Sound is a guaranteed chapter in AMU — 1-2 questions every year. Echo numericals, frequency-wavelength relationship, ultrasound applications, and the human ear are the most tested areas. The formulas are easy — v = fλ is your best friend. This expanded guide covers wave types, parameters, speed of sound, reflection, Doppler effect, ultrasound, hearing range, ear, 15+ PYQs, 15+ practice problems, and common mistakes. Let's go.
1) What is Sound? — Production and Propagation
Sound is a form of energy that produces the sensation of hearing in our ears. It is produced by vibrating objects. When an object vibrates, it disturbs the particles of the surrounding medium, and this disturbance travels as a sound wave.
1.1 Sound is a Mechanical Wave
- Sound requires a medium (solid, liquid, or gas) to travel — it cannot travel through vacuum
- Sound travels as longitudinal waves in air (compressions and rarefactions)
- Sound waves are mechanical waves — they need particles to vibrate for propagation
1.2 The Vacuum Experiment (Bell Jar Experiment)
An electric bell is placed inside a glass jar. Air is gradually removed using a vacuum pump. As the air is removed, the sound of the bell becomes fainter and fainter, and when the jar is evacuated (empty of air), no sound is heard. But the bell's hammer can still be seen striking the gong. This proves that sound needs a material medium to travel.
🔥 AMU PYQ: Sound waves are:
(A) Transverse (B) Longitudinal (C) Both (D) Neither
Ans: B (sound travels as longitudinal waves in air)
🔥 AMU PYQ: Sound cannot travel through:
(A) Air (B) Water (C) Vacuum (D) Steel
Ans: C (sound needs a medium)
(A) Transverse (B) Longitudinal (C) Both (D) Neither
Ans: B (sound travels as longitudinal waves in air)
🔥 AMU PYQ: Sound cannot travel through:
(A) Air (B) Water (C) Vacuum (D) Steel
Ans: C (sound needs a medium)
2) Wave Motion — Types and Classification
A wave is a disturbance that travels through a medium, transferring energy from one point to another without the permanent transfer of matter.
2.1 Types of Waves
| Classification Basis | Type | Description | Examples |
|---|---|---|---|
| Based on medium requirement | Mechanical waves | Need a medium to travel | Sound waves, water waves, waves on a string |
| Electromagnetic waves | Do NOT need a medium; can travel in vacuum | Light, radio waves, X-rays, microwaves | |
| Based on particle motion | Transverse waves | Particles vibrate perpendicular to wave direction | Light, string waves, water surface waves |
| Longitudinal waves | Particles vibrate parallel to wave direction | Sound waves in air, spring (slinky) waves | |
| Based on energy transfer | Progressive waves | Transfer energy from one point to another | Sound, light, water waves |
| Standing/Stationary waves | Energy is confined; no net transfer | Vibrating string fixed at both ends |
2.2 Transverse Waves — In Detail
- Particles of the medium vibrate perpendicular to the direction of wave propagation
- Characterized by crests (highest points) and troughs (lowest points)
- Distance between two successive crests = one wavelength (λ)
- Can travel through solids and liquids surfaces (not through bulk fluids generally)
- Examples: Light waves, ripples on water surface, waves on a guitar string
2.3 Longitudinal Waves — In Detail
- Particles of the medium vibrate parallel to the direction of wave propagation
- Characterized by compressions (C) — regions of high pressure/density and rarefactions (R) — regions of low pressure/density
- Distance between two successive compressions = one wavelength (λ)
- Can travel through solids, liquids, and gases
- Examples: Sound waves in air, seismic P-waves, waves in a slinky spring
🟡 Difference between transverse and longitudinal:
Transverse: Particles vibrate ⟂ to wave direction. Has crests & troughs.
Longitudinal: Particles vibrate ∥ to wave direction. Has compressions & rarefactions.
Key memory: "Transverse = up-down motion (think: TRANSverse = ACROSS). Longitudinal = back-forth motion (think: along the length)."
Transverse: Particles vibrate ⟂ to wave direction. Has crests & troughs.
Longitudinal: Particles vibrate ∥ to wave direction. Has compressions & rarefactions.
Key memory: "Transverse = up-down motion (think: TRANSverse = ACROSS). Longitudinal = back-forth motion (think: along the length)."
3) Wave Parameters — The Language of Waves
To describe a wave completely, we need five key parameters. These form the vocabulary of wave motion.
The Master Equation:
v = f × λv = speed of wave (m/s)f = frequency (Hz)λ = wavelength (m)T = 1/f and f = 1/Tω = 2πf = 2π/T (angular frequency — for advanced use)
| Parameter | Symbol | Definition | SI Unit |
|---|---|---|---|
| Amplitude | A | Maximum displacement of a particle from its mean position. Determines loudness (energy) of sound. | metre (m) |
| Wavelength | λ (lambda) | Distance between two successive compressions or rarefactions (longitudinal) or two crests (transverse). | metre (m) |
| Frequency | f (or ν) | Number of complete oscillations per second. Determines pitch of sound. | Hertz (Hz) = s⁻¹ |
| Time Period | T | Time taken for one complete oscillation (one cycle). | second (s) |
| Speed / Velocity | v | Distance travelled by the wave per second. v = fλ = λ/T. | m/s |
🟡 Important relationships to memorize:
• v = fλ → Most important formula of this chapter
• f = 1/T → Frequency and time period are inverses
• v = λ/T → Direct from the above two
• If frequency increases → wavelength decreases (at constant speed)
• Higher frequency → higher pitch (shrill sound)
• Larger amplitude → louder sound (higher intensity)
• v = fλ → Most important formula of this chapter
• f = 1/T → Frequency and time period are inverses
• v = λ/T → Direct from the above two
• If frequency increases → wavelength decreases (at constant speed)
• Higher frequency → higher pitch (shrill sound)
• Larger amplitude → louder sound (higher intensity)
🔥 AMU PYQ: If the frequency of a wave is 100 Hz and its speed is 340 m/s, its wavelength is:
(A) 3.4 m (B) 34 m (C) 0.34 m (D) 3400 m
Ans: A (λ = v/f = 340/100 = 3.4 m)
🔥 AMU PYQ: If the time period of a wave is 0.05 s, its frequency is:
(A) 5 Hz (B) 10 Hz (C) 20 Hz (D) 50 Hz
Ans: C (f = 1/0.05 = 20 Hz)
(A) 3.4 m (B) 34 m (C) 0.34 m (D) 3400 m
Ans: A (λ = v/f = 340/100 = 3.4 m)
🔥 AMU PYQ: If the time period of a wave is 0.05 s, its frequency is:
(A) 5 Hz (B) 10 Hz (C) 20 Hz (D) 50 Hz
Ans: C (f = 1/0.05 = 20 Hz)
4) Speed of Sound — Factors and Values
The speed of sound is not constant — it varies depending on the medium, temperature, density, and other factors. Understanding these variations is important for AMU.
4.1 Speed in Different Media
Speed of sound in air (0°C) ≈ 330 m/sSpeed of sound in air (25°C) ≈ 346 m/sSpeed of sound in water ≈ 1500 m/sSpeed of sound in steel ≈ 5000 m/sSpeed of sound in glass ≈ 4500 m/sSpeed of light ≈ 3 × 10⁸ m/s (way faster!)
| Medium | Speed (m/s) | Trend |
|---|---|---|
| Air (0°C) | 330 | Slowest in gases |
| Air (25°C) | 346 | Faster at higher temperature |
| Water (fresh) | ~1500 | Faster in liquids (~4.5× air) |
| Seawater | ~1530 | Slightly faster than fresh water (higher density) |
| Steel | ~5000 | Fastest in solids (~15× air) |
| Glass | ~4500 | Very fast in hard solids |
4.2 Factors Affecting Speed of Sound
| Factor | Effect | Explanation |
|---|---|---|
| Temperature (T) | v ∝ √T | Higher temperature → molecules move faster → sound travels faster. In air, speed increases by about 0.6 m/s for every 1°C rise. |
| Density (ρ) | v ∝ 1/√ρ | For gases at same temperature: lighter gases transmit sound faster (e.g., sound in hydrogen is faster than in air). For solids: speed is also affected by elasticity. |
| Humidity | Increases speed | Moist air is less dense than dry air → sound travels faster in humid air. That's why sound carries farther on a humid day. |
| Pressure (of gas) | No effect (at fixed T) | For an ideal gas, speed of sound does NOT depend on pressure (density changes proportionally). |
| Medium state | v_solid > v_liquid > v_gas | Solids have the highest elasticity → fastest sound propagation. Gases have least elasticity → slowest. |
4.3 Formula for Speed of Sound in Gases (Theoretical)
v = √(γRT / M)γ = adiabatic index (≈ 1.4 for diatomic gases like N₂, O₂)R = universal gas constant (8.314 J/mol·K)T = absolute temperature (K)M = molar mass of the gas (kg/mol)For air at 0°C (273 K): v = √(1.4 × 8.314 × 273 / 0.029) ≈ 330 m/s
🟡 Key facts for AMU:
• Speed of sound: Solids > Liquids > Gases
• Speed in air at room temp ≈ 340 m/s (memorize this)
• Speed increases with temperature: 0.6 m/s per °C rise
• Sound travels about 4.5× faster in water than in air
• Light is ~9,00,000× faster than sound → lightning before thunder!
• Speed of sound: Solids > Liquids > Gases
• Speed in air at room temp ≈ 340 m/s (memorize this)
• Speed increases with temperature: 0.6 m/s per °C rise
• Sound travels about 4.5× faster in water than in air
• Light is ~9,00,000× faster than sound → lightning before thunder!
🔥 AMU PYQ: In which medium does sound travel fastest?
(A) Air (B) Water (C) Vacuum (D) Iron
Ans: D (solids have highest speed of sound)
🔥 AMU PYQ: Why do we see lightning before we hear thunder?
Ans: Light travels at 3 × 10⁸ m/s while sound travels at only ~340 m/s. Light reaches us almost instantly while sound takes time.
(A) Air (B) Water (C) Vacuum (D) Iron
Ans: D (solids have highest speed of sound)
🔥 AMU PYQ: Why do we see lightning before we hear thunder?
Ans: Light travels at 3 × 10⁸ m/s while sound travels at only ~340 m/s. Light reaches us almost instantly while sound takes time.
5) Reflection of Sound — Echo and Reverberation
Like light, sound also bounces back when it hits a hard, smooth surface. This phenomenon is called reflection of sound. The laws of reflection for sound are the same as for light: the angle of incidence equals the angle of reflection.
5.1 Echo — The Reflected Sound
An echo is the repetition of sound heard after the sound is reflected from a distant obstacle. The time delay between the original sound and the echo depends on the distance to the obstacle.
Distance calculation for echo:
Distance (d) = (v × t) / 2v = speed of sound, t = time interval between sound and echoFactor of 2 because sound travels to obstacle AND back
Numerical — Echo distance:
A man shouts near a cliff and hears the echo after 2 seconds. Speed of sound = 340 m/s. Find distance to cliff.
Solution: d = (340 × 2)/2 = 340 m
A man shouts near a cliff and hears the echo after 2 seconds. Speed of sound = 340 m/s. Find distance to cliff.
Solution: d = (340 × 2)/2 = 340 m
5.2 Condition for Hearing an Echo
The human ear can distinguish two sounds only if they are separated by at least 0.1 seconds. If the reflected sound reaches our ear within 0.1 s, it merges with the original sound (not perceived as a separate echo).
Minimum distance for echo:
d_min = (v × 0.1) / 2 = (340 × 0.1) / 2 = 17 m (commonly remembered as 17.2 m at 344 m/s)
So the obstacle must be at least 17 m away for a distinct echo to be heard.
d_min = (v × 0.1) / 2 = (340 × 0.1) / 2 = 17 m (commonly remembered as 17.2 m at 344 m/s)
So the obstacle must be at least 17 m away for a distinct echo to be heard.
5.3 Reverberation
Reverberation is the persistence of sound in an enclosed space due to multiple reflections from walls, ceiling, and floor. If the reflected sound persists for too long (> 0.1 s), it interferes with the next sound, making it unclear.
🟡 Controlling reverberation:
• In auditoriums and concert halls, carpets, curtains, and sound-absorbing materials are used to reduce reverberation
• Acoustic tiles on ceilings absorb excess sound
• Cushioned seats absorb sound rather than reflecting it
• Too much absorption makes the hall "dead" (dull); too little makes it "live" (echoey)
• Optimal reverberation time is important for good acoustics
• In auditoriums and concert halls, carpets, curtains, and sound-absorbing materials are used to reduce reverberation
• Acoustic tiles on ceilings absorb excess sound
• Cushioned seats absorb sound rather than reflecting it
• Too much absorption makes the hall "dead" (dull); too little makes it "live" (echoey)
• Optimal reverberation time is important for good acoustics
5.4 Uses of Multiple Reflections of Sound
- Megaphones / Loudspeakers / Horns: The conical shape directs sound forward without spreading, increasing its range
- Stethoscope: Sound of the heartbeat reaches the doctor's ears through multiple reflections inside the tube
- Sound Boards: Curved surfaces placed behind the stage in concert halls reflect sound toward the audience
- Ear Trumpet: Olden-day hearing aid — a conical tube collects sound and directs it into the ear
- Whispering Gallery: In domed buildings (like St. Paul's Cathedral), sound reflects along the curved walls, allowing a whisper to be heard across the gallery
5.5 SONAR — Sound Navigation and Ranging
SONAR is a technique that uses ultrasonic waves and their reflection (echo) to detect and locate objects underwater.
Working of SONAR:
1. A transmitter sends out a pulse of ultrasonic sound waves
2. The waves travel through water and strike an object (submarine, iceberg, shipwreck, fish school)
3. The waves are reflected as an echo
4. A receiver detects the echo
5. The time interval between sending and receiving is measured
6. Distance = (v × t)/2, where v = speed of sound in water (~1500 m/s)
Applications: Navigation, detecting submarines, locating fish schools, mapping ocean floors
1. A transmitter sends out a pulse of ultrasonic sound waves
2. The waves travel through water and strike an object (submarine, iceberg, shipwreck, fish school)
3. The waves are reflected as an echo
4. A receiver detects the echo
5. The time interval between sending and receiving is measured
6. Distance = (v × t)/2, where v = speed of sound in water (~1500 m/s)
Applications: Navigation, detecting submarines, locating fish schools, mapping ocean floors
🔥 AMU PYQ: A man claps near a tower and hears the echo after 4 seconds. Speed of sound = 344 m/s. Distance between man and tower is:
Ans: d = (344 × 4)/2 = 688 m
🔥 AMU PYQ: SONAR works on the principle of:
(A) Reflection of light (B) Reflection of sound (C) Refraction of light (D) Echo of sound
Ans: D (SONAR uses echo of ultrasonic waves)
🔥 AMU PYQ: Carpets and curtains are used in auditoriums to:
(A) Increase loudness (B) Reduce reverberation (C) Increase pitch (D) Amplify sound
Ans: B (they absorb sound, reducing echoes / reverberation)
Ans: d = (344 × 4)/2 = 688 m
🔥 AMU PYQ: SONAR works on the principle of:
(A) Reflection of light (B) Reflection of sound (C) Refraction of light (D) Echo of sound
Ans: D (SONAR uses echo of ultrasonic waves)
🔥 AMU PYQ: Carpets and curtains are used in auditoriums to:
(A) Increase loudness (B) Reduce reverberation (C) Increase pitch (D) Amplify sound
Ans: B (they absorb sound, reducing echoes / reverberation)
6) Characteristics of Sound — Loudness, Pitch, Quality
Every sound we hear can be described by three characteristics:
| Characteristic | Depends on | Description | Unit |
|---|---|---|---|
| Loudness | Amplitude (A) | Larger amplitude → more energy → louder sound. Also depends on the sensitivity of the ear and distance from source. | decibel (dB) |
| Pitch (Shrillness) | Frequency (f) | Higher frequency → higher pitch (shrill sound). Low frequency → low pitch (grave sound). Example: Woman's voice has higher pitch than man's voice. | Hertz (Hz) [the characteristic itself is perceived, not measured in Hz] |
| Quality / Timbre | Waveform (presence of overtones/harmonics) | Distinguishes two sounds of the same pitch and loudness produced by different instruments. E.g., same note played on a piano vs flute sound different because of their unique waveforms. | — (subjective) |
🟡 Key relationships to remember:
• Amplitude ↑ → Loudness ↑ (Loudness ∝ A²)
• Frequency ↑ → Pitch ↑ (shrill sound)
• Frequency ↓ → Pitch ↓ (deep, bass sound)
• Intensity is the physical quantity related to loudness (measured in W/m²)
• Loudness is the sensation (subjective), intensity is the physical measurement (objective)
• Amplitude ↑ → Loudness ↑ (Loudness ∝ A²)
• Frequency ↑ → Pitch ↑ (shrill sound)
• Frequency ↓ → Pitch ↓ (deep, bass sound)
• Intensity is the physical quantity related to loudness (measured in W/m²)
• Loudness is the sensation (subjective), intensity is the physical measurement (objective)
🔥 AMU PYQ: Loudness of sound depends on:
(A) Frequency (B) Amplitude (C) Wavelength (D) Speed
Ans: B (amplitude determines loudness)
🔥 AMU PYQ: Pitch of sound depends on:
(A) Amplitude (B) Frequency (C) Wavelength (D) Velocity
Ans: B (frequency determines pitch)
🔥 AMU PYQ: Which of the following is not a characteristic of sound?
(A) Loudness (B) Pitch (C) Quality (D) Weight
Ans: D (sound has loudness, pitch, and quality)
(A) Frequency (B) Amplitude (C) Wavelength (D) Speed
Ans: B (amplitude determines loudness)
🔥 AMU PYQ: Pitch of sound depends on:
(A) Amplitude (B) Frequency (C) Wavelength (D) Velocity
Ans: B (frequency determines pitch)
🔥 AMU PYQ: Which of the following is not a characteristic of sound?
(A) Loudness (B) Pitch (C) Quality (D) Weight
Ans: D (sound has loudness, pitch, and quality)
7) Range of Hearing — Infrasonic, Audible, Ultrasonic
The human ear can detect sound waves only within a specific frequency range. Sounds outside this range are inaudible to humans.
| Category | Frequency Range | Can Humans Hear? | Sources / Examples | Animals that can hear/produce |
|---|---|---|---|---|
| Infrasonic | < 20 Hz | No | Earthquakes, volcanoes, thunder, ocean waves, elephants, whales (communication over long distances) | Elephants, whales, rhinoceroses |
| Audible (Sonic) | 20 Hz – 20,000 Hz (20 kHz) | Yes ✓ | Speech, music, most everyday sounds (birds, engines, conversation) | Most mammals, birds |
| Ultrasonic | > 20,000 Hz (20 kHz) | No | Bats (echolocation up to 100 kHz), dolphins (up to 150 kHz), medical imaging | Bats, dolphins, dogs, cats, mice, rats |
🟡 Important:
• The audible range varies with age — children can sometimes hear up to 25 kHz, but the upper limit decreases with age
• Dogs can hear up to ~40 kHz → dog whistles use ultrasound
• Bats use ultrasound for navigation — they emit ultrasonic squeaks and detect the echoes (echolocation)
• Elephants use infrasound to communicate over several kilometers
• The audible range varies with age — children can sometimes hear up to 25 kHz, but the upper limit decreases with age
• Dogs can hear up to ~40 kHz → dog whistles use ultrasound
• Bats use ultrasound for navigation — they emit ultrasonic squeaks and detect the echoes (echolocation)
• Elephants use infrasound to communicate over several kilometers
🔥 AMU PYQ: The audible range of sound for humans is:
(A) 20–200 Hz (B) 20–2000 Hz (C) 20–20,000 Hz (D) 200–20,000 Hz
Ans: C (20 Hz to 20 kHz)
🔥 AMU PYQ: Which animals can hear ultrasonic sounds?
(A) Dogs (B) Bats (C) Dolphins (D) All of these
Ans: D
🔥 AMU PYQ: Which of the following produces infrasonic sound?
(A) Mosquito (B) Earthquake (C) Dog (D) Bat
Ans: B (earthquakes produce infrasonic <20 Hz)
(A) 20–200 Hz (B) 20–2000 Hz (C) 20–20,000 Hz (D) 200–20,000 Hz
Ans: C (20 Hz to 20 kHz)
🔥 AMU PYQ: Which animals can hear ultrasonic sounds?
(A) Dogs (B) Bats (C) Dolphins (D) All of these
Ans: D
🔥 AMU PYQ: Which of the following produces infrasonic sound?
(A) Mosquito (B) Earthquake (C) Dog (D) Bat
Ans: B (earthquakes produce infrasonic <20 Hz)
8) Ultrasound — Applications ⭐ (AMU Favourite)
Ultrasound refers to sound waves with frequency above 20,000 Hz (20 kHz). These waves have high energy and can be focused into narrow beams. They have numerous practical applications.
🟡 Properties of ultrasound that make it useful:
• High frequency → short wavelength → can be directed as a narrow beam
• High energy → can penetrate materials
• Reflects from boundaries between different media
• Can be focused into a small area → high intensity
• High frequency → short wavelength → can be directed as a narrow beam
• High energy → can penetrate materials
• Reflects from boundaries between different media
• Can be focused into a small area → high intensity
8.1 Applications of Ultrasound
| Application | Details |
|---|---|
| SONAR (Navigation) | Ultrasonic waves are sent underwater. The time for the echo gives the distance to objects (submarines, icebergs, seabed, fish schools). |
| Medical Imaging (Sonography) | Ultrasound waves are sent into the body. The reflections from different tissues create an image. Used to view foetus development, detect tumors, gallstones, etc. Safe (no radiation like X-rays). |
| Breaking Kidney Stones (Lithotripsy) | High-intensity ultrasound waves are focused on kidney stones. The intense vibrations break the stones into small pieces that can pass out naturally — no surgery needed! |
| Cleaning (Ultrasonic Cleaners) | Objects (jewellery, watches, surgical instruments, delicate parts) are placed in a liquid. Ultrasonic waves create rapid vibrations that dislodge dirt and grease from hard-to-reach crevices. |
| Detection of Flaws / Cracks (NDT) | Ultrasonic waves are sent through metal blocks, bridges, aircraft parts. Any internal crack or flaw reflects the waves differently, allowing detection without cutting the material (Non-Destructive Testing). |
| Drilling and Cutting | Ultrasonic drills can cut extremely hard materials (glass, ceramics, gems) by vibrating an abrasive tool at ultrasonic frequencies. |
| Echolocation in Animals | Bats emit ultrasonic squeaks (up to 100 kHz) and listen to echoes to navigate and catch insects in complete darkness. Dolphins use similar echolocation. |
| Inspection of Concrete | Ultrasonic pulses through concrete structures can detect voids, cracks, or honeycombing in buildings and bridges. |
🔥 AMU PYQ: Ultrasonic waves are used in which medical technique?
Ans: Sonography (ultrasound imaging)
🔥 AMU PYQ: Bats use which type of sound for navigation?
Ans: Ultrasonic waves (echolocation)
🔥 AMU PYQ: Which of the following is an application of ultrasound?
(A) SONAR (B) X-ray (C) MRI (D) ECG
Ans: A (SONAR uses ultrasound)
Ans: Sonography (ultrasound imaging)
🔥 AMU PYQ: Bats use which type of sound for navigation?
Ans: Ultrasonic waves (echolocation)
🔥 AMU PYQ: Which of the following is an application of ultrasound?
(A) SONAR (B) X-ray (C) MRI (D) ECG
Ans: A (SONAR uses ultrasound)
9) Doppler Effect — The Change in Frequency
Have you noticed how the siren of an ambulance sounds higher-pitched as it approaches you, and then lower-pitched as it moves away? This is the Doppler Effect.
Doppler Effect: The apparent change in frequency (or pitch) of a sound when there is relative motion between the source and the observer.
When source moves TOWARD observer: Apparent frequency increases → pitch sounds higher
When source moves AWAY from observer: Apparent frequency decreases → pitch sounds lower
When source moves TOWARD observer: Apparent frequency increases → pitch sounds higher
When source moves AWAY from observer: Apparent frequency decreases → pitch sounds lower
9.1 Examples of Doppler Effect
- Ambulance siren: approaches → higher pitch; passes and moves away → lower pitch
- Train horn: same effect — pitch seems to drop as the train passes
- Race cars: The engine sound changes as a car zooms past
- Aircraft: The sound of a jet engine shifts noticeably
🟡 Remember for AMU:
• Source approaching → f_apparent > f_actual (higher pitch)
• Source receding → f_apparent < f_actual (lower pitch)
• The Doppler Effect applies to ALL waves (sound, light, water)
• For light, it's called "redshift" (moving away) and "blueshift" (moving toward)
• Source approaching → f_apparent > f_actual (higher pitch)
• Source receding → f_apparent < f_actual (lower pitch)
• The Doppler Effect applies to ALL waves (sound, light, water)
• For light, it's called "redshift" (moving away) and "blueshift" (moving toward)
🔥 AMU PYQ: The apparent change in frequency of sound due to relative motion between source and observer is called:
(A) Reverberation (B) Doppler Effect (C) Echo (D) Interference
Ans: B
🔥 AMU PYQ: When an ambulance approaches you, the siren sounds:
(A) Louder (B) Quieter (C) Higher pitch (D) Lower pitch
Ans: C (frequency increases → higher pitch)
(A) Reverberation (B) Doppler Effect (C) Echo (D) Interference
Ans: B
🔥 AMU PYQ: When an ambulance approaches you, the siren sounds:
(A) Louder (B) Quieter (C) Higher pitch (D) Lower pitch
Ans: C (frequency increases → higher pitch)
10) Structure of Human Ear
The human ear is a remarkable organ that converts sound (mechanical vibrations in the air) into electrical signals that the brain can interpret. It has three main parts:
| Part | Components | Function |
|---|---|---|
| 1. Outer Ear | Pinna (external visible part) + Ear Canal (auditory canal) | Pinna collects sound waves from the surroundings and directs them into the ear canal. The ear canal channels the sound to the eardrum. |
| 2. Middle Ear | Eardrum (tympanic membrane) + 3 small bones (Ossicles): Hammer (Malleus), Anvil (Incus), Stirrup (Stapes) | Sound waves strike the eardrum, causing it to vibrate. The ossicles amplify these vibrations (~20×) and transmit them to the inner ear. The middle ear is air-filled. |
| 3. Inner Ear | Cochlea (snail-shaped, fluid-filled) + Auditory nerve | The cochlea converts the mechanical vibrations into electrical signals. The auditory nerve carries these signals to the brain, which interprets them as sound. |
🟡 Ear in AMU — what to remember:
• Pinna → collects sound
• Eardrum → vibrates in response to sound
• Ossicles (Hammer, Anvil, Stirrup) → amplify vibrations
• Cochlea → converts vibrations to electrical signals (most important!)
• Auditory nerve → carries signal to brain
• Pinna → collects sound
• Eardrum → vibrates in response to sound
• Ossicles (Hammer, Anvil, Stirrup) → amplify vibrations
• Cochlea → converts vibrations to electrical signals (most important!)
• Auditory nerve → carries signal to brain
🔥 AMU PYQ: The part of the ear that converts sound vibrations into electrical signals is:
(A) Eardrum (B) Cochlea (C) Hammer (D) Pinna
Ans: B (cochlea)
🔥 AMU PYQ: Which part of the ear amplifies sound vibrations?
(A) Pinna (B) Ear canal (C) Ossicles (D) Cochlea
Ans: C (hammer, anvil, stirrup amplify vibrations)
(A) Eardrum (B) Cochlea (C) Hammer (D) Pinna
Ans: B (cochlea)
🔥 AMU PYQ: Which part of the ear amplifies sound vibrations?
(A) Pinna (B) Ear canal (C) Ossicles (D) Cochlea
Ans: C (hammer, anvil, stirrup amplify vibrations)
11) PYQ Battle — AMU Past Questions on Sound
Q1. A wave has frequency 50 Hz and wavelength 4 m. Its speed is:
Ans: v = fλ = 50 × 4 = 200 m/s
Ans: v = fλ = 50 × 4 = 200 m/s
Q2. A person hears an echo after 3 s. Speed of sound = 340 m/s. Distance from obstacle:
Ans: d = (340 × 3)/2 = 510 m
Ans: d = (340 × 3)/2 = 510 m
Q3. Which of the following is not a characteristic of sound?
(A) Loudness (B) Pitch (C) Quality (D) Weight
Ans: D
(A) Loudness (B) Pitch (C) Quality (D) Weight
Ans: D
Q4. The speed of sound is maximum in which medium?
(A) Air (B) Water (C) Steel (D) Vacuum
Ans: C
(A) Air (B) Water (C) Steel (D) Vacuum
Ans: C
Q5. What is the SI unit of frequency?
(A) m/s (B) Hz (C) s (D) m
Ans: B (Hertz)
(A) m/s (B) Hz (C) s (D) m
Ans: B (Hertz)
Q6. Sound of which frequency is audible to humans?
(A) 5 Hz (B) 50 Hz (C) 25,000 Hz (D) 50,000 Hz
Ans: B (5 Hz is infrasonic, 25k & 50k are ultrasonic)
(A) 5 Hz (B) 50 Hz (C) 25,000 Hz (D) 50,000 Hz
Ans: B (5 Hz is infrasonic, 25k & 50k are ultrasonic)
Q7. Which part of the ear amplifies sound vibrations?
(A) Pinna (B) Ear canal (C) Ossicles (D) Cochlea
Ans: C
(A) Pinna (B) Ear canal (C) Ossicles (D) Cochlea
Ans: C
Q8. The time period of a wave is 0.02 s. Its frequency is:
Ans: f = 1/0.02 = 50 Hz
Ans: f = 1/0.02 = 50 Hz
Q9. Which of the following produces infrasonic sound?
(A) Mosquito (B) Earthquake (C) Dog (D) Bat
Ans: B
(A) Mosquito (B) Earthquake (C) Dog (D) Bat
Ans: B
Q10. Carpets and curtains are used in auditoriums to:
(A) Increase loudness (B) Reduce reverberation (C) Increase pitch (D) Amplify sound
Ans: B
(A) Increase loudness (B) Reduce reverberation (C) Increase pitch (D) Amplify sound
Ans: B
Q11. The frequency of a wave is 200 Hz and its speed is 340 m/s. Find the wavelength.
Ans: λ = v/f = 340/200 = 1.7 m
Ans: λ = v/f = 340/200 = 1.7 m
Q12. A ship sends ultrasound waves to the seabed. The echo is received after 4 s. If speed in water = 1500 m/s, depth of sea is:
Ans: d = (1500 × 4)/2 = 3000 m = 3 km
Ans: d = (1500 × 4)/2 = 3000 m = 3 km
Q13. What type of wave is a sound wave in air?
(A) Transverse (B) Longitudinal (C) Electromagnetic (D) Stationary
Ans: B
(A) Transverse (B) Longitudinal (C) Electromagnetic (D) Stationary
Ans: B
Q14. The minimum distance from an obstacle to hear an echo (v = 340 m/s) is about:
(A) 17 m (B) 34 m (C) 170 m (D) 340 m
Ans: A (d_min = 340 × 0.1 / 2 = 17 m)
(A) 17 m (B) 34 m (C) 170 m (D) 340 m
Ans: A (d_min = 340 × 0.1 / 2 = 17 m)
Q15. Doppler effect is observed when there is relative motion between:
(A) Source and observer (B) Two sources (C) Two observers (D) None
Ans: A
(A) Source and observer (B) Two sources (C) Two observers (D) None
Ans: A
Q16. The unit of loudness is:
(A) Hz (B) m/s (C) dB (D) W
Ans: C (decibel)
(A) Hz (B) m/s (C) dB (D) W
Ans: C (decibel)
12) Practice Problems — 15+ Numericals with Solutions
Practice 1: A wave has frequency 25 Hz and wavelength 2 m. Find its speed.
Solution: v = fλ = 25 × 2 = 50 m/s
Solution: v = fλ = 25 × 2 = 50 m/s
Practice 2: Sound travels at 340 m/s. If a wave has λ = 0.5 m, find its frequency.
Solution: f = v/λ = 340/0.5 = 680 Hz
Solution: f = v/λ = 340/0.5 = 680 Hz
Practice 3: A source produces 600 waves in 1 minute. Find frequency and time period.
Solution: f = 600/60 = 10 Hz. T = 1/10 = 0.1 s
Solution: f = 600/60 = 10 Hz. T = 1/10 = 0.1 s
Practice 4: A person hears thunder 5 seconds after seeing lightning. Speed of sound = 340 m/s. How far away is the storm?
Solution: d = v × t = 340 × 5 = 1700 m = 1.7 km (no factor of 2 — sound travels directly from lightning to observer)
Solution: d = v × t = 340 × 5 = 1700 m = 1.7 km (no factor of 2 — sound travels directly from lightning to observer)
Practice 5: A boat sends an ultrasonic signal to the seabed and receives the echo after 0.6 s. Speed of sound in water = 1500 m/s. Find depth of sea.
Solution: d = (1500 × 0.6)/2 = 900/2 = 450 m
Solution: d = (1500 × 0.6)/2 = 900/2 = 450 m
Practice 6: What is the wavelength of sound of frequency 256 Hz in air at 0°C (v = 330 m/s)?
Solution: λ = 330/256 ≈ 1.29 m
Solution: λ = 330/256 ≈ 1.29 m
Practice 7: A tuning fork produces 512 Hz sound. If v = 340 m/s, find wavelength.
Solution: λ = 340/512 ≈ 0.664 m
Solution: λ = 340/512 ≈ 0.664 m
Practice 8: A wave travels 900 m in 3 seconds. If its frequency is 150 Hz, find wavelength.
Solution: v = 900/3 = 300 m/s. λ = 300/150 = 2 m
Solution: v = 900/3 = 300 m/s. λ = 300/150 = 2 m
Practice 9: The audible range of a person is 30 Hz to 16000 Hz. Find the corresponding wavelength range (v = 340 m/s).
Solution: λ_min = 340/16000 = 0.02125 m. λ_max = 340/30 = 11.33 m. Range: 0.021 m to 11.33 m
Solution: λ_min = 340/16000 = 0.02125 m. λ_max = 340/30 = 11.33 m. Range: 0.021 m to 11.33 m
Practice 10: A SONAR device sends ultrasound and receives echo after 8 s. Speed in seawater = 1530 m/s. Find distance of object.
Solution: d = (1530 × 8)/2 = 12240/2 = 6120 m = 6.12 km
Solution: d = (1530 × 8)/2 = 12240/2 = 6120 m = 6.12 km
Practice 11: The speed of sound in air at 0°C is 330 m/s. What will be the speed at 30°C? (Hint: speed increases by 0.6 m/s per °C)
Solution: v = 330 + (0.6 × 30) = 330 + 18 = 348 m/s
Solution: v = 330 + (0.6 × 30) = 330 + 18 = 348 m/s
Practice 12: A bat emits ultrasonic sound of frequency 50 kHz. If speed in air = 340 m/s, find wavelength.
Solution: λ = 340 / 50000 = 0.0068 m = 6.8 mm
Solution: λ = 340 / 50000 = 0.0068 m = 6.8 mm
Practice 13: A man stands 340 m from a cliff. He shouts and hears the echo after how much time? (v = 340 m/s)
Solution: t = 2d/v = (2 × 340)/340 = 2 s
Solution: t = 2d/v = (2 × 340)/340 = 2 s
Practice 14: A wave has amplitude 0.02 m, frequency 100 Hz, and speed 340 m/s. Find: (a) λ (b) T (c) angular frequency ω.
Solution: (a) λ = 340/100 = 3.4 m. (b) T = 1/100 = 0.01 s. (c) ω = 2πf = 2 × 3.14 × 100 = 628 rad/s
Solution: (a) λ = 340/100 = 3.4 m. (b) T = 1/100 = 0.01 s. (c) ω = 2πf = 2 × 3.14 × 100 = 628 rad/s
Practice 15: An observer is 1020 m from a cliff. The observer fires a gun. After how long will the echo be heard? (v = 340 m/s)
Solution: t = 2d/v = (2 × 1020)/340 = 2040/340 = 6 s
Solution: t = 2d/v = (2 × 1020)/340 = 2040/340 = 6 s
Practice 16: A vibrating body completes 6000 oscillations in 2 minutes. Find its frequency and time period.
Solution: f = 6000/120 = 50 Hz. T = 1/50 = 0.02 s
Solution: f = 6000/120 = 50 Hz. T = 1/50 = 0.02 s
13) Common Mistakes & Tricks
⚠️ Top common mistakes in AMU exam:
Mistake 1: Forgetting the factor of 2 in echo numericals
→ d = (v × t)/2, NOT v × t. Sound travels to obstacle AND back.
Mistake 2: Confusing transverse vs longitudinal
→ Sound = longitudinal (compressions/rarefactions). Light = transverse (crests/troughs). Don't mix them up!
Mistake 3: Loudness vs Pitch confusion
→ Loudness depends on AMPLITUDE. Pitch depends on FREQUENCY. These are distinct!
Mistake 4: Units mix-up
→ Frequency = Hz, wavelength = m, speed = m/s, loudness = dB. Don't assign wrong units.
Mistake 5: Using f = 1/T incorrectly
→ T is in seconds and f in Hz. If T = 0.5 s, f = 2 Hz. If T = 2 s, f = 0.5 Hz. They are inverses.
Mistake 6: Speed of sound vs speed of light
→ Sound ≈ 340 m/s. Light ≈ 3 × 10⁸ m/s. Don't use 340 m/s for light problems!
Mistake 7: Infrasonic vs Ultrasonic
→ Infra = below (below 20 Hz), Ultra = above (above 20 kHz). Easy to swap these.
Mistake 1: Forgetting the factor of 2 in echo numericals
→ d = (v × t)/2, NOT v × t. Sound travels to obstacle AND back.
Mistake 2: Confusing transverse vs longitudinal
→ Sound = longitudinal (compressions/rarefactions). Light = transverse (crests/troughs). Don't mix them up!
Mistake 3: Loudness vs Pitch confusion
→ Loudness depends on AMPLITUDE. Pitch depends on FREQUENCY. These are distinct!
Mistake 4: Units mix-up
→ Frequency = Hz, wavelength = m, speed = m/s, loudness = dB. Don't assign wrong units.
Mistake 5: Using f = 1/T incorrectly
→ T is in seconds and f in Hz. If T = 0.5 s, f = 2 Hz. If T = 2 s, f = 0.5 Hz. They are inverses.
Mistake 6: Speed of sound vs speed of light
→ Sound ≈ 340 m/s. Light ≈ 3 × 10⁸ m/s. Don't use 340 m/s for light problems!
Mistake 7: Infrasonic vs Ultrasonic
→ Infra = below (below 20 Hz), Ultra = above (above 20 kHz). Easy to swap these.
✅ Tricks to score full marks:
• v = fλ is the king formula — write it at the top of your rough sheet
• In echo problems, ALWAYS check: does the time given include the return trip? If yes, divide by 2.
• If echo from a cliff: d = (v×t)/2
• If sound travels one way (like lightning to observer): d = v × t (no division)
• For minimum echo distance: d_min = v × 0.1 / 2 ≈ 17 m (memorize 17 m)
• Audible range: 20 Hz to 20,000 Hz — just memorize this!
• Speed in air ≈ 340 m/s at room temp — use this unless another value is given
• In MCQs, read the unit: if answer is in Hz, it's asking about frequency; if in m, it's wavelength
• v = fλ is the king formula — write it at the top of your rough sheet
• In echo problems, ALWAYS check: does the time given include the return trip? If yes, divide by 2.
• If echo from a cliff: d = (v×t)/2
• If sound travels one way (like lightning to observer): d = v × t (no division)
• For minimum echo distance: d_min = v × 0.1 / 2 ≈ 17 m (memorize 17 m)
• Audible range: 20 Hz to 20,000 Hz — just memorize this!
• Speed in air ≈ 340 m/s at room temp — use this unless another value is given
• In MCQs, read the unit: if answer is in Hz, it's asking about frequency; if in m, it's wavelength
14) Important Formulae — At a Glance
| Quantity | Formula | SI Unit |
|---|---|---|
| Wave speed | v = fλ | m/s |
| Frequency | f = 1/T | Hz (s⁻¹) |
| Time period | T = 1/f | s |
| Angular frequency | ω = 2πf | rad/s |
| Echo distance | d = (v × t) / 2 | m |
| Min distance for echo | d ≥ v × 0.1 / 2 | m |
| Speed of sound in air (0°C) | ≈ 330 m/s | m/s |
| Speed of sound in air (25°C) | ≈ 346 m/s | m/s |
| Speed of sound in water | ≈ 1500 m/s | m/s |
| Speed of sound in steel | ≈ 5000 m/s | m/s |
| Temperature effect | v_t = v₀ + 0.6t | m/s |
| Loudness | ∝ A² (amplitude squared) | dB |
| SONAR depth | d = (v × t) / 2 | m |
| Audible range (humans) | 20 Hz — 20,000 Hz | Hz |
15) Quick Revision — Power-packed One-liners
- Sound is a mechanical, longitudinal wave — needs a medium; cannot travel in vacuum
- Transverse waves: particles ⟂ to wave direction (crest/trough). Long. waves: particles ∥ (compression/rarefaction)
- v = fλ — MOST IMPORTANT formula. Also f = 1/T and v = λ/T
- Speed of sound: Solids > Liquids > Gases
- Speed in air ≈ 340 m/s (increases 0.6 m/s per °C rise in temperature)
- Speed in water ≈ 1500 m/s, in steel ≈ 5000 m/s
- Sound is faster in humid air and at higher temperatures
- Audible range (human): 20 Hz — 20,000 Hz
- Infrasonic (< 20 Hz): earthquakes, volcanoes. Ultrasonic (> 20 kHz): bats, SONAR
- Echo = reflected sound. d = (v×t)/2. Min distance for echo ≈ 17 m
- Reverberation = persistence of sound due to multiple reflections (reduced by carpets, curtains)
- SONAR uses ultrasound for underwater detection
- Loudness → amplitude (dB). Pitch → frequency (Hz). Quality → waveform (timbre)
- Doppler Effect: Apparent frequency change due to relative motion. Approaching → higher pitch. Receding → lower pitch
- Human ear: Outer (collects) → Middle (amplifies via ossicles) → Inner (cochlea converts to electrical signals)
- Ultrasound applications: SONAR, sonography, lithotripsy, cleaning, flaw detection, drilling
✅ Sound & Waves — COMPLETE. v = fλ and echo numericals are the top-scoring areas. Master these formulas and you've got this chapter locked. Next up: Structure of Atom (Chemistry)!