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Lesson 7 of 8

Waves · Lesson 7 of 8

Beats and Musical Sound

“Derive beat frequency, use beats to tune strings, and connect harmonic mixtures with musical timbre and resonant objects.”

Learning Objectives

• Describe beats as slow variations of amplitude and intensity from nearby frequencies. • Derive the average oscillation frequency and the beat frequency without a factor-of-two error. • Use a controlled change in tension to resolve the two possible frequencies inferred from beats. • Explain why instruments can share a fundamental frequency yet sound different.

A note can grow and fade without either source changing

Sound two nearly matched tuning forks together. Instead of a steady loudness, you hear repeated increases and decreases: beats. Each fork can vibrate steadily, but the two displacements at your ear do not keep a constant relative phase because their frequencies differ slightly. Sometimes they reinforce; later they oppose. The relative phase then wraps through another full cycle and the loudness pattern repeats. The fast oscillation is heard as a note, while the slow variation is heard as waxing and waning loudness.

The frequencies should be close enough that the slow envelope can be distinguished from the rapid vibration. Comparable amplitudes make the effect easy to hear. Exactly equal amplitudes can cancel fully at minima; unequal amplitudes leave a nonzero minimum. Widely separated sound frequencies need not be heard as clear slow beats. The mathematical sum remains valid, but the particular slow-modulation interpretation has a limited regime.

Separate the fast oscillation from its slow envelope

Fix the listening location so we need not carry x through the calculation. Choose equal amplitude a and a convenient common phase: s₁ = a cos(ω₁t) and s₂ = a cos(ω₂t). Let ω₁ > ω₂ for now. Superposition gives s = a[cos(ω₁t) + cos(ω₂t)]. Using cos U + cos V = 2 cos[(U − V)/2] cos[(U + V)/2] separates the slow difference from the fast average.

The beat waveformLaTeX
The average angular frequency is ωₐ = (ω₁ + ω₂)/2; the signed envelope coefficient varies with ωᵦ = (ω₁ − ω₂)/2.

If |ω₁ − ω₂| is much smaller than either frequency, the first cosine changes slowly while the second completes many cycles. The rapid oscillation has frequency fₐ = (f₁ + f₂)/2. Its slowly varying physical amplitude is 2a|cos[(ω₁ − ω₂)t/2]|. The absolute value matters: a negative prefactor reverses the rapid oscillation’s phase, but does not produce a negative loudness or amplitude.

Why beat frequency is the full difference

Loud maxima occur when the envelope cosine equals either +1 or −1. Therefore the amplitude magnitude repeats when its argument changes by π, not only when the signed cosine completes a full 2π cycle. Write |ω₁ − ω₂|Tᵦ/2 = π. Rearranging gives Tᵦ = 2π/|ω₁ − ω₂| = 1/|f₁ − f₂|. There are two loudness maxima during one full period of the signed envelope cosine. This removes the common but incorrect factor of two.

Beat period and frequencyLaTeX
Use a nonnegative difference regardless of which source has the larger frequency. The slow-envelope interpretation assumes nearby frequencies.

For a linear acoustic wave, mean intensity is related to the square of the oscillation amplitude. Averaging over the rapid vibration while the envelope changes slowly gives intensity proportional to cos²[(ω₁ − ω₂)t/2]. Its repetition rate is again |f₁ − f₂|. Beats are a modulation of intensity, not a series of occasions on which the source starts and stops. If the frequencies become exactly equal, the relative phase stops drifting and the reinforcement or cancellation is stationary rather than beating.

Reading a beat graph

Problem
Combine two equal-amplitude signals at 11 Hz and 9 Hz at the same listening point. Describe the resulting waveform and mark successive maxima.

  1. 1.The individual angular frequencies are 22π and 18π rad s⁻¹. Their sum is s = 2a cos(2πt) cos(20πt).
  2. 2.The fast oscillation is at 10 Hz. The signed envelope cosine is at 1 Hz, but its magnitude reaches a maximum every 0.5 s.
  3. 3.The beat frequency is |11 − 9| = 2 Hz. Maxima occur at t = 0, 0.5, 1.0,… s; zero-amplitude minima occur at t = 0.25, 0.75,… s.
  4. 4.On a graph of s against t, draw the rapid oscillations inside the upper and lower bounds ±2a|cos(2πt)|. The bounds touch zero at the destructive-interference times.
Beats come from a drifting relative phase11 Hz signal9 Hz signalSum: 10 Hz fast oscillation, 2 Hz beats00.250.50.751t / sLoud maxima every 0.5 s; amplitude minima at 0.25 s and 0.75 s.
Beat waveform— Exact curves for equal amplitudes at 11 and 9 Hz; orange bounds show the physical envelope.

Tuning uses both the beat rate and the direction of change

A beat rate alone gives two possibilities. If one string has frequency 427 Hz and the beat rate is 5 Hz, the other could be 422 Hz or 432 Hz. To decide, perturb a known parameter. For a fixed length and linear density, a string’s natural frequencies are proportional to √F. Raising tension raises frequency. Observe whether that makes the frequency gap grow or shrink, using a sufficiently small adjustment so it does not cross the reference frequency without being noticed.

Finding which sitar string is lower

Problem
Strings A and B produce 5 Hz beats. A has frequency 427 Hz. A slight increase in B’s tension reduces the beat rate to 3 Hz. Find B’s original frequency.

  1. 1.Initially |fᴮ − 427| = 5, so fᴮ is either 422 or 432 Hz.
  2. 2.Increasing B’s tension raises fᴮ. If it began at 432 Hz, this would take it farther above 427 and increase the beat rate.
  3. 3.The observed decrease therefore places B initially below A under the stated small-adjustment interpretation. Its original frequency was 427 − 5 = 422 Hz.
  4. 4.The post-adjustment gap is 3 Hz. Without further tracking, its magnitude alone does not tell us whether the adjusted string remained below 427 or crossed above it; the original lower-frequency conclusion is what the experiment establishes.
Reducing tension instead of increasing it

Problem
String A initially vibrates at 324 Hz and makes 6 Hz beats with B. Reducing A’s tension slightly reduces the beat rate to 3 Hz. Find B’s frequency.

  1. 1.The initial candidates are fᴮ = 318 Hz and 330 Hz.
  2. 2.Reducing A’s tension lowers fᴬ. If B were 330 Hz, lowering A from 324 Hz would increase the gap, contrary to the observed decrease.
  3. 3.B must be the lower-frequency string at 318 Hz. Moving A down towards it reduces the gap. No exact new tension is needed to resolve the original frequency.

Pitch, harmonic mixture and timbre

An ideal normal mode produces one sinusoidal frequency, but a played instrument generally excites a mixture of modes. Two instruments can share the same fundamental, and hence the same nominal pitch, while differing in the amplitudes and phases of their overtones and in the way the sound begins and fades. These differences contribute to timbre, the quality that lets us distinguish a violin from a sitar. The point of bowing or plucking, the instrument body and its coupling to air all affect the mixture. One frequency number alone does not specify the complete time waveform.

The connection extends beyond strings and air columns. Resonant stone pillars can produce distinct notes when tapped because their elasticity, density, geometry and mechanical support determine their vibrations. The musical-pillar example in the Nellaiappar temple in Tamil Nadu illustrates this physical idea; related examples occur at Hampi and other southern Indian sites. The source groups pillars by their musical role: Shruti for notes or swaras, Gana Thoongal for tunes associated with ragas, and Laya Thoongal for rhythmic response. A stone does not need to be a conventional manufactured instrument to have resonant modes.

Sound can also carry information about obstacles

A bat emits ultrasound and analyses returning echoes. A simple travel-time estimate gives obstacle distance d = vΔt/2 because the sound must go out and back. Differences in arrival direction and timing at the ears help locate the obstacle. The strengths and frequency patterns of returned waves depend on how the target scatters sound; wavelength relative to object size affects the available detail. Thus echoes can carry clues to direction, size and nature as well as distance. This is information transmitted by the wave pattern, not a flow of air from the obstacle to the animal.

Quick Quiz

Quiz

Quick check

Equal-amplitude sounds at 440 Hz and 444 Hz produce a beat rate of

Explanation

The beat rate is the magnitude of their frequency difference.

Quick check

In 2a cos(2πt) cos(20πt), the beat frequency is

Explanation

Both positive and negative envelope peaks give a loudness maximum.

Quick check

A 500 Hz reference makes 4 Hz beats with B. Raising B’s tension slightly reduces the beat rate. B initially has frequency

Explanation

Raising a lower frequency towards the reference reduces the gap.

Quick check

Unequal-amplitude nearby waves generally have beat minima that

Explanation

Opposite phase cancels only equal amplitudes completely.

Quick check

A violin and sitar play the same fundamental but sound different mainly because

Explanation

Timbre depends on more than the fundamental frequency.

Practice Problems

Practice Problems
  1. Starting with s₁ = a cos(ω₁t) and s₂ = a cos(ω₂t), derive the product form of their sum. Identify the average oscillation frequency and explain the absolute value needed for physical amplitude.
  2. Explain why the beat frequency is |f₁ − f₂| rather than half that value. Use maxima at both signs of the envelope cosine and confirm the result from its squared magnitude.
  3. For equal-amplitude 11 Hz and 9 Hz signals, write the resultant equation and sketch one second of the waveform with envelope bounds. Mark maxima and minima in loudness.
  4. A 427 Hz sitar string makes 5 Hz beats with a second string. Raising the second string’s tension slightly reduces the beat rate to 3 Hz. Find its original frequency and explain why the upper candidate is rejected.
  5. String A at 324 Hz gives 6 Hz beats with B. Reducing A’s tension slightly reduces the beat rate to 3 Hz. Determine B’s original frequency, explaining the direction of the frequency change.
  6. Explain why equal fundamental frequencies do not make violin and sitar notes identical. Relate your answer to mode mixtures and the start and decay of a note.
  7. A bat hears an echo 0.040 s after emitting ultrasound in air at 340 m s⁻¹. Estimate the reflector distance and explain what additional echo features can convey direction or object information.

Key Takeaways

Key Takeaways

• Nearby frequencies make a fast oscillation with a slowly varying amplitude envelope. • Beat frequency is the absolute difference |f₁ − f₂|; counting only positive signed-envelope peaks misses half the loud maxima. • A small known tension change identifies which side of a reference frequency an out-of-tune string occupies. • Harmonic mixtures and temporal envelopes help determine timbre; resonant behaviour also appears in air columns, membranes and stone objects.