A guitarist presses a string against the 12th fret, exactly halving the string's effective length while tension and mass per unit length remain unchanged. What happens to the fundamental frequency?
AIt doubles — halving L means f₁ = v/(2L) doubles, raising pitch by one octave
BIt stays the same — the physical properties of the string (tension, mass) haven't changed
CIt halves — the shorter string has less distance to vibrate so it is slower
DIt increases by a factor of √2 — because frequency scales as the square root of length
Wave speed v = √(T/μ) depends only on tension and mass per unit length, both unchanged. The fundamental frequency f₁ = v/(2L). Halving L while keeping v fixed doubles f₁. This is a one-octave increase — exactly the octave, because the second harmonic of the open string (f₂ = 2f₁) equals the fundamental of the half-length string. This is why the 12th fret is exactly at the midpoint of the string, and it is why every fret has a precise mathematical location based on the twelfth root of 2.
Question 2 Multiple Choice
A violinist tightens a string (increases tension T) while its length and mass per unit length remain unchanged. Which statement correctly describes the effect on the string's resonant frequencies?
AOnly the fundamental frequency increases; higher harmonics remain unchanged because they depend only on string length
BAll harmonic frequencies f_n = nv/(2L) increase proportionally, because v = √(T/μ) increases and all harmonics scale with wave speed
CWave speed decreases with higher tension because the string resists deformation more strongly
DTimbre changes but pitch stays the same — tension affects only which harmonics are amplified, not their frequencies
All normal mode frequencies are f_n = nv/(2L). Since n, L are fixed, all frequencies scale directly with v = √(T/μ). Increasing T increases v, which proportionally increases every harmonic frequency. The fundamental and all overtones rise together, maintaining their integer-multiple ratios (the harmonic series is preserved). Tuning a string by tightening it raises the pitch of all its harmonics simultaneously — the string sounds higher but retains its tonal character. Option C is incorrect: higher tension increases restoring force, which increases wave speed, not decreases it.
Question 3 True / False
When a guitar string is plucked, multiple normal modes are excited simultaneously, and the relative amplitudes of those harmonics — not just the fundamental — determine the timbre of the resulting sound.
TTrue
FFalse
Answer: True
Plucking at a particular point on the string excites a superposition of normal modes. The fundamental determines the perceived pitch, but the relative amplitudes of the harmonics shape the tonal quality (timbre). A violin and a guitar playing the same pitch differ primarily in which harmonics are emphasized. Plucking near the bridge excites more high harmonics (brighter, harsher sound); plucking near the midpoint suppresses the second harmonic and above (warmer sound). The Fourier analysis of the initial displacement determines the harmonic content — a core reason why Fourier decomposition is so musically relevant.
Question 4 True / False
A string fixed at both ends can sustain standing waves at wavelengths λₙ = L/n, where L is the string length and n = 1, 2, 3, ..., because each wavelength fits an integer number of full waves into the string.
TTrue
FFalse
Answer: False
The correct condition is λₙ = 2L/n, not L/n. The fixed endpoints must be nodes (zero displacement), and the constraint is that an integer number of *half-wavelengths* must fit within L: L = nλₙ/2, giving λₙ = 2L/n. For the fundamental (n = 1), the half-wavelength equals L, so the full wavelength λ₁ = 2L is *twice* the string length. The string contains exactly half a wave at the fundamental. Stating λₙ = L/n would mean fitting full wavelengths — this would give the correct positions for nodes but would require nodes at places other than the endpoints, violating the boundary conditions for n = 1.
Question 5 Short Answer
Explain why a string fixed at both ends can only sustain certain discrete frequencies rather than vibrating at any frequency.
Think about your answer, then reveal below.
Model answer: The fixed endpoints impose boundary conditions: displacement must be zero at both ends at all times (these points cannot move). A standing wave can persist only if both endpoints are nodes. This geometric constraint limits which wavelengths fit: only wavelengths λₙ = 2L/n (for integer n) place nodes at both x = 0 and x = L. All other wavelengths produce a displacement at one or both endpoints, which cannot persist — the driving and reflected waves destructively interfere and cancel. Since frequency is f = v/λ and wave speed v is fixed by the string's tension and mass, each allowed wavelength corresponds to exactly one frequency f_n = nv/(2L). The discreteness is a consequence of the boundary conditions, not of any inherent property of waves.
This is a physical realization of an eigenvalue problem: the boundary conditions select a discrete spectrum of allowed modes from a continuous infinity of possible waves. The same mathematics describes quantum particle in a box (infinite square well), modes of electromagnetic cavities, and acoustic resonances in organ pipes — any wave equation on a bounded domain with fixed-endpoint conditions produces a discrete spectrum of normal modes.