Sound — Print · Grade 12 Science

Chapter: Waves, Sound & Optics

Sound

Learning goals

Readings for this unit

Before this unit

Mechanical waves and longitudinal compression.

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Start with something to explain

Why can a louder tone have the same pitch?

Illustrative evidence case. A vibrating source changes amplitude while its measured cycle interval stays fixed.

Sound requires a material medium and local pressure changesSound requires a material medium and local pressure changesDense regions: compressions; sparse regions: rarefactionssound propagationAir molecules oscillate locally along this direction.
Original selected-particle compression schematic, not a literal molecular photograph or a density calibration.

Notice and wonder

Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.

Why can a louder tone have the same pitch?

Your first model

Read its time axis and equilibrium position before comparing amplitudes.

Build the language from the idea

longitudinal wave
A wave in which particles vibrate back and forth in the same direction the wave travels, creating compressions and rarefactions.
pitch
How high or low a sound is perceived; determined by the sound wave's frequency.
loudness
How intense a sound seems; determined by the wave's amplitude and measured on the logarithmic decibel (dB) scale.
medium
The matter — solid, liquid, or gas — that a mechanical wave such as sound must travel through.
Doppler effect
The change in observed frequency (pitch) when a wave source and observer move relative to each other.
resonance
The large-amplitude vibration that occurs when a system is driven at its natural frequency.
standing wave
A stationary wave pattern of fixed nodes and antinodes formed when a wave reflects and interferes with itself, as on a guitar string.

Sound is a longitudinal wave, and that single fact explains most of its behavior. Instead of the up-and-down motion of a rope, the particles carrying a sound wave vibrate back and forth along the same line the wave travels, squeezing into compressions where they crowd together and stretching into rarefactions where they thin out. A struck tuning fork, a vibrating speaker cone, or your own vocal cords launches this pattern of pressure ripples outward. Because sound is nothing more than particles jostling their neighbors, it absolutely requires a medium — which is why space is silent, and why the booming explosions in science-fiction films could never actually reach your ears across a vacuum.

Sound requires a material medium and local pressure changesSound requires a material medium and local pressure changesDense regions: compressions; sparse regions: rarefactionssound propagationAir molecules oscillate locally along this direction.
Original selected-particle compression schematic, not a literal molecular photograph or a density calibration.

Sound speed depends on the medium’s elastic properties, density, and conditions such as temperature. For small disturbances in a suitable approximately nondispersive model, it is not set by source loudness or pitch. Room-temperature air is often approximated near 343 m/s; use the value stated for a particular task. Many solids transmit sound faster than air, but closely spaced particles alone are not a sufficient explanation, and material properties must be considered.

Pitch is perception strongly related to frequency for simple tones. Intensity is sound power per area, while perceived loudness also depends on frequency, duration and listener. A sound-intensity level uses decibels: β=10log₁₀(I/I₀), with a stated reference. A 10 dB difference means tenfold intensity, not tenfold perceived loudness. A pressure-amplitude ratio uses a different logarithmic factor under matched impedance conditions. Keep the physical and perceptual quantities distinct.

Pitch, intensity and perceived loudness are distinctPitch, intensity and perceived loudness are distinctQuantityPhysical meaningPerception connectionFrequencyCycles per secondA major influence on pitchIntensityPower per areaOne influence on loudnessSpectrumComponent frequenciesContributes to timbrePerception also depends on frequency, duration and listener.
Original distinction table; it is not an individual hearing assessment.

The Doppler effect compares emitted and received frequency. For a stationary receiver in still air and a subsonic source moving directly toward it, fobs=fsource v/(v−vs); recession uses v+vs. Source and receiver velocities are relative to the medium and the formula depends on geometry. Vacuum-light shifts use relativistic relations rather than simply substituting c into a sound formula; cosmic expansion adds another interpretive framework.

Standing-wave patterns can arise through suitable overlapping incident and reflected waves. Nodes have zero displacement amplitude and antinodes larger amplitude in the ideal pattern. Boundaries determine allowed modes. Driving near a natural frequency can produce resonance, while damping and coupling limit real amplitudes. A string and an air column have different displacement and pressure boundary conditions, so their harmonic patterns should not be transferred without checking those conditions.

Musical instruments combine source vibration, boundary-supported modes and energy loss. Walls transmit, reflect and absorb sound according to material, thickness, openings and frequency; a single “poor medium” label is insufficient. Sonar and biological echolocation use sound travel, while radar uses electromagnetic waves. Medical ultrasound uses acoustic interactions with tissues under controlled professional equipment and protocols; this course uses diagrams and supplied records rather than requiring exposure experiments.

Travel time and spreading require different assumptionsTravel time and spreading require different assumptionsAt2r:4× areaI becomes I/4Echo distance uses one-way time, not total return time.
Original spreading schematic; circles are 2D cross-sections of spherical wavefronts.

Model check: Sound requires a medium; pitch and loudness measure different aspects. Round-trip 170 m; one-way distance 85 m.

Build your understanding

Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.

Part 1 · Use the medium to explain sound propagation

Sound is a mechanical disturbance. In air it propagates mainly through longitudinal pressure and density variations: compressions are locally denser/higher-pressure regions and rarefactions are less dense regions. Molecules exchange forces with neighbors and oscillate locally rather than streaming from a speaker to your ear as a parcel. An ordinary vacuum cannot carry airborne sound because it lacks this material interaction chain.

Sound requires a material medium and local pressure changesSound requires a material medium and local pressure changesDense regions: compressions; sparse regions: rarefactionssound propagationAir molecules oscillate locally along this direction.
Original selected-particle compression schematic, not a literal molecular photograph or a density calibration.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What important limit separates sound from an electromagnetic wave?

Visual reasoning — after your attempt

Sound needs a material medium; electromagnetic waves can propagate through vacuum.

Ordinary sound is a mechanical disturbance requiring matter. In a simple fluid model, compressions and rarefactions propagate while particles oscillate locally. Speed depends on elastic properties, density, and conditions such as temperature; particle spacing alone is not a sufficient explanation.

A vacuum does not transmit ordinary sound. Many solids transmit it faster than air, but use the value supplied for the actual material rather than a universal ranking.

Pause and explain

Why is sound speed not determined by source loudness alone?

Check your reasoning after answering

The medium and its physical conditions govern the linear propagation model; louder excitation does not by itself specify a different speed.

Part 2 · Distinguish pitch, intensity, and loudness

Pitch is a perceptual attribute strongly related to frequency for simple tones. Intensity is acoustic power per area, while loudness is perception and depends on frequency, duration and listener. Timbre distinguishes sounds with different spectral or temporal structure even at similar pitch. Do not call a higher-amplitude trace “higher pitch” when its period is unchanged, and do not assume equal physical intensity always produces equal perceived loudness.

Pitch, intensity and perceived loudness are distinctPitch, intensity and perceived loudness are distinctQuantityPhysical meaningPerception connectionFrequencyCycles per secondA major influence on pitchIntensityPower per areaOne influence on loudnessSpectrumComponent frequenciesContributes to timbrePerception also depends on frequency, duration and listener.
Original distinction table; it is not an individual hearing assessment.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

Can two tones have the same frequency but different intensity?

Visual reasoning — after your attempt

Yes. Their oscillation rates can match while amplitudes and power per area differ.

Frequency is strongly associated with perceived pitch, while intensity is physical power per area. Loudness also depends on frequency and listener response. Amplitude and intensity relationships need a stated model. A high-pitch sound need not be intense, and an intense sound need not be high-pitched.

For comparable waves in a linear medium, intensity is often proportional to amplitude squared. This does not make a loudness judgment a precise intensity measurement.

Pause and explain

Can frequency alone rank sound intensity?

Check your reasoning after answering

No. Frequency and intensity are different variables; source power, distance, and propagation also matter.

Part 3 · Interpret the decibel ratio

A sound-intensity level is β=10log₁₀(I/I₀) for a stated reference intensity I₀. Differences in decibels compare ratios: a 20 dB increase means 100 times the intensity, not 20 times. Decibels are not a direct linear loudness scale. The plotted points show selected calculated ratios on a linear horizontal axis; between-point connections are a guide rather than an exact smooth logarithmic curve.

A decibel difference represents a logarithmic intensity ratioA decibel difference represents a logarithmic intensity ratioLevel difference 10 log₁₀(I/I₀) (dB)010203002505007501000Intensity ratio I/I₀ (linear axis)10× intensity → +10 dB;100× → +20 dB.
Calculated selected intensity-ratio points; not measured sound exposure or a safety threshold.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

What intensity ratio corresponds to a 30 dB increase?

Visual reasoning — after your attempt

10^(30/10)=1000, provided the levels use the same intensity reference.

Sound intensity level uses β=10 log₁₀(I/I₀) with a stated reference. A 10 dB difference corresponds to a tenfold intensity ratio, while 20 dB corresponds to a hundredfold ratio. Decibels are logarithmic; do not add intensity levels as if they were ordinary powers.

The same level difference does not establish an identical perceived-loudness change for every listener or frequency. Keep the physical ratio and perception distinct.

Pause and explain

What does a 3 dB increase approximately mean?

Check your reasoning after answering

It is about a factor of two in intensity because 10 log₁₀2≈3.01 dB; it is not simply three times the intensity.

Part 4 · Use travel-time and spreading models carefully

For an echo, a pulse travels to a reflector and back, so distance is vΔt/2 when the same sound speed applies both ways. For an ideal point source spreading uniformly without absorption, intensity falls as 1/r² because the same power crosses a growing spherical area. Real rooms have reflections, directional sources and absorption; the inverse-square comparison is not a universal rule for every microphone position indoors.

Travel time and spreading require different assumptionsTravel time and spreading require different assumptionsAt2r:4× areaI becomes I/4Echo distance uses one-way time, not total return time.
Original spreading schematic; circles are 2D cross-sections of spherical wavefronts.

Read the visual model

Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.

An echo returns after 0.40 s at 340 m/s. How far away is the reflector?

Visual reasoning — after your attempt

d=340×0.40/2=68 m in the stated still-air model.

An echo time covers a round trip when the signal returns to the source. For an ideal isotropic point source without reflections or absorption, intensity decreases as 1/r². Enclosed spaces and directional sources can differ. Specify the propagation path before applying either shortcut.

An echo returning after 0.50 s at 340 m/s corresponds to about 85 m one-way distance, not 170 m.

Pause and explain

When can doubling distance reduce intensity to one-quarter?

Check your reasoning after answering

Under the stated point-source geometric-spreading model with comparable conditions, without major absorption, reflections, or source-direction effects.

Apply the whole unit · Interpret an echo and intensity comparison

Illustrative echo returns after 0.400 s in air with sound speed 340 m/s. A separate ideal point-source record has intensity I at 2.00 m and no significant absorption or reflections.

Your task

Find reflector distance, predict intensity at 4.00 m, and calculate the intensity-level change. Explain why these are different measurements and why the spreading model may fail indoors.

What your response should show

  • Define the system, quantities, units, and relevant conditions before selecting a model.
  • Show intermediate calculations or a labeled evidence account, including signs or denominators.
  • Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
  • Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
  1. Round-trip distance=340×0.400=136 m; reflector distance=68.0 m.

  2. At 4.00 m the intensity is I/4.

  3. Level change=10 log₁₀(1/4)≈−6.02 dB.

The echo calculation needs the return-path assumption; the intensity calculation needs a source-geometry assumption. Neither is a hearing assessment. Walls, directionality, multiple reflected paths, and background sound can make indoor records differ from ideal inverse-square spreading.

Scientific references and source-model foundations

Parent / teacher teaching notes

Use supplied quiet observations or records. Ask why echo distance is halved and why decibels require an intensity ratio rather than a linear subtraction of powers. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.

An animated longitudinal wave showing compressions and rarefactions moving through a line of particles.

Watch particles compress and spread apart as a longitudinal sound wave passes through them.

The key idea

Sound needs a medium. Frequency strongly influences pitch, while intensity and loudness differ. Decibel ratios, Doppler shifts and standing waves require stated conditions.

Distance from Speed and Time

d = v × t

v is the speed of sound (about 343 m/s in air); t is time in seconds; d is distance in meters. For an echo, the sound covers 2d, so d = v × t ÷ 2.

Doppler Effect (moving source)

f′ = f × v ÷ (v ∓ vₛ)

f′ is the observed frequency; f is the source frequency; v is the speed of sound; vₛ is the source's speed. Use the minus sign when the source approaches, the plus sign when it recedes.

Fundamental Frequency (resonance)

f₁ = v ÷ (2L)

f₁ is the lowest resonant (fundamental) frequency of a string fixed at both ends or an open pipe of length L; v is the wave speed. Higher harmonics are whole-number multiples of f₁.

Worked examples

Example 1

A sound wave travels through air at 343 m/s with a frequency of 686 Hz. Find its wavelength.
  1. Rearrange v = f × λ to solve for wavelength: λ = v ÷ f.
  2. λ = 343 ÷ 686.
  3. = 0.5.
Answer: 0.5 m

Example 2 · Worked application

Two tones have frequencies 440 and 444 Hz. Find the simple beat frequency.
  1. Beat frequency is the absolute frequency difference.
  2. 444−440=4 Hz.
  3. The amplitude envelope repeats about four times per second.
Answer: 4 Hz.
Check the reasoning

Beat frequency is distinct from the approximately 442 Hz pitch region.

Example 3

A siren emitting 300 Hz moves toward a stationary listener at 40 m/s. Using the speed of sound as 340 m/s, what frequency does the listener hear?
  1. The source is approaching, so use the minus sign: f′ = f × v ÷ (v − vₛ).
  2. f′ = 300 × 340 ÷ (340 − 40) = 300 × 340 ÷ 300.
  3. The 300 s cancel, leaving f′ = 340.
Answer: 340 Hz (higher than 300 Hz, as expected for an approaching source)

Example 4 · Find and fix the mistake

Context: Two tones have frequencies 440 and 444 Hz. Find the simple beat frequency. A student claims: “A louder tone necessarily has higher frequency.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Beat frequency is distinct from the approximately 442 Hz pitch region.
  2. Loudness depends on amplitude and perception; frequency primarily influences pitch.
Answer: Loudness depends on amplitude and perception; frequency primarily influences pitch.
Check the reasoning

Beat frequency is distinct from the approximately 442 Hz pitch region.

Example 5 · Reason from the visual model

A decibel difference represents a logarithmic intensity ratioA decibel difference represents a logarithmic intensity ratioLevel difference 10 log₁₀(I/I₀) (dB)010203002505007501000Intensity ratio I/I₀ (linear axis)10× intensity → +10 dB;100× → +20 dB.
Calculated selected intensity-ratio points; not measured sound exposure or a safety threshold.
What intensity ratio corresponds to a 30 dB increase?
  1. A sound-intensity level is β=10log₁₀(I/I₀) for a stated reference intensity I₀. Differences in decibels compare ratios: a 20 dB increase means 100 times the intensity, not 20 times. Decibels are not a direct linear loudness scale. The plotted points show selected calculated ratios on a linear horizontal axis; between-point connections are a guide rather than an exact smooth logarithmic curve.
  2. 10^(30/10)=1000, provided the levels use the same intensity reference.
Answer: 10^(30/10)=1000, provided the levels use the same intensity reference.
Check the reasoning

Calculated selected intensity-ratio points; not measured sound exposure or a safety threshold.

Avoid these mistakes

Check your understanding

An echo returns after.50 s at 340 m/s. Find wall distance.

Discuss the answer

Round-trip 170 m; one-way distance 85 m.

Return to the phenomenon

Return to the initial case. Why can a louder tone have the same pitch? Revise your drawing or explanation using two named ideas and one stated limitation.

Compare explanations after your attempt

Amplitude/intensity and frequency describe different quantities; perceived loudness is not identical to either one alone.

What should we investigate next?

Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.

Investigate, calculate and revise

Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.

Activity 1 · Return to the opening case

Assigned reading and preparation

Use the opening phenomenon, unit definitions and explained Parts 1–4.

Your task

Return to the initial case. Why can a louder tone have the same pitch? Revise your drawing or explanation using two named ideas and one stated limitation.

Success criteria

  • Use the stated observations and appropriate units.
  • Connect a labeled model with the physical/environmental mechanism.
  • State assumptions and an evidence limitation.
  • Explain a substantive revision or next test.
Hint

Keep object/material identity, material transfer and energy account distinct.

Response guidance — discuss after your attempt

Amplitude/intensity and frequency describe different quantities; perceived loudness is not identical to either one alone.

Activity 2 · Transfer to a changed case

Assigned reading and preparation

Cover the worked answer and use the explained models, relationships and notebook.

Your task

Two sound intensity levels differ by 15 dB. Find their intensity ratio. A separate echo returns in 0.30 s at 340 m/s; find reflector distance.

Success criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.
Hint

Name known quantities and conditions, show intermediate steps and check conservation.

Response guidance — discuss after your attempt

The intensity ratio is 10^1.5≈31.6. Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.

Words to know

longitudinal wave
A wave in which particles vibrate back and forth in the same direction the wave travels, creating compressions and rarefactions.
pitch
How high or low a sound is perceived; determined by the sound wave's frequency.
loudness
How intense a sound seems; determined by the wave's amplitude and measured on the logarithmic decibel (dB) scale.
medium
The matter — solid, liquid, or gas — that a mechanical wave such as sound must travel through.
Doppler effect
The change in observed frequency (pitch) when a wave source and observer move relative to each other.
resonance
The large-amplitude vibration that occurs when a system is driven at its natural frequency.
standing wave
A stationary wave pattern of fixed nodes and antinodes formed when a wave reflects and interferes with itself, as on a guitar string.

Sound is faster in steel than air — and instant in neither

A common misconception is that sound is effectively instantaneous; in fact it travels at a definite, medium-dependent speed. In room-temperature air that is about 343 m/s, but in water roughly 1,480 m/s and in solid steel around 5,900 m/s — because tightly packed particles hand off the vibration far faster. That is why railroad workers in old films pressed an ear to the rail: the steel delivers the sound of an approaching train sooner and farther than the air. Temperature matters too — warmer air carries sound a little faster, since its molecules move more quickly.

Christian Doppler, Buys Ballot, and a trainful of trumpeters

In 1842 the Austrian physicist Christian Doppler predicted that a moving source should shift the pitch a listener hears. Three years later the Dutch scientist Buys Ballot staged one of science's most charming tests: he had trumpeters play a sustained note aboard an open railway car speeding past musicians with perfect pitch, who confirmed the note rose as the car approached and fell as it receded — exactly as Doppler foretold. The same effect, applied to light, later let astronomers discover that distant galaxies are rushing away, evidence for an expanding universe.

Decibels are logarithmic: why 'twice as loud' isn't twice the number

Loudness is measured on the decibel (dB) scale, and — like pH — it is logarithmic, not linear. Every 10 dB marks a tenfold jump in sound intensity, so a 60 dB conversation and a 110 dB rock concert differ by 50 dB, which is 10⁵ — about 100,000 times more intense, not 'roughly twice.' This is why loud sound damages hearing so much faster than people expect, and why safety limits on headphones and workplaces are written in decibels: a few extra dB hides a huge jump in the actual energy hitting your eardrum.

Resonance can shatter a glass — and rattle a bridge

Every object has natural frequencies at which it 'wants' to vibrate. Drive it at one and the amplitude builds enormously — resonance. A trained singer sustaining a note at a wine glass's natural frequency can feed it energy until it shatters; a child pumping a swing in rhythm climbs higher and higher on the same principle. Engineers respect resonance too: the 1940 collapse of the Tacoma Narrows Bridge, whose deck twisted itself apart in a moderate wind, is the textbook cautionary tale (its cause is better described as aeroelastic flutter, but it made resonance famous).

Echolocation and ultrasound: seeing with sound

Sound above 20,000 Hz — ultrasound — is inaudible to humans but immensely useful. Bats and dolphins echolocate, emitting rapid ultrasonic pulses and reading the returning echoes to map size, distance, and even the texture of prey in total darkness. We borrowed the trick: sonar lets ships and submarines 'see' the seafloor, and medical ultrasound images a developing fetus or a beating heart safely, using no radiation at all — just the timing and strength of reflected pressure waves, the very same physics that makes an echo bounce off a canyon wall.

In the real world

Doppler weather radar bounces microwaves off raindrops and reads the frequency shift of the return to measure how fast a storm's winds are moving toward or away from the station.

Exit check

An echo returns after.50 s at 340 m/s. Find wall distance.

Check your explanation after answering

Round-trip 170 m; one-way distance 85 m.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

Using the speed of sound as 340 m/s, how far away is a storm if thunder is heard 3 seconds after the lightning flash?
Answer and explanation

1020 m

distance = speed × time = 340 × 3 = 1,020 m.

Problem 2

The pitch of a sound is determined by its…
  1. amplitude
  2. loudness
  3. the medium's color
  4. frequency
Answer and explanation

frequency

Pitch depends on frequency — higher frequency means higher pitch.

Problem 3

As an ambulance moves away from you, the siren's pitch drops because…
  1. the waves stretch out behind it, lowering the frequency you hear
  2. the sound doubles in speed
  3. the pitch rises as the waves compress
  4. the pitch stays exactly the same
Answer and explanation

the waves stretch out behind it, lowering the frequency you hear

For a receding source, the wavefronts stretch out, so the observed wavelength grows and the frequency (pitch) drops — the Doppler effect.

Problem 4

A sound wave in air has a wavelength of 0.68 m. Using the speed of sound as 340 m/s, find its frequency.
Answer and explanation

500 Hz

f = v ÷ λ = 340 ÷ 0.68 = 500 Hz.

Problem 5

Explain what resonance is, and give one everyday example of it.
Answer and explanation

Resonance is the large build-up of vibration that happens when a system is driven at its natural frequency. Pushing a swing in time with its natural back-and-forth rhythm makes it go higher and higher; other examples include a singer shattering a wine glass by matching its natural frequency, or a guitar string producing a loud clear note.

Problem 6

A train sounds an 800 Hz whistle while moving away from you at 60 m/s. Using the speed of sound as 340 m/s, what frequency do you hear? (f′ = f × v ÷ (v + vₛ))
Answer and explanation

680 Hz

The source recedes, so f′ = f × v ÷ (v + vₛ) = 800 × 340 ÷ (340 + 60) = 800 × 340 ÷ 400 = 800 × 0.85 = 680 Hz.

Problem 7

An echo returns in 0.20 s with sound speed 340 m/s. Find wall distance. Show the reasoning, not just the conclusion.
Hints

Beat frequency is distinct from the approximately 442 Hz pitch region.

Answer and explanation

340×0.20/2=34 m, including the round trip.

Problem 8

Why does sound speed change between materials?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Elastic and inertial properties of the medium affect propagation; speed is not fixed by loudness alone.

Problem 9

A decibel difference represents a logarithmic intensity ratioA decibel difference represents a logarithmic intensity ratioLevel difference 10 log₁₀(I/I₀) (dB)010203002505007501000Intensity ratio I/I₀ (linear axis)10× intensity → +10 dB;100× → +20 dB.
Calculated selected intensity-ratio points; not measured sound exposure or a safety threshold.
Interpret the supplied visual. What intensity ratio corresponds to a 30 dB increase?
Answer and explanation

10^(30/10)=1000, provided the levels use the same intensity reference.

A sound-intensity level is β=10log₁₀(I/I₀) for a stated reference intensity I₀. Differences in decibels compare ratios: a 20 dB increase means 100 times the intensity, not 20 times. Decibels are not a direct linear loudness scale. The plotted points show selected calculated ratios on a linear horizontal axis; between-point connections are a guide rather than an exact smooth logarithmic curve.

10^(30/10)=1000, provided the levels use the same intensity reference.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

Two sound intensity levels differ by 15 dB. Find their intensity ratio. A separate echo returns in 0.30 s at 340 m/s; find reflector distance.
Answer and explanation

The intensity ratio is 10^1.5≈31.6. Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.

The intensity ratio is 10^1.5≈31.6.

Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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