What is the difference between a standing and a travelling wave?

A travelling wave transfers energy along its direction. A standing wave has fixed nodes and antinodes and transfers no net energy.

Why must there be a node at each end of a fixed string?

The ends are clamped and cannot move, so their displacement is always zero — the definition of a node.

Why are only certain frequencies allowed?

Only wavelengths that fit a whole number of half-wavelengths between the fixed ends satisfy the node condition at both ends.

How does tightening a string raise its pitch?

Greater tension increases the wave speed T/, and with the length fixed, a higher speed means a higher frequency.

Back to course

63, 67–69WavesCore

Standing Waves

Pluck a guitar string and the wave does not travel along it — it seems to stand still, vibrating in fixed loops. That $\textbf{standing wave}$ is two travelling waves moving in opposite directions, superposing to leave points that never move and points that move the most.

35 min Topic 5 · Waves and the Particle Nature of Light

What you'll be able to do

Explain how a standing wave forms from two travelling waves
Identify nodes and antinodes and state their spacing
Find the frequencies of harmonics on a string fixed at both ends
Use $v = \sqrt{T/\mu}$ to relate wave speed to tension and mass per unit length

How standing waves form

A standing wave forms when two waves of the $\textbf{same frequency and amplitude}$ travel in $\textbf{opposite directions}$ and superpose — typically a wave and its reflection.

At some points the two waves always cancel: these are $\textbf{nodes}$ , with zero displacement at all times. Midway between, they always reinforce: these are $\textbf{antinodes}$ , with maximum amplitude.

Unlike a travelling wave, a standing wave transfers no net energy along the string. Energy is stored, oscillating between kinetic and potential forms.

\text{node–node spacing} = \dfrac{\lambda}{2}

Adjacent nodes are half a wavelength apart.

Harmonics on a string

A string fixed at both ends must have a node at each end. The simplest pattern — the $\textbf{fundamental}$ or first harmonic — is a single loop, so the string length is half a wavelength: $L = \lambda/2$ .

The next pattern has two loops ( $L = \lambda$ ), then three ( $L = 3\lambda/2$ ), and so on. In general the $n$ th harmonic has $n$ loops and $L = n\lambda/2$ .

Since the wave speed on the string is fixed, the frequencies are whole-number multiples of the fundamental.

L = \dfrac{n\lambda}{2}

Allowed wavelengths.

f_{n} = \dfrac{nv}{2L} = nf_{1}

Harmonic frequencies.

Example 1A 0.65 m string has wave speed

286\ \mathrm{m\,s^{-1}}

. Find its fundamental frequency and third harmonic.

f_{1} = \dfrac{v}{2L} = \dfrac{286}{1.30} = 220

Hz.

f_{3} = 3 \times 220

Answer

f_{1} = 220\ \mathrm{Hz}

;

f_{3} = 660\ \mathrm{Hz}

Tip — Count the loops in a diagram: that number is the harmonic number $n$ , and each loop is half a wavelength.

Wave speed on a string

The speed of a transverse wave on a string depends on the tension $T$ and the $\textbf{mass per unit length}$ $\mu$ : $v = \sqrt{T/\mu}$ .

Higher tension makes the string restore itself faster, raising the speed; a heavier string responds more sluggishly, lowering it.

Combining with the fundamental gives $f_{1} = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ — the three ways to tune a string: shorter, tighter or lighter all raise the pitch.

v = \sqrt{\dfrac{T}{\mu}}

Wave speed on a string.

f_{1} = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

Fundamental frequency.

Example 1A string of mass per unit length

4.0\times10^{-3}\ \mathrm{kg\,m^{-1}}

is under tension 64 N. Find the wave speed.

v = \sqrt{\dfrac{64}{4.0\times10^{-3}}} = \sqrt{16000}

Answer

v = 126\ \mathrm{m\,s^{-1}}

The required practical

In the practical, a string passes over a pulley with masses hanging from it to set the tension, and a vibration generator drives one end. The frequency is adjusted until clear standing-wave patterns appear.

Varying the length or the tension and recording the resonant frequency tests the relationships above. A graph of $f$ against $\sqrt{T}$ should be a straight line through the origin with gradient $\dfrac{1}{2L\sqrt{\mu}}$ .

Mass per unit length is found by weighing a measured length of the string. Tension is the weight of the hanging masses, $mg$ .

Tip — Plotting $f$ against $\sqrt{T}$ rather than $T$ produces a straight line, which makes the relationship easy to confirm and the gradient easy to use.

Equation recap

\lambda/2

Node–node spacing.

L = n\lambda/2

Allowed modes.

f_{n} = nf_{1}

Harmonics.

v = \sqrt{T/\mu}

Speed on a string.

f_{1} = \tfrac{1}{2L}\sqrt{T/\mu}

Fundamental.

Common mistakes to avoid

Saying adjacent nodes are one wavelength apart.

They are half a wavelength apart.

Using

v = T/\mu

without the square root.

It is

v = \sqrt{T/\mu}

Claiming standing waves transfer energy along the string.

No net energy is transferred; it is stored in the oscillation.

Using the mass of the whole string for

\mu

\mu

is mass per unit length.

Key takeaways

Standing waves form from two identical waves travelling in opposite directions.
Nodes have zero displacement; antinodes maximum; adjacent nodes are $\lambda/2$ apart.
On a fixed string, $L = n\lambda/2$ and $f_{n} = nf_{1}$ .
Wave speed on a string is $v = \sqrt{T/\mu}$ .
Shorter, tighter or lighter strings have higher fundamental frequencies.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 11 lessons