Here’s your concise guide to Class 9 Science Chapter 12: Sound! Perfect for Class 9 students, parents, and teachers prepping for the 2025 CBSE exams, this post covers in-text questions and answers, exercise solutions, and detailed notes on sound’s nature and properties. Let’s explore Class 9 Science Chapter 12 questions, answers, and notes!
Class 9 Science Chapter 12 Notes
1. Nature of Sound
- Definition: Sound is a mechanical wave produced by vibrating objects, requiring a medium (solid, liquid, gas) to travel.
- Type: Longitudinal wave—particles vibrate parallel to wave direction.
2. Propagation
- Travels as compressions (high pressure) and rarefactions (low pressure) in a medium.
- Speed: Varies—fastest in solids (e.g., steel: ~5000 m/s), slower in liquids (water: ~1480 m/s), slowest in air (~343 m/s at 20°C).
3. Characteristics
- Frequency (f): Vibrations per second (Hz); determines pitch.
- Amplitude: Wave height; affects loudness.
- Wavelength (λ): Distance between consecutive compressions (m).
- Speed (v): v = f × λ.
4. Human Hearing
- Range: 20 Hz to 20,000 Hz (20 kHz).
- Ultrasound: >20 kHz (e.g., used in medical imaging).
- Infrasound: <20 Hz (e.g., elephant calls).
5. Reflection
- Echo: Sound bouncing off surfaces; requires ~0.1 s delay (17 m distance in air).
- Reverberation: Multiple reflections causing prolonged sound.
6. Applications
- Sonar (distance measurement), musical instruments, noise control.
These notes outline the chapter—use them for the Q&As!
In-Text Questions and Answers: Page 174
Question 1: Explain comprehensively how sound is produced and how it propagates through a medium, including the role of vibrations and the nature of the wave involved.
Answer:
Sound is produced when an object vibrates, disturbing the surrounding medium’s particles to create a mechanical wave that propagates outward. For instance, striking a tuning fork makes its prongs oscillate, pushing air particles back and forth. These vibrations generate alternating regions of high pressure (compressions) and low pressure (rarefactions). Sound is a longitudinal wave, meaning particle displacement occurs parallel to the wave’s travel direction—unlike transverse waves (e.g., light), where motion is perpendicular.
Propagation happens as vibrating particles transfer energy to neighbors: in air, a drumbeat compresses nearby air, which compresses further layers, carrying the sound. In solids like a metal rod, vibrations travel faster due to tightly packed particles; in water, it’s intermediate. The wave requires a medium—sound can’t travel in a vacuum (e.g., space) because there’s no matter to vibrate. This process, from a guitar string to your ear, shows sound as energy transfer via particle motion, with frequency and amplitude shaping its pitch and loudness, illustrating its mechanical nature in everyday contexts like speech or music.
Question 2: Discuss in detail why sound needs a material medium to travel, and provide examples contrasting its behavior in different states of matter.
Answer:
Sound needs a material medium—solid, liquid, or gas—to travel because it’s a mechanical wave relying on particle vibration and collision to propagate energy. Unlike electromagnetic waves (e.g., light), which move through a vacuum via oscillating fields, sound requires matter to transmit the compressions and rarefactions formed by a vibrating source.
In a vacuum, with no particles, sound cannot propagate—astronauts use radios, not air, to communicate in space. In solids like steel, sound travels fastest (~5000 m/s) because particles are closely packed and rigid, transferring vibrations efficiently (e.g., hearing a train through rails before air). In liquids like water (~1480 m/s), looser particle arrangement slows it, but it’s still effective—whales communicate underwater over kilometers. In gases like air (~343 m/s), sparse, free-moving particles make it slowest, as seen when thunder lags lightning. These differences highlight sound’s dependence on a medium’s density and elasticity, with tighter particle bonds enhancing speed and range, critical for applications like sonar or stethoscopes.
Question 3: Analyze extensively why the speed of sound varies in different media, including the physical properties affecting it and examples from real-world scenarios.
Answer:
The speed of sound varies in different media due to the medium’s physical properties: elasticity (ability to regain shape) and density (mass per unit volume). In solids, high elasticity and moderate density allow rapid vibration transfer—sound in steel (~5000 m/s) outpaces air because steel’s stiffness lets particles snap back quickly despite higher mass. In liquids, lower elasticity and density (water: ~1480 m/s) slow it compared to solids, but denser packing than gases boosts speed over air. In gases, low density and elasticity (air: ~343 m/s) mean particles move freely, delaying energy transfer.
Temperature also affects gases—sound speeds up in warmer air (e.g., 360 m/s at 40°C) as particles move faster. Examples: a train’s rumble reaches you faster through tracks (solid) than air; underwater explosions are heard sooner by divers than surface listeners; a shout in a helium balloon sounds high-pitched and fast (~1000 m/s) due to helium’s low density. These variations stem from how easily and quickly particles transmit vibrations, shaping sound’s behavior in engineering (e.g., sonar) and nature (e.g., animal calls).
In-Text Questions and Answers: Page 179
Question 1: Provide a thorough explanation of how humans perceive sound, including the role of the ear and the range of frequencies detectable.
Answer:
Humans perceive sound through a process where vibrating objects generate pressure waves that travel through air to the ear, which converts them into electrical signals for the brain. When a bell rings, air particles compress and rarefy, reaching the outer ear (pinna), which funnels waves into the ear canal. The eardrum vibrates, passing motion via three tiny bones (hammer, anvil, stirrup) to the cochlea—a fluid-filled spiral lined with hair cells.
These cells bend with fluid movement, triggering nerve impulses sent via the auditory nerve to the brain, interpreted as sound. The detectable frequency range is 20 Hz to 20,000 Hz (20 kHz)—low frequencies (e.g., bass drums) feel deep, high ones (e.g., whistles) shrill. Below 20 Hz (infrasound) or above 20 kHz (ultrasound), we can’t hear, though animals like elephants (infrasound) or bats (ultrasound) can. Age reduces this range—older adults may not hear above 12 kHz. This system lets us enjoy music or detect danger, linking physics to biology seamlessly.
Question 2: Elaborate in depth on the difference between pitch and loudness, including how they relate to frequency and amplitude, with examples.
Answer:
Pitch and loudness are distinct sound properties tied to frequency and amplitude, respectively, shaping how we hear. Pitch is the perceived highness or lowness of sound, determined by frequency (f, Hz)—higher frequency means higher pitch. A 400 Hz tone (e.g., a flute) sounds higher than a 100 Hz tone (e.g., a bass guitar) because more vibrations per second tighten the cochlea’s response. Loudness is the perceived intensity, governed by amplitude—larger vibrations (more energy) make louder sounds.
A drum hit hard (high amplitude) is louder than a soft tap, though frequency may stay constant. Example: a child’s shout (high frequency, moderate amplitude) has high pitch but may not be loud; a lion’s roar (low frequency, high amplitude) is low-pitched yet deafening. Frequency and amplitude interact—a 1000 Hz tone at low amplitude is a faint high note, but cranked up, it’s piercing. This distinction explains musical scales (pitch) and volume controls (loudness), blending physics with auditory experience.
Exercise Questions and Answers: Page 184
Question 1: A sound wave has a frequency of 500 Hz and a wavelength of 0.68 m. Calculate in detail the speed of the sound wave, including steps and its significance.
Answer:
To calculate the speed (v) of a sound wave with frequency (f) 500 Hz and wavelength (λ) 0.68 m:
- Formula: v = f × λ.
- Step 1: Substitute values: v = 500 × 0.68.
- Step 2: Multiply: 500 × 0.68 = 340.
- Speed = 340 m/s.
This matches sound’s speed in air (~343 m/s at 20°C), confirming typical conditions. The significance lies in understanding wave propagation—340 m/s means sound travels 340 m in 1 s, useful for timing echoes or designing audio systems. It’s slower than in water or steel, reflecting air’s properties, key for applications like weather forecasting or acoustics.
Question 2: A stone dropped into a well produces an echo heard 2 seconds later. If sound speed is 340 m/s, compute in detail the depth of the well, explaining all steps.
Answer:
To find the well’s depth (d):
- Total time (t) = 2 s (stone falling + echo returning).
- Sound speed (v) = 340 m/s.
- Stone falls: s = ½gt², g = 9.8 m/s²; total distance = d.
- Echo time: t_sound = 2d/v (down and up).
- Total time: t_fall + t_sound = 2. Let t_fall = t, then t_sound = 2 – t.
- d = ½ × 9.8 × t²; t_sound = 2d/340 = 2 – t.
- Substitute: 2 × (4.9t²) / 340 = 2 – t. Simplify: 0.0288t² + t – 2 = 0.
- Solve quadratic: t ≈ 1.94 s (positive root).
- Depth: d = ½ × 9.8 × (1.94)² ≈ 18.4 m.
Depth = 18.4 m. This measures well depth using sound’s travel time, practical for real-world surveying.
Class 9 Science Chapter 12: Sound unravels wave mechanics for your 2025 CBSE exams. With these notes and Q&As, you’re set to succeed. Explore NCERT solutions for practice, and comment your doubts—we’re here!
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