Wavelengths of three notes

Problem

Rank from shortest to longest the wavelengths in air of A$_3$ ($220\;\mathrm{Hz}$), A$_4$ ($440\;\mathrm{Hz}$), A$_5$ ($880\;\mathrm{Hz}$). Use $v = 343\;\mathrm{m/s}$.

Solution

Principle. At a fixed speed of sound, $\lambda = v/f$: wavelength is inversely proportional to frequency. The higher the note (greater frequency), the shorter the wavelength.

Calculation.

$\lambda_{5} = \frac{343}{880} \approx 0{,}39\;\mathrm{m}$

$\lambda_{4} = \frac{343}{440} \approx 0{,}78\;\mathrm{m}$

$\lambda_{3} = \frac{343}{220} \approx 1{,}56\;\mathrm{m}$

From shortest to longest:

$\ev{\lambda_{\text{la}_5} < \lambda_{\text{la}_4} < \lambda_{\text{la}_3}}$

Every octave jump (frequency doubled) halves the wavelength.

Linked atoms

Topics: Onde Concepts: Periodo e frequenza Skills: Ragionamento di ranking