A string fixed at both ends has length 0.80\ m and the wave speed is 200\ m s^-1. Find the fundamental frequency. [2 marks]

SingaporePhysicsSyllabus dot point

How do two waves travelling in opposite directions form a stationary wave with fixed nodes and antinodes?

Explain the formation of stationary waves by superposition, identify nodes and antinodes, and apply the conditions for stationary waves on strings and in air columns

A focused answer to the H2 Physics learning outcome on stationary waves. Their formation by superposition of two opposite progressive waves, nodes and antinodes, and the harmonic conditions on strings and in open and closed air columns.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to explain how stationary (standing) waves form by superposition, to identify nodes and antinodes, and to apply the harmonic conditions for stationary waves on strings and in air columns. Stationary waves underlie all musical instruments and many resonance experiments.

The answer

Formation of a stationary wave

A stationary wave forms when two progressive waves of the same frequency, wavelength and amplitude travel in opposite directions and superpose. In practice this happens when a wave reflects off a boundary and overlaps the incoming wave.

By the principle of superposition:

at points where the two waves are always in antiphase, they cancel permanently, giving zero displacement (a node),
at points where they are always in phase, they reinforce, giving maximum displacement oscillation (an antinode).

Nodes and antinodes

The spacing between adjacent nodes (or adjacent antinodes) is half a wavelength:

\text{node-to-node distance} = \frac{\lambda}{2}

A node and its neighbouring antinode are a quarter wavelength apart. All points between two adjacent nodes oscillate in phase; the amplitude varies with position from zero at the nodes to maximum at the antinodes.

Stationary versus progressive waves

A stationary wave does not transfer energy along the medium (the energy is stored, oscillating between kinetic and potential), it has fixed nodes and antinodes, and its amplitude varies with position. A progressive wave transfers energy, has no fixed nodes, and every point oscillates with the same amplitude.

Strings fixed at both ends

A string fixed at both ends must have a node at each end. The allowed wavelengths are:

L = \frac{n\lambda}{2} \implies \lambda_n = \frac{2L}{n}, \quad f_n = \frac{nv}{2L}, \quad n = 1, 2, 3, \dots

The fundamental ( $n = 1$ ) has $\lambda = 2L$ ; the harmonics are integer multiples of the fundamental frequency.

Air columns

For a pipe, the closed end is a displacement node and the open end is (approximately) an antinode:

Open at both ends: antinodes at both ends, fundamental $\lambda = 2L$ , all harmonics present, $f_n = \dfrac{nv}{2L}$ .
Closed at one end: node at the closed end, antinode at the open end, fundamental $\lambda = 4L$ , only odd harmonics present, $f_n = \dfrac{nv}{4L}$ with $n = 1, 3, 5, \dots$

Worked example

A pipe closed at one end has length $0.85\ \text{m}$ . The speed of sound in air is $340\ \text{m s}^{-1}$ . Find (a) the fundamental frequency and (b) the frequency of the next harmonic present.

Step 1: Set up the fundamental for a closed pipe

A pipe closed at one end has a node at the closed end and an antinode at the open end, so the length is a quarter wavelength in the fundamental: $\lambda_1 = 4L = 4 \times 0.85 = 3.4\ \text{m}$ .

Step 2: Find the fundamental frequency

f_1 = \frac{v}{\lambda_1} = \frac{340}{3.4} = 100\ \text{Hz}

Step 3: Identify the next harmonic present

A closed pipe supports only odd harmonics, so the next is the third harmonic, $n = 3$ .

Step 4: Find its frequency

f_3 = 3f_1 = 3 \times 100 = 300\ \text{Hz}

The fundamental is $100\ \text{Hz}$ and the next harmonic present is $300\ \text{Hz}$ (the third), because even harmonics are absent in a closed pipe.

Examples in context

Example 1. A guitar string. Plucking a guitar string sets up a stationary wave with nodes at the fixed ends. Pressing a fret shortens the vibrating length $L$ , raising the fundamental frequency $f_1 = v/2L$ and so the pitch. The integer harmonics give the note its characteristic timbre.

Example 2. A bottle as a closed pipe. Blowing across the top of a partly filled bottle excites a stationary wave in the air column, which acts as a pipe closed at the water surface. Adding water shortens the air column, raising the fundamental frequency $f_1 = v/4L$ and the pitch, a everyday demonstration of the closed-pipe condition.

Try this

Q1. State two differences between a stationary wave and a progressive wave. [2 marks]

Cue. Stationary wave: no net energy transfer, fixed nodes and antinodes, amplitude varies with position. Progressive wave: transfers energy, no fixed nodes, uniform amplitude.

Q2. A string fixed at both ends has length $0.80\ \text{m}$ and the wave speed is $200\ \text{m s}^{-1}$ . Find the fundamental frequency. [2 marks]

Cue. $\lambda = 2L = 1.6\ \text{m}$ ; $f_1 = v/\lambda = 200/1.6 = 125\ \text{Hz}$ .

Q3. Explain why a pipe closed at one end produces only odd harmonics. [2 marks]

Cue. It must have a node at the closed end and an antinode at the open end; only standing waves with an odd number of quarter wavelengths satisfy both boundary conditions, giving $n = 1, 3, 5, \dots$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA string of length

0.60\ \text{m}

is fixed at both ends and vibrates in its fundamental mode at

256\ \text{Hz}

. (a) Find the wavelength of the fundamental. (b) Find the wave speed on the string. (c) State the frequency of the third harmonic.

Show worked answer →

(a) For a string fixed at both ends, the fundamental has a node at each end and one antinode in the middle, so the length is half a wavelength: $\lambda = 2L = 2 \times 0.60 = 1.2\ \text{m}$ .

(b) Wave speed: $v = f\lambda = 256 \times 1.2 = 307\ \text{m s}^{-1}$ .

(c) Harmonics of a fixed string are integer multiples of the fundamental, so the third harmonic is $3 \times 256 = 768\ \text{Hz}$ .

Markers reward $\lambda = 2L$ for the fundamental on a string fixed at both ends, the wave speed from $v = f\lambda$ , and the third harmonic as three times the fundamental.

Original4 marks(a) Explain how a stationary wave is formed on a stretched string. (b) State two differences between a stationary wave and a progressive wave.

Show worked answer →

(a) A stationary wave forms when two progressive waves of the same frequency and amplitude travel in opposite directions and superpose. On a string this happens when the wave reflects from a fixed end and overlaps the incident wave. At points where they are always in antiphase the displacement is permanently zero (nodes); where they are always in phase the displacement oscillates with maximum amplitude (antinodes).

(b) Any two of: a stationary wave stores energy and does not transfer it along the medium, whereas a progressive wave transfers energy; a stationary wave has fixed nodes and antinodes, whereas every point on a progressive wave oscillates with the same amplitude; the amplitude varies with position in a stationary wave but is uniform in a progressive wave; all points between adjacent nodes oscillate in phase in a stationary wave.

Markers reward the formation by superposition of two oppositely travelling identical waves with reflection, and two valid distinguishing features.

What this dot point is asking

The answer

Formation of a stationary wave

Nodes and antinodes

Stationary versus progressive waves

Strings fixed at both ends

Air columns

Examples in context

Try this

Exam-style practice questions

Related dot points