- Sound is a mechanical longitudinal wave — it needs a material medium and travels as compressions and rarefactions.
- Key relation: v = fλ — speed = frequency × wavelength.
- Speed of sound in air at 25 °C ≈ 343 m/s (NCERT uses 344 m/s for echo problems).
- Speed: solids > liquids > gases (steel ~5100 m/s, water ~1500 m/s, air ~343 m/s).
- Echo: minimum distance of reflecting surface from source = 17.2 m (at 344 m/s, persistence of hearing 0.1 s).
- Human hearing range: 20 Hz to 20,000 Hz. Below = infrasound; above = ultrasound.
- Ultrasound applications: medical imaging, cleaning delicate parts, echocardiography, flaw detection.
- SONAR measures depth using time of echo: d = v × t / 2.
- Board weightage: ~5 marks/year — typically one numerical (echo/SONAR) and one short answer (ultrasound uses or reverberation).
1. Production and Propagation of Sound
Sound is produced by vibrating objects. When a body vibrates, it disturbs the surrounding medium and creates a wave that travels outward. Examples:
- Plucking a stretched rubber band — the band vibrates and you can both see and hear the vibration.
- Striking a school bell — feel the bell surface; it vibrates when ringing and stops when you press your palm on it.
- Tuning fork — two prongs vibrate when struck; hold it near your ear or dip it in water to see ripples forming on the water surface.
- Human voice — vocal cords in the larynx vibrate when air passes over them.
- Drum — the stretched membrane (diaphragm) vibrates when hit.
Key point: Once the vibration stops, sound also stops. The vibrating source must have a material medium (solid, liquid, or gas) to transfer energy to. Sound cannot travel through vacuum.
NCERT Demonstration — Bell in a vacuum jar: Place an electric bell inside a glass jar connected to a vacuum pump. As air is pumped out, the sound becomes fainter and eventually disappears, even though the hammer is still striking the bell. This proves conclusively that sound needs a material medium to travel. Once air is let back in, the sound returns.
The medium particles themselves do not travel — they oscillate about their mean positions and pass the disturbance to the next particle. The energy of the vibration travels, not the medium itself. This is why sound is a wave, not a bulk flow of air.
Strike a tuning fork on a rubber pad and bring it close to your ear. You hear a continuous hum. Now touch the handle lightly to a table — the table begins to vibrate and the sound becomes louder (the table acts as a sounding board). Dip the vibrating prong in a beaker of water — the water splashes, proving the prong is vibrating rapidly.
2. Sound as a Longitudinal Wave — Compressions and Rarefactions
When a vibrating object (e.g., a tuning fork prong) moves forward, it pushes the air particles ahead, creating a region of high pressure and high density called a compression (C). When the prong moves back, it creates a region of low pressure and low density called a rarefaction (R).
These compressions and rarefactions travel outward one after another, forming a longitudinal wave. In a longitudinal wave, the particles of the medium vibrate parallel to (along) the direction of wave propagation — unlike transverse waves (e.g., water ripples, waves on a string) where particles vibrate perpendicular to the direction of travel.
The pattern of the wave: ... C R C R C R C R ... spreads outward from the source at the speed of sound.
Sound is always a longitudinal wave — in air, water, or solids. It cannot be polarised (polarisation is only possible for transverse waves).
Slinky analogy: Push one end of a slinky coil forward and back. Bunching of coils = compression; spreading out of coils = rarefaction. The coils move to-and-fro along the direction of the push — that is longitudinal motion. The disturbance travels forward even though individual coils return to their original positions.
Place a long stretched slinky coil on a smooth floor. Ask a friend to hold one end. Give the other end a sharp push and pull. You will see compressions (regions where coils crowd together) and rarefactions (regions where coils spread apart) travelling along the slinky. This is exactly how a sound wave travels through air.
3. Wave Parameters — Frequency, Time Period, Wavelength, Amplitude
A sound wave is completely described by four parameters:
Frequency (f) — number of complete oscillations (vibrations) per second. Unit: hertz (Hz) = 1 cycle per second. A tuning fork marked 256 Hz makes 256 complete vibrations every second. Higher frequency = higher pitch (more shrill).
Time Period (T) — time taken for one complete oscillation (one full compression + rarefaction cycle). Unit: second (s).
T = 1 / f and f = 1 / T
Example: f = 256 Hz ⇒ T = 1/256 s ≈ 0.0039 s.
Wavelength (λ) — the distance between two consecutive compressions (or two consecutive rarefactions), i.e., the length of one complete wave cycle. Also = the distance the wave travels in one time period. Unit: metre (m).
Amplitude (A) — maximum displacement of a medium particle from its rest (equilibrium) position. Unit: metre (m); for sound waves, amplitude is related to the pressure change (loudness). Larger amplitude = louder sound = more energy carried. Amplitude does NOT affect pitch.
Summary — What each parameter controls:
| Parameter | Symbol | Unit | What it determines |
|---|---|---|---|
| Frequency | f | Hz | Pitch (shrillness) of sound |
| Time period | T | s | T = 1/f; inversely related to f |
| Wavelength | λ | m | Distance of one wave cycle |
| Amplitude | A | m (or Pa) | Loudness (intensity) of sound |
4. Speed of Sound — Formula v = fλ
In one complete time period T, the wave travels exactly one wavelength λ. So:
Speed = Distance / Time
v = λ / T
Since f = 1/T, we get v = f × λ
Units: v in m/s, f in Hz (= 1/s), λ in m. Consistent: (1/s) × m = m/s.
Important: The speed of sound in a given medium at a given temperature is fixed. If the source's frequency changes, the wavelength changes proportionally so that v stays the same. E.g., if f doubles, λ halves (and v stays constant).
Question: A sound wave has a frequency of 2 kHz and wavelength 35 cm. How long will it take to travel 1.5 km?
Solution:
Given: f = 2 kHz = 2000 Hz, λ = 35 cm = 0.35 m
Step 1: Speed v = f × λ = 2000 × 0.35 = 700 m/s
Step 2: Distance = 1.5 km = 1500 m
Step 3: Time = Distance / Speed = 1500 / 700 = 2.14 s
Question: A guitar string vibrates at 660 Hz. The speed of sound in air is 330 m/s. Find the wavelength of the sound produced.
Solution:
λ = v / f = 330 / 660 = 0.5 m
5. Speed of Sound in Different Media
Sound travels fastest through solids and slowest through gases because the restoring forces between particles (elasticity) are strongest in solids, allowing vibrations to pass more quickly.
Rule: Speed of sound — Solids > Liquids > Gases
Approximate speeds at 25 °C (NCERT values):
- Steel: 5120 m/s
- Aluminium: ~6420 m/s
- Sea water: 1531 m/s
- Water (distilled): 1498 m/s
- Air: 346 m/s (also 343 m/s; NCERT echo problems use 344 m/s)
- Hydrogen gas: ~1284 m/s (gases vary with molecular mass and temperature)
Effect of temperature: Speed in air increases with temperature — at 0 °C it is 331 m/s; at 25 °C it is about 346 m/s. Approximately 0.6 m/s increase per 1 °C rise.
Practical consequence: You can hear a train approaching by pressing your ear to the steel rail — sound in steel (5120 m/s) arrives far ahead of the sound through air (343 m/s). This was how railway workers detected approaching trains before the days of modern signalling.
Note: Speed of sound depends on the medium and its temperature, not on frequency or amplitude of the sound source.
6. Reflection of Sound — Laws of Reflection
Sound bounces back when it strikes a hard, rigid surface (wall, cliff, building, hillside). This is called reflection of sound. The same two laws that apply to reflection of light also apply to reflection of sound:
First Law: The angle of incidence equals the angle of reflection — both measured from the normal (perpendicular) to the surface at the point of incidence.
Angle of incidence = Angle of reflection
Second Law: The incident sound wave, the reflected sound wave, and the normal at the point of incidence all lie in the same plane.
Good reflectors of sound: Hard, smooth, large flat surfaces — concrete walls, rocky cliffs, tall buildings, metal sheets.
Poor reflectors (good absorbers): Soft, porous, rough surfaces — curtains, carpets, foam, acoustic tiles, human bodies.
Take two long cardboard tubes. Place them at an angle toward a smooth vertical wall (like a hard-bound book). Put a ticking clock at the opening of one tube (incident tube). Place your ear at the opening of the other tube (reflected tube). Adjust the angles until you hear the ticking loudest. Measure the angles with a protractor — you will find the angle of incidence = angle of reflection, just as in light reflection.
7. Echo — Minimum Distance 17.2 m
An echo is the reflected sound that is heard distinctly as a repetition of the original sound. For the brain to perceive the echo as a separate sound from the original, the reflected sound must arrive at least 0.1 second after the direct sound. This 0.1 s is the persistence of hearing (the minimum time gap the human ear needs to distinguish two separate sounds).
Derivation of minimum echo distance:
Speed of sound in air = 344 m/s (NCERT standard value for echo problems)
In 0.1 s, total distance sound travels = 344 × 0.1 = 34.4 m
This is the total path: from source to wall AND back to ear.
Minimum distance of reflector = 34.4 / 2 = 17.2 m
If the reflector is closer than 17.2 m: the reflected sound reaches the ear in less than 0.1 s and merges with the original sound, causing reverberation instead of a distinct echo.
Everyday examples:
- Shouting in a large, empty hall — a far wall reflects your voice as a distinct echo.
- Yodelling in mountains — cliffs reflect the sound repeatedly, producing multiple echoes.
- Outdoors, there are no nearby reflectors, so echoes require distant walls or cliffs.
Question: A person clapped his hands near a cliff and heard the echo after 4 s. What is the distance of the cliff from the person if the speed of sound is 346 m/s?
Solution:
Total distance sound travels = Speed × Time = 346 × 4 = 1384 m
This is the round-trip distance (to cliff and back).
Distance to cliff = 1384 / 2 = 692 m
Question: What is the minimum distance a person must stand from a wall to hear a distinct echo, if the speed of sound in air is 340 m/s?
Solution:
In 0.1 s, sound travels = 340 × 0.1 = 34 m (total, to and back)
Minimum distance = 34 / 2 = 17 m
(NCERT uses 344 m/s giving 17.2 m; always check which speed value a problem gives.)
8. Reverberation and Its Control
Reverberation is the persistence of sound caused by rapid, repeated reflections of sound from the walls, ceiling, and floor of an enclosed space. Because the reflections come back quickly (the space is small), they overlap with the original sound and with each other.
Echo vs. Reverberation:
- Echo: distinct repetition of original sound; reflecting surface far away (≥17.2 m); gap ≥ 0.1 s between sounds.
- Reverberation: prolonged, blurred sound in closed rooms; reflections arrive too quickly to be heard separately; overlapping makes speech unclear.
Problems: Too much reverberation in auditoria, lecture halls, and cinemas makes words run into each other and become unintelligible. Also a problem in recording studios where unwanted reflections distort recordings.
How to reduce/control reverberation (NCERT):
- Line walls and ceiling with sound-absorbing materials: compressed fibreboard, rough plaster, special acoustic tiles.
- Use heavy curtains and thick carpets — they absorb sound instead of reflecting it.
- Padded seats in auditoria — chosen to absorb sound even when unoccupied, so the hall doesn't sound very different when empty vs. full of an audience.
- Architectural design — curved surfaces to reflect sound uniformly toward the audience; avoid parallel walls (which cause repeated back-and-forth reflections).
- The presence of an audience itself helps — human clothing and bodies absorb a significant amount of sound.
9. Uses of Multiple Reflection of Sound
Multiple reflections of sound are deliberately put to use in many practical devices and architectural designs:
- Megaphone / Loudspeaker horn / Bullhorn: A funnel-shaped or conical tube. Sound enters the narrow end and is channelled by repeated reflections off the inner walls, so it travels mainly in one direction with little sideways spreading. Energy is concentrated forward. Used by coaches, traffic police, tour guides, and cheerleaders to project voices over a large crowd.
- Ear trumpet (hearing aid): A funnel-shaped instrument that collects sound over a large area and concentrates it into the ear canal. Historically used before electronic hearing aids existed. The principle is the same: multiple reflections from the inner walls direct sound into the ear.
- Stethoscope: Sound from the patient's heartbeat or breathing travels through the rubber (or plastic) tube by multiple reflections off the walls of the tube and reaches the doctor's ears clearly. The tube prevents the sound from spreading out and getting lost.
- Curved ceilings in concert halls and auditoria: Carefully designed curved ceilings act as acoustic reflectors. Sound from the stage strikes the curved ceiling and is reflected back toward the audience, ensuring everyone hears clearly even at the back.
- Whispering gallery: A circular or elliptical hall (e.g., St. Paul's Cathedral, London; Gol Gumbaz, Bijapur) where a whisper at one focus point is carried by repeated reflections along the curved wall and heard clearly at the other focal point, across the room. The curved wall acts as a series of reflectors.
10. Range of Hearing
The human ear can detect sound only within a specific frequency range:
Human hearing range: 20 Hz to 20,000 Hz (= 20 kHz)
- Sounds with frequency below 20 Hz: called Infrasound — inaudible to humans.
- Sounds with frequency above 20,000 Hz: called Ultrasound — inaudible to humans.
The audible range shrinks with age — elderly people typically cannot hear frequencies above ~15,000 Hz.
Infrasound — examples from nature:
- Rhinoceroses communicate using infrasound at about 5 Hz.
- Elephants produce infrasound rumbles (14–35 Hz) that can travel several kilometres through the ground.
- Whales communicate with infrasound across ocean basins.
- Earthquakes generate infrasound — many animals (dogs, elephants, birds) sense these precursor vibrations and behave strangely before tremors begin.
- The pendulum of a very large clock swings so slowly its frequency is below 20 Hz — infrasonic.
Ultrasound — examples from nature:
- Bats: Navigate and hunt in darkness using echolocation — they emit ultrasonic pulses (typically 50–200 kHz) and use the returning echo to pinpoint insects and obstacles.
- Dolphins: Use ultrasound for echolocation in water to navigate and find fish.
- Dogs: Can hear up to ~50,000 Hz — a "dog whistle" is ultrasonic for humans but clearly audible to dogs.
- Porpoises, some moths (which can detect a bat's ultrasound and take evasive action), and certain rodents also use or detect ultrasound.
11. Ultrasound — Properties and Applications
Ultrasonic waves (frequency > 20,000 Hz) have very high frequencies and correspondingly short wavelengths. Their key physical properties make them useful:
- Short wavelength ⇒ can detect very small features or defects.
- Travel in straight lines and penetrate deep into materials.
- Reflected at boundaries between media of different densities.
- Can pass through soft tissue safely (unlike X-rays which cause ionisation).
NCERT Applications of Ultrasound:
1. Industrial cleaning: Objects (jewellery, electronic components, watch mechanisms, surgical instruments) are placed in a cleaning solution. Ultrasonic vibrations (transducer emitting ultrasound into the liquid) cause rapid oscillations that shake loose dirt, grease, soot, and chemical residues — even from parts with intricate shapes that a brush cannot reach. The process is fast, thorough, and non-damaging to delicate components.
2. Detecting cracks and flaws in metal blocks (Non-Destructive Testing, NDT): Ultrasonic waves are directed into the metal casting or railway track. At a flaw (crack, air pocket, or impurity), some waves are reflected back. A detector (usually on the same side as the transmitter) picks up this early reflection, revealing the location and size of the defect — without cutting or damaging the metal. Widely used for aircraft parts, railway tracks, ship hulls, and nuclear reactor vessels.
3. Echocardiography: Ultrasonic waves are reflected from different surfaces of the heart (valves, walls). The returning echoes are used to construct a moving image of the heart's chambers and valves. This helps diagnose valve defects, blockages, and other heart conditions without surgery or radiation.
4. Ultrasonography (abdominal/obstetric ultrasound scan): A probe (transducer) placed on the skin emits ultrasonic pulses. These reflect at boundaries between different tissues (liver, kidney, foetus, etc.) and the time delay gives distance information. A computer assembles the reflected signals into a 2-D image on a screen. Used to monitor foetal growth, detect tumours, check internal organs. Completely safe — no ionising radiation.
5. Breaking kidney stones and gall bladder stones: High-intensity focused ultrasonic waves are aimed at the kidney stone. The intense vibration causes the stone to break into fine, sand-like grains that are then flushed out by urine. This technique (lithotripsy) avoids surgical incision entirely.
Bats emit rapid ultrasonic pulses as they fly. Each pulse reflects from insects, walls, or trees and returns as an echo. The bat's brain calculates the distance from the time taken for the echo to arrive, and the direction from which ear the echo is louder. This gives the bat a precise "acoustic picture" of its surroundings, allowing it to navigate in complete darkness and catch fast-moving prey. Some moths have evolved ultrasound-sensitive ears to detect approaching bats and take evasive action.
12. SONAR — Sound Navigation and Ranging
SONAR stands for Sound Navigation And Ranging. It is a device that uses ultrasonic waves to measure distances, detect submerged objects, and map the ocean floor — in water, where light cannot travel more than a few metres.
Working of SONAR:
- A transducer (a device that can both emit and receive ultrasonic waves) is mounted on the bottom of a ship or submarine.
- The transducer transmits a beam of ultrasonic waves downward (or in a chosen direction).
- The waves hit an object (ocean floor, submarine, iceberg, school of fish, shipwreck) and are reflected back as an echo.
- The same transducer (acting now as a receiver) detects the returning echo.
- The electronics measure the time interval t between transmission and reception.
SONAR depth formula:
If v = speed of sound in sea water, t = total time for echo to return:
d = (v × t) / 2
We divide by 2 because the sound travels from the ship to the object (distance d) AND back (distance d) — so total distance = 2d.
Uses of SONAR:
- Measuring the depth of the sea (ocean bathymetry / echo-sounding).
- Locating enemy submarines and icebergs (which can be 90% below the water surface — a hazard for ships, as the Titanic disaster showed).
- Finding sunken ships and wreckage.
- Detecting shoals of fish for commercial fishing.
- Mapping the contours of the ocean floor (producing detailed seafloor maps).
Question: A SONAR device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Solution:
Given: t = 5 s (total round-trip time), d = 3625 m (one way)
Total distance = 2d = 2 × 3625 = 7250 m
Speed v = Total distance / Total time = 7250 / 5 = 1450 m/s
Question: A ship sends out an ultrasound pulse. The echo from the sea bed is received 3.5 s later. If the speed of sound in sea water is 1531 m/s, find the depth of the sea bed.
Solution:
d = (v × t) / 2 = (1531 × 3.5) / 2 = 5358.5 / 2 = 2679.25 m
Question: A SONAR device transmits ultrasound of frequency 5 × 10^5 Hz. If the speed of sound in sea water is 1500 m/s, find the wavelength of this ultrasound.
Solution:
λ = v / f = 1500 / (5 × 10^5) = 3 × 10^-3 m = 3 mm
13. Quick Comparison and Common Mistakes
Infrasound vs. Audible vs. Ultrasound:
| Type | Frequency Range | Heard by humans? | Key examples |
|---|---|---|---|
| Infrasound | Below 20 Hz | No | Earthquakes, elephants, rhinos, whales |
| Audible sound | 20 Hz to 20,000 Hz | Yes | Speech, music, instruments |
| Ultrasound | Above 20,000 Hz | No | Bats, dolphins, medical imaging, SONAR |
Common mistakes to avoid:
- Calling sound a transverse wave — sound is always longitudinal in all media.
- Confusing echo (distinct, far reflector) with reverberation (overlapping, enclosed space).
- In SONAR problems: t is the total (round-trip) time; d = vt/2 (one way). Never write d = vt.
- Saying amplitude affects pitch — amplitude affects loudness; frequency affects pitch.
- Forgetting that speed of sound depends on medium and temperature, not on source frequency.
- Saying ultrasound is >2000 Hz — the correct NCERT threshold is >20,000 Hz (20 kHz).
- Using the wrong echo speed: NCERT standard is 344 m/s ⇒ min. distance = 17.2 m. Always use the value given in the problem.
- Thinking that sound speeds up as it goes from a cold room to a warm room — true! Higher temperature = faster speed in gas.
- Steel
- Water
- Vacuum
- Air
- Perpendicular to the direction of wave propagation
- Parallel to the direction of wave propagation
- In circular paths
- In random directions
- 512 s
- 1/512 s
- 0.512 s
- 5120 s
- Doubles
- Remains the same
- Halves
- Becomes four times
- Air at 0 °C
- Water
- Vacuum
- Steel
- 34.4 m
- 344 m
- 17.2 m
- 8.6 m
- Absorption of sound by the walls
- Repeated reflections of sound in an enclosed space
- A single echo from a cliff 2 km away
- Diffraction of sound around corners
- 0 Hz to 2,000 Hz
- 20 Hz to 20,000 Hz
- 200 Hz to 2,00,000 Hz
- 2 Hz to 20,000 Hz
- Infrared radiation
- Radio waves
- Ultrasonic sound waves
- Visible light
- 6.8 m
- 3.4 m
- 0.68 m
- 13.6 m
Echo: An echo is the sound heard after reflection from a distant hard surface. For it to be heard distinctly, the reflected sound must reach the ear at least 0.1 s after the original sound (persistence of hearing).
Derivation: In 0.1 s, sound travels = 344 × 0.1 = 34.4 m (total, to reflector and back). So minimum distance of reflector = 34.4 / 2 = 17.2 m.
Wave equation: v = f × λ
Speed v = 500 × 0.66 = 330 m/s
Time to travel 1 km: t = 1000 / 330 = 3.03 s
d = (v × t) / 2 = (1500 × 2) / 2 = 3000 / 2 = 1500 m
(a) Three uses of ultrasound:
1. Medical imaging (ultrasonography) — imaging of internal organs and foetus without harmful radiation.
2. Industrial cleaning — shaking dirt off delicate parts, jewellery, surgical instruments placed in cleaning solution.
3. Non-destructive testing — detecting cracks and flaws inside metal castings and railway tracks.
(b) Echo vs. Reverberation:
Echo: heard as a distinct, separate repetition; reflector is far (≥17.2 m); time gap ≥ 0.1 s.
Reverberation: heard as a prolonged, overlapping sound; occurs in enclosed spaces; reflections are too rapid to be individually distinguished.
In a hall, the walls, ceiling, and floor repeatedly reflect sound back to the listener — this repeated reflection is reverberation, which adds energy to the direct sound, making it seem louder. In open air, sound spreads out in all directions and there are no reflectors nearby, so energy disperses rapidly.
Acoustic design of auditoria: Walls and ceilings are lined with sound-absorbing materials (fibreboard, rough plaster). Heavy curtains and carpets are used. Seats are padded. The ceiling and walls are curved or angled to direct reflected sound toward the audience uniformly. These measures control reverberation without making the hall too "dead" (over-absorbed).
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