← Class 9 Science
Production Medium Waves Characteristics Speed Pitch & Loudness Reflection Echo Reverberation Uses Range Ultrasound SONAR Ear Exercises Summary
Class 9 · NCERT Science · Ch 10

🔊 Sound Waves

Characteristics & Applications (ध्वनि)

🔊
🔨 1. Production of Sound (ध्वनि का उत्पादन)

Sound is produced by vibrating objects. When an object vibrates, it moves back and forth rapidly, disturbing the particles of the medium around it. These disturbances travel as waves and reach our ears, where we perceive them as sound.

Every sound you hear — a guitar string, a drum, your voice, a bell — comes from something vibrating. Stop the vibration, and the sound stops too.

🎸 Guitar String

Plucking the string makes it vibrate. The vibration produces sound.

🗣️ Human Voice

Vocal cords in the larynx vibrate when air from the lungs passes through them.

🔔 Bell

When struck, the metal body of the bell vibrates and produces sound.

🥁 Drum

Striking the membrane makes it vibrate, creating sound waves in air.
🎨 How Sound Travels Through Air

Watch how a vibrating object pushes air particles, creating compressions (C) and rarefactions (R):

✅ NCERT Activity 10.1

Feel the Vibration in Your Throat

Aim: To show that sound is produced by vibration.

Do this: Place your hand on your throat. Now speak or hum a song. Feel your throat with your fingers.

👤

Observation: You can feel vibrations in your throat when you speak or hum.

Conclusion: The vocal cords in our throat vibrate to produce sound.

✅ NCERT Activity 10.2

Tuning Fork in Water

Aim: To show that a vibrating object can cause effects in the surrounding medium.

Do this: Strike a tuning fork on a rubber pad. Quickly touch the prongs to the surface of water in a beaker.

Observation: Water splashes when the vibrating prongs touch the surface.

Conclusion: The vibrating tuning fork transfers energy to the water, causing splashes. This proves the fork is vibrating.

✅ NCERT Activity 10.3

Tuning Fork & Suspended Ball

Aim: To show that a vibrating tuning fork can push a light object.

Do this: Suspend a table-tennis ball from a stand. Touch a vibrating tuning fork to the ball.

Observation: The ball bounces away from the vibrating tuning fork.

Conclusion: The vibrating fork pushes the ball, proving that vibration produces a force that can displace nearby objects.

✅ NCERT Activity 10.4

Vibrating Ruler

Do this: Hold a ruler firmly at the edge of a table so part of it extends beyond the edge. Press the free end down and release.

Observation: The free end vibrates up and down, producing a humming sound. Shorter free length → faster vibration → higher pitch.

❓ In-Text Questions (Page 162)

Q1
How does the sound produced by a vibrating object in a medium reach your ear?
When an object vibrates, it creates compressions (regions of high pressure) and rarefactions (regions of low pressure) in the surrounding medium. These disturbances travel outward as a longitudinal wave. When these waves enter our ear, they cause the eardrum to vibrate, and we hear sound. The particles of the medium do not travel — only the disturbance (energy) travels from the source to the ear.
🌐 2. Sound Needs a Material Medium

Sound waves are mechanical waves — they need a material medium (solid, liquid, or gas) to propagate. Unlike light, sound cannot travel through a vacuum.

The particles of the medium vibrate and pass the disturbance along to their neighbours. No particles = no sound transmission.

✅ NCERT Activity 10.5 — Bell Jar Experiment

Sound Cannot Travel Through Vacuum

Do this: Place an electric bell inside a glass bell jar connected to a vacuum pump. Switch on the bell. Now gradually pump out the air.

Observation: As air is pumped out, the sound becomes fainter and fainter. In a perfect vacuum, no sound is heard at all. When air is let back in, the sound returns.

Conclusion: Sound needs a material medium to travel. It cannot travel through a vacuum.

💡 Remember: Astronauts on the Moon cannot talk to each other directly because there is no atmosphere (no medium). They use radio waves (electromagnetic waves), which do not need a medium.

❓ In-Text Questions (Page 163)

Q1
Explain how sound is produced by your school bell.
When the school bell is struck by the hammer, the metallic body of the bell begins to vibrate. These vibrations create compressions and rarefactions in the surrounding air, which travel as sound waves and reach our ears.
Q2
Why are sound waves called mechanical waves?
Sound waves are called mechanical waves because they need a material medium (solid, liquid, or gas) for their propagation. They cannot travel through a vacuum. The particles of the medium vibrate to transfer the sound energy.
Q3
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
No. The moon has no atmosphere (no air). Since sound needs a material medium to travel, and there is no medium on the moon, we cannot hear any sound produced by our friend on the moon.
3. Sound as a Longitudinal Wave

Sound travels as a longitudinal wave. In a longitudinal wave, the particles of the medium vibrate parallel (back and forth) to the direction of wave propagation.

This creates alternating regions of:

Compression (संपीडन)

Region where particles are close together. High-pressure, high-density zone.

Rarefaction (विरलन)

Region where particles are far apart. Low-pressure, low-density zone.
✅ NCERT Activity 10.6 — Slinky Longitudinal Wave

Longitudinal Wave on a Slinky

Do this: Stretch a slinky on a table. Push one end forward and pull it back. Watch the compression travel along the slinky.

Observation: A compression (bunch of coils close together) moves along the slinky, followed by rarefactions (coils far apart). Each particle oscillates back and forth — it doesn't travel with the wave.

Conclusion: This is a longitudinal wave — the coils (particles) vibrate back and forth in the same direction as the wave travels. Sound in air behaves exactly like this.

✅ NCERT Activity 10.7 — Slinky Transverse Wave

Transverse Wave on a Slinky (for contrast)

Do this: Stretch the slinky. Move one end up and down (sideways). Watch the wave travel along.

Observation: Crests (peaks) and troughs (valleys) move along the slinky.

Conclusion: This is a transverse wave — the coils vibrate perpendicular to the direction of wave travel. Light waves are transverse; sound waves are NOT.

How Sound Travels Through Air — Particle Energy Transfer

Watch how a vibrating speaker pushes air molecules, creating compressions (C) and rarefactions (R). Particles don’t travel — only energy does!

Side-by-Side: Longitudinal vs Transverse Wave

Compare how particles move in each type. Both waves carry energy rightward, but particle motion differs!

PropertyLongitudinal WaveTransverse Wave
Particle motionParallel to wave directionPerpendicular to wave direction
FeaturesCompressions & rarefactionsCrests & troughs
ExampleSound in air, slinky push-pullLight, water waves, slinky up-down
Needs medium?Yes (mechanical)Not always (EM waves don’t)
📈 4. Characteristics of a Sound Wave

A sound wave can be described by these key properties:

λ Wavelength (λ)

Distance between two consecutive compressions or rarefactions. Unit: metre (m)

𝑓 Frequency (𝑓 or ν)

Number of vibrations per second. Unit: hertz (Hz). 1 Hz = 1 vibration/s

T Time Period (T)

Time for one complete vibration. T = 1/f. Unit: second (s)

A Amplitude

Maximum displacement of a particle from its rest position. Determines loudness.
Key Relations
f = 1 / T     and     v = f × λ
v = speed of sound, f = frequency, λ = wavelength, T = time period
Frequency & Time Period — Visual Explanation

Watch how a vibrating particle creates waves. Frequency = how many cycles per second. Time Period = time for one complete cycle.


T = 0.500 s   |   f = 2 Hz   |   f × T = 1 (always!)
🎨 Interactive Wave Explorer

Drag the sliders to see how frequency and amplitude change the wave shape:



λ = px  |  T = s  |  Pitch: Medium  |  Loudness: Medium

❓ In-Text Questions (Page 166)

Q1
Which wave property determines (a) loudness, (b) pitch?
(a) Loudness is determined by the amplitude of the sound wave. Greater amplitude = louder sound.
(b) Pitch is determined by the frequency of the sound wave. Higher frequency = higher pitch.
Q2
Guess which sound has a higher pitch: guitar or car horn?
A guitar generally has a higher pitch than a car horn. Guitar strings vibrate at a higher frequency compared to the lower frequency vibrations produced by a car horn.
Q3
What are wavelength, frequency, time period and amplitude of a sound wave?
Wavelength (λ): The distance between two consecutive compressions or two consecutive rarefactions. SI unit: metre (m).
Frequency (f): The number of oscillations per unit time. SI unit: hertz (Hz).
Time Period (T): The time taken for one complete oscillation. T = 1/f. SI unit: second (s).
Amplitude (A): The maximum displacement of the medium particle from its mean (rest) position. It determines the loudness of sound.
Q4
How are the wavelength and frequency of a sound wave related to its speed?
Speed, wavelength and frequency are related by the equation:
v = f × λ
where v = speed of sound (m/s), f = frequency (Hz), λ = wavelength (m).
This means: Speed = Frequency × Wavelength.
Q5
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Given: f = 220 Hz, v = 440 m/s
Formula: v = f × λ ⇒ λ = v / f
λ = 440 / 220 = 2 m
Q6
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
The time interval between successive compressions is equal to the time period (T).
T = 1/f = 1/500 = 0.002 s
Note: The distance of the listener does not affect the time period. T depends only on frequency.
🚀 5. Speed of Sound

The speed of sound depends on the properties of the medium. It is fastest in solids, slower in liquids, and slowest in gases.

Speed of Sound
v = f × λ
In air at 20°C, v ≈ 344 m/s  |  In water: ~1498 m/s  |  In steel: ~5960 m/s
📊 Speed Comparison in Different Media
See how particle arrangement affects sound speed. Tap a speaker to send a pulse — watch how fast energy transfers through each medium.
Why? In solids, particles are very close and tightly bound — they transfer vibrations quickly. In gases, particles are far apart and transfer energy slowly.

🌡️ Temperature Effect

Speed of sound in air increases with temperature. At 0°C: 331 m/s; at 20°C: 344 m/s.

💧 Humidity Effect

Sound travels faster in humid air than dry air because water vapour is lighter than nitrogen and oxygen.
MediumStateSpeed (m/s) at 25°C
Air (0°C)Gas331
Air (20°C)Gas344
Water (25°C)Liquid1498
Sea waterLiquid1531
IronSolid5130
AluminiumSolid6420
Speed of Sound — Bar Chart for Quick Memory

❓ In-Text Questions (Page 167)

Q1
In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature, and why?
Sound travels fastest in iron. This is because iron is a solid where particles are closely packed and strongly bonded. The closeness of particles allows vibrations to be transmitted much more quickly than in liquids or gases. Order: Iron > Water > Air.
🎵 6. Pitch, Loudness & Quality (Timbre)

🎼 Pitch (तीक्ष्णता)

How high or low a sound appears. Depends on frequency. Higher frequency = higher pitch (e.g., whistle). Lower frequency = lower pitch (e.g., drum).

🔊 Loudness (प्रबलता)

How loud or soft a sound appears. Depends on amplitude. Greater amplitude = louder sound. Unit of loudness: decibel (dB).

🎸 Quality / Timbre

The characteristic that distinguishes two sounds of same pitch and loudness. Depends on the waveform. It’s why a sitar sounds different from a flute even at the same note.
🎧 Pitch Demo — Drag to Change Frequency & HEAR IT!
Low pitch
🔈 Loudness Demo — Drag to Change Amplitude & HEAR IT!
Medium
Sound SourceLoudness (dB)
Normal breathing10 dB
Whisper20 dB
Normal conversation60 dB
Busy traffic70 dB
Factory / loud music90–100 dB
Jet plane at 30 m150 dB
⚠️ Health Warning: Prolonged exposure to sound above 80 dB can cause hearing damage. Always protect your ears from very loud sounds!
🔄 7. Reflection of Sound

Sound waves obey the laws of reflection, just like light:

Law 1

The angle of incidence equals the angle of reflection.
∠i = ∠r

Law 2

The incident wave, the reflected wave, and the normal to the reflecting surface all lie in the same plane.

Sound reflects best off hard, smooth surfaces (walls, cliffs, polished metal). Soft, rough surfaces absorb sound.

✅ NCERT Activity 10.8 — Reflection of Sound

Two Tubes & a Reflecting Surface

Setup: Place two long tubes at angles against a smooth wall. Put a ticking clock at the end of one tube. Listen through the other tube.

Observation: The sound is heard clearly when the angle of the listening tube equals the angle of the source tube from the wall (∠i = ∠r).

Conclusion: Sound follows the laws of reflection, similar to light.

📣 8. Echo (प्रतिध्वनि)

An echo is the repetition of a sound caused by the reflection of sound waves from a hard surface (like a cliff or building).

For an echo to be heard distinctly, the reflected sound must reach the ear at least 0.1 seconds after the original sound (this is the persistence of hearing).

Minimum Distance for Echo
d = v × t / 2 = 344 × 0.1 / 2 = 17.2 m
The reflecting surface must be at least 17.2 m away (at 20°C in air) for an echo to be heard.
Echo Animation — Watch Sound Reflect Off a Cliff

A person shouts towards a cliff. The sound travels, hits the wall, reflects back, and is heard as an echo after a delay.


🧮 Echo Calculator

🔎 Will You Hear an Echo?

❓ In-Text Questions (Page 168)

Q1
An echo returned in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Given: t = 3 s, v = 342 m/s
Formula: d = v × t / 2
d = 342 × 3 / 2 = 1026 / 2 = 513 m
The reflecting surface is 513 m away from the source.
🏢 9. Reverberation

Reverberation is the persistence of sound in an enclosed space after the original sound has stopped. It is caused by repeated reflections of sound from walls, ceiling, and floor.

Excessive reverberation makes speech unclear. Concert halls and auditoriums must be carefully designed to reduce it.

Reverberation — Sound bouncing in a room

Click "Clap" to produce a sound pulse. Watch it reflect off walls multiple times, creating reverberation.

Methods to Reduce Reverberation

🧳 Soft Materials

Curtains, carpets, and seat cushions absorb sound and reduce reflections.

🪏 Acoustic Panels

Walls and ceilings are covered with sound-absorbing materials like compressed fibreboard or rough plaster.

🏠 Curved Ceilings

Curved surfaces in halls help distribute sound evenly and prevent focussing of reflected sound.
📣 10. Uses of Multiple Reflection of Sound

📣 Megaphone / Horn

Cone shape reflects sound waves inward repeatedly, focusing them in one direction. Sound travels farther without spreading out.

🩺 Stethoscope

Sound of heartbeat is reflected multiple times inside the tube and reaches the doctor’s ears without loss.

🎭 Soundboard

Curved board placed behind speakers on a stage. It reflects sound towards the audience, making the speaker audible to a large crowd.

🏛️ Concert Halls

Curved ceilings and walls are designed to reflect sound uniformly so every seat gets clear audio. This is acoustic design.

❓ In-Text Questions (Page 169)

Q1
Why are the ceilings of concert halls curved?
The ceilings of concert halls are curved so that sound produced on stage is reflected uniformly towards the audience. The curved surface spreads the reflected sound evenly across the hall, ensuring that listeners in all parts of the hall can hear clearly. This also helps reduce reverberation by directing sound downward rather than allowing it to bounce repeatedly.
👂 11. Range of Hearing

The human ear can hear sounds in the frequency range of 20 Hz to 20,000 Hz. This is called the audible range.

🚨 Infrasound (< 20 Hz)

Sound below 20 Hz. Humans cannot hear it. Produced by earthquakes, volcanoes, elephants, whales. Used in seismology.

👂 Audible Sound (20–20,000 Hz)

The range humans can hear. Speech: 500–5000 Hz. Music: 20–15,000 Hz. Hearing range decreases with age.

🔬 Ultrasound (> 20,000 Hz)

Sound above 20,000 Hz. Humans cannot hear it. Produced by bats, dolphins, dog whistles. Has many applications.
Infrasound
Audible Range
Ultrasound
0 Hz
20 Hz
20,000 Hz
>20 kHz
🐸 Elephants hear infrasound  |  🦇 Bats & 🐬 Dolphins use ultrasound

❓ In-Text Questions (Page 170)

Q1
What is the audible range of the average human ear?
The audible range of the average human ear is 20 Hz to 20,000 Hz (20 kHz).
Q2
What is the range of frequencies associated with (a) Infrasound? (b) Ultrasound?
(a) Infrasound: Frequencies below 20 Hz.
(b) Ultrasound: Frequencies above 20,000 Hz (20 kHz).
🔬 12. Applications of Ultrasound

Ultrasound (frequency > 20,000 Hz) has many practical uses because it can be directed in well-defined beams and has high energy:

💧 Cleaning

Ultrasonic waves dislodge dirt and grease from hard-to-reach parts of objects like jewellery, electronic components, surgical instruments.

🔬 Detecting Flaws

Ultrasound is passed through metal blocks. If there is a crack or flaw, the wave reflects back differently, revealing the defect. Used in industries to test quality.

🩺 Medical Imaging (Ultrasonography)

Ultrasound waves are used to get images of internal organs. Commonly used to examine a developing baby (foetus) during pregnancy. Safe — no harmful radiation.

💢 Breaking Kidney Stones

High-energy ultrasound waves (lithotripsy) are focused on kidney stones to break them into tiny pieces that can pass out naturally.
🎨 Ultrasound Applications Animation
🚢 13. SONAR

SONAR stands for Sound Navigation And Ranging. It uses ultrasound waves to measure distances, find objects underwater, and map the ocean floor.

How SONAR Works

1. A transmitter on the ship sends ultrasound pulses towards the ocean floor.
2. The pulses reflect off the ocean floor (or any object) and return to the ship.
3. A receiver on the ship detects the reflected signal.
4. The time taken is used to calculate the depth.

SONAR Depth Formula
d = v × t / 2
d = depth, v = speed of sound in water, t = total time for echo to return. Divide by 2 because the sound travels down AND back up.
🚢 SONAR Calculator

🔎 Calculate Ocean Depth

❓ In-Text Questions (Page 171)

Q1
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Given: t = 1.02 s, v = 1531 m/s
Formula: d = v × t / 2
d = 1531 × 1.02 / 2 = 1561.62 / 2 = 780.81 m
The cliff is approximately 780.81 m away from the submarine.
👂 14. Structure of the Human Ear

The human ear is a remarkable organ that converts sound waves (mechanical energy) into electrical signals that the brain interprets as sound. It has three main parts:

👂 Outer Ear

Pinna (auricle): The visible, funnel-shaped part. It collects sound waves from the surroundings and directs them into the ear canal.
Ear canal: A tube that carries sound waves to the eardrum.

🔄 Middle Ear

Eardrum (tympanic membrane): A thin membrane that vibrates when sound waves hit it.
Ossicles: Three tiny bones — hammer (malleus), anvil (incus), stirrup (stapes) — that amplify the vibrations and transmit them to the inner ear. The middle ear amplifies sound about 20 times.

🧠 Inner Ear

Cochlea: A coiled, fluid-filled tube with tiny hair cells. The vibrations create waves in the fluid, which bend the hair cells, generating electrical signals.
Auditory nerve: Carries these electrical signals to the brain, which interprets them as sound.
👂 Interactive Ear Diagram — Click each part to learn more:
👈 Click on any part of the ear diagram or the buttons above to see its function.
How We Hear — Step by Step

1. Pinna collects sound waves → 2. Waves travel through ear canal → 3. Hit eardrum, making it vibrate → 4. Ossicles amplify vibrations (~20x) → 5. Vibrations reach cochlea, where fluid waves bend hair cells → 6. Hair cells convert vibrations to electrical signals → 7. Auditory nerve sends signals to brain → 8. Brain interprets signals as sound!

📝 NCERT Exercises

Complete end-of-chapter exercises with step-by-step solutions. Try answering first, then click to check!

Q1
What is sound and how is it produced?
Sound is a form of energy that produces a sensation of hearing in our ears. It is produced by vibrating objects. When an object vibrates, it creates disturbances (compressions and rarefactions) in the surrounding medium, which travel as waves and reach our ears.
Q2
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
When a vibrating object (e.g., a tuning fork prong) moves forward, it pushes the air molecules ahead, creating a region of high pressure (compression). When it moves backward, it leaves a region of low pressure (rarefaction). These alternate compressions and rarefactions spread outward through the air as a longitudinal wave.

Diagram: C — R — C — R — C → (wave direction)
C = Compression (particles close), R = Rarefaction (particles far apart)
Q3
Cite an experiment to show that sound needs a material medium for its propagation.
Bell Jar Experiment: Place an electric bell inside a bell jar connected to a vacuum pump. Switch on the bell — sound is heard clearly. Now gradually pump out the air. As the air is removed, the sound becomes fainter and fainter. In a near-perfect vacuum, almost no sound is heard, even though the bell is still vibrating (we can see it). When air is let back in, the sound returns. This proves sound needs a material medium to propagate.
Q4
Why is sound wave called a longitudinal wave?
Sound is called a longitudinal wave because the particles of the medium vibrate back and forth in the same direction (parallel) as the wave travels. This creates alternating compressions and rarefactions along the direction of propagation.
Q5
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Quality (Timbre) of sound helps us identify a person by their voice. Each person’s vocal cords produce a unique waveform, giving their voice a distinct quality or timbre, even when the pitch and loudness may be similar.
Q6
Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen. Why?
The speed of light (3 × 10⁸ m/s) is much greater than the speed of sound (~344 m/s). Light travels almost instantaneously, so we see the flash immediately. Sound takes much longer to cover the same distance, so we hear the thunder a few seconds later.
Q7
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m/s.
For 20 Hz:
λ = v/f = 344/20 = 17.2 m

For 20,000 Hz:
λ = v/f = 344/20000 = 0.0172 m = 1.72 cm

The audible wavelength range is from 1.72 cm to 17.2 m.
Q8
Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the other child. (Speed of sound in air = 346 m/s, in aluminium = 6420 m/s)
Let the length of the rod be L.
Time in air: t₁ = L / 346
Time in aluminium: t₂ = L / 6420

Ratio: t₁ / t₂ = (L/346) / (L/6420) = 6420 / 346 = 18.55 : 1

Sound in air takes about 18.55 times longer than in aluminium.
Q9
The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Given: f = 100 Hz (= 100 vibrations per second)
1 minute = 60 seconds
Total vibrations = 100 × 60 = 6000 vibrations
Q10
Does sound follow the same laws of reflection as light does? Explain.
Yes. Sound follows the same two laws of reflection as light:
(i) The angle of incidence is equal to the angle of reflection (∠i = ∠r).
(ii) The incident wave, the reflected wave, and the normal at the point of incidence all lie in the same plane.
This can be verified using the two-tube experiment with a smooth reflecting surface.
Q11
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production be d. Find the minimum value of d to hear an echo if the speed of sound is v and the persistence of hearing is 1/10 s.
Total distance travelled by sound = 2d (going + coming back)
Time for echo ≥ 1/10 s (persistence of hearing)

2d = v × t
2d = v × (1/10)
d = v / 20

For v = 344 m/s: d = 344/20 = 17.2 m

The reflecting surface must be at least v/20 metres away.
Q12
Why are the ceilings of concert halls curved?
The ceilings of concert halls are curved so that sound produced on stage is reflected and spread uniformly across the hall. The curved shape acts like a concave reflector that directs sound towards the audience area, ensuring even distribution. This also helps reduce excessive reverberation and echoes.
Q13
What is the audible range of the average human ear?
The audible range of the average human ear is from 20 Hz to 20,000 Hz (20 kHz). Sounds below 20 Hz are called infrasound, and those above 20,000 Hz are called ultrasound.
Q14
What is the range of frequencies associated with (a) Infrasound? (b) Ultrasound?
(a) Infrasound: Frequencies below 20 Hz.
(b) Ultrasound: Frequencies above 20,000 Hz (20 kHz).
Q15
A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Given: t = 1.02 s, v = 1531 m/s
Formula: d = v × t / 2
d = 1531 × 1.02 / 2
d = 1561.62 / 2
d = 780.81 m

The cliff is approximately 780.81 m away from the submarine.
Q16
What is sound? What type of wave is it?
Sound is a form of energy that produces a sensation of hearing. It is produced by vibrating objects and travels through a material medium as a longitudinal wave (mechanical wave), consisting of alternating compressions and rarefactions.
Q17
A sound wave has a frequency of 2 kHz and wave length 35 cm. How long will it take to travel 1.5 km?
Given: f = 2 kHz = 2000 Hz, λ = 35 cm = 0.35 m, distance = 1.5 km = 1500 m

Step 1: Find speed: v = f × λ = 2000 × 0.35 = 700 m/s

Step 2: Find time: t = distance / v = 1500 / 700 = 2.14 s (approx.)
Q18
Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Given: f = 220 Hz, v = 440 m/s
Formula: λ = v / f = 440 / 220 = 2 m
What You Have Learnt
  • Sound is produced by vibrating objects.
  • Sound requires a material medium (solid, liquid, or gas) to travel. It cannot travel through vacuum.
  • Sound travels as a longitudinal wave with compressions and rarefactions.
  • In a transverse wave, particles vibrate perpendicular to the wave direction (crests and troughs).
  • The important characteristics of a sound wave are wavelength (λ), frequency (f), time period (T), and amplitude (A).
  • Speed of sound v = f × λ. Frequency f = 1/T.
  • Sound travels fastest in solids, slower in liquids, and slowest in gases.
  • Speed of sound increases with temperature.
  • Pitch depends on frequency; loudness depends on amplitude; quality/timbre depends on waveform.
  • Sound obeys the laws of reflection: angle of incidence = angle of reflection.
  • An echo is heard when the reflecting surface is at least 17.2 m away (at 20°C).
  • Reverberation is repeated reflection of sound in enclosed spaces. It can be reduced using soft, sound-absorbing materials.
  • Audible range: 20 Hz – 20,000 Hz. Below 20 Hz: infrasound. Above 20 kHz: ultrasound.
  • Ultrasound is used in cleaning, detecting flaws, medical imaging (ultrasonography), SONAR, and breaking kidney stones.
  • SONAR (Sound Navigation And Ranging) uses ultrasound to find distances under water. d = v×t/2.
  • The human ear has three parts: outer ear (pinna, canal), middle ear (eardrum, ossicles), inner ear (cochlea, auditory nerve).
📑 Formula Sheet
Speed of Sound
v = f × λ
v (m/s), f (Hz), λ (m)
Time Period
T = 1 / f
T (s), f (Hz)
Echo Distance
d = v × t / 2
Min d = 17.2 m (in air at 20°C)
SONAR Depth
d = v × t / 2
v = speed in water, t = echo time
💡 Quick Memory Aid:
v = fλ → “Very Fast Lambda”
• Pitch = Frequency, Loudness = Amplitude, Timbre = Waveform
• Speed order: Solid > Liquid > Gas
• Echo minimum distance: 17.2 m (remember: 17.2 = 344 × 0.1 / 2)
• Audible range: 20 Hz – 20 kHz