Senior 6 practice book · Unit 1 of 10

Sound Waves

50 questions that complete the Senior 6 quiz for this unit: 25 core and 25 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

When two sounds cancel at a quiet spot, their energy is lost.Interference only moves energy around: what is missing at the quiet spots appears at the loud spots, so the total energy is conserved.
An echo and reverberation are the same thing.An echo is a reflection heard separately (more than about 0.1 s later); reverberation is many quick reflections that merge and make the sound last longer.
A source moving towards you at steady speed gives a pitch that keeps rising as it comes nearer.At constant speed the heard frequency is constant and higher than normal; it drops suddenly to a constant lower value as the source passes.
If you run towards a sound source, the wavelength of the sound gets shorter.Only a moving source changes the wavelength in the air; a moving listener meets more wavefronts per second, so the frequency heard rises while the wavelength stays the same.
Ultrasound is a very loud sound.Ultrasound is sound above 20 kHz, too high in frequency for humans to hear; it can be very quiet or very intense.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Nature of sound: longitudinal mechanical wave, compressions and rarefactions, need for a medium
Speed of sound in solids, liquids and gases; effect of temperature (v ∝ √T)
Characteristics of sound: pitch, loudness, quality and the audible range
Echoes and timing methods for measuring distance and the speed of sound
Reverberation, absorption and the acoustics of halls; noise control
Interference of sound from two coherent sources (path difference)
Diffraction and refraction of sound
Beats and their use in tuning
Stationary waves on strings: nodes, antinodes, harmonics, v = √(T/μ)
Stationary waves in open and closed pipes; resonance tube and end correction
Sound intensity, inverse-square law and the decibel scale
Doppler effect for a moving source
Doppler effect for a moving observer, both moving and reflected sound
Ultrasound and infrasound: properties and applications (sonar, medical scans, cleaning)

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 25 harder ones are marked ★ Challenge.

1True or false
At a quiet spot in front of two loudspeakers that are in phase, the sound energy is destroyed.
Show answer
Answer: False
Energy is redistributed: less arrives at the quiet spots and more at the loud spots, so the total is conserved.
Common mistakeThinking energy is destroyed breaks conservation of energy; destructive interference only changes where the energy goes.
2Fill in the blank · ★ Challenge
When the distance from a point source of sound is doubled, the intensity level falls by about ______ dB.
Show answer
Answer: 6
Intensity falls to ¼, and 10 log 4 ≈ 6 dB.
Common mistakeWriting 3 dB uses a factor of 2 in intensity; doubling the distance gives a factor of 4 because I ∝ 1/r².
3True or false · ★ Challenge
Because of the end correction, the real fundamental frequency of a closed pipe is slightly higher than v/4L.
Show answer
Answer: False
The antinode lies a little beyond the open end, so the effective length L + e is longer and f = v/4(L + e) is slightly lower.
Common mistakeThinking the frequency is higher reverses the effect: a longer effective column means a longer wavelength and a lower frequency.
4True or false
Sound waves change direction when they pass from a layer of cool air into a layer of warmer air.
Show answer
Answer: True
The speed changes between the layers, so the waves refract.
Common mistakeThinking only light refracts is wrong: refraction happens to every kind of wave whose speed changes between media.
5Multiple choice · ★ Challenge
At the centre of a compression in a sound wave in air, what are the displacement of the air particles and the pressure?
1. ADisplacement maximum, pressure maximum
2. BDisplacement zero, pressure maximum
3. CDisplacement zero, pressure minimum
4. DDisplacement maximum, pressure normal
Show answer
Answer: B. Displacement zero, pressure maximum
Particles on both sides are displaced towards the centre of a compression, so the particle there is not displaced but the pressure is highest; displacement and pressure are a quarter-cycle out of step.
Common mistakeChoosing 'displacement maximum' assumes the most crowded place is where particles move most; in fact the crowding comes from neighbours moving towards a particle that itself stays put.
6True or false
When a listener runs towards a stationary sound source, the wavelength of the sound in the air becomes shorter.
Show answer
Answer: False
The source is at rest, so the wavefronts in the air keep the same spacing; the listener simply meets them more often.
Common mistakeThinking the wavelength shortens copies the moving-source picture; only the motion of the source changes the spacing of the wavefronts.
7True or false
If the air pressure on a mountain top falls while the temperature stays the same, the speed of sound there becomes smaller.
Show answer
Answer: False
In a gas, pressure and density change in the same proportion, so the speed of sound depends on temperature, not on pressure.
Common mistakeLinking lower pressure to slower sound ignores that the density falls by the same factor, so the ratio that sets the speed is unchanged.
8Multiple choice · ★ Challenge
The ear can separate two sounds only if they arrive at least 0.10 s apart. What is the smallest distance from a wall at which a clap gives a distinct echo (v = 340 m/s)?
1. A8.5 m
2. B34 m
3. C3.4 m
4. D17 m
Show answer
Answer: D. 17 m
In 0.10 s sound travels 340 × 0.10 = 34 m; this is there and back, so the wall must be at least 34 ÷ 2 = 17 m away.
Common mistake34 m forgets that the sound covers the distance twice; 8.5 m halves the distance twice.
9Multiple choice · ★ Challenge
A stretched wire fixed at both ends is found to resonate at 150 Hz, 225 Hz and 300 Hz, which are three neighbouring resonant frequencies. What is its fundamental frequency?
1. A150 Hz
2. B37.5 Hz
3. C75 Hz
4. D50 Hz
Show answer
Answer: C. 75 Hz
On a string all harmonics occur, so neighbouring resonances differ by f₁: 225 − 150 = 75 Hz (150, 225 and 300 Hz are the 2nd, 3rd and 4th harmonics).
Common mistakeChoosing 150 Hz assumes the lowest frequency measured is the fundamental; the difference between neighbouring resonances gives f₁.
10True or false
As a car drives straight towards you at a steady speed sounding its horn, the pitch you hear keeps rising until the car reaches you.
Show answer
Answer: False
At constant speed the heard frequency is constant (and higher than the true one); it drops suddenly to a lower constant value as the car passes.
Common mistakeThinking the pitch rises steadily confuses increasing loudness as the car comes nearer with increasing frequency.
11Multiple choice · ★ Challenge
A worker using a brick-cutting machine wears ear defenders that lower the sound level at her ears by 30 dB. What percentage of the sound intensity reaches her ears?
1. A0.1 %
2. B3.3 %
3. C33 %
4. D1 %
Show answer
Answer: A. 0.1 %
Δβ = 10 log(I₁/I₂) = 30 dB, so I₁/I₂ = 10³ and only 1/1000 = 0.1 % of the intensity gets through.
Common mistake3.3 % (1/30) assumes decibels are proportional to intensity; every 10 dB is a factor of 10 in intensity, so 30 dB is a factor of 1000.
12Fill in the blank
On a stationary wave, the distance between two neighbouring nodes is ______ of a wavelength.
Show answer
Answer: half
Nodes are λ/2 apart, with an antinode midway between them.
Common mistakeAnswering 'one' confuses node-to-node spacing with a full wavelength, which spans two loops.
13Fill in the blank
Two sound sources with the same frequency and a constant phase difference are called ______ sources.
Show answer
Answer: coherent
Coherent sources are needed for a steady interference pattern.
Common mistakeWriting 'identical' is not enough: two identical-looking speakers on different generators are not coherent because their phase difference drifts.
14Multiple choice · ★ Challenge
A bat flies at 10 m/s towards a cave wall, sending out 40.0 kHz pulses (v = 340 m/s). What frequency does the bat receive in the echo from the wall?
1. A41.2 kHz
2. B37.7 kHz
3. C40.0 kHz
4. D42.4 kHz
Show answer
Answer: D. 42.4 kHz
The wall receives f₁ = 40.0 × 340/330 ≈ 41.2 kHz and reflects it as a stationary source; the moving bat hears f₂ = 41.2 × 350/340 ≈ 42.4 kHz (in one step: 40.0 × 350/330).
Common mistake41.2 kHz stops after the first shift; the bat is also a moving listener for the echo, so a second shift must be applied.
15Fill in the blank
The way sound lingers in a hall after the source stops, because of many reflections from the walls, floor and ceiling, is called ______.
Show answer
Answer: reverberation
Repeated reflections that arrive too quickly to be heard separately prolong the sound: this is reverberation.
Common mistakeAnswering 'echo' is not right here: an echo is a single reflection heard separately, not the lingering of the sound.
16Multiple choice · ★ Challenge
A classroom doorway is 0.85 m wide. Sound spreads out most strongly through it when the wavelength is about the same as the width. Which frequency is that (v = 340 m/s)?
1. A289 Hz
2. B800 Hz
3. C400 Hz
4. D200 Hz
Show answer
Answer: C. 400 Hz
f = v/λ = 340 ÷ 0.85 = 400 Hz.
Common mistake289 Hz comes from multiplying 340 by 0.85; use f = v/λ, which divides the speed by the wavelength.
17Multiple choice
What is the difference between an echo and reverberation?
1. AAn echo is reflected sound; reverberation is sound that has passed through a wall
2. BAn echo comes from a soft, absorbing surface, while reverberation always comes from a hard, smooth one such as concrete
3. CAn echo is heard as a separate repeat; in reverberation many quick reflections merge and prolong the sound
4. DThere is no difference; the two words mean the same thing
Show answer
Answer: C. An echo is heard as a separate repeat; in reverberation many quick reflections merge and prolong the sound
Reflections arriving within about 0.1 s cannot be separated by the ear, so they merge with the original and make it last longer (reverberation).
Common mistakeSaying they are the same ignores the time delay: only a reflection more than about 0.1 s later is heard as a distinct echo.
18Multiple choice · ★ Challenge
An older teacher can hear sounds up to only 12 kHz. What is the shortest wavelength in air that she can hear (v = 340 m/s)?
1. A35 m
2. B0.28 cm
3. C28 cm
4. D2.8 cm
Show answer
Answer: D. 2.8 cm
λ = v/f = 340 ÷ 12 000 = 0.028 m = 2.8 cm; the highest frequency gives the shortest wavelength.
Common mistake35 m comes from f/v = 12 000 ÷ 340, which has the formula upside down; check that m/s ÷ Hz gives metres.
19Fill in the blank
A scanner probe contains a piezoelectric ______ that turns electrical oscillations into ultrasound and turns returning echoes back into electrical signals.
Show answer
Answer: transducer
A transducer changes energy from one form to another; the piezoelectric crystal both sends and receives the ultrasound.
Common mistakeAnswering 'microphone' gives only half the job: the same crystal acts as both the emitter and the receiver of ultrasound.
20Multiple choice · ★ Challenge
Fork X gives 3 beats per second with a 384 Hz fork and 5 beats per second with a 392 Hz fork. What is the frequency of X?
1. A381 Hz
2. B387 Hz
3. C389 Hz
4. D397 Hz
Show answer
Answer: B. 387 Hz
From the first fork X = 381 or 387 Hz; from the second X = 387 or 397 Hz. The only value in both lists is 387 Hz.
Common mistakePicking 381 Hz uses only the first fork; each beat reading allows two values, and you need both readings to choose one.
21Fill in the blank
A note of 1024 Hz is one ______ above a note of 512 Hz, because its frequency is exactly double.
Show answer
Answer: octave
Two notes whose frequencies are in the ratio 2 : 1 are an octave apart.
Common mistakeWriting 'harmonic' mixes two ideas: 1024 Hz is the second harmonic of 512 Hz, but the musical interval is called an octave.
22Multiple choice · ★ Challenge
Two cars approach each other on a straight road, each at 20 m/s. Car A sounds a 400 Hz horn (v = 340 m/s). What frequency does the driver of car B hear?
1. A425 Hz
2. B356 Hz
3. C400 Hz
4. D450 Hz
Show answer
Answer: D. 450 Hz
Both moving towards each other: f' = f(v + v_o)/(v − v_s) = 400 × 360 ÷ 320 = 450 Hz.
Common mistake425 Hz allows only for the moving source; the moving listener also meets the waves faster, which raises the frequency further.
23Multiple choice
Why are pregnant women at a health centre scanned with ultrasound rather than X-rays?
1. AUltrasound travels faster through the body than X-rays
2. BUltrasound is not ionising, so it does not damage the cells of the baby
3. CUltrasound can pass through bone but X-rays cannot
4. DX-ray machines cannot show soft tissue at all, but ultrasound can show bone better
Show answer
Answer: B. Ultrasound is not ionising, so it does not damage the cells of the baby
At the power used in scanning, ultrasound causes no ionisation, so it is safe for the developing baby; X-rays can damage DNA.
Common mistakeChoosing 'faster' is wrong: X-rays travel at the speed of light, far faster than ultrasound; the reason is safety, not speed.
24Short answer · ★ Challenge
Explain why a cyclist riding towards a stationary siren hears a higher frequency, and why this differs from the case where the siren moves towards a stationary cyclist.
Show answer
Model answer: When the cyclist moves towards the siren, the wavelength in the air is unchanged, but she meets the wavefronts more often (relative speed v + v_o), so more waves reach her per second. When the siren moves, it squashes the waves in front of it, shortening the wavelength itself. The two cases give different formulae and slightly different frequencies for the same speed.
Common mistakeSaying both cases are identical ignores that only a moving source changes the wavelength; the formulae f(v + v_o)/v and fv/(v − v_s) give different answers.
25Multiple choice
At a wedding in a compound, guests behind a building hear the deep drums clearly but the high whistles only faintly. Why?
1. AThe long-wavelength drum sound diffracts more around the building
2. BHigh-frequency sound travels more slowly in air
3. CThe drums are louder, and a louder sound always bends further round obstacles
4. DWhistle sound is reflected back by the air
Show answer
Answer: A. The long-wavelength drum sound diffracts more around the building
Diffraction is greatest when the wavelength is similar to or larger than the obstacle; low notes have long wavelengths (several metres) and spread round the building.
Common mistakeChoosing 'high frequency travels more slowly' is wrong: all audible frequencies travel at the same speed in air; the difference is the amount of diffraction.
26Short answer · ★ Challenge
On a still, clear night, people in a village hear a distant road much more clearly than in the afternoon. At night the ground cools and the air near it is colder than the air above. Explain the effect.
Show answer
Model answer: Sound travels faster in warmer air, so at night the upper part of each wavefront moves faster than the lower part. The sound refracts (bends) back down towards the ground instead of escaping upwards, so more of it reaches listeners far away.
Common mistakeSaying sound travels faster in cold air reverses the effect; v ∝ √T, so the warmer layer above is the faster one and the waves bend down.
27Multiple choice
Tuning forks of 256 Hz and 260 Hz are sounded together. What does a listener hear?
1. AA 4 Hz note too low to hear
2. BTwo separate, steady notes of 256 Hz and 260 Hz
3. CA note of about 258 Hz whose loudness rises and falls 4 times each second
4. DA 516 Hz note, the sum of the two, whose loudness pulses 4 times each second
Show answer
Answer: C. A note of about 258 Hz whose loudness rises and falls 4 times each second
The ear hears the average frequency, (256 + 260)/2 = 258 Hz, with its loudness swelling and fading at the beat frequency 260 − 256 = 4 Hz.
Common mistakeThinking a 4 Hz note is heard confuses the beat frequency with the frequency of the sound; the 4 Hz is how often the loudness rises and falls.
28Short answer · ★ Challenge
A new church hall in Kigali has bare concrete walls, a tiled floor and metal chairs. Worshippers complain that the preacher's words are blurred. Explain the cause and suggest two practical improvements.
Show answer
Model answer: Hard, smooth surfaces reflect most of the sound, so many reflections arrive soon after the direct sound and the hall has a long reverberation time; syllables overlap and blur. Improvements: hang curtains or fit acoustic (porous) panels on the walls and ceiling, add soft seat cushions or carpets; these absorb sound and shorten the reverberation.
Common mistakeSuggesting a louder amplifier makes it worse: it increases the reflected sound as much as the direct sound, so the words stay blurred.
29Multiple choice
A learner blows across the top of a bottle and hears a note. Water is then poured in. What happens to the note and why?
1. APitch falls, because the water makes the bottle heavier
2. BPitch rises, because the air column is shorter
3. CPitch falls, because the air column is shorter
4. DPitch stays the same, because only the air vibrates
Show answer
Answer: B. Pitch rises, because the air column is shorter
The bottle acts as a closed pipe, f = v/4L; a shorter air column gives a higher frequency.
Common mistakeChoosing 'heavier bottle' mixes up the vibrating object: it is the air column, not the glass, that sets the note.
30Multiple choice · ★ Challenge
The speed of sound in air is 331 m/s at 0 °C and is proportional to √T, with T in kelvin. At what air temperature is the speed of sound 10 % greater than at 0 °C?
1. A27 °C
2. B330 °C
3. C57 °C
4. D300 °C
Show answer
Answer: C. 57 °C
v ∝ √T so T = 273 × 1.10² = 273 × 1.21 ≈ 330 K, which is 330 − 273 = 57 °C.
Common mistake27 °C comes from taking T ∝ v (273 × 1.10 = 300 K); because v ∝ √T, the kelvin temperature must rise by the square of the factor.
31Multiple choice
The threshold of pain is about 120 dB. What sound intensity is this? (I₀ = 10⁻¹² W/m²)
1. A120 W/m²
2. B1.0 W/m²
3. C10¹² W/m²
4. D12 W/m²
Show answer
Answer: B. 1.0 W/m²
I = I₀ × 10^β/10 = 10⁻¹² × 10¹² = 1 W/m².
Common mistake10¹² W/m² forgets to multiply by I₀; 120 W/m² treats the decibel scale as if it were linear.
32Multiple choice · ★ Challenge
Two neighbouring resonant frequencies of a pipe are 210 Hz and 350 Hz, and no resonance lies between them. What is the fundamental frequency, and is the pipe open or closed at one end?
1. A140 Hz, open
2. B70 Hz, open
3. C140 Hz, closed
4. D70 Hz, closed
Show answer
Answer: D. 70 Hz, closed
The gap 350 − 210 = 140 Hz. If the pipe were open, f₁ = 140 Hz and 210 Hz would not be a multiple of it. For a closed pipe neighbouring odd harmonics differ by 2f₁, so f₁ = 70 Hz; 210 = 3 × 70 and 350 = 5 × 70.
Common mistakeChoosing '140 Hz, open' takes the gap as f₁ without checking: 210 Hz is not a whole-number multiple of 140 Hz.
33Multiple choice
A minibus passenger travels at 25 m/s straight towards a church bell ringing at 510 Hz (v = 340 m/s). What frequency does the passenger hear?
1. A473 Hz
2. B510 Hz
3. C548 Hz
4. D551 Hz
Show answer
Answer: C. 548 Hz
Moving observer: f' = f(v + v_o)/v = 510 × 365 ÷ 340 ≈ 548 Hz.
Common mistake551 Hz uses the moving-source formula f v/(v − v_s); when the listener moves, the speed is added on top, f(v + v_o)/v.
34Multiple choice · ★ Challenge
A learner walks across in front of two loudspeakers that are in phase. At the first quiet spot beside the central loud spot, the path difference to the speakers is 0.17 m (v = 340 m/s). What is the frequency of the sound?
1. A1000 Hz
2. B2000 Hz
3. C500 Hz
4. D58 Hz
Show answer
Answer: A. 1000 Hz
At the first quiet spot the path difference is λ/2, so λ = 2 × 0.17 = 0.34 m and f = v/λ = 340 ÷ 0.34 = 1000 Hz.
Common mistake2000 Hz takes 0.17 m as a full wavelength; the first minimum is where one wave is half a wavelength behind the other.
35Multiple choice
An echo sounder on a boat on Lake Kivu receives the echo from the lake bed 0.60 s after sending a pulse. The speed of sound in the water is 1500 m/s. How deep is the lake at that point?
1. A450 m
2. B900 m
3. C102 m
4. D225 m
Show answer
Answer: A. 450 m
The pulse travels down and back: depth = vt/2 = 1500 × 0.60 ÷ 2 = 450 m.
Common mistake900 m forgets that the 0.60 s covers the trip down and back up; 102 m uses the speed of sound in air instead of water.
36Short answer · ★ Challenge
Explain why no beats are heard when notes of 300 Hz and 350 Hz are played together.
Show answer
Model answer: The beat frequency would be 350 − 300 = 50 Hz. The ear cannot follow loudness changes faster than about 15–20 per second, so instead of throbbing beats it hears two separate notes (or a rough combined sound).
Common mistakeSaying beats need the same frequency is wrong: beats need close but different frequencies; here the difference is too large for the ear to follow.
37Multiple choice
A 500 Hz siren moves at 40 m/s straight towards a listener (v = 340 m/s). What is the wavelength of the sound in the air between the siren and the listener?
1. A0.68 m
2. B0.76 m
3. C0.08 m
4. D0.60 m
Show answer
Answer: D. 0.60 m
Each second the siren sends 500 waves into a distance of 340 − 40 = 300 m, so λ = 300 ÷ 500 = 0.60 m.
Common mistake0.68 m is the wavelength from a stationary siren; 0.76 m uses v + vs, which applies behind the siren, not in front.
38Short answer · ★ Challenge
Two loudspeakers are connected to two separate signal generators, both set to 500 Hz. A learner expects fixed loud and quiet spots in the room, but they keep moving. Explain why, and say how to obtain a steady pattern.
Show answer
Model answer: Two separate generators are never exactly the same frequency and their phase difference keeps changing, so the sources are not coherent and the loud and quiet places drift (slow beats). Connecting both speakers to the same generator makes them coherent with a constant phase difference, giving a steady pattern.
Common mistakeBlaming room echoes misses the point: a stable interference pattern needs coherent sources with a fixed phase difference.
39Short answer
An electric bell rings inside a glass bell jar. Describe what is heard as the air is slowly pumped out, and explain why a faint sound may still be heard at the end.
Show answer
Model answer: The sound gets quieter and quieter as the air is removed, because sound needs particles of a medium to pass on the vibrations. A faint sound remains because a little air is left in the jar and some vibration passes through the bell's supports and the jar itself.
Common mistakeSaying the bell stops vibrating is wrong: the bell can still be seen striking; only the medium that carries the sound is removed.
40Short answer · ★ Challenge
In a resonance-tube experiment, a learner uses only the first resonance length L₁ and the formula v = 4fL₁. Explain why her value of v is too low, and describe how the method can be improved.
Show answer
Model answer: The antinode forms slightly above the open end, so the effective column is L₁ + e and λ = 4(L₁ + e); ignoring e makes λ and v too small. Measure the second resonance L₂ as well and use λ = 2(L₂ − L₁), which cancels the end correction; repeat with several forks and plot a graph.
Common mistakeBlaming the tuning fork is wrong: the systematic error comes from the end correction, which the two-resonance method removes.
41Multiple choice · ★ Challenge
A doctor chooses a 3.5 MHz probe to look at a patient's liver. Sound travels at 1540 m/s in soft tissue. How long is one wave of this ultrasound inside the body?
1. A0.097 mm
2. B2.3 mm
3. C0.44 mm
4. D0.88 mm
Show answer
Answer: C. 0.44 mm
λ = v/f = 1540 ÷ (3.5 × 10⁶) = 4.4 × 10⁻⁴ m = 0.44 mm; such a short wavelength lets the scan show fine detail.
Common mistake0.097 mm uses the speed of sound in air (340 m/s); in tissue sound travels much faster, 1540 m/s.
42Short answer
Give two uses of ultrasound outside hospitals, and explain why ultrasound rather than audible sound is used to find small cracks inside metal pipes.
Show answer
Model answer: Uses: finding fish shoals or the depth of a lake by sonar; cleaning jewellery or delicate parts in a liquid bath; measuring thickness or flaws in metal. Ultrasound has a very short wavelength, so it reflects from tiny cracks and forms a narrow beam that can be aimed; audible sound would diffract round small flaws.
Common mistakeSaying ultrasound is used 'because it is louder' is wrong: the advantage is its short wavelength, which gives reflections from small objects and little spreading.
43Multiple choice
Two loudspeakers driven in phase by the same generator emit 680 Hz (v = 340 m/s). Point P is 3.00 m from one speaker and 3.75 m from the other. What does a listener at P hear?
1. AA loud sound, because the path difference is a whole number of wavelengths
2. BA loud sound, because both speakers send the same frequency
3. CA quiet sound, because the waves have travelled different distances and lost energy
4. DA quiet sound, because the path difference is an odd number of half-wavelengths
Show answer
Answer: D. A quiet sound, because the path difference is an odd number of half-wavelengths
λ = 340 ÷ 680 = 0.50 m; path difference = 3.75 − 3.00 = 0.75 m = 1.5λ, so the waves arrive in antiphase and cancel (destructive interference).
Common mistakeChoosing 'loud, whole number of wavelengths' comes from not dividing 0.75 m by λ; 0.75 ÷ 0.50 = 1.5, an odd number of half-wavelengths.
44Short answer · ★ Challenge
Learners measure the sound level near a small water pump: 80 dB at 2 m, 74 dB at 4 m and 68 dB at 8 m. Show that the data agree with an inverse-square law and predict the level at 16 m.
Show answer
Model answer: If I ∝ 1/r², doubling r divides I by 4, a change of 10 log 4 ≈ 6 dB. The readings fall by 6 dB for each doubling of distance, as expected. At 16 m: 68 − 6 = 62 dB.
Common mistakeExpecting the level to halve to 40 dB at 4 m treats decibels as proportional to intensity; a quarter of the intensity is only 6 dB less.
45Multiple choice
Two notes are shown on the same oscilloscope settings. Trace Q has the same height as trace P but shows twice as many waves across the screen. Compared with P, note Q is:
1. AHigher in pitch, with the same loudness
2. BLouder, with the same pitch
3. CLower in pitch, with exactly the same loudness
4. DHigher in pitch and louder
Show answer
Answer: A. Higher in pitch, with the same loudness
Equal heights mean equal amplitudes (same loudness); more waves on the same time-base means a higher frequency, so a higher pitch.
Common mistakeChoosing 'louder' confuses the number of waves with the height of the trace; loudness is shown by the height only.
46Short answer · ★ Challenge
A worker strikes one end of a steel rail 510 m long. A learner with an ear on the other end hears two sounds, one through the rail and one through the air, 1.4 s apart. Taking v(air) = 340 m/s, calculate the speed of sound in the steel.
Show answer
Model answer: Time through air = 510 ÷ 340 = 1.5 s. Time through steel = 1.5 − 1.4 = 0.10 s. v(steel) = 510 ÷ 0.10 = 5100 m/s.
Common mistakeDividing 510 m by 1.4 s treats the time gap as the travel time in steel; the gap is the difference between the two travel times.
47Multiple choice
During a storm over Huye, thunder is heard 4.5 s after the lightning flash is seen (v = 340 m/s). About how far away did the lightning strike?
1. A0.77 km
2. B76 m
3. C3.1 km
4. D1.5 km
Show answer
Answer: D. 1.5 km
Light arrives almost instantly, so the delay is the sound's travel time: d = vt = 340 × 4.5 ≈ 1530 m ≈ 1.5 km.
Common mistake0.77 km halves the distance as if this were an echo; thunder travels only one way from the strike to the listener.
48Short answer · ★ Challenge
A learner presses a ukulele string at different places and records the vibrating length L for each note: 256 Hz, 0.60 m; 320 Hz, 0.48 m; 384 Hz, 0.40 m; 512 Hz, 0.30 m. Show that the data fit f ∝ 1/L and predict L for a 480 Hz note.
Show answer
Model answer: f × L = 256 × 0.60 = 153.6, 320 × 0.48 = 153.6, 384 × 0.40 = 153.6, 512 × 0.30 = 153.6: the product is constant, so f ∝ 1/L. For 480 Hz: L = 153.6 ÷ 480 = 0.32 m.
Common mistakeChecking f ÷ L instead of f × L would test direct proportion; for an inverse relationship the product must stay constant.
49Multiple choice
Two guitar strings have the same length and tension, but string B has 4 times the mass per unit length of string A. How does the fundamental frequency of B compare with that of A?
1. AHalf as high
2. BA quarter as high
3. CTwice as high
4. DThe same
Show answer
Answer: A. Half as high
v = √(T/μ): 4 times μ halves v, and with the same length f = v/2L also halves.
Common mistake'A quarter as high' forgets the square root; f ∝ 1/√μ, so 4 times μ gives ½ the frequency.
50Short answer · ★ Challenge
A learner stands 50 m from a tall wall and claps steadily so that each clap is made exactly as the echo of the previous one arrives. A friend times 30 clap intervals as 9.0 s. Calculate the speed of sound, and give one way to make the result more reliable.
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Model answer: Time for one interval = 9.0 ÷ 30 = 0.30 s, during which sound travels 2 × 50 = 100 m. v = 100 ÷ 0.30 ≈ 333 m/s. To improve: time more intervals, repeat and average, or stand further from the wall so the timing error is a smaller fraction.
Common mistakeUsing 50 m instead of 100 m halves the answer; in each interval the sound goes to the wall and back.