Beats — Question 10

Question 10

Three equally spaced tones create more than one envelope peak height

Three equal pressure amplitudes $P$ at frequencies $440$, $442$ and $444\,\mathrm{Hz}$ start in phase. Consider the slowly varying signed factor after extracting the central $442\,\mathrm{Hz}$ oscillation.

1. Derive that factor and find its zeros during $0<t<0.50\,\mathrm s$.

2. Compare the envelope magnitudes at $t=0$ and $t=0.25\,\mathrm s$. Explain why counting every local loudness peak as an identical two-tone beat is misleading.

Question 10 - Solution

Three equally spaced tones create more than one envelope peak height

(a) Method and calculation

Pair the outer frequencies:$$p=P\cos(884\pi t)+2P\cos(4\pi t)\cos(884\pi t)$$$$=P[1+2\cos(4\pi t)]\cos(884\pi t).$$The slow factor vanishes when $\cos(4\pi t)=-1/2$, giving $t=1/6$ and $1/3\,\mathrm s$ in the stated interval.

(b) Interpretation and further calculation

At $t=0$ the envelope magnitude is $3P$. At $t=0.25\,\mathrm s$ it is $|P(1-2)|=P$. Thus the corresponding carrier-averaged intensity peaks differ by a factor of nine. The full envelope repeats after $0.50\,\mathrm s$ but contains a weaker intervening peak; it is not the simple equal-peak pattern of two equal tones.

Physical check

The alternating zero intervals are $1/6$ and $1/3\,\mathrm s$ across repeated cycles, confirming that a single identical-beat picture is insufficient.