Waves

Sound: Standing Waves, Resonance, Harmonics, and Intensity

5th Year · 6th Year (Leaving Cert)

  • By the end of this lesson students will be able to describe the formation of standing waves and identify nodes and antinodes.
  • By the end of this lesson students will be able to explain the phenomenon of resonance and its applications.
  • By the end of this lesson students will be able to calculate the frequencies and wavelengths of harmonics in vibrating strings, open pipes, and closed pipes.
  • By the end of this lesson students will be able to define sound intensity and apply the decibel scale to compare sound levels (HL).
  • By the end of this lesson students will be able to solve problems involving the speed of sound, frequency, wavelength, and harmonic relationships.

Key concepts

Standing Waves (Stationary Waves)

A standing wave is formed when two waves of the same frequency, amplitude, and wavelength travelling in opposite directions superpose. This typically occurs when a wave reflects back on itself and interferes with the incident wave. In a standing wave, certain points (nodes) remain permanently at rest, while other points (antinodes) vibrate with maximum amplitude. Energy is not transferred along the medium in a standing wave.

Nodes and Antinodes

Nodes are points on a standing wave where the displacement is always zero (no vibration). Antinodes are points on a standing wave where the displacement is maximum (maximum vibration). The distance between two consecutive nodes or two consecutive antinodes is half a wavelength (λ/2). The distance between a node and an adjacent antinode is a quarter wavelength (λ/4).

Resonance

Resonance occurs when a vibrating system is subjected to a periodic force at a frequency equal or very close to its own natural frequency. When this happens, the system absorbs energy efficiently, leading to a rapid increase in the amplitude of its oscillations. Examples include a child on a swing, a tuning fork resonating with another, or the shattering of a glass by a specific sound frequency.

Harmonics in Vibrating Strings

When a string fixed at both ends vibrates, it can form standing waves. The ends of the string must be nodes. The simplest mode of vibration is the fundamental frequency (or first harmonic), where the string vibrates with one antinode in the middle. Higher modes of vibration are called overtones or harmonics. The nth harmonic has n antinodes and (n+1) nodes. The speed of a wave on a string depends on the tension (T) and its mass per unit length (μ).

v = √(T/μ) Fundamental frequency (1st harmonic, n=1): f₁ = v / (2L) Wavelength of nth harmonic: λ_n = 2L / n Frequency of nth harmonic: f_n = n * (v / 2L) = n * f₁ (where L is the length of the string, v is wave speed, n = 1, 2, 3, ...)
Harmonics in Open Pipes

An open pipe is open at both ends. When air in an open pipe vibrates, standing waves are formed. The open ends of the pipe must be antinodes (as air particles can move freely there). The simplest mode of vibration is the fundamental frequency (or first harmonic), with an antinode at each end and one node in the centre. All harmonics (odd and even multiples of the fundamental) are present.

Fundamental frequency (1st harmonic, n=1): f₁ = v / (2L) Wavelength of nth harmonic: λ_n = 2L / n Frequency of nth harmonic: f_n = n * (v / 2L) = n * f₁ (where L is the effective length of the pipe, v is speed of sound, n = 1, 2, 3, ...)
Harmonics in Closed Pipes

A closed pipe is open at one end and closed at the other. The open end must be an antinode, and the closed end must be a node (as air particles cannot move there). The simplest mode of vibration is the fundamental frequency (or first harmonic), with a node at the closed end and an antinode at the open end. Only odd harmonics are present in a closed pipe.

Fundamental frequency (1st harmonic, n=1): f₁ = v / (4L) Wavelength of nth harmonic: λ_n = 4L / (2n-1) Frequency of nth harmonic: f_n = (2n-1) * (v / 4L) = (2n-1) * f₁ (where L is the effective length of the pipe, v is speed of sound, n = 1, 2, 3, ... for the 1st, 3rd, 5th harmonic etc.)
Intensity of Sound

The intensity of sound (I) is defined as the rate at which sound energy is transmitted per unit area perpendicular to the direction of propagation. It is the power (P) carried by the sound wave per unit area (A). Intensity is measured in watts per square metre (W m⁻²). The intensity of sound decreases with the square of the distance from the source (assuming an isotropic source in a non-absorbing medium).

I = P / A
Decibel Scale (HL)

The decibel (dB) scale is a logarithmic scale used to measure sound intensity level (L) because the human ear perceives sound over a very wide range of intensities. It compares the intensity of a sound (I) to a reference intensity (I₀), which is the threshold of human hearing (I₀ = 1.0 × 10⁻¹² W m⁻²).

L = 10 log₁₀ (I / I₀) (where L is the sound intensity level in decibels (dB), I is the sound intensity in W m⁻², and I₀ is the reference intensity, 1.0 × 10⁻¹² W m⁻²)

Key facts to remember

  • 1Standing waves are formed by the superposition of two identical waves travelling in opposite directions.
  • 2Nodes are points of zero displacement; antinodes are points of maximum displacement.
  • 3Resonance occurs when the driving frequency matches the natural frequency, leading to maximum amplitude.
  • 4For a vibrating string fixed at both ends, the ends are nodes. All harmonics (f, 2f, 3f...) are present.
  • 5For an open pipe, both ends are antinodes. All harmonics (f, 2f, 3f...) are present.
  • 6For a closed pipe, the closed end is a node and the open end is an antinode. Only odd harmonics (f, 3f, 5f...) are present.
  • 7The speed of sound in air is approximately 340 m s⁻¹ at room temperature.
  • 8Sound intensity (I) is power per unit area (W m⁻²).
  • 9(HL) The decibel scale is logarithmic, comparing sound intensity to a reference intensity I₀ = 1.0 × 10⁻¹² W m⁻².

Worked examples

Example 1

A string of length 0.80 m is fixed at both ends. When plucked, it vibrates at its fundamental frequency of 250 Hz. Calculate: (i) The speed of the wave on the string. (ii) The frequency of its third harmonic.

I(i) For the fundamental frequency (first harmonic) of a string, the wavelength is λ₁ = 2L.
IIGiven L = 0.80 m, so λ₁ = 2 * 0.80 m = 1.60 m.
IIIThe relationship between wave speed (v), frequency (f), and wavelength (λ) is v = fλ.
IVv = f₁ * λ₁ = 250 Hz * 1.60 m = 400 m s⁻¹.
V(ii) The frequency of the nth harmonic for a string is f_n = n * f₁.
VIFor the third harmonic, n = 3.
VIIf₃ = 3 * f₁ = 3 * 250 Hz = 750 Hz.

Answer

(i) Speed of wave = 400 m s⁻¹ (ii) Frequency of third harmonic = 750 Hz

Remember that for a string fixed at both ends, all harmonics (1st, 2nd, 3rd, etc.) are possible.

Example 2

A closed organ pipe has a length of 0.60 m. The speed of sound in air is 340 m s⁻¹. Calculate: (i) The fundamental frequency of the pipe. (ii) The frequency of its first overtone.

I(i) For a closed pipe, the fundamental frequency (1st harmonic) corresponds to a wavelength of λ₁ = 4L.
IIGiven L = 0.60 m, so λ₁ = 4 * 0.60 m = 2.40 m.
IIIUsing v = fλ, we can find the fundamental frequency f₁ = v / λ₁.
IVf₁ = 340 m s⁻¹ / 2.40 m = 141.67 Hz (or 142 Hz to 3 significant figures).
V(ii) For a closed pipe, only odd harmonics are present. The first overtone is the 3rd harmonic.
VIThe frequency of the nth harmonic for a closed pipe is f_n = (2n-1) * f₁.
VIIFor the 3rd harmonic (n=2 in the (2n-1) formula, or simply 3 * f₁), f₃ = 3 * f₁.
VIIIf₃ = 3 * 141.67 Hz = 425.01 Hz (or 425 Hz to 3 significant figures).

Answer

(i) Fundamental frequency = 142 Hz (ii) Frequency of first overtone = 425 Hz

Be careful with 'first overtone' for closed pipes; it refers to the 3rd harmonic, not the 2nd.

Example 3

(HL) The sound intensity level of a rock concert is 110 dB. Calculate the intensity of the sound in W m⁻².

IThe formula for sound intensity level is L = 10 log₁₀ (I / I₀).
IIGiven L = 110 dB and I₀ = 1.0 × 10⁻¹² W m⁻².
IIISubstitute the values into the formula: 110 = 10 log₁₀ (I / (1.0 × 10⁻¹²)).
IVDivide both sides by 10: 11 = log₁₀ (I / (1.0 × 10⁻¹²)).
VTo remove the logarithm, raise 10 to the power of both sides: 10¹¹ = I / (1.0 × 10⁻¹²).
VISolve for I: I = 10¹¹ * (1.0 × 10⁻¹²).
VIII = 1.0 × 10⁻¹ W m⁻² or 0.1 W m⁻².

Answer

The intensity of the sound is 0.1 W m⁻².

Ensure you are comfortable with logarithmic calculations and the use of the reference intensity I₀.

Common mistakes

  • Confusing the number of nodes/antinodes with the harmonic number for different systems (strings vs. pipes).
  • Incorrectly identifying the 'first overtone' for closed pipes (it's the 3rd harmonic, not the 2nd).
  • Using the wrong formula for wavelength or frequency for open vs. closed pipes.
  • Errors in logarithmic calculations when converting between intensity and decibels (HL).
  • Forgetting that the speed of sound (v) is constant for a given medium and temperature, but the speed of a wave on a string depends on tension and mass per unit length.

Exam tips

  • Always draw diagrams for standing wave problems (strings, open pipes, closed pipes) to visualise nodes and antinodes and determine the wavelength in terms of pipe/string length.
  • Clearly state all formulae used and show all steps in calculations. Include units at each step where appropriate.
  • For HL questions involving the decibel scale, be proficient with logarithms and anti-logarithms on your calculator. Remember I₀ is a constant.
  • Pay close attention to whether a pipe is 'open' or 'closed' as this fundamentally changes the harmonic series and formulae.

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