Standing waves and resonance

Lesson C.4 Standing Waves and Resonance

Strings, Air Columns, Resonance and Damping

This lesson explains how standing waves form in strings and air columns, how boundary conditions determine their allowed frequencies, and how resonance and damping affect oscillating systems.

Learning objectives

Explain how standing waves form through superposition and reflection.
Identify nodes and antinodes and describe their spacing.
Determine allowed wavelengths and frequencies for strings and pipes.
Distinguish displacement patterns from pressure patterns in air columns.
Explain resonance using natural and driving frequencies.
Compare light, critical and heavy damping.

1. How standing waves form

A standing wave forms when two waves of the same frequency, wavelength and amplitude travel in opposite directions and superpose. This commonly occurs when an incident wave overlaps with its reflection.

The resulting pattern has fixed positions of zero oscillation amplitude and fixed positions of maximum oscillation amplitude. The pattern does not travel along the medium.

Node: A point where the displacement remains zero throughout the oscillation.

Antinode: A point where the oscillation amplitude is greatest.

An antinode moves back and forth through equilibrium. Its displacement is not always at its maximum value.

Reflection at a fixed end

When a transverse wave on a string reflects from a fixed end, the reflected displacement is inverted. This corresponds to a phase change of $\pi$ radians, or $180^\circ$ . The fixed end must remain a displacement node.

Spacing and phase

The distance between adjacent nodes is $\frac{\lambda}{2}$ .
The distance between adjacent antinodes is $\frac{\lambda}{2}$ .
The distance between a node and its nearest antinode is $\frac{\lambda}{4}$ .
Points between the same pair of adjacent nodes oscillate in phase.
Points in neighbouring loops oscillate in antiphase.

Energy transfer: An ideal standing wave has no net energy transfer along the medium. Its two component travelling waves carry equal energy in opposite directions. Real driven systems still require energy input to replace dissipated energy.

2. Standing waves on a string fixed at both ends

A string fixed at both ends must have a node at each end. Its length must contain a whole number of half-wavelengths.

$\displaystyle L=\frac{n\lambda_n}{2}$

$\displaystyle \lambda_n=\frac{2L}{n}$

$\displaystyle f_n=\frac{v}{\lambda_n}=\frac{nv}{2L}$

$\displaystyle n=1,2,3,\ldots$

Here, $L$ is the string length, $v$ is the wave speed, and $n$ is the harmonic number.

The fundamental and higher harmonics

The fundamental frequency is the lowest allowed frequency. For a string fixed at both ends:

$\displaystyle f_1=\frac{v}{2L}$

First harmonic: one loop, with an antinode at the centre.
Second harmonic: two loops, with a node at the centre.
Third harmonic: three loops.

$\displaystyle f_n=nf_1$

An overtone is an allowed frequency above the fundamental. For this string, the first overtone is the second harmonic, and the second overtone is the third harmonic.

Worked example 1 — String harmonics

A string of length $0.80\,\mathrm{m}$ is fixed at both ends. Waves travel along it at $120\,\mathrm{m\,s^{-1}}$ .

(a) Calculate the fundamental frequency.
(b) Calculate the third-harmonic frequency.

Solution:

(a)

$\displaystyle f_1=\frac{v}{2L}=\frac{120}{2(0.80)}=75\,\mathrm{Hz}$

(b)

$\displaystyle f_3=3f_1=3(75)=225\,\mathrm{Hz}$

Answers: $75\,\mathrm{Hz}$ and $225\,\mathrm{Hz}$ .

Wave speed on a stretched string

The speed of transverse waves on an ideal stretched string is:

$\displaystyle v=\sqrt{\frac{F_T}{\mu}}$

$F_T$ : tension in the string, measured in newtons.
$\mu$ : mass per unit length, measured in $\mathrm{kg\,m^{-1}}$ .

Combining this with the harmonic-frequency equation gives:

$\displaystyle f_n=\frac{n}{2L}\sqrt{\frac{F_T}{\mu}}$

Increasing tension increases wave speed and harmonic frequencies.
Increasing mass per unit length decreases wave speed and frequencies.
Increasing string length decreases frequencies if tension and mass per unit length remain constant.

3. Standing waves in air columns

Sound waves in a pipe are longitudinal waves. Air particles oscillate parallel to the pipe’s length. The standing-wave pattern depends on whether each end is open or closed.

Boundary conditions

Open end: approximately a displacement antinode and a pressure node. Air can move, while the pressure remains close to atmospheric pressure.

Closed end: a displacement node and a pressure antinode. Air cannot move through the rigid boundary, but pressure variations can be large.

The pressure and displacement patterns have their nodes and antinodes at different positions. A displacement node corresponds to a pressure antinode, and vice versa.

In the following equations, end corrections are neglected: the effective length of the air column is taken to be the pipe length.

4. Pipe open at both ends

Both ends are displacement antinodes. The fundamental contains half a wavelength within the pipe.

$\displaystyle L=\frac{n\lambda_n}{2}$

$\displaystyle \lambda_n=\frac{2L}{n}$

$\displaystyle f_n=\frac{nv}{2L}$

$\displaystyle n=1,2,3,\ldots$

All positive integer harmonics are allowed. The frequency equations have the same form as those for a string fixed at both ends, although the displacement boundary conditions differ.

Here, $v$ is the speed of sound in air. Use the value supplied in the question.

5. Pipe closed at one end

The closed end is a displacement node, while the open end is a displacement antinode. The shortest pattern that fits contains one quarter of a wavelength.

$\displaystyle L=\frac{\lambda_1}{4}$

$\displaystyle \lambda_1=4L$

$\displaystyle f_1=\frac{v}{4L}$

Additional allowed patterns contain three, five, seven and further odd numbers of quarter-wavelengths. Therefore, only odd harmonics are allowed in the ideal model.

$\displaystyle L=\frac{n\lambda_n}{4}$

$\displaystyle \lambda_n=\frac{4L}{n}$

$\displaystyle f_n=\frac{nv}{4L}$

$\displaystyle n=1,3,5,7,\ldots$

In this notation, $n$ is the harmonic number and takes only odd values.

The first overtone is the third harmonic. The second overtone is the fifth harmonic.

Worked example 2 — Open–closed pipe

A pipe of length $0.60\,\mathrm{m}$ is closed at one end and open at the other. The speed of sound is $340\,\mathrm{m\,s^{-1}}$ .

(a) Calculate its fundamental frequency.
(b) Calculate its next allowed resonant frequency.

Solution:

(a)

$\displaystyle f_1=\frac{340}{4(0.60)}=141.7\,\mathrm{Hz}$

$\displaystyle f_1\approx142\,\mathrm{Hz}$

(b) The next allowed frequency is the third harmonic:

$\displaystyle f_3=3f_1=3\left(\frac{340}{2.40}\right)=425\,\mathrm{Hz}$

Answers: Approximately $142\,\mathrm{Hz}$ and $425\,\mathrm{Hz}$ .

Exam tip

First identify whether the question describes displacement or pressure. Then apply the correct boundary conditions. Label the pattern before selecting an equation.

6. Natural frequency and forced oscillations

A system’s natural frequency is the frequency at which it oscillates freely after being displaced and released, in the absence of significant damping.

A forced oscillation occurs when an external periodic force drives the system. After initial transients have decayed, the system oscillates at the driving frequency.

Mass–spring system

$\displaystyle T=2\pi\sqrt{\frac{m}{k}}$

$\displaystyle f_0=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Here, $m$ is the oscillating mass and $k$ is the spring constant. The ideal model assumes the spring obeys Hooke’s law.

Simple pendulum

$\displaystyle T=2\pi\sqrt{\frac{L}{g}}$

$\displaystyle f_0=\frac{1}{2\pi}\sqrt{\frac{g}{L}}$

These pendulum equations apply to small angular displacements. Here, $L$ is the pendulum length and $g$ is the gravitational field strength.

7. Resonance

Resonance is the large response of a system driven at a frequency near one of its natural frequencies. For a lightly damped oscillator, its displacement amplitude is greatest when the driving frequency is close to its natural frequency.

$\displaystyle f_{\mathrm{drive}}\approx f_0$

The driving force supplies energy to the system. Near resonance, it can transfer energy efficiently, producing a large oscillation amplitude.

For a lightly damped oscillator near resonance, the driving force is approximately in phase with the oscillator’s velocity. It therefore does positive work efficiently.

In a real system, the amplitude does not increase indefinitely. A steady response occurs when the average energy supplied by the driver equals the average energy dissipated.

Examples

A swing gains a large amplitude when pushed at an appropriate frequency.
Air columns in musical instruments resonate at allowed frequencies.
A vibrating string can excite the body of an instrument and surrounding air.
Structures may respond strongly to periodic forces near their natural frequencies.

Worked example 3 — Spring resonance

A mass of $0.50\,\mathrm{kg}$ is attached to a spring with spring constant $80\,\mathrm{N\,m^{-1}}$ .

(a) Calculate the natural frequency.
(b) Describe the response if the lightly damped system is driven near this frequency.

Solution:

(a)

$\displaystyle f_0=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

$\displaystyle f_0=\frac{1}{2\pi}\sqrt{\frac{80}{0.50}}$

$\displaystyle f_0=2.01\,\mathrm{Hz}\approx2.0\,\mathrm{Hz}$

(b) Driving the system near $2.0\,\mathrm{Hz}$ produces resonance and a large amplitude, limited by damping.

8. Damping

Damping is the dissipation of an oscillator’s mechanical energy through resistive effects such as friction, air resistance and internal losses. In a freely oscillating system, damping reduces the amplitude with time.

The energy is transferred to the surroundings or converted into other forms, usually internal energy. Total energy remains conserved.

Comparison of light, critical and heavy damping

Light damping — underdamped motion

The system continues to oscillate about equilibrium, but successive maximum displacements decrease. For linear damping, the amplitude envelope decays exponentially.

Critical damping

For a system displaced and released from rest, critical damping gives the fastest return towards equilibrium without oscillation.

Heavy damping — overdamped motion

The system returns towards equilibrium without oscillating, but more slowly than a critically damped system.

Reading damping graphs

A graph that repeatedly crosses equilibrium with decreasing peaks shows underdamped motion. A non-oscillating return may be critically damped or overdamped; compare the return times to distinguish them.

These graphs normally show displacement against time. The amplitude is the envelope of the oscillating peaks.

Application: Vehicle suspension

Shock absorbers dissipate energy and reduce repeated bouncing after a vehicle passes over a bump. A simplified design aim is a rapid settling response with little oscillation, often discussed using critical damping.

9. How damping affects resonance

For the same driving-force amplitude, increasing damping changes the displacement response:

The maximum amplitude decreases.
A distinct resonance peak becomes broader and less sharp.
The frequency of maximum displacement can shift below the undamped natural frequency.
With sufficiently strong damping, a distinct displacement resonance peak disappears.

Optional extension: Linear damping

A common model uses a damping force proportional to velocity:

$\displaystyle F_d=-bu$

Here, $b$ is the damping coefficient and $u$ is the instantaneous velocity. The negative sign shows that the force opposes the motion.

For underdamped free motion in this model, the amplitude envelope is:

$\displaystyle A(t)=A_0e^{-bt/(2m)}$

Focus first on explaining the qualitative changes to displacement–time graphs and resonance curves.

10. Practice questions

Question 1 — Nodes and wavelength

Adjacent nodes on a standing wave are separated by $0.15\,\mathrm{m}$ . Calculate:

(a) the wavelength;
(b) the distance between a node and its nearest antinode.

Show answer

(a) $\lambda=2(0.15)=0.30\,\mathrm{m}$ .

(b) $\frac{\lambda}{4}=0.075\,\mathrm{m}$ .

Question 2 — String harmonics

A string fixed at both ends has length $1.20\,\mathrm{m}$ . Its third-harmonic frequency is $150\,\mathrm{Hz}$ .

(a) Find the fundamental frequency.
(b) Calculate the wave speed.
(c) State the number of nodes, including the endpoints, in the third harmonic.

Show answer

(a) $f_1=\frac{150}{3}=50.0\,\mathrm{Hz}$ .

(b) $v=2Lf_1=2(1.20)(50.0)=120\,\mathrm{m\,s^{-1}}$ .

(c) Four nodes: two endpoints and two internal nodes.

Question 3 — Comparing pipes

Two pipes each have length $0.85\,\mathrm{m}$ . One is open at both ends; the other is closed at one end. The speed of sound is $340\,\mathrm{m\,s^{-1}}$ .

(a) Calculate the fundamental frequency of each pipe.
(b) Find the first-overtone frequency of each pipe.

Show answer

(a) Open–open: $f_1=\frac{340}{2(0.85)}=200\,\mathrm{Hz}$ .

Open–closed: $f_1=\frac{340}{4(0.85)}=100\,\mathrm{Hz}$ .

(b) Open–open: first overtone $=2f_1=400\,\mathrm{Hz}$ .

Open–closed: first overtone $=3f_1=300\,\mathrm{Hz}$ .

Question 4 — Resonance and damping

An oscillator has a natural frequency of $3.0\,\mathrm{Hz}$ . It is driven by a periodic force whose amplitude remains constant.

(a) Describe the response as the driving frequency approaches $3.0\,\mathrm{Hz}$ when damping is light.
(b) Explain how increasing damping changes the resonance peak.
(c) Distinguish critical damping from overdamping for a system displaced and released from rest.

Show answer

(a) The amplitude increases towards a large maximum near the natural frequency because the system resonates.

(b) The peak becomes lower and broader because energy is dissipated more rapidly. Its position can shift to a lower driving frequency.

(c) Both return without oscillating. Critical damping gives the fastest such return; an overdamped system returns more slowly.

Summary

Standing waves form from equal-amplitude waves travelling in opposite directions.
Nodes have zero amplitude; antinodes have maximum amplitude.
Adjacent nodes are separated by $\frac{\lambda}{2}$ .
Fixed–fixed string: $f_n=\frac{nv}{2L}$ , where $n=1,2,3,\ldots$ .
Open–open pipe: $f_n=\frac{nv}{2L}$ , where $n=1,2,3,\ldots$ .
Open–closed pipe: $f_n=\frac{nv}{4L}$ , where $n=1,3,5,\ldots$ .
An open end is a displacement antinode and a pressure node.
A closed end is a displacement node and a pressure antinode.
Resonance produces a large response near a natural frequency.
Damping dissipates mechanical energy and limits resonance amplitude.

The Creator Who Sustains All Things

Memory verse — Colossians 1:17 (KJV):
“And he is before all things, and by him all things consist.”

Creation: Order, sound and music

Standing waves help us understand the sounds produced by strings and air columns in musical instruments. Their allowed frequencies follow patterns that we can investigate and describe mathematically. As Christians, we receive the created world and our ability to explore it as gifts from God.

Colossians 1:16–17 teaches that all things were created through Jesus Christ and for Him, and that in Him all things hold together. Hebrews 1:2–3 also presents the Son as the One through whom God made the universe and who sustains all things by His powerful word. The physical processes we study belong to His creation.

Music gives us one way to respond in gratitude. Psalm 33:3 encourages singing and skilful music in praise to God. Learning how instruments work can deepen our appreciation of the opportunities we have to worship Him with creativity and care.

The Fall: Our need for restoration

Genesis 3 describes humanity’s disobedience and broken fellowship with God. Sin affects our relationships, choices and use of our gifts. We need forgiveness and a restored relationship with our Creator. First John 1:8–9 calls us to acknowledge our sin and come honestly to God.

Redemption: Peace through Jesus Christ

Colossians 1:19–22 teaches that reconciliation comes through Jesus Christ and the blood of His cross. The Creator entered His creation to rescue us. Through His death and resurrection, He offers forgiveness, peace with God and new life.

We can turn to Jesus in repentance and faith, trusting His grace. Our response includes using our knowledge, abilities and creativity to honour Him and serve others.

Restoration: A life of worship

Revelation 21:3–5 points to God’s promised restoration: His people will dwell with Him, and sorrow and death will pass away. As we look forward to that hope, we can worship Him today through prayer, music, honest work and love for others.

A personal invitation: Receive Jesus Christ as your personal God, Saviour and Lord. Read and study the Bible, spend time in prayer, and connect with other believers in a local church so that you can grow in faith and in your knowledge of God.

Reflection: How can you use your learning, musical abilities or other gifts to praise God and encourage someone this week?

Prayer: Lord Jesus, thank You for creating and sustaining all things. Thank You for the gifts of sound, music and understanding. Forgive my sins and restore my relationship with You. Help me study faithfully, use my gifts wisely, and grow through Your Word, prayer and fellowship with other believers. Amen.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Standing waves and resonance

Lesson C.4 Standing Waves and Resonance

Learning objectives

1. How standing waves form

Reflection at a fixed end

Spacing and phase

2. Standing waves on a string fixed at both ends

The fundamental and higher harmonics

Worked example 1 — String harmonics

Wave speed on a stretched string

3. Standing waves in air columns

Boundary conditions

4. Pipe open at both ends

5. Pipe closed at one end

Worked example 2 — Open–closed pipe

Exam tip

6. Natural frequency and forced oscillations

Mass–spring system

Simple pendulum

7. Resonance

Examples

Worked example 3 — Spring resonance

8. Damping

Light damping — underdamped motion

Critical damping

Heavy damping — overdamped motion

Reading damping graphs

Application: Vehicle suspension

9. How damping affects resonance

Optional extension: Linear damping

10. Practice questions

Question 1 — Nodes and wavelength

Question 2 — String harmonics

Question 3 — Comparing pipes

Question 4 — Resonance and damping

Summary

The Creator Who Sustains All Things

Creation: Order, sound and music

The Fall: Our need for restoration

Redemption: Peace through Jesus Christ

Restoration: A life of worship

Related

Leave A Reply Cancel reply

Company

Links

Support

Recommend

alphy school publishers

Customise Consent Preferences

Log in with your account

Modal title