GRADE 9 · MECHANICAL WAVES AND SOUND
Sound, echo and resonance
Pitch, loudness, timbre, frequency ranges and sound reflection.
Read the full lesson theory here without an account. In the interactive lesson, the same theory is presented as cards, followed by questions with feedback and practice problems. Your progress is saved.
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?
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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
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
Sound frequency
\nuThe frequency of vibration of the source and sound wave.
Unit: hertz · \mathrm{Гц}
Distance to obstacle
sOne-way distance from the source to the sound-reflecting obstacle.
Unit: metre · \mathrm{м}
Lesson formulas
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?
Initial values
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.