Beat Frequency from Two Close Tones

When two sound waves with slightly different frequencies combine, you hear a single tone whose amplitude pulses at a rate equal to the difference between the two frequencies. Those pulses are called beats, and they're how piano tuners get strings into perfect agreement.

The Math

Use the sum-to-product identity:

$\sin(\omega_1 t) + \sin(\omega_2 t) = 2\cos\!\left(\dfrac{\omega_1 - \omega_2}{2}\,t\right)\sin\!\left(\dfrac{\omega_1 + \omega_2}{2}\,t\right)$

The result has two factors: a slowly varying envelope at frequency $|\omega_1 - \omega_2|/2$, and a fast oscillation at the average frequency $(\omega_1 + \omega_2)/2$. The envelope is what your ear interprets as "loudness pulsing" — the beat.

The beat frequency (how many loudness peaks per second) is twice the envelope frequency, because $|\cos|$ has two peaks per cycle:

$f_{\text{beat}} = |f_1 - f_2|$

Step-by-Step Solution

Given: Two waves $\sin(10\,t)$ and $\sin(10.5\,t)$ — angular frequencies 10 and 10.5 rad/s.

Find: The beat frequency in Hz.

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Step 1 — Convert angular frequencies to ordinary frequencies.

$f_1 = \dfrac{\omega_1}{2\pi} = \dfrac{10}{2\pi} \approx 1.5915\;\text{Hz}$

$f_2 = \dfrac{\omega_2}{2\pi} = \dfrac{10.5}{2\pi} \approx 1.6711\;\text{Hz}$

Step 2 — Compute the beat frequency.

$f_{\text{beat}} = |f_1 - f_2| = \dfrac{|\omega_1 - \omega_2|}{2\pi} = \dfrac{0.5}{2\pi} \approx 0.0796\;\text{Hz}$

That's about one beat every 12.6 seconds — pretty slow.

Step 3 — Compute the average (carrier) frequency.

$\omega_{\text{avg}} = \dfrac{\omega_1 + \omega_2}{2} = \dfrac{20.5}{2} = 10.25\;\text{rad/s}$

$f_{\text{avg}} = \dfrac{10.25}{2\pi} \approx 1.6313\;\text{Hz}$

This is the fast oscillation that fills in the envelope.

Step 4 — Apply the sum-to-product identity to verify.

$\sin(10\,t) + \sin(10.5\,t) = 2\cos\!\left(\dfrac{10.5 - 10}{2}\,t\right)\sin\!\left(\dfrac{10.5 + 10}{2}\,t\right)$

$= 2\cos(0.25\,t)\sin(10.25\,t)$

The envelope is $\pm 2\cos(0.25 t)$, oscillating slowly. The carrier $\sin(10.25 t)$ fills it in fast.

Step 5 — Period of the envelope.

The envelope $\cos(0.25\,t)$ has period $T_{\text{env}} = 2\pi/0.25 = 8\pi \approx 25.13$ seconds. But since the amplitude (loudness) follows $|\cos(0.25 t)|$, we hear two beats per envelope cycle — so the beat period is $4\pi \approx 12.57$ seconds.

That matches $f_{\text{beat}} \approx 0.0796\;\text{Hz}$ exactly: $1/0.0796 \approx 12.57\;\text{s}$ per beat. ✓

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Answer:

$\boxed{\;f_{\text{beat}} = |f_1 - f_2| = \dfrac{0.5}{2\pi} \approx 0.0796\;\text{Hz}\;}$

That's approximately one beat every 12.57 seconds. The combined wave looks like a fast oscillation (period $\approx 0.6$ s) inside a slowly modulating envelope (period $\approx 25$ s).

Try It

• Adjust the frequency difference widget — wider differences give faster beats.
• The dashed envelope is plotted on top of the sum so you can see it modulating.
• Audio analogy: try setting a difference of 1 Hz — that's roughly the beat rate when two adjacent strings on a piano are slightly out of tune.