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

This study guide follows the IB Physics guide, first assessment 2025, and covers section C.4 (standing waves and resonance), which is taught to both SL and HL. The guide adds no higher-level content to C.4, so everything here applies at both levels. It sits within topic C, wave behaviour. You can check your progress against the printable checklist and the IB Physics hub. Condensed recall is in the revision notes and exam-style questions are in the practice set. For the wider course map see the IB DP Physics syllabus guide, and to find out where you stand today try the free diagnostics.

What this unit covers

Guide section	What you must be able to do	Level
C.4 (1)	Explain how a standing wave forms from two identical waves travelling in opposite directions	SL and HL
C.4 (2)	Describe nodes and antinodes, and compare the amplitude and phase of different points on a standing wave	SL and HL
C.4 (3)	Describe and use standing wave patterns in strings and pipes, including fixed, free, open and closed ends	SL and HL
C.4 (4)	Explain resonance, natural frequency, and how amplitude depends on driving frequency	SL and HL
C.4 (5)	Explain how damping changes the maximum amplitude and the resonant frequency	SL and HL
C.4 (6)	Describe what light, critical and heavy damping do to the motion	SL and HL

The numbers in brackets are the order of the bullet points under “Understandings” in the guide. Resonance in real structures and instruments is used throughout as the context for the damping ideas.

1. How a standing wave forms

A travelling wave carries energy along the medium. A standing wave does not. It is the superposition of two waves with the same frequency, wavelength and amplitude, travelling in opposite directions. In a string or pipe the second wave is usually the reflection of the first, and the two keep overlapping at every point.

By the principle of superposition, the displacement at any point is the sum of the two displacements there. At some points the two waves are always in antiphase, so they cancel at every instant. These points are nodes. Halfway between nodes, the two waves are always in phase and add to give the largest oscillation. These points are antinodes.

Key contrasts with a travelling wave:

• The pattern stays in place. The nodes and antinodes do not move along the medium.
• No net energy is transferred along the medium. Energy is stored in the oscillation.
• The amplitude depends on position. In a travelling wave every point reaches the same maximum displacement.

Adjacent nodes are half a wavelength apart, and so are adjacent antinodes. A node and the next antinode are a quarter of a wavelength apart. These distances are the most useful facts in the topic.

2. Amplitude and phase along a standing wave

Let the antinode amplitude be A. A point at distance x from a node has amplitude A sin(2πx/λ). This gives zero at a node and A at an antinode. Every point oscillates at the same frequency as the original waves.

Phase rule. All points between one pair of adjacent nodes (one loop) oscillate in phase. Points in neighbouring loops are in antiphase, a phase difference of π. Points two loops apart are in phase again. Points in the same loop reach maximum displacement together, and all pass through equilibrium together, though their maximum displacements differ.

Worked example 1. The antinode amplitude of a standing wave is 6.0 mm. Find the amplitude of a point one-twelfth of a wavelength from a node, and describe the amplitude of each of the two original travelling waves.

• Phase angle: 2πx/λ = 2π × (1/12) = π/6.
• Amplitude: 6.0 × sin(π/6) = 6.0 × 0.5 = 3.0 mm.
• At an antinode the two waves add, so each has amplitude 6.0 ÷ 2 = 3.0 mm.

Both answers are 3.0 mm, but they describe different things: one is a point on the standing wave, the other is the amplitude of each travelling wave.

3. Standing waves in strings and pipes

The pattern that forms depends on the boundary conditions, the way the medium is constrained at its ends. Only wavelengths that fit those conditions can form a stable standing wave.

String fixed at both ends

Both ends are nodes. The simplest pattern is one loop, with a node at each end and an antinode in the middle. The length is half a wavelength. Higher patterns add loops. For the nth harmonic the string holds n half-wavelengths.

• λ = 2L/n
• f = n v / (2L) = n f₁, where v is the wave speed in the string and f₁ is the frequency of the first harmonic.

The first harmonic is the lowest frequency that fits. The second harmonic has twice that frequency, and so on. Counting nodes, the nth harmonic has n + 1 nodes (ends included) and n antinodes.

The speed v is set by the string: a tighter string, or a lighter one per unit length, gives a higher speed and therefore higher frequencies for the same length.

Worked example 2. A string of length 0.80 m has wave speed 240 m/s. Find the frequencies of the first three harmonics and the wavelength of the third.

• f₁ = v/2L = 240 ÷ 1.60 = 150 Hz.
• f₂ = 300 Hz and f₃ = 450 Hz.
• λ₃ = 2L/3 = 1.60 ÷ 3 = 0.533 m.

Pipe open at both ends

Air at an open end can move freely, so each open end is a displacement antinode. The simplest pattern has an antinode at each end and a node in the middle, so L = λ/2. The harmonic series is the same as for the string.

• f = n v / (2L), with n = 1, 2, 3, …

Pipe closed at one end

The closed end has air that cannot move, a displacement node. The open end is an antinode. The simplest pattern has L = λ/4. The next pattern that fits has L = 3λ/4, then 5λ/4, so only odd multiples of the first harmonic occur.

• f = n v / (4L), with n = 1, 3, 5, …

Here “third harmonic” means three times the first harmonic frequency. The second harmonic does not exist in this pipe.

Other boundary conditions

A string can also have a free end, such as a loop that slides on a rod, which is an antinode. A string fixed at one end and free at the other behaves like the closed pipe: L = λ/4, odd n only, f = n v / (4L). A string free at both ends has an antinode at each end and behaves like the fixed string, with f = n v / (2L). A pipe closed at both ends has a displacement node at each end, so it also follows f = n v / (2L), with n = 1, 2, 3, …

Worked example 3. Air in pipes of length 0.60 m has speed of sound 340 m/s. Find the first harmonic of an open pipe and of a closed pipe, and the next two harmonics of the closed pipe.

• Open: f₁ = 340 ÷ (2 × 0.60) = 283.3, so 283 Hz.
• Closed: f₁ = 340 ÷ (4 × 0.60) = 141.7, so 142 Hz.
• Closed: f₃ = 3 × 141.7 = 425 Hz, and f₅ = 5 × 141.7 = 708 Hz.

The closed pipe sounds an octave lower than the open pipe of the same length. An octave is a frequency ratio of 2.

Worked example 4. A pipe closed at one end sounds its third harmonic at 1020 Hz in air where the speed of sound is 340 m/s. Find its length.

• f₃ = 3v/(4L), so L = 3v/(4 f₃).
• L = (3 × 340) ÷ (4 × 1020) = 1020 ÷ 4080 = 0.25 m.

Using a resonance tube. A tuning fork above a tube whose air column length can be changed gives loud sound whenever the column resonates. Successive resonances are half a wavelength apart in column length. If the fork is 440 Hz and successive resonances are 0.390 m apart, then λ = 0.780 m and v = 440 × 0.780 = 343 m s⁻¹.

4. Resonance and driving frequency

Every oscillating system has one or more natural frequencies, at which it oscillates freely once disturbed. A driving force is a periodic force applied from outside, with its own driving frequency.

When the driving frequency is far from the natural frequency, the system responds weakly. As the driving frequency approaches the natural frequency, the amplitude of oscillation grows. At or very near the natural frequency the amplitude is largest. This is resonance. The graph of amplitude against driving frequency is a peak centred close to the natural frequency.

In the steady state the system oscillates at the driving frequency, not at its natural frequency. Resonance happens because the driving force then keeps adding energy in step with the motion, so the energy stored builds up until losses balance the energy input.

Standing waves in strings and pipes are examples. A string or air column has a set of natural frequencies, the harmonics. Driving it at one of them produces a strong standing wave. That is why the tuning fork above the tube works.

5. Damping and its effect on resonance

Damping is a resistive force, such as friction or air resistance, that removes energy from an oscillation. The amplitude falls over time unless energy is supplied.

On the amplitude against driving frequency graph, increasing damping:

• lowers the maximum amplitude;
• broadens the peak, so the response is spread over a wider range of frequencies;
• moves the peak to a slightly lower frequency than the natural frequency.

With very light damping the resonant frequency is almost exactly the natural frequency and the peak is tall and narrow.

6. Light, critical and heavy damping

Consider a system displaced and then released, with no driving force.

Type	What happens	Typical displacement-time shape
Light damping	Oscillates many times, amplitude decreasing gradually	Oscillation inside a decaying envelope
Critical damping	Returns to equilibrium in the shortest time without oscillating	One smooth curve to zero, no overshoot
Heavy damping	Returns to equilibrium without oscillating, but slowly	A slow curve to zero, taking longer than critical

Critical damping is the borderline: any less and the system overshoots, any more and it is slower. This is why a vehicle suspension or a door closer is designed close to critical damping.

Useful and harmful resonance. Resonance is wanted in musical instruments, where the body or air column reinforces particular notes, and in tuning circuits. It is a danger in bridges and buildings, where wind, footsteps or ground motion at a natural frequency could build up large oscillations. Engineers add damping, or change the stiffness or mass, to move natural frequencies away from the expected driving frequencies.

Common errors

• Saying a standing wave “carries energy along the string”. It does not transfer energy along the medium.
• Placing the nodes of a closed pipe wrongly. The closed end is a displacement node and the open end an antinode.
• Using f = nv/2L for a closed pipe. It needs 4L and odd n.
• Numbering harmonics from a higher mode. The first harmonic is the lowest frequency.
• Saying points in different loops are “out of phase” without giving the phase difference π, or saying all points on a standing wave are in phase.
• Claiming damping has no effect on the frequency of the peak, or that heavy damping makes the amplitude larger.
• Describing critical damping as “no damping”.

Next steps

Condense this with the revision notes, then test yourself with the practice questions. For exam technique see IB DP Physics exam preparation.

Official syllabus

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