude is permanently zero [1]. An antinode is a point where the amplitude is a maximum [1]. [2] Examiner insight: Say amplitude, not displacement, and say “permanently” or “at all times” for the node; a point that is merely momentarily at rest does not describe a node.

2. Closed pipe: λ = 4L = 1.80 m [1]. f₁ = v/λ = 340 ÷ 1.80 = 189 Hz [1]. Only odd harmonics occur, so the next two are 3 × 189 = 567 Hz and 5 × 189 = 944 Hz [1]. [3] Examiner insight: Giving 378 Hz as the next frequency, by treating the pipe as open, is a common slip; justify the odd-harmonic series from the node at the closed end.

3. (a) Four half-wavelengths fit: λ = 2L/4 = 0.25 m [1]. (b) v = fλ = 640 × 0.25 = 160 m s⁻¹ [1]. (c) f₁ = f₄ ÷ 4 [1] = 160 Hz [1]. [4] Examiner insight: Write the substitution for each part, and in part (c) divide the given 640 Hz by 4 rather than recalculating from your own earlier answers; clear working makes your reasoning easy to follow.

4. (a) Phase angle = 2π × 0.04 ÷ 0.48 = π/6 [1]. Amplitude = 5.0 × sin(π/6) = 2.5 mm [1]. (b) Q is 0.04 m from the node at 0.24 m, so its amplitude is 2.5 mm [1]. P and Q lie in adjacent loops, on either side of the node at 0.24 m [1], so the phase difference is π (antiphase) [1]. [5] Examiner insight: Both amplitude and phase are asked for, so give the numerical phase difference as well as the word antiphase; reasoning from loops is clearer than quoting a sign from a formula.

5. The wave travels along the string and reflects at the fixed end [1]. The reflected wave travels in the opposite direction with the same frequency and amplitude [1]. The incident and reflected waves superpose [1]. Where they are always in antiphase there are nodes, and where always in phase there are antinodes, so the pattern stays in place; only wavelengths that fit the fixed ends give a stable pattern [1]. [4] Examiner insight: Name the two waves that superpose and say they are identical and opposite; saying “waves bounce back and forth” without superposition is incomplete.

6. (a) Successive resonances differ in length by half a wavelength [1]. (b) λ = 2 × (0.505 − 0.165) = 0.680 m [1]. v = fλ = 512 × 0.680 = 348 m s⁻¹ [1]. (c) Next resonance: 0.505 + 0.340 [1] = 0.845 m [1]. [5] Examiner insight: Treating the first length alone as a quarter-wavelength is risky; stating that you use the difference between successive resonances shows a sound method.

7. (a) Open: f₁ = 340 ÷ (2 × 0.85) = 200 Hz [1]. Closed: f₁ = 340 ÷ (4 × 0.85) = 100 Hz [1]. (b) Open pipe: 200, 400, 600 Hz [1]. Closed pipe: 100, 300, 500 Hz (odd multiples only) [1]. (c) For the third harmonic, λ = 4L/3 = 1.13 m [1]. Displacement nodes are at the closed end and at λ/2 = 0.57 m from it; the open end is an antinode [1]. [6] Examiner insight: Distance from which end must be stated; quote the position relative to the closed end and note the antinode at the open end to avoid ambiguity.

8. (a) 4.0 Hz [1]. (b) Heavier damping gives a lower peak amplitude [1] and a broader response; its peak is also at a slightly lower frequency, about 3.5 Hz rather than 4.0 Hz [1]. (c) 31 ÷ 7.5 = 4.1 [1]. (d) Damping dissipates energy each cycle [1]. At the peak the energy supplied by the driver per cycle equals the energy dissipated per cycle, so the amplitude stops growing, and more damping means this balance is reached at a smaller amplitude [1]. [6] Examiner insight: “Differences” questions want comparisons, not two separate descriptions, so use words such as lower, broader and shifted; the peak shift is only supported by the table to the resolution of the data.

9. (a) Critical damping [1]. (b) Light damping: the door would oscillate about the closed position with amplitude decreasing gradually [1]. Critical damping: displacement falls to zero in the shortest time with no oscillation [1]. Heavy damping: no oscillation, but the displacement falls to zero more slowly than for critical damping [1]. [4] Examiner insight: A comparison of times is needed for critical and heavy damping; saying both “do not oscillate” will not distinguish them.

10. (a) The footsteps act as a periodic driving force at a frequency close to the natural frequency of the bridge [1]. Resonance occurs, so energy is transferred to the bridge efficiently and the amplitude grows large [1]. (b) Increase the damping (for example with energy-absorbing dampers) [1]. Change the stiffness or mass of the bridge so its natural frequency is moved away from the stepping frequency [1]. [4] Examiner insight: Link each suggestion to the physics of resonance; “make the bridge stronger” does not address the driving and natural frequencies.

11. (a) λ = 2L = 1.30 m [1]. v = fλ = 220 × 1.30 = 286 m s⁻¹ [1]. (b) L = v/(4f) = 340 ÷ (4 × 220) [1] = 0.386 m [1]. (c) Pipe third harmonic = 3 × 220 = 660 Hz [1]. String third harmonic = 3 × 220 = 660 Hz, so they match [1]. (d) The open pipe of the same length has first harmonic 340 ÷ (2 × 0.386) = 440 Hz [1]. This equals the second harmonic of the string, so the string can drive it at resonance at that frequency [1]. [8] Examiner insight: A “show that” needs both values calculated and compared; stating that the harmonics are equal without evidence does not demonstrate the match.

12. (a) Fixed end is a node and free end an antinode, so L = λ/4 and λ = 2.4 m [1]. f₁ = 90 ÷ 2.4 = 37.5 Hz [1]. Only odd harmonics occur [1]. (b) Both ends are displacement nodes, so L = λ/2 and λ = 1.00 m [1]. f₁ = 340 ÷ 1.00 = 340 Hz [1]. [5] Examiner insight: Decide node or antinode at each end first; the free end of a string is an antinode, unlike a fixed end.

Where marks are usually lost

• Using the open-pipe formula for a pipe closed at one end, or allowing even harmonics there.
• Numbering harmonics from a higher mode instead of the lowest frequency.
• Treating the length of a resonance tube as one wavelength, rather than using successive resonances half a wavelength apart.
• Stating damping effects without comparing: the peak is lower, broader and slightly lower in frequency.
• Failing to distinguish critical from heavy damping, which both avoid oscillation.

Next steps

• Recap with the revision notes.
• Re-read the study guide.
• Return to the IB Physics hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. Section C.4, standing waves and resonance.