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How do we describe a sound wave, and how do amplitude, frequency, period and wavelength relate to what we hear?

Sound as a longitudinal pressure wave: amplitude and loudness, frequency and pitch, period, wavelength and the wave equation, and the audible frequency range.

A focused answer to the Edexcel 9MT0 principles of sound, covering sound as a longitudinal pressure wave, amplitude and loudness, frequency and pitch, period, wavelength, the wave equation and the audible range.

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What this dot point is asking

Edexcel wants you to describe sound as a longitudinal pressure wave and to use the core wave quantities precisely: amplitude (and its link to loudness), frequency (and its link to pitch), period, wavelength, and the wave equation, plus the audible frequency range. These quantities are the vocabulary you will use to analyse every recording and to make every production decision, so they must be second nature.

The answer

Sound as a longitudinal pressure wave

Because sound is a pressure variation, it needs a medium and cannot travel through a vacuum. When we draw a sound wave as a smooth curve, that graph is pressure (or the equivalent voltage from a microphone) plotted against time. The curve is a convenient transverse picture of a wave that is physically longitudinal.

Amplitude and loudness

In a digital audio workstation the amplitude of a waveform is shown as its height in the editor, and the level meters read how large that amplitude is at any moment. Clipping happens when the amplitude exceeds the maximum the system can represent, flattening the peaks and adding harsh distortion.

Frequency and pitch

The note A above middle C is standardised at 440440 Hz (concert pitch). Musical intervals are ratios, not differences: the octave is 2:12:1, the perfect fifth is close to 3:23:2. This is why pitch is perceived logarithmically, and why a synthesiser's pitch control and a graphic EQ both use octave-based spacing.

Period, wavelength and the wave equation

A low bass note at 4040 Hz has a wavelength of λ=34040=8.5\lambda = \frac{340}{40} = 8.5 m, which is why bass is hard to control acoustically in a small room and why subwoofer placement matters. A 1010 kHz cymbal harmonic has a wavelength of only 3.43.4 cm, so it is highly directional and easily shadowed.

The audible range

The human ear responds to roughly 2020 Hz to 2020 kHz. Below 2020 Hz is infrasound (felt rather than heard); above 2020 kHz is ultrasound. The upper limit drops with age and exposure to loud sound, so an adult may hear only to 1515 kHz or less. This range is why audio systems are designed around a 2020 Hz to 2020 kHz bandwidth and why CD audio samples fast enough to capture it.

Examples in context

When you set a high-pass filter at 8080 Hz to clean up a vocal, you are removing energy whose wavelength (over 44 m) is too long to be useful and that only muddies the mix. When you choose a sample rate, you are choosing how many times per second to measure the amplitude, fast enough to capture frequencies up to 2020 kHz. When you tune an oscillator to 440440 Hz, you are setting the cycle rate that the ear reads as the note A. Every one of these decisions rests on the relationship between amplitude, frequency and wavelength.

Try this

Q1. State what amplitude and frequency each control in a sound we hear. [2 marks]

  • Cue. Amplitude controls loudness; frequency controls pitch.

Q2. A note has a period of 4.04.0 ms. Find its frequency. [2 marks]

  • Cue. f=1T=10.0040=250f = \frac{1}{T} = \frac{1}{0.0040} = 250 Hz.

Q3. Calculate the wavelength of a 170170 Hz note if the speed of sound is 340340 m per second. [2 marks]

  • Cue. λ=vf=340170=2.0\lambda = \frac{v}{f} = \frac{340}{170} = 2.0 m.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 9MT0/03 20193 marksA sustained synthesiser note has a fundamental frequency of 110110 Hz. Calculate its period, and state how the pitch would change if the frequency were doubled to 220220 Hz.
Show worked answer →

The period is the reciprocal of the frequency: T=1f=1110=9.1×103T = \frac{1}{f} = \frac{1}{110} = 9.1 \times 10^{-3} s, which is about 9.19.1 ms.

Doubling the frequency from 110110 Hz to 220220 Hz raises the pitch by exactly one octave, because an octave is a 2:12:1 frequency ratio.

Markers reward the formula T=1fT = \frac{1}{f}, the value about 9.19.1 ms, and the statement that the pitch rises by an octave (not simply "it gets higher").

Edexcel 9MT0/03 20224 marksExplain the difference between the amplitude and the frequency of a sound wave, stating what each one controls in the sound we hear, and describe the audible frequency range for a typical young listener.
Show worked answer →

Amplitude is the maximum change in air pressure from the rest (atmospheric) value as the wave passes; it controls the loudness of the sound, because a larger pressure variation carries more energy and is perceived as louder. Frequency is the number of complete pressure cycles per second, measured in hertz; it controls the perceived pitch, because a higher frequency is heard as a higher note.

The audible range for a typical young listener is roughly 2020 Hz to 2020 kHz. Below 2020 Hz is infrasound and above 2020 kHz is ultrasound, both inaudible, and the upper limit falls with age and noise exposure.

Markers reward amplitude linked to loudness, frequency linked to pitch, the units (hertz), and the 2020 Hz to 2020 kHz range.

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