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GRADE 9 · MECHANICAL WAVES AND SOUND

Sound, echo and resonance

Pitch, loudness, timbre, frequency ranges and sound reflection.

24 minutes6 theory cards
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Understanding step by step

A flute and a guitar play the same note, yet we easily recognise the instruments. One can be quiet, the other loud. Which different wave properties lie behind a sound's pitch, loudness and distinctive character?

FluteGuitarΔp, ПаΔp, Па-101-10102,5502,55t, мсt, мсT = 2,5 мсT = 2,5 мсSchematic signals of one note
These schematic flute and guitar signals share a fundamental frequency of 400 Hz and period of 2.5 ms. Different overtone contributions change the pressure waveform. This illustrates timbre differences rather than recordings of specific instruments; real timbre also depends on how the sound changes over time.

Source and medium

A guitar string moves back and forth and acts on the air. Pressure changes arise: compressions and rarefactions pass to neighbouring layers as a longitudinal sound wave. The air does not fly as a whole from string to ear; a disturbance reaches the ear. The source sets the oscillation frequencies, and the medium's properties set propagation speed. Imagine removing air and all other sound-conducting media between source and listener: ordinary sound would not arrive.

Source → air → earDiaphragmSoundearCompressions and rarefactionsSource sets frequency; medium sets speedOrdinary sound cannot travel through a vacuum
The diaphragm oscillates left and right, creating alternating compressions and rarefactions in air. They travel to the ear as a longitudinal wave. Stripes represent air sections, not transverse oscillations. The source sets frequency; the medium determines propagation speed.

Pitch

Play 220 Hz, then 440 Hz. In the second case, oscillations repeat twice as often and the perceived fundamental pitch is higher. Frequency ν is measured in hertz and is related to pitch. A higher note need not be louder: a high note can be quiet and a low note loud. For a complex sound, hearing identifies a fundamental and a combination of components, so one frequency cannot fully describe the sound.

Same interval: 10 мсΔp, Па-11Δp, Па-110510t, мсν = 220 Гцν = 440 ГцEqual amplitudes, different frequencies
Two sinusoidal pressure signals are shown over the same 10 ms. Their frequencies, 220 and 440 Hz, differ by a factor of two; both amplitudes are 1 Pa. The lower plot has higher pitch. Increasing frequency alone does not mean greater intensity or loudness.

Loudness and intensity

Keep frequency the same but increase the string's oscillation amplitude. Other things being equal, the wave carries more energy: intensity increases, meaning more energy passes through unit area per unit time. The sound is usually perceived as louder, but loudness is a sensation that also depends on frequency and hearing sensitivity. Two sounds of equal intensity need not seem equally loud. Oscillating more strongly and more often are different changes.

Same medium and frequencyΔp, Па-11Δp, Па-220510t, мсν = 200 ГцAmplitude ×2I₂ = 4I₁; I = E/(St), Вт/м²
Compare two travelling sinusoidal sound waves in the same medium at 200 Hz. The lower pressure amplitude is twice as large: 2 Pa instead of 1 Pa, so its average intensity is four times as large. In I = E/(St), E is energy, S area and t time. Perceived loudness also depends on frequency and hearing.

Where timbre comes from

Why does the guitar still not sound like a flute on the same note? An instrument usually produces a fundamental frequency and higher components—overtones. Their relative strengths, the onset and decay of the sound shape its timbre. Imagine removing the overtones: the sound approaches a pure tone and loses some of its recognisable character. Timbre describes the sound's composition and development, not just its fundamental frequency or amplitude.

Sound АSound БA/A₁10A/A₁1020060010002006001000ν, Гцν, ГцSame fundamental: 200 ГцTimbre also depends on spectral changes
Two illustrative spectra have the same 200 Hz fundamental but different relative amplitudes of harmonics at 400, 600, 800 and 1000 Hz. Bar heights are relative to each sound's first harmonic. Different composition and its evolution over time affect timbre.

Limits of hearing

The ear does not respond equally to all frequencies. The approximate human audible range is 20 Hz to 20 kHz, with actual limits varying between people and with age. Below 20 Hz is infrasound; above 20 kHz is ultrasound. These names are defined by frequency. Making an audible note quieter does not turn it into infrasound, and loud ultrasound does not become ordinary audible sound just because its amplitude is large.

InfraAudibleUltra20 Гц20 кГцFrequency increases →
The usual human audible range is approximately 20 Hz to 20 kHz, with individual limits. Below it is infrasound; above it, ultrasound.

A resonator strengthens the response

A string without an instrument body sounds different. The body and the air inside receive energy from the source and can respond strongly near their natural frequencies: they resonate. They help transfer energy into the surrounding air; different frequencies are enhanced differently, changing timbre. A resonator does not create energy from nothing: the musician supplies it, and the connections between parts determine how it becomes sound.

String → bridge → body → airStringBridgeInto the airBody and air insideStrong response near natural frequencies
The vibrating string rests on a bridge that transfers its action to the body. The body and the air inside respond especially strongly near their natural frequencies and help radiate sound. The oscillation source supplies energy; the resonator changes transfer efficiency and the spectrum.

An echo returns

Clap briefly in front of a distant wall. Part of the wave reflects and returns. With sufficient delay, the ear distinguishes a second signal—an echo—instead of a continuation merged with the first. The measured time t covers the whole route: to the wall and back to the listener. Moving the wall farther away increases the delay. Sound reflection reveals distance even when the obstacle is hard to see.

Sound reflects from the wallv = constt — outward and backSourceand receiveroutwardreturnsEnough delay makes a separate echo audible
A short sound travels from the source to the front face of the wall and returns along the same route to the nearby receiver. Two arrows on one line show opposite directions. The measured time includes both journeys; with sufficient delay, the reflection is heard separately.

Halve the path length

The source and receiver are close together, and sound speed v is constant along the route. Both legs are then approximately s: 2s = vt, giving s = vt/2. Double the delay and the distance doubles; at zero delay the model gives zero distance. Use metres for s and seconds for t. Ultrasonic echolocation uses the same principle; if source and receiver were separated, the path geometry would need a separate calculation.

One segment travelled twicev = constt — outward and backSourceand receiversss2s = vt → s = vt/2
Source and receiver are close together. On normal reflection from a wall, sound travels distance s to it and the same distance back. At constant speed v, the full path is 2s = vt, so s = vt/2. Time t runs from emission to echo return; measure s in metres and t in seconds.

A distant wall

An echo returns 0.6 s after a short signal. Take sound speed as 340 m/s. Find the distance to the wall.

Echo from a distant wallv = 340 м/сt = 0,6 с — both journeysSourceand receiveroutwardreturns — ?Find distance s to the wall
Source and listener are close together. A short signal and its echo are separated by 0.6 s; take sound speed as 340 m/s. Measure s from the sending/receiving point to the wall's front face. Sound travels this segment twice.

Let's look at an example · 1

Between the signal and its echo, sound travels to the wall and back. Thus 0.6 s cannot be treated as the time for only one leg.

Count both journeysv = 340 м/сt = 0,6 с — both journeysSourceand receiversssFull path: s + s = 2s
The same source, receiver and wall. Blue and green arrows refer to the same segment of length s. The full sound path is s + s = 2s, and 0.6 s is the time for the whole journey.

Let's look at an example · 2

The full path is vt = 340 · 0,6 = 204 м. With source and listener close together, half this path leads to the wall.

Calculate the full pathv = 340 м/сt = 0,6 с — both journeysSourceand receiversss2s = vt = 340 · 0,6 = 204 м
Keep the same route from source to wall and back. Multiplying 340 m/s by the full time of 0.6 s gives 204 m. This is the sum of two journeys along s, not the distance to the wall.

Let's look at an example · 3

s = vt/2 = 204/2 = 102 м. Check: 102/340 = 0,3 с one way, and 0.6 s for the complete round trip.

Halve the full pathv = 340 м/сt = 0,6 с — both journeysSourceand receiveroutwardreturns = 102 мs = 204/2 = 102 м
The distance from the original sending point to the same wall face is now labelled s = 102 м. Sound travels this segment twice, so divide its total path of 204 m by two.

Check the result

The wall is 102 m away. During the measured 0.6 s, sound travels twice as far—204 m—including the return journey.

Key point · 1

Pitch, loudness and timbre concern different sound properties: frequency, perception of intensity and the composition of oscillations.

Key point · 2

Infrasound and ultrasound are defined by frequency, not by how strongly the source oscillates.

Key point · 3

An echo measures round-trip time; with source and receiver close together, s = vt/2.

What comes next

Next, light: it also carries energy but does not require air.

Basic Concepts

3 concepts

Pitch, loudness and timbre

Pitch is mainly related to frequency, loudness to intensity, and timbre to the sound's spectral composition.

The same note played by different instruments differs in timbre because their overtones differ.

Infrasound and ultrasound

Infrasound has a frequency below the human audible range, and ultrasound has a frequency above it.

Ultrasound is used in diagnosis and echolocation, with attention to intensity and exposure safety.

Sound reflection and echo

An echo is a sound that returns after reflection with a noticeable delay.

The distance to the obstacle is equal to half the path of the sound during the time between the signal and the echo.

Physical quantities

2 quantities

Sound frequency

\nu

The frequency of vibration of the source and sound wave.

Unit: hertz · \mathrm{Гц}

Distance to obstacle

s

One-way distance from the source to the sound-reflecting obstacle.

Unit: metre · \mathrm{м}

Lesson formulas

1 formula

Distance from an echo

s=\frac{vt}{2}

The distance to the obstacle is equal to half the path of sound during its propagation there and back.

Test the relationship

Make a prediction first. The button changes the stated quantity; then compare the result and explore other values.

Wave and medium motion

λ = v/ν; y = A sin[2π(x/λ − νt)]

Harmonic transverse wave in a uniform medium without damping or dispersion. The marked medium point has fixed x = 2 m and moves only transversely.

At v = 4 m/s, frequency increases from 1 to 2 Hz. How will wavelength change?

Your prediction

Initial values

0-0.622-0.314060.3180.62y, мx, м
y

Calculated result

λ
4 м
T
1 с
y(2 м)
0 м
Explaining the result

A crest travels right while the medium point moves up and down. At a fixed wave speed, doubling frequency halves the spacing between crests.

Related formulas

The same formulas in the reference, with symbols and units.

Problems on this topic

Statements, answer options and the walkthrough.