Beats — Question 1

Question 1

The audible beat rate is twice the signed-envelope frequency

Two equal pressure amplitudes $P$ at a point oscillate at $f_1=440\,\mathrm{Hz}$ and $f_2=444\,\mathrm{Hz}$ with zero initial phase difference.

1. Factor the total pressure into a rapid oscillation and a slow factor.

2. Find the repetition time of loudness maxima and explain why the beat rate is $4\,\mathrm{Hz}$ rather than $2\,\mathrm{Hz}$.

Question 1 - Solution

The audible beat rate is twice the signed-envelope frequency

(a) Method and calculation

The cosine-sum identity gives$$p=P\cos(2\pi f_1t)+P\cos(2\pi f_2t)$$$$=2P\cos[\pi(f_2-f_1)t]\cos[2\pi(f_1+f_2)t/2]$$$$=2P\cos(4\pi t)\cos(884\pi t).$$The carrier frequency is $442\,\mathrm{Hz}$ and the signed slow factor has frequency $2\,\mathrm{Hz}$.

(b) Interpretation and further calculation

The amplitude magnitude is $2P|\cos(4\pi t)|$, so a positive and a negative extremum of the signed factor both give a loudness maximum. Successive maxima occur at$$t_m=m/4\,\mathrm s,\qquad T_b=0.250\,\mathrm s,\qquad f_b=4\,\mathrm{Hz}.$$Carrier-averaged intensity is proportional to $\cos^2(4\pi t)$, with the same $4\,\mathrm{Hz}$ repetition.

Physical check

The result counts envelope-magnitude peaks, not every zero crossing of the acoustic pressure.