A guitar string can only ring at certain notes, not whatever pitch you like. That is because a string fixed at both ends must have a node at each end, which allows only a special set of wavelengths to fit, the harmonics. Two equations, $\lambda_n = 2L/n$ and $f_n = nv/2L$, give every allowed wavelength and frequency, and explain why one string produces a whole family of musical tones.
A string is fixed at both ends. The two ends cannot move, so each must be a node of any standing wave on it.
Before reading: what is the longest possible standing wave (the fundamental) that fits a string of length $L$ with a node at each end, how does its wavelength compare to $L$? Write your prediction and one sentence of reasoning.
Nodes at both ends allow only a whole number of loops
Because both ends of the string are tied down, a standing wave must have a node at each end, and only wavelengths that fit a whole number of loops between those two nodes are allowed.
A string fixed at both ends forces a node at each end. Each loop of a standing wave is half a wavelength, so the string length must hold a whole number $n$ of half-wavelengths: $L = n\tfrac{\lambda_n}{2}$. Rearranging gives the allowed wavelengths:
$$\lambda_n = \frac{2L}{n} \qquad (n = 1, 2, 3, \dots)$$
A string fixed at both ends has a node at each end, so it holds a whole number of half-wavelengths: $L = n\tfrac{\lambda_n}{2}$, giving the allowed wavelengths $\lambda_n = \dfrac{2L}{n}$ for $n = 1, 2, 3, \dots$ The fundamental ($n=1$) has $\lambda_1 = 2L$.
Pause, copy the highlighted wavelength rule into your book before moving on.
Did you get this? True or false: for a string fixed at both ends, the fundamental wavelength is $\lambda_1 = 2L$.
Combining the wavelengths with v = fλ to get the harmonic frequencies
We just saw that only $\lambda_n = 2L/n$ fits. That raises a question: what frequencies do these wavelengths correspond to? This card answers it → substitute into the wave equation $v = f\lambda$.
Each allowed wavelength maps to one allowed frequency through the wave equation, giving a ladder of harmonics that are whole-number multiples of the fundamental.
Using $v = f\lambda$ with $\lambda_n = 2L/n$ gives the allowed frequencies:
$$f_n = \frac{nv}{2L} \qquad (n = 1, 2, 3, \dots)$$
So $f_2 = 2f_1$, $f_3 = 3f_1$, and so on, the harmonics are integer multiples of the fundamental $f_1 = \dfrac{v}{2L}$.
A guitar string of length $0.65\ \text{m}$ carries waves travelling at $260\ \text{m s}^{-1}$. Find the wavelength and frequency of its third harmonic ($n = 3$).
For a string fixed at both ends, the allowed frequencies are $f_n = \dfrac{nv}{2L}$, integer multiples of the fundamental $f_1 = \dfrac{v}{2L}$. Each harmonic pairs with $\lambda_n = \dfrac{2L}{n}$, consistent with $v = f\lambda$.
Add the highlighted frequency formula to your notes before the check below.
Quick check: A string's fundamental frequency is $120\ \text{Hz}$. What is the frequency of its second harmonic?
Pattern A, Calculate then pattern-spot
A string of length $0.80\ \text{m}$ carries waves at $320\ \text{m s}^{-1}$. In your book: (1) calculate the fundamental frequency $f_1$; (2) calculate $f_2$ and $f_3$ using $f_n = nv/2L$; (3) state in one sentence the pattern linking $f_1$, $f_2$ and $f_3$.
Odd one out: Three of these are true for a string fixed at both ends. Which one is the odd one out?
Dot point: Analyse the frequency and wavelength of standing waves in strings fixed at both ends using λ_n = 2L/n and f_n = nv/2L
Why length, and the standing-wave rules, set the pitch of a string
We just learned the harmonic formulas. That raises a question: how do these explain the guitar trick from the hook, halving the length to jump an octave? This card answers it → shortening $L$ raises every frequency in proportion.
Because frequency depends inversely on length, pressing a string at its midpoint halves the vibrating length and doubles every harmonic frequency, exactly one octave higher.
From $f_n = \dfrac{nv}{2L}$, the frequency is inversely proportional to the length $L$ (for fixed wave speed). So:
This is why a string fixed at both ends can produce a whole musical scale: each note is just a different allowed standing wave, and the length, tension and harmonic number together fix its frequency.
Since $f_n = \dfrac{nv}{2L}$, frequency is inversely proportional to length: halving $L$ doubles every harmonic (one octave up). The fundamental sets the pitch and the higher harmonics give timbre; changing tension changes $v$ and shifts all frequencies.
Pause, copy the highlighted length–frequency relationship into your book before the check below.
Last check: Decide if the statement is true or false.
For a string with fixed wave speed, halving its vibrating length doubles its fundamental frequency.
Fill the gaps: Complete the description of standing waves in strings, then press Check.
A string fixed at both ends must have a at each end, so only certain wavelengths fit. The allowed wavelengths are $\lambda_n = 2L/n$, where $n$ is the number. The lowest allowed frequency, with $n = 1$, is called the . The allowed frequencies are whole-number of that lowest one.
Dot point: Analyse the frequency and wavelength of standing waves in strings fixed at both ends using λ_n = 2L/n and f_n = nv/2L
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A string of length $0.50\ \text{m}$ is fixed at both ends and carries waves at $200\ \text{m s}^{-1}$. Calculate the wavelength and frequency of the fundamental, and explain why only certain frequencies produce a standing wave.
1 mark: $\lambda_1 = 2L = 1.0\ \text{m}$ · 1 mark: $f_1 = v/2L = 200\ \text{Hz}$ · 1 mark: node required at each fixed end · 1 mark: only whole numbers of half-wavelengths fit, giving discrete harmonics
ApplyBand 5(3 marks) 2. A string of length $1.2\ \text{m}$ is fixed at both ends, and waves travel along it at $60\ \text{m s}^{-1}$. Calculate the wavelength and frequency of the fundamental, and the frequency of the third harmonic.
1 mark: $\lambda_1 = 2L = 2.4\ \text{m}$ · 1 mark: $f_1 = v/\lambda_1 = 25\ \text{Hz}$ · 1 mark: $f_3 = 3f_1 = 75\ \text{Hz}$
Dot point: Analyse the frequency and wavelength of standing waves in strings fixed at both ends using λ_n = 2L/n and f_n = nv/2L
AnalyseBand 6(3 marks) 3. Explain why a string fixed at both ends can only support certain wavelengths, and derive the relationship $\lambda_n = 2L/n$ from that condition.
1 mark: both fixed ends must be nodes, since they cannot move · 1 mark: nodes are half a wavelength apart, so a whole number of half-wavelengths must fit into the length $L$ · 1 mark: $L = n\lambda/2$ rearranges to $\lambda_n = 2L/n$
Dot point: Analyse the frequency and wavelength of standing waves in strings fixed at both ends using λ_n = 2L/n and f_n = nv/2L
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): For a string fixed at both ends, the fundamental ($n = 1$) has wavelength $\lambda_1 = 2L = 2 \times 0.50 = 1.0\ \text{m}$. Its frequency is $f_1 = \dfrac{v}{2L} = \dfrac{200}{2 \times 0.50} = 200\ \text{Hz}$. Only certain frequencies produce a standing wave because each fixed end must be a node, so a whole number of half-wavelengths must fit exactly into the length ($L = n\tfrac{\lambda}{2}$). This means only the discrete set of wavelengths $\lambda_n = 2L/n$, and the corresponding frequencies $f_n = nv/2L$, can form a stable standing wave; any other frequency does not fit the boundary condition and does not resonate.
Q2 (3 marks): For a string fixed at both ends the allowed wavelengths are $\lambda_n = 2L/n$. For the fundamental, $n = 1$, so $\lambda_1 = 2L = 2 \times 1.2 = 2.4\ \text{m}$. Its frequency follows from the wave equation: $f_1 = v/\lambda_1 = 60/2.4 = 25\ \text{Hz}$. For the third harmonic, $n = 3$, and $f_n = nv/2L = (3 \times 60)/(2 \times 1.2) = 180/2.4 = 75\ \text{Hz}$. Equivalently, the allowed frequencies are whole-number multiples of the fundamental, so $f_3 = 3f_1 = 3 \times 25 = 75\ \text{Hz}$.
Q3 (3 marks): The two ends of the string are clamped, so they physically cannot move, and every standing wave the string supports must therefore have a node at each end. On a standing wave, adjacent nodes are separated by half a wavelength. For a pattern to have a node at both ends, the length of the string must contain a whole number of half-wavelengths, with no part-fraction left over; any wavelength that does not satisfy this would require the end to move, which is impossible, and the pattern is not sustained. Writing that condition for $n$ half-wavelengths in a string of length $L$ gives $L = n\lambda/2$, where $n = 1, 2, 3, \ldots$ is the harmonic number. Rearranging for the wavelength gives $\lambda_n = \mathbf{2L/n}$. Combining this with the wave equation $v = f\lambda$ gives the corresponding allowed frequencies $f_n = v/\lambda_n = nv/2L$, which are whole-number multiples of the fundamental $f_1 = v/2L$.
You can now calculate harmonic wavelengths with $\lambda_n = 2L/n$, harmonic frequencies with $f_n = nv/2L$, and explain why a string only sounds at discrete pitches. The final lesson of this focus area turns to a moving source, the Doppler effect and its frequency shift.
↩ Back to focus-area hubAt the start you predicted the longest standing wave on a string of length $L$. The fundamental is one loop, which is half a wavelength, so the wavelength is $\lambda_1 = 2L$, twice the string length, not equal to it. If you guessed $\lambda_1 = L$, revisit Card 1, one loop spans only $\tfrac{\lambda}{2}$.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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