Sound — NCERT Solutions, AI Tutor & Practice

Understanding: Sound is a form of energy that produces the sensation of hearing in our ears.

How it is produced: Sound is produced by vibrating objects. When an object vibrates, it moves back and forth, pushing the particles of the surrounding medium (like air).

Example: When we strike a tuning fork or pluck a stretched rubber band, it vibrates. These vibrations disturb the nearby air particles, and the disturbance travels outward as sound.

Conclusion: No vibration means no sound. If you touch a ringing bell, you can feel it vibrating; stopping the vibration stops the sound.

Understanding: Sound is a mechanical wave, which means it needs a material medium (solid, liquid or gas) to travel because it propagates by making the medium's particles vibrate.

Reason: A vacuum has no particles. With nothing to carry the vibration, sound cannot propagate through it.

Bell jar experiment:

1. An electric bell is suspended inside an airtight glass bell jar connected to a vacuum pump.
2. When switched on, the bell is clearly heard.
3. As air is gradually pumped out, the sound becomes fainter and fainter.
4. When almost all air is removed, the sound is (nearly) inaudible even though the hammer is still seen striking the gong.

Conclusion: This proves that sound requires a medium and cannot travel through vacuum.

Longitudinal wave: The particles of the medium vibrate parallel (back and forth) to the direction of wave propagation. It consists of compressions (regions of high density/pressure) and rarefactions (regions of low density/pressure).

Transverse wave: The particles of the medium vibrate perpendicular to the direction of wave propagation. It consists of crests and troughs.

| Feature | Longitudinal | Transverse | |---|---|---| | Particle motion | Parallel to wave | Perpendicular to wave | | Parts | Compressions & rarefactions | Crests & troughs | | Example | Sound in air | Ripples on water |

Conclusion: A sound wave is a longitudinal wave because the air particles vibrate along the direction in which the sound travels.

Frequency ($\nu$): The number of complete oscillations (vibrations) per unit time. SI unit: hertz (Hz).

Time period ($T$): The time taken for one complete oscillation. SI unit: second (s).

Amplitude ($A$): The maximum displacement of a particle of the medium from its mean position. It determines the loudness.

Wavelength ($\lambda$): The distance between two consecutive compressions or two consecutive rarefactions (i.e. one complete wave). SI unit: metre (m).

Relation: $\nu = \frac{1}{T}$

Frequency and time period are reciprocals of each other.

Given: Frequency $\nu = 500\ \text{Hz}$, wavelength $\lambda = 0.66\ \text{m}$.

Formula: $v = \nu \times \lambda$

Substitution: $v = 500 \times 0.66 = 330\ \text{m/s}$

Time to travel 1.32 km = 1320 m: $t = \frac{\text{distance}}{\text{speed}} = \frac{1320}{330} = 4\ \text{s}$

Answer: Speed $= 330\ \text{m/s}$ and time taken $= 4\ \text{s}$.

Loudness: The characteristic by which we distinguish a loud sound from a faint one.

• It depends on the amplitude of the wave.
• Larger amplitude → louder sound; smaller amplitude → fainter sound.

Pitch: The characteristic that helps us tell a shrill (high) sound from a flat (low) sound.

• It depends on the frequency of the wave.
• Higher frequency → higher pitch (shrill); lower frequency → lower pitch (flat).

Example: A woman's voice usually has a higher pitch (higher frequency) than a man's, while shouting increases loudness (amplitude).

Conclusion: Loudness ↔ amplitude; Pitch ↔ frequency.

Echo: The repetition of a sound caused by the reflection of sound waves from a hard surface (like a wall or cliff) is called an echo.

Condition: The human ear can distinguish two sounds only if they reach it at an interval of at least 0.1 s.

Calculation: In time $t = 0.1\ \text{s}$ at speed $v = 344\ \text{m/s}$, the total distance travelled by sound (to the wall and back): $d = v \times t = 344 \times 0.1 = 34.4\ \text{m}$

This is the to-and-fro distance, so the minimum distance from the reflecting surface: $\text{distance} = \frac{34.4}{2} = 17.2\ \text{m}$

Answer: The minimum distance to hear a distinct echo is 17.2 m.

Given: Total time $t = 3.42\ \text{s}$, speed $v = 1531\ \text{m/s}$.

This technique is called SONAR (echo-ranging).

Total distance travelled by the ultrasound: $D = v \times t = 1531 \times 3.42 = 5236.02\ \text{m}$

Since this is the down-and-up (to-and-fro) distance, the depth of the sea: $\text{depth} = \frac{D}{2} = \frac{5236.02}{2} = 2618.01\ \text{m}$

Answer: The depth of the sea is approximately 2618 m.

Audible range: A normal human ear can hear sounds of frequency between 20 Hz and 20,000 Hz.

Infrasound (infrasonic): Sound of frequency below 20 Hz.

• Example: Sounds produced by elephants and whales for communication; some animals (like rhinos) can detect infrasound produced before earthquakes.

Ultrasound (ultrasonic): Sound of frequency above 20,000 Hz.

• Example: Bats and dolphins use ultrasound for navigation; it is used in medical scans (ultrasonography), cleaning parts, and detecting cracks in metals.

Conclusion: Both lie outside the human audible range but are used by many animals and in technology.

SONAR stands for SOund NAvigation And Ranging.

Principle: It works on the reflection of ultrasound waves (echo-ranging).

Working:

1. A transmitter fitted to a ship sends out ultrasonic waves into the water.
2. These waves travel through water and strike an object (seabed, submarine, iceberg, or a shoal of fish).
3. The reflected waves (echo) are picked up by a detector, which converts them into electrical signals.
4. By measuring the time $t$ between sending and receiving, and knowing the speed of sound in water $v$, the distance is found using: $2d = v \times t \quad\Rightarrow\quad d = \frac{v \times t}{2}$

Uses: Measuring the depth of the sea, and locating underwater objects such as submarines, sunken ships, icebergs and fish.

Conclusion: SONAR uses the echo of ultrasound to 'see' underwater.

Understanding: The human ear converts pressure variations (sound waves) into electrical signals sent to the brain. It has three parts: outer, middle and inner ear.

Analysis of the path of sound:

1. Pinna (outer ear): Collects the sound waves from the surroundings.
2. Auditory canal: Guides the sound to the eardrum.
3. Eardrum (tympanic membrane): A thin membrane that vibrates as compressions and rarefactions strike it — pressing in during a compression and out during a rarefaction.
4. Middle ear (three bones – hammer, anvil, stirrup): These ossicles amplify the vibrations and pass them to the inner ear.
5. Cochlea (inner ear): Converts the pressure variations into electrical signals via the auditory nerve.
6. Auditory nerve: Carries these signals to the brain, which interprets them as sound.

Conclusion: The ear acts as a chain — collect → vibrate → amplify → convert → interpret.

Given: Time for echo $t = 2\ \text{s}$, speed $v = 340\ \text{m/s}$.

Total distance travelled by sound: $D = v \times t = 340 \times 2 = 680\ \text{m}$

The sound travels to the cliff and back, so this is twice the distance of the cliff: $\text{distance of cliff} = \frac{D}{2} = \frac{680}{2} = 340\ \text{m}$

Answer: The cliff is 340 m away from the person.