Beats — Question 6

Question 6

Slow beats can compare different harmonics of two instruments

Instrument A has fundamental $200\,\mathrm{Hz}$. Its third harmonic and the second harmonic of instrument B produce $3.0\,\mathrm{Hz}$ beats. Increasing the fundamental of B by $0.50\,\mathrm{Hz}$ raises this beat rate to $4.0\,\mathrm{Hz}$. Harmonics are ideal integer multiples.

1. Infer the original fundamental of B.

2. Find the beat difference between their fundamentals and explain why the observed slow modulation belongs to selected harmonics.

Question 6 - Solution

Slow beats can compare different harmonics of two instruments

(a) Method and calculation

The selected A harmonic is $3(200)=600\,\mathrm{Hz}$. The B second harmonic is $600\pm3$, giving $f_B=298.5$ or $301.5\,\mathrm{Hz}$. Raising its fundamental by $0.50$ raises the second harmonic by $1.0\,\mathrm{Hz}$. The beat then becomes $2$ or $4$, so$$f_B=301.5\,\mathrm{Hz}.$$

(b) Interpretation and further calculation

The fundamental frequency separation is$$|301.5-200|=101.5\,\mathrm{Hz}.$$It is the nearly coincident $600$ and $603\,\mathrm{Hz}$ components that give the $3\,\mathrm{Hz}$ slow envelope. A complex musical sound can contain beat rates arising from different harmonic pairs; they do not all equal the difference of fundamentals.

Physical check

The tuning perturbation of an $n$th harmonic is $n$ times the fundamental perturbation, as used for the second harmonic here.