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Physics · Ch 11 — Waves

Fundamental Frequency and Overtones

11.8.4

Fundamental Frequency and Overtones

Clamping a string rigidly at both its ends, at x=0x=0 and x=Lx=L, and setting it vibrating (as when a guitar string is plucked) can only produce standing waves whose displacement is forced to vanish at both boundaries, y(x=0,t)=0y(x=0,t)=0 and y(x=L,t)=0y(x=L,t)=0. Since consecutive nodes of any stationary wave pattern are spaced λn/2\lambda_n/2 apart, fitting this pattern exactly between the two fixed boundaries requires n(λn/2)=Ln(\lambda_n/2)=L for some positive integer nn, giving the set of allowed (quantised) wavelengths λn=2L/n\lambda_n=2L/n -- so not every wavelength can form a valid standing wave on this string; only this discrete set is permitted by the boundary conditions. The corresponding natural (allowed) frequencies follow directly from fn=v/λn=nv/(2L)f_n=v/\lambda_n=nv/(2L). The lowest of these, corresponding to n=1n=1, is called the fundamental frequency, f1=v/(2L)f_1=v/(2L); because every higher allowed frequency turns out to be an exact integer multiple of this fundamental, fn=nf1f_n=nf_1, the full sequence of natural frequencies forms a complete harmonic series, f1:f2:f3:…=1:2:3:…f_1:f_2:f_3:\ldots=1:2:3:\ldots. The second natural frequency, $f_2 …

Misc Example 11.22First four harmonics of a plucked guitar string

Worked out. A guitar string of length 80 cm and mass 0.32 g is stretched under a tension of 80 N and plucked, and the task is to find its first four lowest natural frequencies. The linear mass density is μ=(0.32×10−3 kg)/(0.8 m)=0.4×10−3 kg/m\mu = (0.32\times10^{-3}\ \text{kg})/(0.8\ \text{m}) = 0.4\times10^{-3}\ \text{kg/m}, giving a wave speed v=T/μ=80/(0.4×10−3)≈447.2 m/sv=\sqrt{T/\mu}=\sqrt{80/(0.4\times10^{-3})}\approx447.2\ \text{m/s}. The fundamental wavelength for a string fixed at both ends is λ1=2L=1.6 m\lambda_1=2L=1.6\ \text{m}, so the fundamental frequency is f1=v/λ1=447.2/1.6≈279.5 Hzf_1=v/\lambda_1=447.2/1.6\approx279.5\ \text{Hz}. Since the string forms a complete harmonic series, the next three natural frequencies are simply integer multiples of this fundamental: f2=2f1≈559 Hzf_2=2f_1\approx559\ \text{Hz}, $f_3=3f_ …