Sound Waves: Characteristics and Applications explains how vibrating objects make sound and why sound needs a medium to travel. It shows sound as a longitudinal mechanical wave of compressions and rarefactions, where energy moves forward while particles only vibrate in place. You learn wavelength, frequency, time period, amplitude, intensity and speed, and the relations Ξ½ = 1/T and v = λν. The chapter ends with pitch, loudness, the audible range, echoes, reverberation, echolocation and sonar, which supply most of the chapter’s short numericals.
Key Concepts
1. Production of Sound: Vibrations
You learnt in Chapter 7 that sound is a form of energy, and that energy only changes from one form to another. So every sound starts with some other form of energy being turned into sound energy.
Activity 10.1 (rubber band on a box): stretch a rubber band across the open side of a cardboard box and pluck it. You hear a sound and see the band vibrating. When the band stops vibrating, the sound stops. Change the tension and the sound changes. Stretch the band between two fingers without the box and the sound is softer.
Conclusion: sound is produced by vibrations. A vibration is the periodic to and fro motion (oscillation) of an object. The object that produces the sound is the source of the sound.
| What vibrates | Example |
|---|---|
| Stretched string | Rubber band, sitar, veena, tanpura |
| Stretched membrane | Tabla, mridangam |
| Air column | Bansuri (flute): the air inside the hollow pipe vibrates |
| Metal object | A struck plate, a bell, a tuning fork |
In most musical instruments more than one part vibrates.
Humans and animals. In humans, sound is produced by the vibration of the vocal cords, tightly stretched muscular flaps inside the voice box, or larynx, in the throat. The tongue, lips, mouth and nasal cavity shape that sound into speech or music. Grasshoppers and crickets produce sound by rubbing their wings or legs.
The tuning fork. A tuning fork is a U-shaped metal bar with a stem, usually made of steel or aluminium. The two sides of the U are the prongs or tines. In Activity 10.2 you hold it by the stem and strike one prong gently on a soft rubber pad (never on a hard surface). Held near the ear, it sounds. Touch a vibrating prong to water and small waves form, which proves the prongs are vibrating. A tuning fork gives a sound of nearly a single frequency.
Sudden loud sounds. In firecrackers and thunder, air or gases heated rapidly expand very fast, sending out a sudden disturbance in air density that we hear as a loud sound pulse. A supersonic aircraft, flying faster than sound, produces a sonic boom.
2. Propagation of Sound: Sound Needs a Medium
The material through which sound propagates is called a medium. It can be a solid, a liquid or a gas.
- Solid (Activity 10.3): press your ear on a desk while a friend gently knocks the other end. You hear the knock through the wood.
- Liquid (Activity 10.4): tap two metal spoons together fully under water in a bucket. You still hear the tap, so the sound travelled through water and then air.
- Gas: almost every sound you hear reaches you through air.
A space with no matter in it is a vacuum. In the vacuum bell jar experiment, an electric bell rings inside a glass jar connected to a vacuum pump. As air is pumped out, the sound becomes fainter. Near vacuum, almost no sound is heard even though the bell is seen ringing. When air is let back in, the sound gradually becomes as loud as before.
Conclusion: sound cannot propagate in vacuum; it needs a material medium. You can still see the bell because light does not need a medium.
Astronauts on a spacewalk. Outer space is a near vacuum. Astronauts cannot directly hear each other speak or hear metal parts clanking. They talk through special devices fitted in their spacesuits.
3. Sound as a Wave: Compressions, Rarefactions and Longitudinal Waves
The slinky analogy (Activity 10.5). Mark one turn of a slinky stretched along a table, with a friend holding the far end. Push your end sharply and pull it back, again and again. Regions where the turns are closer together and regions where they are more spread out travel along the slinky. The marked turn does not travel with them. It only moves back and forth about its own position of rest, parallel to the direction of the disturbance.
The piston in a tube. Picture a long tube of air with a piston at one end, open at the other. The oscillating piston is a simple model of a sound source.
- Piston at rest: the air has a uniform density, the average density.
- Piston moves forward: it pushes nearby air particles forward and squeezes them, raising the density above average. This region is a compression (C). The particles collide with those ahead, which collide with their neighbours, so the compression moves forward while the particles themselves do not travel with it.
- Piston moves back: the air near it becomes less dense. A region with density lower than average is a rarefaction (R). It also moves ahead through collisions.
- Piston keeps oscillating: compressions and rarefactions are produced alternately and travel away from the source.
Definition: a sound wave is the disturbance consisting of a series of alternating compressions and rarefactions propagating through a medium, without the actual flow of the particles of the medium. The direction in which it travels is the direction of propagation.
Longitudinal waves. In sound, the particles vibrate back and forth parallel to the direction of propagation. Such waves are longitudinal waves.
Mechanical waves. Waves that need a material medium to propagate are mechanical waves. Sound is a mechanical wave.
| Feature | Longitudinal wave | Transverse wave |
|---|---|---|
| Particle vibration | Parallel to the direction of propagation | Perpendicular to the direction of propagation |
| Examples in the chapter | Sound; the longitudinal seismic waves, first to be detected by seismographs | Light, which travels even through vacuum |
Not all waves are mechanical. Light can travel through vacuum, which is why sunlight reaches the Earth. In open air, a small source sends compressions and rarefactions out in all directions as spherical waves.
4. Graph of a Sound Wave: Wavelength, Frequency and Time Period
At one instant, the density of the medium rises and falls again and again with distance from the source. On a graph with distance on the x-axis and density on the y-axis, the average density is a horizontal dashed line.
- In a compression the density rises above average. The highest point is the crest (maximum density).
- In a rarefaction it falls below average. The lowest point is the trough (minimum density).
| Quantity | Definition (NCERT) | Symbol | SI unit |
|---|---|---|---|
| Wavelength | The distance between two consecutive crests or two consecutive troughs | λ (lambda) | metre (m) |
| One complete oscillation | The change in density at a fixed point from maximum to minimum and back to maximum (or vice versa) | - | - |
| Frequency | The number of density oscillations at a fixed point per unit time | ν (nu) | s−1, called hertz (Hz) |
| Time period | The time taken for one complete density oscillation at a fixed point | T | second (s) |
Relation (Eq. 10.1): ν = 1 / T. A shorter time period means a higher frequency. A 50 Hz wave has T = 1/50 = 0.02 s; a wave with T = 0.001 s has ν = 1000 Hz = 1 kHz.
- Crest to next crest is one wavelength; crest to next trough is half a wavelength.
- One compression plus one rarefaction make one wavelength.
- More crests in the same distance means a shorter wavelength.
Everyday sounds contain a mixture of many frequencies. A nearly single-frequency sound comes from a tuning fork or oral whistling. In Activity 10.7, a frequency app shows that when you sing Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa, the frequency is lowest for Sa and rises for each following note. Each note has its own distinct frequency.
5. Energy of Sound, Amplitude and Intensity
Activity 10.6 (dancing grains). Stretch a rubber sheet over a wide-mouthed container and sprinkle rice or salt on it. Make a loud sound nearby without touching it. The grains jump: sound travelled through air, made the sheet vibrate, and the sheet moved the grains.
So sound is a form of energy. The vibrating source transfers energy to the medium, and particles pass it on by vibrating and colliding. In a sound wave, energy is transferred while the particles stay near their mean positions. Particles are always moving randomly because of thermal energy; a passing wave only increases their vibration for a while.
- A microphone converts sound energy into electrical energy: sound makes a thin diaphragm vibrate, and the vibrations become an electrical signal.
- A speaker does the opposite: an electrical signal vibrates a cone or diaphragm, which produces sound.
Amplitude is the maximum change in the density of air in a compression (or a rarefaction) compared to the average density. On the graph it is the height of a crest above the average line. A wave with a larger amplitude carries more energy: strike the plate harder in Activity 10.6 and the grains jump higher.
Intensity of sound is the amount of sound energy passing through a unit area perpendicular to the direction of propagation in a unit time. As sound spreads out, the same energy covers a larger area, so intensity decreases with distance. Sounds with a larger initial amplitude travel further before fading away.
6. Speed of Sound
The speed of sound is the distance which a point on the wave, such as a crest (or a trough), travels in unit time.
Deriving v = λ × ν: a crest moves one wavelength λ in one time period T. Speed = distance / time, so v = λ / T. Since ν = 1/T, v = λ × ν (Eq. 10.2). Units: m × s−1 = m s−1.
Speed depends on the medium: fastest in solids, slower in liquids, slowest in gases. Sound travels about 4 to 5 times faster in water and typically 15 to 20 times faster in solids than in air.
| State | Medium | Approximate speed at 15 °C (NCERT Table 10.1) |
|---|---|---|
| Solid | Steel | 5000 m s−1 |
| Liquid | Water | 1500 m s−1 |
| Gas | Air | 340 m s−1 |
Temperature and humidity. In air, the speed increases as temperature or humidity increases: about 331 m s−1 in dry air at 0 °C and nearly 344 m s−1 at 22 °C. Questions use 340, 343, 344 or 345 m s−1 for air; use the value given.
Frequency does not change the speed. In media such as air, the speed depends only on the medium, and neither on the source nor on the frequency. If the frequency changes, the wavelength changes instead: λ = v/ν, so doubling the frequency halves the wavelength.
Lightning and thunder. Both happen together, but light (about 300,000 km s−1) reaches you almost instantly while sound is far slower. The delay × speed of sound gives the distance to the strike.
7. Human Perception of Sound: Pitch, Loudness and the Audible Range
Frequency, amplitude and speed are measurable physical properties. How we experience sound is subjective and is described by pitch and loudness, which have no simple relation with the physical properties.
Pitch is how frequency is perceived by humans. A whistle or siren is shrill (high pitch); thunder or an aircraft rumble is deep (low pitch). In general, higher frequency means higher pitch, although the exact relation is complicated. During adolescence, boys’ vocal cords lengthen and thicken and vibrate less frequently, deepening their voice. Similarly, a thin rubber band vibrates faster than a thick one, giving a higher frequency and a shorter time period.
Loudness is how humans perceive the amplitude. Larger amplitude sounds louder. Loudness decreases as you move away from the source. Intensity is a measurable quantity, while loudness depends on the listener’s hearing ability.
| Physical property | What we perceive | Rule of thumb |
|---|---|---|
| Frequency | Pitch | Higher frequency, higher pitch |
| Amplitude | Loudness | Larger amplitude, louder sound |
| Pattern and intensity of overtones | Timbre | Tells a flute from a tabla on the same note |
Audible range: humans hear 20 Hz to 20,000 Hz (20 kHz). It varies from person to person and decreases with age.
- Infrasonic waves: below 20 Hz. Elephants can detect them.
- Ultrasonic waves: above 20 kHz. Dogs, cats, bats and dolphins can detect them.
Decibels and noise. Loudness is commonly measured in decibels (dB): rustling leaves are a few dB, normal conversation about 60 dB, and firecrackers can exceed 100 dB. A small rise in dB means a large rise in intensity. Unwanted or harmful sound is noise. Long exposure to loud sound can affect health and sleep and cause hearing loss. Hearing aids contain a microphone, an amplifier and a speaker.
Tone, note, timbre, octave. A tone has a single frequency (tuning fork, whistling). A musical note combines a lowest frequency, the fundamental, with higher frequencies called overtones. Timbre is why a flute, ektara and tabla sound different on the same note at the same loudness; it depends on the shape, material and construction of the instrument. An octave is the interval between two notes where one has double the frequency of the other, such as 200 Hz and 400 Hz.
8. Reflection of Sound: Echo, Reverberation, Ultrasound and Sonar
Sound waves bounce off obstacles such as solids or liquids: this is reflection of sound. It follows the same laws as light. The incident and reflected directions make equal angles with the normal at the point of incidence, and all three lie in the same plane.
Echo. Shout near a cliff or down a long corridor and you hear your voice again after a short time. This reflected sound is an echo.
The 0.1 s rule. Two sounds are heard separately only if they reach us at least 0.1 s apart. In a small room, reflections arrive too quickly, so no echo is heard.
Minimum echo distance (340 m s−1): in 0.1 s sound travels 340 × 0.1 = 34 m. That is to the wall and back, so the surface must be at least 34/2 = 17 m away.
Echoes are strongest from hard, smooth surfaces. Curtains absorb sound and rough surfaces scatter it, so echoes from them are unclear.
Echo formula: the sound covers 2d in time t, so d = (v × t)/2.
Reverberation. In a large hall, multiple reflections from the walls make sound persist after the source stops. This occurs when reflections arrive with a time difference of less than 0.05 s. Auditoriums are designed to have desirable reverberation so everyone hears clearly. Sound-absorbing panels, upholstered chairs, curtains and other soft, porous surfaces reduce unwanted reverberation, which would otherwise make the sound garbled.
| Point | Echo | Reverberation |
|---|---|---|
| Reflections | A distinct reflection heard as a repeat | Multiple reflections in a large hall |
| Time gap | At least 0.1 s after the original | Less than 0.05 s apart |
| What you hear | The sound again, separately | The sound lingering after the source stops |
Uses of infrasonic and ultrasonic waves.
| Wave | Applications given in the chapter |
|---|---|
| Infrasonic (below 20 Hz) | Detecting earthquakes, volcanic eruptions and severe storms, since these waves travel long distances through air and the Earth |
| Ultrasonic (above 20 kHz) | Imaging internal organs without surgery (ultrasonography); breaking kidney stones; ultrasonic welding and cleaning delicate machine parts; detecting defects inside metal blocks; locating objects from reflected waves |
Echolocation. Most bats send out short bursts of ultrasonic waves that reflect from nearby objects. By sensing the echoes, the bat finds obstacles and prey in the dark. The ability to locate objects using reflected sound waves is echolocation. Dolphins, whales and some birds also use it.
Sonar (sound navigation and ranging). Ships send ultrasonic waves into water and analyse the reflected waves to find the distance, direction and speed of underwater objects such as submarines or shipwrecks. Depth = (speed in water × return time)/2. A car’s parking sensor works the same way in air, using ultrasound of about 40 kHz.
Formula and Reaction Sheet
This is a physics chapter, so the sheet lists formulas and fixed values. There are no chemical reactions.
| Quantity or rule | Formula or value | Notes |
|---|---|---|
| Frequency from a count | ν = number of oscillations / time taken | Hz (s−1) |
| Frequency and time period | ν = 1/T, T = 1/ν | Inversely related |
| Wave equation | v = λ/T = λ × ν | λ = v/ν, ν = v/λ |
| One-way travel | d = v × t | Thunder, a knock on a fence |
| Echo, sonar, echolocation | d = (v × t)/2 | t = time to go and come back |
| Hearing two sounds separately | Gap of at least 0.1 s | Minimum echo distance 17 m in air (340 m s−1) |
| Reverberation | Reflections less than 0.05 s apart | Sound persists |
| Audible range | 20 Hz to 20 kHz | 1 kHz = 1000 Hz |
| Speeds at 15 °C | Steel 5000, water 1500, air 340 | m s−1 |
| Speed in dry air | About 331 at 0 °C, nearly 344 at 22 °C | Rises with temperature and humidity |
| Wavelength from a graph | Crest to next crest = λ; crest to next trough = λ/2 | Read along the distance axis |
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Important Definitions
- Vibration: the periodic to and fro motion (oscillation) of an object.
- Medium: the material through which sound propagates.
- Vacuum: a space where there is no medium (matter).
- Compression (C): a region of the medium with a higher density than the average density.
- Rarefaction (R): a region of the medium with a lower density than the average density.
- Sound wave: the disturbance consisting of a series of alternating compressions and rarefactions propagating through a medium, without the actual flow of the particles of the medium.
- Longitudinal wave: a wave in which the particles vibrate in a direction parallel to the direction of wave propagation.
- Transverse wave: a wave in which the particles vibrate perpendicular to the direction of wave propagation.
- Mechanical wave: a wave that requires a material medium for propagation.
- Wavelength (λ): the distance between two consecutive crests or two consecutive troughs.
- Frequency (ν): the number of density oscillations at a fixed point per unit time.
- Time period (T): the time taken for one complete density oscillation at a fixed point.
- Amplitude: the maximum change in air density in a compression (or a rarefaction) compared to the average density.
- Intensity of sound: the amount of sound energy passing through a unit area perpendicular to the direction of propagation of the sound wave in a unit time.
- Speed of sound: the distance which a point on a wave, such as a crest (or a trough), travels in unit time.
- Pitch: how frequency is perceived by humans. Loudness: how humans perceive amplitude.
- Infrasonic and ultrasonic waves: sound waves with frequency below 20 Hz and above 20 kHz respectively.
- Echo: sound heard again after reflecting from the hard surface of a distant object.
- Reverberation: the persistence of sound after the source stops, caused by multiple reflections.
- Echolocation: the ability to locate objects using reflected sound waves.
Solved Examples (NCERT-Based)
Example 1: Frequency and time period from a count (NCERT Example 10.1)
Question: If there are 10 density oscillations in 2 seconds at a given position, then calculate the (i) frequency of sound wave, and (ii) its time period.
Solution: (i) Frequency = 10 / 2 s = 5 Hz. (ii) Time for one oscillation = 2 s / 10 = 0.2 s. Check: 1/0.2 = 5 Hz.
Example 2: Wavelengths at the edges of human hearing (NCERT Example 10.2)
Question: Human hearing roughly spans 20 Hz to 20 kHz. What are the corresponding wavelengths in air for these two frequencies? Use the speed of sound in air as 344 m s−1.
Solution: λ = v / ν.
- (i) ν = 20 Hz: λ = 344 / 20 = 17.2 m.
- (ii) ν = 20000 Hz: λ = 344 / 20000 = 0.0172 m = 1.72 cm.
Example 3: Distance of a lightning strike (NCERT Example 10.3)
Question: During a thunderstorm, lightning is seen before thunder is heard because sound travels much slower than light. If the time delay between seeing the lightning flash and hearing the thunder is measured to be 5 s, estimate the distance to the lightning strike. Use the speed of sound in air as 340 m s−1. Assume that light (speed = 300000 km s−1) reaches you almost instantaneously.
Solution: One-way travel, so d = v × t = 340 × 5 = 1700 m. Lightning struck about 1.7 km away.
Example 4: Frequency and time period in steel (adapted from NCERT Example 10.4)
Question: A sound wave in steel has a wavelength of 50 m (read from the textbook graph). Find its frequency and time period if the speed of sound in steel is 5000 m s−1.
Solution: ν = v / λ = 5000 / 50 = 100 Hz. T = 1/ν = 1/100 = 0.01 s.
Example 5: Distance from a wall by echo (NCERT Example 10.5)
Question: You clap in an empty corridor and hear an echo after 0.5 s. If the speed of sound in air is 340 m s−1, calculate your distance from the wall.
Solution: Sound travels to the wall and back. d = (v × t)/2 = (340 × 0.5)/2 = 85 m.
Example 6: Naval sonar (NCERT Example 10.6)
Question: A naval sonar signal sent into seawater returns after 0.90 s. The speed of sound in seawater is 1530 m s−1. How far is the object?
Solution: Time to reach the object = 0.90/2 = 0.45 s. Distance = 1530 × 0.45 = 688.5 m.
Example 7: Two sounds along a steel fence (NCERT Pause and Ponder 11)
Question: Two friends are standing along a steel fence at a distance of 340 m from each other. Gunjan places her ear over the fence and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air given in Table 10.1, calculate the time difference between the sound that reached Gunjan through the air and the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least 0.1 s to be heard separately.)
Solution:
- Through air: 340 / 340 = 1 s.
- Through steel: 340 / 5000 = 0.068 s.
- Difference = 1 − 0.068 = 0.932 s.
0.932 s is more than 0.1 s, so yes, she hears two separate knocks: first through the steel, then through the air.
Example 8: Minimum distance for a 0.2 s echo (NCERT Pause and Ponder 12)
Question: An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 m s−1.
Solution: Path in 0.2 s = 343 × 0.2 = 68.6 m, there and back. Minimum distance = 68.6/2 = 34.3 m.
Example 9: Parking sensor (NCERT Exercise 11)
Question: A vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be 345 m s−1.
Solution: Total path = 2 × 1.2 = 2.4 m. t = 2.4 / 345 ≈ 0.00696 s ≈ 0.007 s (about 7 milliseconds). The 40 kHz value only tells you the wave is ultrasonic; it is not needed.
Example 10: Thunder on a cold day (NCERT Exercise 12)
Question: The speed of sound in air is about 331 m s−1 at 0 °C and nearly 344 m s−1 at 22 °C. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 °C to 0 °C? Assume that all other conditions remain unchanged.
Solution: At 22 °C: 1720 / 344 = 5.00 s. At 0 °C: 1720 / 331 ≈ 5.20 s. Extra time ≈ 0.2 s, because sound is slower in colder air.
Competency-Based Questions (with answers)
1. Case-based: Repairs on a spacewalk
Two astronauts are repairing the arm of a space station during a spacewalk. One drops a metal spanner, which clangs against the frame in front of both of them.
(a) Do they hear the clang directly? No. Space is a near vacuum and sound needs a material medium.
(b) Can they see the spanner hit the frame? Yes. Light travels through vacuum.
(c) How do they talk to each other? Through special devices fitted into their spacesuits.
(d) Which experiment shows the same idea? The vacuum bell jar: the ringing fades as air is pumped out while the bell is still seen ringing.
2. Assertion-Reason (NCERT Pause and Ponder 3)
Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out.
Reason (R): Sound requires a medium to travel.
(i) Both A and R are true, but R is not the correct explanation of A. (ii) Both A and R are true, and R is the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.
Answer: (ii). With most of the air removed there is almost no medium to carry the sound, so R explains A.
3. Assertion-Reason (NCERT Pause and Ponder 4)
Assertion (A): Compressions and rarefactions move through the medium.
Reason (R): Individual particles of the medium continuously move forward with the wave.
Answer: (iii) A is true, but R is false. The compressions and rarefactions travel; the particles only oscillate about their mean positions, like the marked turn of the slinky.
4. Assertion-Reason
Assertion (A): An echo is not heard when you clap in a small room.
Reason (R): The reflected sound from the walls reaches you less than 0.1 s after the original sound.
Answer: Both A and R are true, and R is the correct explanation of A. The walls are much closer than 17 m, so the reflection arrives too quickly to be heard separately.
5. Competency MCQs (NCERT Exercises 1 to 3)
(a) Which observation best supports the idea that sound is a mechanical wave? (i) Sound shows reflection (ii) Sound needs a medium to propagate (iii) Sound has frequency (iv) Sound carries energy.
Answer: (ii). A mechanical wave is one that needs a material medium. Light also shows reflection and carries energy, and light travels through vacuum, so only (ii) singles out a mechanical wave.
(b) For a sound wave propagating in a medium, increasing its frequency will increase its (i) wavelength (ii) speed (iii) number of compressions per second (iv) time period.
Answer: (iii). Speed stays the same; wavelength and time period both fall.
(c) If 20 compressions pass a point in 4 seconds, the frequency is (i) 80 Hz (ii) 5 Hz (iii) 10 Hz (iv) 0.2 Hz.
Answer: (ii). ν = 20/4 = 5 Hz.
6. Source-based: Speed of sound in different media (NCERT Table 10.1)
Approximate speed of sound at 15 °C: steel 5000 m s−1, water 1500 m s−1, air 340 m s−1.
(a) Speed in water relative to air? 1500/340 ≈ 4.4, which fits “about 4 to 5 times faster”.
(b) Speed in steel relative to water? 5000/1500 ≈ 3.3.
(c) A 500 Hz sound passes from air into water. What happens to its wavelength? In air λ = 340/500 = 0.68 m; in water λ = 1500/500 = 3 m. The wavelength increases because the speed increases while the frequency stays that of the source.
7. Case-based: Designing a school auditorium
A new school hall has bare concrete walls, a high hard ceiling and hard chairs. Students at the back say the principal’s voice sounds garbled and seems to linger after each sentence.
(a) Name the effect. Reverberation: multiple reflections make the sound persist after the speaker stops.
(b) Suggest three changes. Sound-absorbing wall panels, upholstered chairs, and curtains or other soft, porous materials.
(c) Should all reflections be removed? No. Halls are designed to have a desirable amount of reverberation so that the whole audience hears clearly.
8. Data-based: Same cliff, air and water
Two identical sources, A in air and B under water, are at the same horizontal distance from the vertical face of a cliff. Both send a sound towards the cliff at the same instant. The echo returns to A after 4.5 times the time it takes to return to B.
(a) Ratio of the speed of sound in air to that in water? Equal distances, so speed is inversely proportional to time: vair/vwater = 1/4.5 = 2 : 9.
(b) If B’s echo takes 0.2 s and sound in water travels at 1500 m s−1, how far is the cliff? d = (1500 × 0.2)/2 = 150 m.
Important Questions for Exams
1-Mark Questions
- Name the part of the human body that vibrates to produce the voice. The vocal cords, inside the larynx.
- What is the SI unit of frequency? Hertz (Hz), the same as s−1.
- What is a compression? A region of the medium with a higher density than the average density.
- State the audible range for humans. 20 Hz to 20,000 Hz (20 kHz).
2-Mark Questions
- If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute? (NCERT Pause and Ponder 8)
20 oscillations per second × 60 s = 1200 oscillations. - A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s−1 find its time period. (NCERT Exercise 9)
ν = 344/3.44 = 100 Hz, so T = 1/100 = 0.01 s. - Distinguish between infrasonic and ultrasonic waves, naming an animal that detects each. Infrasonic: below 20 Hz, detected by elephants. Ultrasonic: above 20 kHz, detected by bats (also dogs, cats, dolphins).
- Why are echoes unclear from curtains or rough walls? Curtains absorb sound, so little is reflected; rough surfaces scatter it in many directions. Hard, smooth surfaces give strong echoes.
3-Mark Questions
- In 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? (NCERT Exercise 8)
(1) Space is a near vacuum and sound needs a medium, so no sound should be heard. (2) Light travels through vacuum, so showing the flash is correct. (3) Even in air, light travels far faster than sound, so the flash would reach an observer first and the sound later. - A sonar signal sent to find the depth of ocean takes 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500 m s−1? (NCERT Pause and Ponder 13)
One-way time = 2 s. Depth = 1500 × 2 = 3000 m. - A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 m s−1 in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located? (NCERT Exercise 10)
One-way time = 2.5 s. Depth = 1525 × 2.5 = 3812.5 m, about 3.8 km. - Show that the minimum distance of a reflecting surface for an echo in air is 17 m (speed 340 m s−1).
Two sounds must be at least 0.1 s apart. In 0.1 s sound travels 34 m, which covers the trip there and back, so the surface must be at least 17 m away.
5-Mark Questions
- Using a piston in a tube of air, explain how sound travels through a medium. Define compression, rarefaction and sound wave, and explain why sound is a longitudinal mechanical wave.
At rest, the air in the tube has a uniform average density. When the piston moves forward it squeezes the nearby air, making a region of higher than average density, a compression. These particles collide with those ahead, so the compression moves forward while each particle stays near its own place. When the piston moves back, the nearby air thins out into a rarefaction, a region of lower than average density, which also moves forward. A continuously oscillating piston sends out alternate compressions and rarefactions. A sound wave is this disturbance propagating through a medium without the actual flow of particles. The particles vibrate parallel to the direction of propagation, so sound is longitudinal. It cannot travel without particles, as the bell jar shows, so it is mechanical. - (a) Define wavelength, frequency and time period. (b) Derive v = λ × ν. (c) A 500 Hz sound travels in air at 340 m s−1. Find its wavelength and time period.
(a) Wavelength: distance between two consecutive crests or troughs. Frequency: number of density oscillations at a fixed point per unit time. Time period: time for one complete density oscillation at a fixed point.
(b) A crest moves one wavelength λ in one time period T, so v = λ/T. Since ν = 1/T, v = λ × ν.
(c) λ = 340/500 = 0.68 m. T = 1/500 = 0.002 s. - (a) What is echolocation and how do bats use it? (b) How does sonar work? (c) A sonar pulse returns from a submarine after 1.2 s; sound in seawater travels at 1500 m s−1. How far away is the submarine?
(a) Echolocation is the ability to locate objects using reflected sound waves. Bats send short bursts of ultrasonic waves and sense the echoes from obstacles and prey to find their positions in the dark.
(b) A ship sends ultrasonic waves into the water and analyses the reflected waves to find the distance, direction and speed of underwater objects.
(c) d = (1500 × 1.2)/2 = 900 m.
Common Mistakes and Examiner Tips
- Mistake: writing that air particles travel from the source to the ear. Fix: particles only vibrate about their mean positions; the disturbance and the energy travel.
- Mistake: forgetting to halve in echo and sonar numericals. Fix: use d = (v × t)/2 for reflected sound. For thunder or a direct knock, use d = v × t.
- Mistake: calling sound a transverse wave. Fix: particles vibrate parallel to propagation, so sound is longitudinal.
- Mistake: saying a higher frequency makes sound travel faster. Fix: in air the speed depends on the medium; a higher frequency gives a shorter wavelength.
- Mistake: measuring wavelength from a crest to the next trough. Fix: that is half a wavelength; use crest to crest or trough to trough.
- Mistake: mixing up kHz and Hz. Fix: convert first: 20 kHz = 20,000 Hz.
- Mistake: linking pitch to amplitude or loudness to frequency. Fix: pitch goes with frequency; loudness goes with amplitude.
- Mistake: writing that sound is fastest in air. Fix: solids > liquids > gases (steel 5000, water 1500, air 340 m s−1).
- Mistake: saying sound slows down in hotter air. Fix: it speeds up with temperature and humidity.
- Mistake: confusing echo and reverberation. Fix: an echo is a distinct repeat at least 0.1 s later; reverberation is lingering sound from multiple reflections less than 0.05 s apart.
- Tip: different questions use 340, 343, 344 or 345 m s−1 for air. Use the value given, and write units in every line.
Quick Revision Points
- Sound is produced by vibrating objects; a vibration is a periodic to and fro motion.
- Human voice: vibrating vocal cords in the larynx.
- Sound travels through solids, liquids and gases, the medium; it cannot travel through vacuum.
- The bell jar experiment shows sound needs a medium; astronauts use devices in their suits.
- Compression: density above average. Rarefaction: density below average.
- A sound wave is a series of alternating compressions and rarefactions; particles do not travel with it.
- Sound is longitudinal and mechanical. Light is transverse and travels through vacuum.
- Sound carries energy; microphones turn sound into electrical energy, speakers do the reverse.
- On a density-distance graph: compression = crest, rarefaction = trough.
- ν = 1/T and v = λ × ν.
- Larger amplitude: more energy, louder sound. Intensity falls with distance.
- Speed: solids > liquids > gases; rises with temperature and humidity in air.
- Pitch depends on frequency, loudness on amplitude, timbre on overtones.
- Audible range 20 Hz to 20 kHz; below is infrasonic, above is ultrasonic.
- Conversation is about 60 dB; noise is unwanted or harmful sound.
- An octave doubles the frequency (200 Hz to 400 Hz).
- Sound obeys the laws of reflection of light.
- Echo needs a gap of at least 0.1 s; minimum distance 17 m in air.
- Echo or sonar distance: d = (v × t)/2.
- Reverberation: multiple reflections less than 0.05 s apart; reduced by soft, porous materials.
- Bats use echolocation; ships use sonar; doctors use ultrasonography.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams (periodic tests, half-yearly and annual papers), and the NCERT textbook gives no mark allocation for it. It mixes explanation with short numericals.
| Question type | What this chapter typically supplies |
|---|---|
| MCQ and one-word | SI units, audible range, pitch and loudness, frequency from a count |
| Assertion-Reason | Sound in vacuum, particles versus disturbance, echo in a small room |
| Short answer | Wave definitions, ν = 1/T and v = λν numericals, echo distance, uses of ultrasound |
| Long answer | Piston model, derivation of v = λν, echolocation and sonar with a numerical |
| Graph and case-based | Labelling C and R, reading wavelength and amplitude, thunder, sonar, auditorium design |
The chapter is the foundation for Class 10 and later physics, where wave quantities such as wavelength, frequency and v = λν are used again.
Class 9 Science Β· Chapter 10 β swipe through all 10 cards to understand the whole chapter.
Sound Comes from Vibrations
Every sound is made by an object vibrating: moving to and fro periodically.
The vibrating object is called the source of the sound.
- Strings, membranes, air columns and metal objects can all vibrate.
- Human voice: vocal cords vibrate inside the larynx.
- A tuning fork’s vibrating prong makes ripples on water.
No Sound in Vacuum
Sound travels through solids, liquids and gases, but never through vacuum.
Light crosses vacuum, so the bell in the jar is still seen ringing.
- Bell jar: pump out the air and the ringing fades away.
- Let the air back in and the sound returns.
- Astronauts on a spacewalk talk through devices in their suits.
Compressions and Rarefactions
A sound wave is a series of high-density and low-density regions moving through a medium.
Particles only vibrate about their mean positions; energy moves on.
- Sound is longitudinal: particles vibrate parallel to propagation.
- Sound is mechanical: it needs a material medium.
- Light is a transverse wave that travels through vacuum.
Wavelength, Frequency, Time Period
On a density-distance graph a compression is a crest and a rarefaction is a trough.
Wavelength Ξ» is crest to next crest; crest to trough is only Ξ»/2.
- Frequency: density oscillations per second, unit hertz (Hz).
- Time period: time for one complete oscillation, unit second.
- 10 oscillations in 2 s gives Ξ½ = 5 Hz and T = 0.2 s.
Amplitude and Intensity
Amplitude is the maximum change in density from the average density.
Intensity is sound energy per unit area per unit time; it falls with distance.
- Strike the plate harder and the grains on the sheet jump higher.
- Microphone: sound energy to electrical energy.
- Speaker: electrical signal to vibrating cone to sound.
Speed of Sound
Speed is the distance a crest travels in unit time.
In air, a change of frequency changes Ξ» while the speed stays the same.
- At 15 Β°C: steel 5000, water 1500, air 340 m/s.
- Air: about 331 m/s at 0 Β°C, nearly 344 m/s at 22 Β°C.
- 20 Hz in air at 344 m/s gives Ξ» = 17.2 m.
Pitch, Loudness, Audible Range
Pitch is perceived frequency; loudness is perceived amplitude.
Below 20 Hz is infrasonic; above 20 kHz is ultrasonic.
- Elephants detect infrasound; bats, dogs and dolphins detect ultrasound.
- Conversation is about 60 dB; firecrackers can exceed 100 dB.
- Timbre lets you tell a flute from a tabla on the same note.
Echo
An echo is sound heard again after reflecting from a distant hard surface.
Two sounds need a gap of at least 0.1 s to be heard separately.
- In air at 340 m/s the surface must be at least 17 m away.
- Clap, echo after 0.5 s: wall is 340 Γ 0.5 / 2 = 85 m away.
- Hard, smooth surfaces give strong echoes; curtains absorb sound.
Reverberation
Multiple reflections in a large hall make sound persist after the source stops.
Halls are designed for a desirable, controlled amount of reverberation.
- Too much reverberation makes speech sound garbled.
- Absorbing panels, upholstered chairs and curtains reduce it.
- Gol Gumbaz, Bijapur: a whisper is heard several times across the dome.
Echolocation and Sonar
Reflected ultrasound is used to locate objects that cannot be seen.
Sonar stands for sound navigation and ranging.
- Bats send ultrasonic bursts and sense echoes from prey.
- Sonar echo after 0.90 s at 1530 m/s: object is 688.5 m away.
- Ultrasound also images organs and breaks kidney stones.
π Practice Sound Waves: Characteristics and Applications - 10 questions
Concept-check questions Β· with answers & explanations
Start βClose β
Why D: Sound is a mechanical wave and cannot propagate without a material medium. Outer space is a near vacuum, so no sound from the explosion can travel to the viewer at all. The flash is fine, because light is not a mechanical wave and travels through vacuum.
Why not A: Light is far faster than sound, so sound can never arrive first even where there is a medium.
Why not B: Air is the carrier of sound, not an obstacle; with no air there is no sound, not a louder one.
Why not C: A delay would be the right correction on Earth, where sound is much slower than light; in space the sound never arrives, so a delay is not the issue.
Remember: In space you can see the flash but never hear the bang.
Why A: Not all waves need a medium. Light is a transverse wave that can travel through vacuum, so it is not a mechanical wave. Sound is a mechanical wave and needs a material medium; with no medium in the near vacuum of space, sound from the Sun cannot propagate at all.
Why not B: Speed is not the issue; without a medium the sound never starts its journey, however long we wait.
Why not C: Frequency decides whether we can hear a sound, not whether it can travel; no sound of any frequency crosses a vacuum.
Why not D: The labels are reversed: sound is longitudinal and light is transverse, and in any case the type of vibration is not what decides whether a medium is needed.
Remember: Light crosses vacuum, sound cannot: only mechanical waves need a medium.
Why D: Intensity is the sound energy passing through unit area, perpendicular to the direction of propagation, per unit time. As the wave spreads out from a small source as spherical waves, the energy released each second must be conserved, but it now passes through a larger area, so each unit of area receives less. Hence intensity falls with distance.
Why not A: With constant speed and frequency, v = Ξ» x Ξ½ fixes the wavelength; it does not change with distance.
Why not B: In a given medium the speed of sound is constant; it does not fall off with distance.
Why not C: Frequency is set by the source and stays the same as the wave travels; a distant siren has the same pitch, only softer.
Remember: Same energy, bigger area, smaller intensity.
Why A: Step 1 (time at 22 Β°C): t = 1720 / 344 = 5.0 s. Step 2 (time at 0 Β°C): t = 1720 / 331 = 5.2 s. Step 3 (extra time): 5.2 – 5.0 = 0.2 s. Colder air carries sound more slowly, so the thunder takes about 0.2 s longer.
Why not B: 5.0 s is the total travel time at 22 Β°C, not the extra time caused by the cooling.
Why not C: 0.02 s is a decimal slip in the subtraction; the real difference is ten times larger.
Why not D: 13 is the difference in speeds (344 – 331) in m/s, which is not a time.
Remember: Speed of sound in air rises with temperature and humidity; colder air means a longer travel time.
Why A: Both statements are true. Longer, thicker vocal cords vibrate less frequently, which lowers the frequency of the sound produced. Since pitch is the perception of frequency, a lower frequency is heard as a deeper voice. R is exactly the physical cause of A.
Why not B: R is not an unrelated fact; the lower vibration frequency is what makes the voice deeper.
Why not C: R is true; the chapter states that the vocal cords lengthen and thicken and vibrate less frequently.
Why not D: A is true; the deepening of boys’ voices in adolescence is the everyday observation being explained.
Remember: Slower vibration = lower frequency = lower pitch = deeper voice.
Why D: Step 1 (same path): both sounds travel the same round-trip distance 2d to the cliff and back. Step 2 (speed = distance / time): vβα΅’α΅£ = 2d / tβα΅’α΅£ and v_water = 2d / t_water, so vβα΅’α΅£ / v_water = t_water / tβα΅’α΅£. Step 3 (substitute): tβα΅’α΅£ = 4.5 t_water, so the ratio is 1 / 4.5, that is 1 : 4.5 (about 2 : 9). Sound is indeed about 4 to 5 times faster in water than in air.
Why not A: 9 : 1 both inverts the ratio and doubles it for the round trip; both sounds make the same round trip, so the factor of 2 cancels, and the slower sound (air) must have the smaller speed.
Why not B: 1 : 2.25 halves the ratio for the round trip; again the factor cancels because both paths are equal.
Why not C: 4.5 : 1 inverts the ratio; the sound that takes longer is the slower one, and that is the sound in air.
Remember: Same distance: the ratio of speeds is the inverse of the ratio of times.
Why A: Both statements are true. Sound propagates by particles of the medium colliding with their neighbours; with almost no particles in outer space there is nothing to carry the disturbance, so no sound reaches the other astronaut. R is exactly the reason for A, which is why astronauts communicate through special devices fitted into their spacesuits.
Why not B: R is not a separate fact; it directly explains why the astronauts hear nothing.
Why not C: R is a true statement, verified by the vacuum bell jar experiment.
Why not D: A is true; there is no medium between the astronauts to carry the sound.
Remember: No medium, no sound: space is silent.
Why C: A is true: the series of compressions and rarefactions is the disturbance that travels from the source through the medium. R is false: the particles only oscillate about their mean positions and do not travel with the wave. In the slinky analogy the marked turn stays near its rest position while the bunched-up region moves along.
Why not A: R is false, so it cannot be the correct explanation of anything.
Why not B: R is false, not merely unrelated.
Why not D: A is true; the density disturbance does propagate, which is what sound is.
Remember: The disturbance travels, the particles stay home.
Why C: Vibrating faster means more oscillations per second, so the frequency is higher. Since Ξ½ = 1/T, a higher frequency means each oscillation takes less time, so the time period is shorter. A shorter time period always corresponds to a higher frequency.
Why not A: Amplitude is about the size of the density change (energy), not about how fast the band vibrates; the frequency clearly changes here.
Why not B: Frequency and time period are inverses; they cannot both increase.
Why not D: Vibrating faster raises the frequency, it does not lower it.
Remember: Faster vibration = higher Ξ½ = shorter T.
Why D: Step 1 (time = distance / speed, air): tβα΅’α΅£ = 340 / 340 = 1.0 s. Step 2 (time = distance / speed, steel): tβββββ = 340 / 5000 = 0.068 s. Step 3 (difference): 1.0 – 0.068 = 0.932 s, about 0.93 s. Since this is more than 0.1 s, she hears two separate knocks, the one through steel first.
Why not A: 1.0 s is the time through air alone; the steel time must be subtracted.
Why not B: 1.07 s adds the two times instead of subtracting them; the difference is what separates the arrivals.
Why not C: 0.07 s is the time through steel alone, not the difference between the two arrivals.
Remember: Same distance, two speeds: find each time, then subtract.
Chapter Navigation
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Frequently Asked Questions
Sound is a mechanical wave. It travels as compressions and rarefactions, which are regions where particles of a medium are squeezed together or spread apart. In a vacuum there are no particles to squeeze or spread, so no wave can form. The vacuum bell jar experiment shows this: as air is pumped out, the ringing fades to almost nothing while the bell is still seen ringing.
In a sound wave the particles of the medium vibrate back and forth parallel to the direction in which the wave travels. Waves in which particles vibrate parallel to the direction of propagation are called longitudinal waves. In a transverse wave, such as light, the vibration is perpendicular to the direction of propagation.
A crest moves one wavelength Ξ» in one time period T, so v = Ξ»/T. Since frequency Ξ½ = 1/T, the speed of sound is v = Ξ» Γ Ξ½. For example, a 500 Hz sound in air at 340 m/s has a wavelength of 340/500 = 0.68 m. In a given medium the speed stays the same, so a higher frequency means a shorter wavelength.
Two sounds are heard separately only if they reach us at least 0.1 s apart. With sound at 340 m/s, the distance covered in 0.1 s is 34 m. That is the path to the reflecting surface and back, so the surface must be at least 17 m away. In a small room reflections return too quickly, so no echo is heard.
An echo is a distinct repeat of a sound reflected from a distant hard surface, heard at least 0.1 s after the original. Reverberation is the persistence of sound after the source stops, caused by multiple reflections in a large hall arriving less than 0.05 s apart. Sound-absorbing panels, upholstered chairs and curtains reduce unwanted reverberation.
Humans hear sounds from 20 Hz to 20 kHz. Sound waves below 20 Hz are infrasonic; elephants can detect them, and they are used to detect earthquakes, volcanic eruptions and severe storms. Sound waves above 20 kHz are ultrasonic; bats, dogs, cats and dolphins can detect them, and they are used in ultrasonography, breaking kidney stones, cleaning machine parts and sonar.
A ship sends an ultrasonic pulse down into the water and measures the time t for the echo to return from the sea bed. The pulse travels down and back, so depth = (speed of sound in water Γ t)/2. For example, if the echo returns after 4 s and sound travels at 1500 m/s in seawater, the depth is 1500 Γ 4/2 = 3000 m.
Pitch is how we perceive the frequency of a sound: a whistle or siren has high frequency and high pitch, while thunder has low pitch. Loudness is how we perceive the amplitude: a larger amplitude sounds louder. Loudness also falls as you move away from the source. Intensity is the measurable quantity, while loudness depends on the listener’s hearing.
