Chapter 10 · Grade 9 Science
Sound Waves:
Characteristics & Applications
Everything you need to know about how sound is produced, how it travels, what shapes it, and how we use it — explained simply.
1. Production of Sound
Sound is all around us — voices, music, thunder, birds. But how exactly is sound made? Every sound comes from something that is vibrating.
Imp What is Vibration?
Vibration is the back-and-forth (to and fro) motion of an object around a fixed position. As long as an object vibrates, it produces sound. When the vibration stops, the sound stops too.When you pluck a guitar string, strike a bell, or blow a flute, the object starts vibrating. This vibrating object is called the source of sound.
🎸 Strings
Vibrating strings of guitar, sitar, violin → produce sound
🥁 Membranes
Vibrating skin of tabla, drum, mridangam → produce sound
🎶 Air Columns
Vibrating air inside flute, bansuri → produce sound
🎙️ Vocal Cords
Vibrating muscular flaps in the larynx (voice box) → human speech
How Humans Produce Sound
Inside your throat is the larynx (voice box). Inside it are the vocal cords — thin, stretchy muscular flaps. When you speak or sing, air from your lungs passes through them, making them vibrate. Your tongue, lips, mouth, and nasal cavity then shape this vibration into words and music.Animal Sound Production
Some animals make sounds in unique ways. Grasshoppers and crickets rub their wings or legs together to produce sound — a process called stridulation.1.1 The Tuning Fork
A tuning fork is a U-shaped metal bar (usually steel or aluminium) with a handle (stem) at the bottom. The two arms of the U are called prongs or tines.
Fig. A tuning fork — prongs vibrate when struck on a rubber pad
When a prong is struck against a rubber pad, it vibrates and produces a nearly single-frequency sound. When the vibrating prong touches a water surface, it creates ripples — confirming that vibration is happening.
Imp Rule to Remember
Sound is produced only when something vibrates. No vibration → No sound.2. Propagation of Sound
Once sound is produced at a source, how does it reach your ears? It travels through a medium.
Imp Medium
A medium is any material (solid, liquid, or gas) through which sound can travel. Sound cannot travel through vacuum (empty space with no matter at all).2.1 Sound Travels Through Solids, Liquids & Gases
Through Solids
Place your ear on a table. Ask someone to scratch the other end. You hear it clearly — sound traveled through the solid wood.
Through Liquids
Tap two spoons together while they are submerged in water. You can still hear the sound — it traveled through water and then air.
Through Gases
We hear speech in everyday life. Sound travels through the air (a gas) from the source to our ears.
Through Vacuum?
❌ Sound CANNOT travel through vacuum. No particles = no vibration transfer = no sound.
2.2 The Vacuum Bell Jar Experiment
Fig. Vacuum Bell Jar — as air is removed, sound becomes fainter until almost silent
When air is pumped out of the jar, the bell can still be seen ringing but is barely heard. When air is allowed back in, the sound returns. This proves that sound needs a medium to propagate.
Imp Space & Astronauts
Outer space is nearly a vacuum. That is why astronauts doing spacewalks cannot hear each other speak directly or hear metal objects clanking. They communicate using radio devices inside their spacesuits.3. Sound Waves — How Sound Travels
Sound doesn't just jump from source to ear. It moves by creating disturbances in the particles of the medium — passing the push forward, one particle to the next.
3.1 Compressions and Rarefactions
Imagine a piston pushing and pulling inside a tube filled with air. This is a model for how a sound source works:
Fig. How compression (C) and rarefaction (R) form and travel in a medium
Compression (C)
When the source pushes forward, nearby air particles are squeezed together. The density increases above average. This is a compression.Rarefaction (R)
When the source pulls back, nearby air particles spread apart. The density decreases below average. This is a rarefaction.Imp Particles Don't Travel — Energy Does
In a sound wave, air particles do NOT flow from source to ear. Each particle only vibrates back and forth around its own position. What travels is the disturbance (energy) — passed from one particle to the next by collisions.Imp Definition of Sound Wave
A sound wave is a disturbance consisting of alternating compressions and rarefactions propagating through a medium, without any actual flow of the medium's particles.3.2 Sound as a Longitudinal Wave
Longitudinal Wave
Particles vibrate parallel (in the same direction) to the wave's direction of travel. Sound is a longitudinal wave.
Transverse Wave
Particles vibrate perpendicular (at 90°) to the wave's direction of travel. Example: light waves.
Imp Mechanical Wave
Sound is a mechanical wave — it needs a material medium to propagate. Light, on the other hand, is a transverse wave that can travel through vacuum. That's why light from the Sun reaches us, but Sun's sound cannot.4. Energy of Sound Waves
Sound carries energy. When a loud sound is made near a stretched membrane, grains placed on it jump — even though nothing is touching the membrane. The sound waves carry energy through air and transfer it to the membrane, making it vibrate.
Imp Energy Transfer in Sound
When a source vibrates, it transfers energy to the surrounding medium. As the sound wave moves through the medium, the vibrating particles collide with their neighbours, passing energy along the wave — all the way to your ear.🎙️ Microphone
Converts sound energy → electrical energy. Sound waves make the diaphragm vibrate, creating an electrical signal.
🔊 Speaker
Converts electrical energy → sound energy. An electrical signal makes the cone/diaphragm vibrate, producing sound.
5. Graphical Representation of a Sound Wave
We can draw a graph of a sound wave by plotting how the density of the medium changes with distance (at a fixed moment in time).
Fig. Graphical representation of a sound wave — density vs. distance. Crests = Compression, Troughs = Rarefaction
Crest
The highest point on the graph — corresponds to a Compression (maximum density).Trough
The lowest point on the graph — corresponds to a Rarefaction (minimum density).6. Characteristics of a Sound Wave
6.1 Wavelength (λ)
Imp Wavelength
The distance between two consecutive crests (or two consecutive troughs) is called the wavelength of the wave.Symbol: λ (Greek letter lambda)
SI Unit: metre (m)
6.2 Frequency (ν)
Imp Frequency
The number of complete density oscillations per second at a fixed point is the frequency.Symbol: ν (Greek letter nu)
SI Unit: Hertz (Hz) = per second (s⁻¹)
6.3 Time Period (T)
Imp Time Period
The time taken for one complete density oscillation at a fixed point.Symbol: T
SI Unit: second (s)
Imp Relation — Frequency & Time Period
ν = 1 / T or T = 1 / ν
Frequency and time period are inversely related. Higher frequency → shorter time period.
6.4 Amplitude
Imp Amplitude
The maximum change in density in a compression (or rarefaction) compared to the average density. A larger amplitude means the wave carries more energy.Fig. Comparison of low and high amplitude sound waves
6.5 Intensity of Sound
Imp Intensity
The amount of sound energy passing through a unit area per unit time, perpendicular to the direction of sound propagation.As sound travels farther from the source, it spreads over a larger area. The same energy now covers more area → intensity decreases with distance.
6.6 Speed of Sound
Imp Speed of Sound
The distance traveled by a point on the wave (like a crest or trough) per unit time.Imp Wave Speed Formula
v = λ × ν
Speed = Wavelength × Frequency
Imp Speed in Different Media
Sound travels at different speeds in different media:- Fastest in solids (e.g. steel: ~5000 m/s) — particles are closest
- Slower in liquids (e.g. water: ~1500 m/s)
- Slowest in gases (e.g. air: ~340 m/s)
Sound travels 4–5 times faster in water and 15–20 times faster in solids compared to air.
| State of Medium | Substance | Speed (at 15°C) |
|---|---|---|
| Solid | Steel | 5000 m/s |
| Liquid | Water | 1500 m/s |
| Gas | Air | 340 m/s |
Effect of Temperature & Humidity
As temperature or humidity increases, the speed of sound in air increases.Speed in dry air: 331 m/s at 0°C and ~344 m/s at 22°C.
6.7 Human Perception of Sound — Pitch & Loudness
🔺 Pitch
How a human perceives frequency. High frequency → high pitch (shrill sounds like a whistle). Low frequency → low pitch (deep sounds like thunder).
🔊 Loudness
How a human perceives amplitude. Large amplitude → loud sound. Small amplitude → soft sound. Measured in decibels (dB).
| Sound | Approx. Level (dB) |
|---|---|
| Rustling leaves | ~10 dB |
| Normal conversation | ~60 dB |
| Firecrackers | > 100 dB |
Intensity vs. Loudness
Intensity is a measurable, objective physical quantity. Loudness is subjective — it depends on the listener's hearing ability. Both are commonly measured in dB, but they are not the same thing.6.8 Human Hearing Range
Infrasound (< 20 Hz)
Below human hearing. Detected by elephants. Used to detect earthquakes, volcanic eruptions, and severe storms (travels very long distances).Ultrasound (> 20,000 Hz)
Above human hearing. Detected by dogs, cats, bats, dolphins. Many important applications in medicine and industry.6.9 Tone, Musical Note & Timbre
Tone
A sound of a single frequency. Produced by a tuning fork or whistling. Simple, pure sound wave.
Musical Note
A combination of frequencies — the lowest (fundamental) + higher overtones. Produced by singing or plucking a string. Rich and complex.
Timbre
The unique quality that lets us tell a flute apart from a violin even when they play the same note. Comes from the pattern of overtones, shape, and material of the instrument.
Octave
The interval between two notes where one has double the frequency of the other. Example: 200 Hz and 400 Hz are one octave apart.
7. Reflection of Sound
Sound waves bounce off solid or liquid surfaces — this is called the reflection of sound. Sound follows the same laws of reflection as light:
Imp Law of Reflection of Sound
The angle of incidence (angle between incoming sound and the normal to the surface) = angle of reflection (angle between reflected sound and the normal). All three — incident direction, normal, and reflected direction — lie in the same plane.7.1 Echo
Imp Echo
When sound reflects off a distant hard surface and returns to the listener, it is heard again after a gap. This distinct repetition is called an echo.For an echo to be heard separately, the reflected sound must reach your ears at least 0.1 seconds after the original sound.
Minimum Distance for an Echo
Distance = (v × t) / 2
v = speed of sound | t = time for echo to return
Minimum echo distance = (340 m/s × 0.1 s) / 2 = 17 metres
Minimum echo distance = (340 m/s × 0.1 s) / 2 = 17 metres
Why Echoes Are Stronger From Some Surfaces
- Hard, smooth surfaces (stone walls, cliffs) → strong echo
- Soft surfaces (curtains, cushions) → absorb sound, weak/no echo
- Rough surfaces → scatter sound in different directions, weak echo
7.2 Reverberation
Imp Reverberation
In a large hall or auditorium, sound undergoes multiple reflections from the walls, ceiling, and floor. These reflections overlap and the sound persists (continues) after the source stops. This is called reverberation.It occurs when sound reflections arrive with a time gap of less than 0.05 s — too quick for the ear to separate.
Reverberation in Auditoriums
Some reverberation is desirable — it enriches music. Too much reverberation garbles sound. Auditoriums use sound-absorbing panels, upholstered chairs, curtains, and curved walls to control reverberation.8. Ultrasonic & Infrasonic Waves — Applications
Ultrasonography
Ultrasound is used to create images of internal organs without surgery (e.g. pregnancy scans).
Kidney Stone Removal
High-frequency ultrasound breaks kidney stones into tiny pieces that pass out naturally.
Industrial Cleaning
Ultrasonic waves clean delicate machine parts and are used in ultrasonic welding.
Flaw Detection
Ultrasound detects hidden cracks or defects inside metal blocks used in construction.
Earthquake Detection
Infrasound (low frequency) is used to detect earthquakes, volcanic eruptions, and severe storms.
Object Location
Ultrasound reflects off objects to determine their distance, direction, and speed (sonar, echolocation).
8.1 Echolocation
Bats are nocturnal and cannot rely on sight in the dark. Most bats emit short bursts of ultrasonic waves. These waves bounce off objects (prey, obstacles) and the reflected echoes return to the bat. By analyzing these echoes, the bat determines the exact position, size, and distance of the object.
Fig. Echolocation by bats — sending ultrasound and receiving echo to locate prey
Other animals that use echolocation: Dolphins Whales Some birds (oilbirds, swiftlets)
8.2 SONAR — Underwater Detection
Imp SONAR
SONAR = Sound Navigation And Ranging. It uses the same principle as echolocation. Ultrasonic waves are sent into water from a ship. The reflected waves are analyzed to find the distance, direction, and speed of underwater objects (submarines, fish schools, shipwrecks, ocean floor).Imp SONAR Distance Formula
Distance = (v × t) / 2
v = speed of sound in water | t = total time for signal to go and return
Exercise Questions & Answers
Multiple Choice Questions (MCQ)
1Which observation best supports the idea that sound is a mechanical wave?
Answer: (ii) A mechanical wave is defined as one that requires a material medium to travel. The fact that sound cannot travel through vacuum (as proven by the bell jar experiment) directly shows it is a mechanical wave.
2For a sound wave propagating in a medium, increasing its frequency will increase its —
Answer: (iii) Speed in a medium stays constant regardless of frequency. If frequency increases: more oscillations per second = more compressions per second. Wavelength actually decreases (v = λν, v constant → λ = v/ν decreases). Time period also decreases (T = 1/ν).
3If 20 compressions pass a point in 4 seconds, the frequency is —
Answer: (ii) 5 Hz
Frequency = number of compressions / time = 20 / 4 = 5 Hz
Frequency = number of compressions / time = 20 / 4 = 5 Hz
4In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation?
Answer: Reverberation.
An echo requires the time gap between the original and reflected sound to be at least 0.1 s so the brain can hear them as separate sounds. Here the gap is only 0.05 s — less than 0.1 s — so the reflected sound merges with the original. This is reverberation, not an echo.
An echo requires the time gap between the original and reflected sound to be at least 0.1 s so the brain can hear them as separate sounds. Here the gap is only 0.05 s — less than 0.1 s — so the reflected sound merges with the original. This is reverberation, not an echo.
5Two sound wave graphs (a) and (b) are shown with the same axis scales. Which has (i) greater wavelength, and (ii) smaller amplitude?
Answer:
(i) Wave (a) has greater wavelength — the crests are farther apart, meaning the distance between two consecutive crests is more.
(ii) Wave (b) has smaller amplitude — the height of the crests (and depth of troughs) from the average density line is less in (b).
(i) Wave (a) has greater wavelength — the crests are farther apart, meaning the distance between two consecutive crests is more.
(ii) Wave (b) has smaller amplitude — the height of the crests (and depth of troughs) from the average density line is less in (b).
6Sound waves from three sources A, B, C are represented in a figure. Frequency of A is maximum and C is minimum. Identify the curves.
Answer:
Higher frequency → shorter wavelength (more waves squeezed in the same distance).
• A = the curve with the shortest wavelength (most compressions per unit distance)
• B = the curve with medium wavelength
• C = the curve with the longest wavelength (fewest compressions per unit distance)
Higher frequency → shorter wavelength (more waves squeezed in the same distance).
• A = the curve with the shortest wavelength (most compressions per unit distance)
• B = the curve with medium wavelength
• C = the curve with the longest wavelength (fewest compressions per unit distance)
7Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
Answer: The graph shows a sinusoidal wave. The crests rise 3 units above the average density line and the troughs dip 3 units below. The distance between two consecutive crests = 4 cm = wavelength.
8In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?
Answer — Two errors:
1. No sound in space: Outer space is nearly a vacuum. Sound is a mechanical wave and needs a medium to travel. In vacuum, there are no particles to carry the sound wave — so no sound should be heard at all.
2. Light and sound together: Even if there were an atmosphere, light travels at 300,000 km/s while sound travels at only ~340 m/s. Light would reach the observer almost instantly, but sound would take much longer. They would never be heard and seen at exactly the same time from any significant distance.
1. No sound in space: Outer space is nearly a vacuum. Sound is a mechanical wave and needs a medium to travel. In vacuum, there are no particles to carry the sound wave — so no sound should be heard at all.
2. Light and sound together: Even if there were an atmosphere, light travels at 300,000 km/s while sound travels at only ~340 m/s. Light would reach the observer almost instantly, but sound would take much longer. They would never be heard and seen at exactly the same time from any significant distance.
9A source produces a sound wave of wavelength 3.44 m. The wave travels with a speed of 344 m/s. Find its time period.
Solution:
v = λ × ν → ν = v / λ = 344 / 3.44 = 100 Hz
T = 1 / ν = 1 / 100 = 0.01 s
The time period of the sound wave is 0.01 seconds.
v = λ × ν → ν = v / λ = 344 / 3.44 = 100 Hz
T = 1 / ν = 1 / 100 = 0.01 s
The time period of the sound wave is 0.01 seconds.
10A ship sent a sonar signal and detected an echo after 5 s. Ultrasonic wave travels at 1525 m/s in seawater. How far down is the wreckage?
Solution:
Time for signal to reach wreckage = 5/2 = 2.5 s
Distance = speed × time = 1525 × 2.5 = 3812.5 m ≈ 3813 m
The wreckage is approximately 3812.5 metres (≈ 3.81 km) below the surface.
Time for signal to reach wreckage = 5/2 = 2.5 s
Distance = speed × time = 1525 × 2.5 = 3812.5 m ≈ 3813 m
The wreckage is approximately 3812.5 metres (≈ 3.81 km) below the surface.
11A parking sensor emits 40 kHz ultrasound. Warning beep starts at 1.2 m from obstacle. Speed of ultrasound in air = 345 m/s. How much time does the ultrasound take to travel to the obstacle and come back?
Solution:
Total distance = 1.2 m (to obstacle) + 1.2 m (back) = 2.4 m
Time = Distance / Speed = 2.4 / 345 = 0.00696 s ≈ 6.96 × 10⁻³ s
The ultrasound takes approximately 0.007 seconds (6.96 ms) to travel to the obstacle and return.
Total distance = 1.2 m (to obstacle) + 1.2 m (back) = 2.4 m
Time = Distance / Speed = 2.4 / 345 = 0.00696 s ≈ 6.96 × 10⁻³ s
The ultrasound takes approximately 0.007 seconds (6.96 ms) to travel to the obstacle and return.
12Speed of sound in air: 331 m/s at 0°C, 344 m/s at 22°C. How much extra time will sound of thunder take to travel 1720 m if temperature drops from 22°C to 0°C?
Solution:
Time at 22°C: t₁ = 1720 / 344 = 5.0 s
Time at 0°C: t₂ = 1720 / 331 ≈ 5.196 s
Extra time = t₂ − t₁ = 5.196 − 5.0 = ≈ 0.196 s ≈ 0.2 s
Sound takes approximately 0.2 seconds more at 0°C than at 22°C to travel the same distance.
Time at 22°C: t₁ = 1720 / 344 = 5.0 s
Time at 0°C: t₂ = 1720 / 331 ≈ 5.196 s
Extra time = t₂ − t₁ = 5.196 − 5.0 = ≈ 0.196 s ≈ 0.2 s
Sound takes approximately 0.2 seconds more at 0°C than at 22°C to travel the same distance.
13A sound wave propagates at 340 m/s. The figure (Fig. 10.32) shows the density variation — wavelength = 8 cm. Calculate wavelength and frequency.
Solution:
Wavelength (λ) = 8 cm = 0.08 m (from graph — distance for one full wave cycle)
Speed (v) = 340 m/s
Frequency (ν) = v / λ = 340 / 0.08 = 4250 Hz
Wavelength = 0.08 m (8 cm), Frequency = 4250 Hz
Wavelength (λ) = 8 cm = 0.08 m (from graph — distance for one full wave cycle)
Speed (v) = 340 m/s
Frequency (ν) = v / λ = 340 / 0.08 = 4250 Hz
Wavelength = 0.08 m (8 cm), Frequency = 4250 Hz
14Two sound waves A and B propagate at 345 m/s. From Fig. 10.33: Wave A has wavelength 5.0 cm, Wave B has wavelength 2.5 cm. Calculate their frequencies.
Solution:
Reading the graph:
• Wave A: one full cycle spans 5.0 cm → λ_A = 5.0 cm = 0.05 m
• Wave B: one full cycle spans 2.5 cm → λ_B = 2.5 cm = 0.025 m
Frequency of A: ν_A = v / λ_A = 345 / 0.05 = 6900 Hz
Frequency of B: ν_B = v / λ_B = 345 / 0.025 = 13,800 Hz
Reading the graph:
• Wave A: one full cycle spans 5.0 cm → λ_A = 5.0 cm = 0.05 m
• Wave B: one full cycle spans 2.5 cm → λ_B = 2.5 cm = 0.025 m
Frequency of A: ν_A = v / λ_A = 345 / 0.05 = 6900 Hz
Frequency of B: ν_B = v / λ_B = 345 / 0.025 = 13,800 Hz
15Sound sources at A (air) and B (water) are equidistant from a cliff. Time for sound to return to A is 4.5 times that of B. What is the ratio of speed of sound in air to water?
Solution:
Both A and B are at the same distance (d) from the cliff, so the total path (to cliff and back) = 2d for both.
Time taken by A (in air): t_A = 2d / v_air
Time taken by B (in water): t_B = 2d / v_water
Given: t_A = 4.5 × t_B
→ 2d / v_air = 4.5 × (2d / v_water)
→ 1 / v_air = 4.5 / v_water
→ v_water / v_air = 4.5
→ v_air : v_water = 1 : 4.5 = 2 : 9
The ratio of speed of sound in air to water = 1 : 4.5 (or 2 : 9).
Both A and B are at the same distance (d) from the cliff, so the total path (to cliff and back) = 2d for both.
Time taken by A (in air): t_A = 2d / v_air
Time taken by B (in water): t_B = 2d / v_water
Given: t_A = 4.5 × t_B
→ 2d / v_air = 4.5 × (2d / v_water)
→ 1 / v_air = 4.5 / v_water
→ v_water / v_air = 4.5
→ v_air : v_water = 1 : 4.5 = 2 : 9
The ratio of speed of sound in air to water = 1 : 4.5 (or 2 : 9).
Pause & Ponder Questions — Answers
P3Assertion: We cannot hear the bell in a closed jar after most air is pumped out. Reason: Sound requires a medium to travel.
Answer: (ii) Both statements are true, and R directly explains A. Because sound needs a medium to travel, removing the air (medium) prevents sound from reaching your ear — which is why you can't hear the bell.
P4Assertion: Compressions and rarefactions move through the medium. Reason: Individual particles of the medium continuously move forward with the wave.
Answer: (iii) Assertion is TRUE — compressions and rarefactions do travel through the medium. But the Reason is FALSE — individual particles do NOT move forward with the wave. They only vibrate back and forth about their own mean positions. It is the disturbance (energy) that moves forward, not the particles.
P5When sound travels from a tuning fork to your ear, what actually reaches your ear?
Answer: (ii) Only energy travels from source to ear. Air particles do not flow — they only vibrate around their own positions, passing energy to their neighbours through collisions.
P7Thin rubber band vs thick rubber band in Activity 10.1 — does the thin one vibrate faster? How do frequency and time period differ?
Answer:
Yes, the thin rubber band vibrates faster than the thick one (under the same tension).
• The thin band has higher frequency → shorter time period (T = 1/ν)
• The thick band has lower frequency → longer time period
This is why thin strings in musical instruments produce higher-pitched sounds.
Yes, the thin rubber band vibrates faster than the thick one (under the same tension).
• The thin band has higher frequency → shorter time period (T = 1/ν)
• The thick band has lower frequency → longer time period
This is why thin strings in musical instruments produce higher-pitched sounds.
P8Frequency of sound wave = 20 Hz. How many oscillations does the piston complete per minute?
Solution:
Frequency = 20 Hz = 20 oscillations per second
In 1 minute (60 seconds): oscillations = 20 × 60 = 1200 oscillations per minute
Frequency = 20 Hz = 20 oscillations per second
In 1 minute (60 seconds): oscillations = 20 × 60 = 1200 oscillations per minute
P9From the graph (Fig. 10.19), the x-axis goes 0 to 4.5 cm. What is half the wavelength?
Solution:
From the graph, one full wave cycle (crest to crest) = 3.0 cm → Wavelength (λ) = 3.0 cm
Half wavelength = λ / 2 = 3.0 / 2 = 1.5 cm
From the graph, one full wave cycle (crest to crest) = 3.0 cm → Wavelength (λ) = 3.0 cm
Half wavelength = λ / 2 = 3.0 / 2 = 1.5 cm
P10Compare speeds: (i) Water vs Air, (ii) Steel vs Water (using Table 10.1)
Solution:
(i) Speed in water / Speed in air = 1500 / 340 ≈ 4.41
→ Sound travels about 4.4 times faster in water than in air.
(ii) Speed in steel / Speed in water = 5000 / 1500 ≈ 3.33
→ Sound travels about 3.3 times faster in steel than in water.
(i) Speed in water / Speed in air = 1500 / 340 ≈ 4.41
→ Sound travels about 4.4 times faster in water than in air.
(ii) Speed in steel / Speed in water = 5000 / 1500 ≈ 3.33
→ Sound travels about 3.3 times faster in steel than in water.
P11Two friends 340 m apart along a steel fence. Gunjan's ear is on the fence. Knocking produces sound. Find time difference between sound via air and via steel. Can she distinguish them?
Solution:
Distance = 340 m
Time via air: t_air = 340 / 340 = 1.000 s
Time via steel: t_steel = 340 / 5000 = 0.068 s
Time difference = 1.000 − 0.068 = 0.932 s
Since 0.932 s > 0.1 s (the minimum gap needed to hear two sounds separately), yes, Gunjan can clearly distinguish between the two sounds — the steel sound arrives much earlier than the air sound.
Distance = 340 m
Time via air: t_air = 340 / 340 = 1.000 s
Time via steel: t_steel = 340 / 5000 = 0.068 s
Time difference = 1.000 − 0.068 = 0.932 s
Since 0.932 s > 0.1 s (the minimum gap needed to hear two sounds separately), yes, Gunjan can clearly distinguish between the two sounds — the steel sound arrives much earlier than the air sound.
P12Echoes must arrive at least 0.2 s after emission. What minimum distance should the reflecting surface be at? (Speed of sound = 343 m/s)
Solution:
Total distance traveled by sound = v × t = 343 × 0.2 = 68.6 m
This is the total round trip (to surface and back).
Minimum distance of reflecting surface = 68.6 / 2 = 34.3 m
Total distance traveled by sound = v × t = 343 × 0.2 = 68.6 m
This is the total round trip (to surface and back).
Minimum distance of reflecting surface = 68.6 / 2 = 34.3 m
P13A sonar signal takes 4 s to return. Speed of sound in seawater = 1500 m/s. What is the depth?
Solution:
Time to reach ocean floor = 4 / 2 = 2 s
Depth = speed × time = 1500 × 2 = 3000 m (3 km)
Time to reach ocean floor = 4 / 2 = 2 s
Depth = speed × time = 1500 × 2 = 3000 m (3 km)
Quick Concept Reference
| Term | Definition | Symbol / Unit |
|---|---|---|
| Wavelength | Distance between two consecutive crests (or troughs) | λ / metre (m) |
| Frequency | Number of complete oscillations per second | ν / Hertz (Hz) |
| Time Period | Time for one complete oscillation | T / second (s) |
| Amplitude | Maximum change in density from average | — / density units |
| Intensity | Sound energy per unit area per unit time | — / W/m² |
| Speed | Distance traveled by wave per unit time | v / m/s |
| Pitch | Human perception of frequency | High/Low |
| Loudness | Human perception of amplitude | dB |
| Echo | Reflected sound heard after ≥ 0.1 s gap | Min. distance: 17 m |
| Reverberation | Multiple reflections with < 0.05 s gap | — |
| Infrasound | Frequency below 20 Hz | < 20 Hz |
| Ultrasound | Frequency above 20,000 Hz | > 20 kHz |