Piano string and timbre

Problem

A piano string of length $1{,}2\;\mathrm{m}$, tension $700\;\mathrm{N}$ and linear density $6\cdot 10^{-3}\;\mathrm{kg/m}$. Find the fundamental frequency and the first two harmonics. Comment on how timbre arises.

Solution

Setup. First the wave speed on the string:

$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{700}{6\cdot 10^{-3}}} = \sqrt{1{,}167\cdot 10^{5}} \approx 342\;\mathrm{m/s}$

Fundamental (string fixed at both ends, $f_n = n\,v/2L$):

$f_1 = \frac{v}{2L} = \frac{342}{2\cdot 1{,}2} = \frac{342}{2{,}4} \approx 142\;\mathrm{Hz}$

First two harmonics.

$f_2 = 2 f_1 \approx 285\;\mathrm{Hz}, \qquad f_3 = 3 f_1 \approx 427\;\mathrm{Hz}$

$\ev{f_1 \approx 142\;\mathrm{Hz},\quad f_2 \approx 285\;\mathrm{Hz},\quad f_3 \approx 427\;\mathrm{Hz}}$

How timbre arises. The string does not vibrate in a single mode, but as a superposition of the fundamental and many harmonics, each with its own amplitude. It is the particular “recipe” of harmonic intensities (the spectrum) that defines the timbre: the same note $f_1$ sounds different on a piano, a violin or a voice precisely because the relative weight of the higher harmonics differs.

Linked atoms

Topics: Onde Concepts: Onde stazionarie Skills: Analisi dimensionale