Navigating through the newly revised CBSE Class 9 Science curriculum (Exploration) requires a thorough physical and mathematical understanding of mechanical wave propagation, compressions and rarefactions, wave characteristics (wavelength, amplitude, time period, frequency, and wave velocity v = f \times \lambda), the distinction between pitch and loudness, speed of sound across different states of matter (solids, liquids, gases) and temperatures, laws of reflection of sound, persistence of hearing (0.1\text{ s}) and minimum distance for a distinct echo (17.2\text{ m}), reverberation control in auditoriums, the human audible spectrum (20\text{ Hz to } 20,000\text{ Hz}), infrasound vs. ultrasound, and industrial/medical applications of ultrasound including SONAR, ultrasonic flaw detection, and echolocation in bats. Chapter 10 of Class 9 Physics, “Sound Waves: Characteristics and Applications”, establishes the wave mechanics governing acoustics. It explores why sound cannot propagate across the vacuum of outer space; explains how density and pressure oscillations travel through elastic media; analyzes the mathematical relationships between frequency, wavelength, and acoustic speed; details the dual-path calculations for echo ranging and underwater depth measurement; and evaluates real-world problem sets. To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE Class 9 evaluation standards.
Every question presented in the official NCERT textbook—ranging from all introductory “Think It Over” sections and in-text “Pause and Ponder” prompts (Pages 184, 188, 192, and 196) to the complete end-of-chapter “Revise, Reflect, Refine” exercises (Questions 1 to 15 on Pages 204–208)—has been solved with exhaustive detail. Numerical problems and graphical interpretations follow a step-by-step box format with explicit variable legends, standard SI unit conversions, and algebraic substitutions using clean plain-text symbols without raw LaTeX tags. Key scoring terms, official CBSE exam tags, and dynamic summary tables have been highlighted to ensure students secure maximum marks in their examinations.
Master Concept & Comparative Summary Tables
1. Sound Wave Characteristics & Physical Quantities
| Physical Quantity | Formal Definition | Standard SI Unit & Symbol | Controlling Perceptual Quality | Key Mathematical Formula |
|---|---|---|---|---|
| Wavelength (\lambda) | The linear distance between two consecutive compressions or two consecutive rarefactions. | Metre (m) | Relates inversely to frequency for a given wave speed. | Wavelength = Speed / Frequency = v / f |
| Frequency (f or \nu) | The total number of complete wave oscillations (or compressions) passing a fixed point per unit time. | Hertz (Hz) or s⁻¹ | Determines the Pitch (Shrillness) of the sound. | Frequency = 1 / Time Period = 1 / T |
| Time Period (T) | The time taken by one complete wave cycle or oscillation to pass a given point. | Second (s) | Determines frequency (T = 1 / f). | Time Period = Total Time / Number of Waves |
| Amplitude (A) | The maximum displacement or density variation of particles from their mean equilibrium position. | Metre (m) or Density units | Determines the Loudness / Intensity of sound (Loudness ∝ Amplitude²). | Measured from central baseline to crest peak. |
| Wave Speed (v) | The distance traversed by a sound wave per unit second in a given elastic medium. | Metres per second (m/s) | Speed depends on medium density, elasticity, and temperature. | Speed = Frequency × Wavelength = f × λ |
2. Frequency Spectrum of Sound Waves
| Acoustic Category | Frequency Range | Natural Sources & Animal Emitters | Key Practical / Technological Applications |
|---|---|---|---|
| Infrasound (Infrasonic Waves) | Below 20\text{ Hz} (< 20\text{ vibrations/second}) | Earthquakes (seismic P-waves), volcanic eruptions, elephants, rhinoceroses, and whales. | Early seismic earthquake warning systems; monitoring volcanic magma movement. |
| Audible Sound Spectrum | 20\text{ Hz} to 20,000\text{ Hz} (20\text{ Hz to } 20\text{ kHz}) | Human vocal cords, musical instruments (flute, guitar, piano), vibrating tuning forks. | Human acoustic communication, speech, music, and audible audio devices. |
| Ultrasound (Ultrasonic Waves) | Above 20,000\text{ Hz} (> 20\text{ kHz}) | Bats, dolphins, porpoises, moths, and specialized piezoelectric quartz crystals. | • SONAR (depth measurement & submarine detection) • Echocardiography & Ultrasonography (medical imaging) • Non-destructive metal flaw detection • Ultrasonic cleaning of spiral tubes and electronic parts. |
3. Comparison: Pitch vs. Loudness
| Basis of Distinction | Pitch (Shrillness / Flatness) | Loudness (Volume / Softness) |
|---|---|---|
| Determining Characteristic | Depends strictly on the Frequency of the sound wave. | Depends strictly on the Amplitude of the sound wave. |
| Physical Effect of Variation | Higher frequency produces a shrill, sharp, high-pitched sound; lower frequency produces a flat, grave, low-pitched sound. | Larger amplitude produces a loud sound (carries more energy); smaller amplitude produces a faint, soft sound. |
| Everyday Example | A woman’s voice, whistle, or mosquito buzz has a high pitch; a man’s deep voice or lion’s roar has a low pitch. | A lion’s roar has high loudness, while a mosquito’s buzz has low loudness. |
NCERT In-Text Questions: “Think It Over”
Page No. 184: Think It Over (Questions 1 & 2)
Question 1 Can astronauts talk to each other and hear the sounds of metal tools clanking directly in space as they do on Earth? [Exam Favorite]
Answer: No, astronauts cannot talk directly or hear the clanking of metal tools in outer space.
- Scientific Reason: Sound is a mechanical longitudinal wave. It propagates through the physical transfer of kinetic energy from one vibrating atom/molecule to the next in an elastic medium (solids, liquids, or gases).
- Outer space is an almost complete vacuum, containing virtually no material particles. In the absence of a transmitting medium, mechanical vibrations cannot propagate.
- How Astronauts Communicate: Astronauts communicate via two-way radio transmitters built into their spacesuits. Radio waves are electromagnetic waves that do not require any material medium and travel freely through the vacuum of space.
Question 2 How do bats use sound to navigate and locate prey in complete darkness at night? [Exam Favorite]
Answer: Bats use a specialized bio-acoustic biological navigation technique known as Echolocation:
- Emission of Ultrasound: While flying, the bat produces high-frequency, inaudible ultrasonic squeaks (> 20,000\text{ Hz}).
- Reflection (Echo): These ultrasonic pulses travel through the air, strike flying insects or obstacles, and reflect back as echoes to the bat’s highly sensitive ears.
- Brain Interpretation: By analyzing the time delay between emission and return, as well as the changes in frequency and intensity of the returning echo, the bat instantly computes the exact distance, direction, size, and flying speed of its prey in total darkness.
NCERT In-Text Questions: “Pause and Ponder”
Page No. 188: Pause and Ponder (Question 1)
Question 1 Why does the flash of lightning appear almost instantaneously during a thunderstorm, whereas the sound of thunder is heard several seconds later? [Exam Favorite]
Answer: This time lag occurs due to the colossal difference in the propagation speeds of light and sound:
- Speed of Light (c): Light travels at an enormous speed of 300,000,000\text{ m/s} (3 \times 10⁸\text{ m/s}) in air. The visual flash of lightning reaches the observer’s eyes almost instantaneously (t \approx 0\text{ seconds}).
- Speed of Sound (v): Sound travels at a much slower speed of approximately 344\text{ m/s} in air at room temperature (22^\circ\text{C}).
- Because sound travels nearly a million times slower than light, the sound waves of thunder take several seconds to travel the exact same atmospheric distance from the cloud to the observer.
Page No. 192: Pause and Ponder (Question 2)
Question 2 Does sound travel faster in solids, liquids, or gases? Explain the physical reason behind this variation. [Exam Favorite]
Answer: Sound travels fastest in solids, slower in liquids, and slowest in gases: Speed in Solids (~5000 m/s in steel) > Speed in Liquids (~1500 m/s in water) > Speed in Gases (~344 m/s in air).
- Physical Reason (Density and Elasticity):
- In solids, atoms and molecules are closely packed in a rigid crystal lattice with strong intermolecular forces. When one particle vibrates, it immediately transfers its mechanical disturbance to its neighboring particles.
- In gases, molecules are spaced far apart with weak intermolecular forces. Particles must travel across relatively large distances before colliding with another particle to transfer the vibration, resulting in a significantly lower acoustic speed.
Page No. 196: Pause and Ponder (Question 3)
Question 3 If the frequency of a musical note played on a violin is increased, what happens to its pitch and what happens to its speed in the room? [Exam Favorite]
Answer:
- Effect on Pitch: The pitch of the sound increases (becomes sharper and more shrill). Pitch is directly proportional to frequency; a higher frequency of string vibration produces a higher-pitched sound.
- Effect on Speed: The speed of the sound in the room remains CONSTANT. The speed of sound in a given medium (air at constant room temperature) depends only on the physical properties of the medium, not on the frequency or amplitude of the source. As frequency increases, the wavelength decreases proportionally (v = f \times \lambda), keeping the wave speed constant.
NCERT Chapter-End Exercises: “Revise, Reflect, Refine” (Pages 204–208)
Question 1 Which observation best supports the idea that sound is a mechanical wave? [Exam Favorite] (i) Sound shows reflection. (ii) Sound needs a medium to propagate. (iii) Sound has frequency. (iv) Sound carries energy.
Answer: Correct Option: (ii) Sound needs a medium to propagate.
- Reasoning: A mechanical wave is defined as a disturbance that propagates through the physical oscillation of particles in an elastic material medium and cannot travel through a vacuum. Observations like reflection, carrying energy, and having frequency are shared by electromagnetic waves (like light), but the strict requirement for a material medium uniquely identifies sound as a mechanical wave.
Question 2 For a sound wave propagating in a medium, increasing its frequency will increase its: [Exam Favorite] (i) wavelength (ii) speed (iii) number of compressions passing a point per second (iv) time period
Answer: Correct Option: (iii) number of compressions passing a point per second.
- Reasoning: By physical definition, the frequency of a wave is the number of complete wave cycles (or compressions) that pass a given point in one second. (Wavelength and time period decrease when frequency increases, while speed remains constant in a uniform medium).
Question 3 If 20 compressions pass a point in 4 seconds, the frequency of the sound wave is: [Exam Favorite] (i) 80 Hz (ii) 5 Hz (iii) 10 Hz (iv) 0.2 Hz
Answer:
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CALCULATION:
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GIVEN DATA:
• Total Number of Compressions (Cycles) = 20
• Total Time Taken (t) = 4 seconds
CALCULATION:
Formula: Frequency (f) = Total Number of Compressions / Total Time
f = 20 / 4 = 5 Hz (5 oscillations per second)
FINAL ANSWER:
Option (ii) 5 Hz is the correct answer.
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Question 4 In a room, the reflected sound reaches the ear 0.05 s after its production. Will a person hear a clear, distinct echo or reverberation? Explain. [Exam Favorite]
Answer: The person will hear Reverberation (prolonged overlapping sound), NOT a distinct echo.
- Scientific Justification (Persistence of Hearing):
- The human brain and auditory system retain the sensation of any heard sound for approximately 0.1\text{ seconds} (1/10\text{th of a second}).
- For a distinct, separate echo to be perceived, the reflected sound wave must reach the ear after a time interval of at least 0.1\text{ seconds} (t \ge 0.1\text{ s}).
- In this room, the reflected wave returns in only 0.05\text{ seconds} (0.05\text{ s} < 0.1\text{ s}). The reflected sound blends and overlaps with the original sound sensation in the brain, resulting in acoustic persistence known as reverberation.
Question 5 Graphs representing two sound waves are shown in Fig. 10.30 (a) and (b). Compare the two waves and determine which wave has: (i) a greater wavelength, and (ii) a smaller amplitude. [Exam Favorite]
Answer:
- (i) Greater Wavelength: Wave (a) has a greater wavelength. The horizontal distance between two consecutive wave crests (or compressions) is visibly larger in wave (a) than in wave (b).
- (ii) Smaller Amplitude: Wave (a) has a smaller amplitude. The vertical height of the crest (maximum displacement from the mean central baseline) is smaller in wave (a) than in wave (b), meaning wave (a) represents a softer sound with lower intensity.
Question 6 The sound waves emitted by three sources A, B, and C are represented graphically. If the frequency of A is maximum and C is minimum, identify the corresponding curves and mark A, B, and C on them. [Exam Favorite]
Answer:
- Curve A (Maximum Frequency): The wave pattern that contains the most densely packed, crowded oscillations within a given horizontal distance.
- Curve B (Intermediate Frequency): The wave pattern that displays a moderate number of wave cycles within the same distance.
- Curve C (Minimum Frequency): The wave pattern that contains the fewest, most widely spaced wave cycles (largest wavelength).
Question 7 Draw a graph to represent a sound wave for which the density amplitude is 3 units and the wavelength is 4 cm. [Exam Favorite]
Answer:
================================================================================ GRAPHICAL PLOT SPECIFICATION (DENSITY-DISTANCE GRAPH): -------------------------------------------------------------------------------- 1. Central Baseline: Horizontal X-axis representing Distance in cm (0 to 12 cm). 2. Vertical Y-axis: Density Variation (from -3 units to +3 units). 3. Amplitude Points: • Peak of Crest (Compression) = +3 units on Y-axis. • Trough (Rarefaction) = -3 units on Y-axis. 4. Wavelength Calibration: • First Compression Crest at X = 1 cm • Next Compression Crest at X = 5 cm (Distance = 5 - 1 = 4 cm = Wavelength λ) • Third Compression Crest at X = 9 cm (Distance = 9 - 5 = 4 cm) 5. Wave Shape: A continuous, smooth symmetrical sinusoidal wave oscillating between +3 and -3 with crest-to-crest distance strictly equal to 4 cm. ================================================================================
Question 8 In a movie, while showing the explosion of a spacecraft in space, a bright flash of light is shown along with a loud booming sound at the same time. Is this scene scientifically correct? Explain. [Exam Favorite]
Answer: No, this depiction is scientifically INCORRECT.
- Explanation:
- Light Flash is Visible: The explosion produces light, which is an electromagnetic wave. Light does not require a material medium and can travel across the vacuum of space, so the flash is visible.
- No Sound Can Be Heard: Sound is a mechanical wave that requires a material medium of particles to propagate. Since outer space is a vacuum, the pressure vibrations cannot travel. The explosion would occur in absolute, complete silence to any nearby observer.
Question 9 A source produces a sound wave of wavelength 3.44 m propagating with a speed of 344 m/s in air. Calculate its frequency and time period. [Exam Favorite]
Answer:
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NUMERICAL SOLUTION (FREQUENCY AND TIME PERIOD):
--------------------------------------------------------------------------------
GIVEN DATA:
• Wavelength (λ) = 3.44 m
• Wave Speed (v) = 344 m/s
STEP 1: Calculating Frequency (f)
Formula: Speed (v) = Frequency (f) × Wavelength (λ)
f = v / λ
f = 344 / 3.44 = 100 Hz
STEP 2: Calculating Time Period (T)
Formula: Time Period (T) = 1 / Frequency (f)
T = 1 / 100 = 0.01 seconds (0.01 s)
FINAL ANSWER:
• Frequency of the wave = 100 Hz
• Time Period of the wave = 0.01 s
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Question 10 A ship searching for a sunken wreck sends an ultrasonic SONAR signal down toward the ocean floor and detects the returning echo after 5 s. If the speed of sound in seawater is 1525 m/s, calculate the depth of the sunken wreck. [Exam Favorite]
Answer:
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NUMERICAL SOLUTION (SONAR ECHO RANGING):
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GIVEN DATA:
• Total Two-Way Time (t) = 5 s
• Speed of sound in seawater (v) = 1525 m/s
CALCULATION:
In echo sounding, the sound travels down to the wreck and returns back up (Distance = 2d).
Formula: 2 × Depth (d) = Speed (v) × Time (t)
d = (v × t) / 2
d = (1525 × 5) / 2
d = 7625 / 2 = 3812.5 m
FINAL ANSWER:
The sunken wreck lies at a depth of 3812.5 metres (approximately 3.81 km).
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Question 11 A vehicle is fitted with an ultrasonic distance sensor as part of its reverse parking assistance system. While reversing, the sensor emits an ultrasonic wave toward a wall 1.2 m away. If the speed of the ultrasonic wave in air is 345 m/s, calculate the time taken by the signal to return to the vehicle sensor. [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (PARKING SENSOR ECHOLOCATION):
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GIVEN DATA:
• One-way distance to wall (d) = 1.2 m
• Total two-way distance = 2 × d = 2 × 1.2 m = 2.4 m
• Speed of ultrasound (v) = 345 m/s
CALCULATION:
Formula: Time (t) = Total Distance / Speed
t = (2 × d) / v
t = 2.4 / 345 ≈ 0.006956 seconds ≈ 0.007 seconds (or 7 milliseconds)
FINAL ANSWER:
The signal returns to the sensor in 0.007 seconds (7 ms).
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Question 12 The speed of sound in air is about 331 m/s at 0°C and nearly 344 m/s at 22°C. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m if the air temperature changes from 22°C to 0°C? [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (TEMPERATURE EFFECT ON SOUND SPEED): -------------------------------------------------------------------------------- GIVEN DATA: • Distance of lightning strike (d) = 1720 m • Speed of sound at 22°C (v₁) = 344 m/s • Speed of sound at 0°C (v₂) = 331 m/s STEP 1: Time taken at 22°C (t₁) t₁ = Distance / v₁ = 1720 / 344 = 5.0 seconds STEP 2: Time taken at 0°C (t₂) t₂ = Distance / v₂ = 1720 / 331 ≈ 5.196 seconds STEP 3: Extra Time Taken (Δt) Δt = t₂ - t₁ = 5.196 s - 5.0 s = 0.196 seconds ≈ 0.20 seconds FINAL ANSWER: The sound of thunder will take approximately 0.20 seconds extra at 0°C. ================================================================================
Question 13 The variation of density of a medium for a sound wave propagating with a speed of 340 m/s is recorded. If two consecutive wave cycles span a distance of 0.08 m (8 cm), calculate the wavelength and frequency of the sound wave. [Exam Favorite]
Answer:
================================================================================
CALCULATION:
--------------------------------------------------------------------------------
GIVEN DATA:
• Speed of sound (v) = 340 m/s
• Distance for 2 wave cycles = 0.08 m (8 cm)
STEP 1: Calculating Wavelength (λ)
Since two complete wave cycles span 0.08 m:
Wavelength (λ) = 0.08 m / 2 = 0.04 m (or 4 cm)
STEP 2: Calculating Frequency (f)
Formula: Speed (v) = Frequency (f) × Wavelength (λ)
f = v / λ
f = 340 / 0.04 = 8500 Hz (8.5 kHz)
FINAL ANSWER:
• Wavelength of the sound wave = 0.04 m (4 cm)
• Frequency of the sound wave = 8500 Hz
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Question 14 The graphical representations of two sound waves A and B propagating at the same speed of 345 m/s show that wave A completes one cycle in 2.5 cm, while wave B completes one cycle in 5.0 cm. What is the wavelength of each? Also, calculate their respective frequencies. [Exam Favorite]
Answer:
================================================================================ CALCULATION: -------------------------------------------------------------------------------- GIVEN DATA: • Speed of both waves (v) = 345 m/s • Wavelength of wave A (λ_A) = 2.5 cm = 0.025 m • Wavelength of wave B (λ_B) = 5.0 cm = 0.050 m STEP 1: Frequency of Wave A (f_A) f_A = v / λ_A = 345 / 0.025 = 13,800 Hz (13.8 kHz) STEP 2: Frequency of Wave B (f_B) f_B = v / λ_B = 345 / 0.050 = 6,900 Hz (6.9 kHz) FINAL ANSWER: • Wave A : Wavelength = 0.025 m (2.5 cm), Frequency = 13,800 Hz • Wave B : Wavelength = 0.050 m (5.0 cm), Frequency = 6,900 Hz ================================================================================
Question 15 Two identical sound sources are placed at A and B, one in air and one submerged in water. Both produce sounds simultaneously that travel to a vertical cliff and reflect back. If the time taken by the sound to return to A (in air) is 4.5 times that of B (in water), what is the ratio between the speeds of sound in air and water? [Exam Favorite]
Answer:
================================================================================ RATIO DERIVATION: -------------------------------------------------------------------------------- Let the one-way distance to the cliff be d (identical for both). • Total distance travelled by each sound = 2d • Speed in air = v_air, Time in air = t_air • Speed in water = v_water, Time in water = t_water RELATION: t_air = 2d / v_air t_water = 2d / v_water GIVEN RATIO: t_air = 4.5 × t_water (2d / v_air) = 4.5 × (2d / v_water) 1 / v_air = 4.5 / v_water v_air / v_water = 1 / 4.5 = 10 / 45 = 2 / 9 FINAL ANSWER: The ratio of the speed of sound in air to that in water is 2 : 9 (or 1 : 4.5). ================================================================================
Frequently Asked Questions (FAQs) – Class 9 Science Chapter 10
Question 1: What is a Sound Wave? Is it transverse or longitudinal in air? [Exam Favorite] Answer: Sound is a mechanical wave produced by vibrating bodies. In air and liquids, sound travels as a Longitudinal Wave, where medium particles oscillate back and forth parallel to the direction of wave propagation, creating alternating regions of high pressure (compressions) and low pressure (rarefactions).
Question 2: What is the audible frequency range for human hearing? [Exam Favorite] Answer: The normal audible frequency range for a healthy human ear is from 20\text{ Hz} to 20,000\text{ Hz} (20\text{ kHz}).
Question 3: What is the minimum distance required from an obstacle to hear a distinct echo in air at 22°C? [Exam Favorite] Answer: Total Distance = Speed × Time = 344 m/s × 0.1 s = 34.4 m. Since sound travels to the obstacle and returns (2d = 34.4 m), the minimum distance is d = 34.4 / 2 = 17.2 metres.
Question 4: What is Reverberation? How is it minimized in auditoriums and cinema halls? [Exam Favorite] Answer: Reverberation is the persistence of sound caused by repeated multiple reflections off walls, ceiling, and floor. It is minimized by covering walls and ceilings with sound-absorbing materials (such as compressed fiberboard, rough plaster, heavy draperies, and cushioned seats).
Question 5: What is SONAR? State its expansion and operating principle. [Exam Favorite] Answer: SONAR stands for SOund Navigation And Ranging. It uses high-frequency ultrasonic waves to detect the depth, distance, and speed of submerged underwater objects (like submarines, icebergs, and shipwrecks) using the echo-ranging formula 2d = v \times t.
Question 6: Why are sound waves called mechanical waves? [Exam Favorite] Answer: Sound waves are called mechanical waves because they require a material medium (solid, liquid, or gas) containing physical particles to transmit energy and cannot travel through a vacuum.
Question 7: How does an increase in temperature affect the speed of sound in air? [Exam Favorite] Answer: The speed of sound in air increases with an increase in temperature because gas molecules move faster and collide more frequently at higher thermal energies (speed increases by roughly 0.6\text{ m/s} for every 1^\circ\text{C} rise in temperature).
Question 8: Differentiate between Loudness and Intensity of sound. [Exam Favorite] Answer:
- Intensity: The amount of sound energy passing per second through a unit area perpendicular to the direction of propagation (measured in \text{Watts/m}²). It is an objective physical quantity.
- Loudness: The physiological sensation of sound perceived by the human ear, which depends on both acoustic intensity and ear sensitivity.
Question 9: What is Ultrasound? Give two major medical applications. [Exam Favorite] Answer: Ultrasound consists of sound waves with frequencies greater than 20,000\text{ Hz} (20\text{ kHz}). Medical applications include: (i) Ultrasonography (imaging internal organs and fetal growth), and (ii) Echocardiography (imaging heart valves and muscle motion).
Question 10: How do ultrasonic waves detect invisible internal cracks in metal blocks? [Exam Favorite] Answer: Ultrasonic pulses are passed through the metal block. If an internal crack or air void is present, the ultrasound beam is reflected back prematurely and detected by sensors, revealing the exact internal defect without damaging the structure.
Question 11: What is a Megaphone or Stethoscope operating on? [Exam Favorite] Answer: Both devices operate on the principle of Multiple Reflection of Sound. In a stethoscope, sound waves from the patient’s heartbeat undergo repeated reflections inside the rubber tube to reach the doctor’s ears with high intensity.
Question 12: Why are the ceilings of concert halls and cinema theatres curved? [Exam Favorite] Answer: Ceilings are curved so that sound waves, after reflecting from the curved surface, are distributed uniformly to all corners of the auditorium, ensuring all audience members hear clearly.
Question 13: An ultrasonic signal sent from a boat returns in 0.8 s. If sound speed in water is 1500 m/s, find the sea depth. [Exam Favorite] Answer: Depth = (Speed × Time) / 2 = (1500 × 0.8) / 2 = 1200 / 2 = 600 metres.
Question 14: What is the relationship between frequency, wavelength, and wave speed? [Exam Favorite] Answer: Wave Speed (v) = Frequency (f) × Wavelength (λ).
Question 15: What is Infrasound? Name two animals that communicate using infrasound. [Exam Favorite] Answer: Infrasound consists of sound waves with frequencies below 20\text{ Hz}. Animals that communicate via infrasound include Elephants and Whales.
Mastering the NCERT Solutions for Class 9 Science Chapter 10 (Exploration), “Sound Waves: Characteristics and Applications”, equips students with the wave equations, echo calculations, acoustic principles, and ultrasound applications required for top performance in CBSE physics evaluations. Review the numerical box solutions, the master summary tables, and the 15 high-yield FAQs above to secure full marks in your examinations.