NCERT Solutions for Class 9 Science Exploration Chapter 10 Sound Waves: Characteristics and Applications

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• Chapter 10 Exercises Solutions
• Very Short Answer Type Questions
• Short Answer Type Questions
• Long Answer Type Questions

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The chapter 10 of 9th Science begins with how sound is produced by vibrating objects, moves into how it propagates as a longitudinal mechanical wave through compressions and rarefactions, and builds up to the mathematical characteristics of sound waves – wavelength, frequency, time period, amplitude and speed. It then explores reflection of sound (echoes and reverberation), followed by a rich section on ultrasonic and infrasonic waves and their applications in SONAR, medical imaging, echolocation and industrial testing.

What makes this chapter particularly special in the 2026–27 edition is its inquiry-based, activity-driven pedagogy — with experiments using rubber bands, tuning forks, slinkies and even mobile apps like Phyphox. It connects physics to Indian science heritage through Sir C. V. Raman’s contributions to acoustics and the acoustic design of instruments like the tabla and mridangam. Cultural references such as the Whistling Village of Meghalaya (Kongthong) and the Whispering Gallery of Gol Gumbaz, Bijapur make the chapter memorable and contextually rich.

Class 9 Science Exploration Chapter 10 Question Answer

(i) Sound shows reflection
(ii) Sound needs a medium to propagate
(iii) Sound has frequency
(iv) Sound carries energy
Answer:
(ii) Sound needs a medium to propagate.
Explanation:
A mechanical wave is defined as a wave that requires a material medium (solid, liquid or gas) to travel. Sound cannot travel in vacuum, which proves that it needs a medium for propagation. Hence, this observation strongly supports that sound is a mechanical wave. The other options are general properties of waves but do not specifically prove that sound is mechanical.

(i) wavelength
(ii) speed
(iii) number of compressions per second
(iv) time period
Answer:
(iii) number of compressions per second.
Explanation:
Frequency is defined as the number of vibrations or compressions per second. Therefore, if frequency increases, the number of compressions per second also increases.

• Wavelength decreases (since v = fλ and speed remains constant in a given medium).
• Speed does not change (it depends on the medium).
• Time period decreases (T = 1/f).

(i) 80 Hz
(ii) 5 Hz
(iii) 10 Hz
(iv) 0.2 Hz
Answer:
(ii) 5 Hz
Explanation:
Frequency is the number of compressions or vibrations passing a point in one second.
Given:
Number of compressions = 20
Time = 4 s
Frequency = Number of compressions / Time
= 20/4 = 5 Hz
Therefore, the correct answer is 5 Hz.

Answer:
It will produce reverberation, not an echo.
Explanation:
For an echo to be heard separately, the reflected sound must reach the ear at least 0.1 second after the original sound.
Given:
Time gap = 0.05 s
Since 0.05 s is less than 0.1 s, the reflected sound will mix with the original sound and will not be heard as a separate echo. This effect is called reverberation.
Therefore, the sound will produce reverberation.

Answer:
(i) Greater wavelength:
Wave (a) has greater wavelength.
Explanation:
Wavelength is the distance between two successive compressions or two successive rarefactions. Since graph (a) shows fewer waves in the same horizontal distance, each wave occupies more space. Therefore, wave (a) has a greater wavelength.

(ii) Smaller amplitude:
Wave (a) has smaller amplitude.
Explanation:
Amplitude is the maximum displacement from the mean position. In the graphs, the height of the crests and depth of the troughs from the mean line are smaller in graph (a) than in graph (b). Therefore, wave (a) has smaller amplitude.

Answer:
Frequency depends on the number of waves present in the same horizontal distance.

• The curve with the greatest number of oscillations has the highest frequency.
• The curve with the smallest number of oscillations has the lowest frequency.

From Fig. 10.31:

• The green curve has the maximum number of oscillations → maximum frequency
• The blue curve has the minimum number of oscillations → minimum frequency
• The red curve has intermediate frequency

Therefore:

• A = Green curve
• B = Red curve
• C = Blue curve

Answer:
To draw the graph:

• Take the horizontal axis as distance
• Take the vertical axis as density
• Mark the mean density line
• The maximum displacement above and below the mean line should be 3 units
• The distance between two successive crests or two successive troughs should be 4 cm

Description of the graph:

• Amplitude = 3 units
• Wavelength = 4 cm

A suitable wave can be drawn with:

• Crest at +3 units
• Trough at −3 units
• Distance from one crest to the next crest = 4 cm.

Answer:
There are two errors in this depiction:
1. Sound cannot travel in space:
Space is nearly a vacuum and sound requires a material medium to travel. Therefore, no sound of explosion should be heard in space.
2. Light and sound cannot reach together:
Even if sound could travel, light travels much faster than sound. Therefore, they would not be seen and heard at the same time.
So, in reality, only the flash of light should be seen, but no sound should be heard.

Answer:
Wavelength, λ = 3.44 m
Speed, v = 344 m s⁻¹
Using the relation: v = νλ
⇒ ν = v / λ
⇒ ν = 344 / 3.44
⇒ ν = 100 Hz
Now, Time period, T = 1 / ν
⇒ T = 1 / 100
= 0.01 s
The time period of the sound wave is 0.01 s.

Answer:
Time for echo = 5 s
Speed of ultrasonic wave = 1525 m s⁻¹
The sound travels to the wreckage and back, so the one-way time is:
Time taken to reach wreckage = 5/2 = 2.5 s
Distance = Speed × Time
Distance = 1525 × 2.5 = 3812.5 m
The wreckage is approximately 3812.5 m below the surface of the ocean.

Answer:
Distance to obstacle = 1.2 m
Speed of ultrasonic wave = 345 m s⁻¹
The wave travels to the obstacle and comes back.
Total distance travelled = 2 × 1.2 = 2.4 m
Time = Distance / Speed
⇒ Time = 2.4/345
= 0.00696 s
The time taken by the ultrasonic wave to travel to the obstacle and come back is about 0.007 s.

Answer:
Distance travelled by sound, d = 1720 m
Speed of sound at 22°C, v₁ = 344 m s⁻¹
Speed of sound at 0°C, v₂ = 331 m s⁻¹
We know, Time = Distance / Speed
Time taken at 22°C:
t₁ = 1720 / 344 = 5 s
Time taken at 0°C:
t₂ = 1720 / 331 ≈ 5.20 s

Extra time taken = t₂ − t₁ = 5.20 − 5.00 = 0.20 s (approximately)
The sound of thunder will take about 0.2 s extra time to travel 1720 m when the temperature changes from 22°C to 0°C.

Answer:
From Fig. 10.32, the marked distance 8 cm covers two complete wavelengths.
Therefore, 2λ = 8 cm
⇒ λ = 8/2 = 4 cm
Converting into metre: λ = 4 cm = 0.04 m
Now, using the wave equation: v = fλ
Given:
v = 340 m s⁻¹
λ = 0.04 m
So, f = v / λ
⇒ f = 340 / 0.04 = 8500 Hz
Wavelength of the sound wave = 4 cm
Frequency of the sound wave = 8500 Hz.

Answer:
Wave A completes more cycles in the same distance, so it has a shorter wavelength.
Wave B completes fewer cycles in the same distance, so it has a longer wavelength.

By observing the graph:

• Wavelength of wave A ≈ 2.5 cm
• Wavelength of wave B ≈ 5.0 cm

Now convert into metre:
For wave A: λ₁ = 2.5 cm = 0.025 m
For wave B: λ₂ = 5.0 cm = 0.05 m
Given speed of both waves: v = 345 m s⁻¹
Using: f = v / λ
For wave A:
f₁ = 345/0.025 = 13800 Hz
For wave B:
f₂ = 345/0.05 = 6900 Hz
Hence, Wavelength of wave A = 2.5 cm and Frequency of wave A = 13,800 Hz.
Wavelength of wave B = 5.0 cm and Frequency of wave B = 6,900 Hz.

Answer:
Source A is in air
Source B is in water
The sound from both sources travels to the cliff and returns
Time taken for sound to return to A is 4.5 times the time taken to return to B
Let:
Distance from source to cliff be d for both A and B
Speed of sound in air = v_air
Speed of sound in water = v_water
Since the sound goes to the cliff and comes back, total distance travelled in each case = 2d
So,
Time taken in air:
t_air = 2d / v_air

Time taken in water:
t_water = 2d / v_water

According to the question:
t_air = 4.5 × t_water
Substituting:
2d/v_air = 4.5 × (2d/v_water)
⇒ 1/v_air = 4.5/v_water [Cancelling 2d from both sides]
⇒ v_water = 4.5 × v_air
Therefore, v_air : v_water = 1 : 4.5
To write in whole numbers: v_air : v_water = 2 : 9
The ratio between the speeds of sound in air and water is 2 : 9.

Since sound takes more time in air and less time in water for the same distance, sound must travel faster in water than in air. The given time ratio directly gives the inverse speed ratio.

NCERT Class 9 Science Exploration Chapter 10 Very Short Answer Type Questions with Explanation.

1. What is the source of sound?
   Answer:
   Any vibrating object is the source of sound. The object that produces sound by its vibrations is called the source.
2. What is vibration?
   Answer:
   Vibration is the periodic to-and-fro motion (oscillation) of an object about its mean position of rest.
3. What is a compression in a sound wave?
   Answer:
   A compression is a region in the medium where the density of particles is higher than the average density, formed when the source moves forward.
4. What is rarefaction in a sound wave?
   Answer:
   Rarefaction is a region where the density of the medium is lower than the average density, formed when the source moves backward.
5. State the SI unit of frequency.
   Answer:
   The SI unit of frequency is hertz (Hz), equivalent to one oscillation per second (s⁻¹).
6. Write the relation between frequency (ν) and time period (T).
   Answer:
   ν = 1/T. Frequency and time period are inversely related; a higher frequency corresponds to a shorter time period.
7. What is the human audible range of sound?
   Answer:
   The human audible range is from 20 Hz to 20,000 Hz (20 kHz). This range varies with age and person.
8. Why can astronauts not hear each other directly during a spacewalk?
   Answer:
   Outer space is a near vacuum — there is no medium. Since sound requires a material medium to propagate, it cannot travel in space.
9. Define the wavelength of a sound wave.
   Answer:
   Wavelength (λ) is the distance between two consecutive crests or two consecutive troughs of a wave. Its SI unit is metre (m).
10. What is the speed of sound in dry air at 0°C and 22°C?
    Answer:
    Speed of sound in dry air is approximately 331 m s⁻¹ at 0°C and about 344 m s⁻¹ at 22°C.
11. What are ultrasonic waves? Give one example of an animal that detects them.
    Answer:
    Ultrasonic waves have frequency above 20 kHz, inaudible to humans. Bats detect and use ultrasonic waves for echolocation.
12. What is echolocation?
    Answer:
    Echolocation is the ability to locate objects by emitting sound waves and detecting the reflected echoes. Bats, dolphins, and whales use it.
13. What is the minimum distance from a reflecting wall to hear an echo?
    Answer:
    The minimum distance is 17 m. Sound must travel 34 m (to wall and back) in at least 0.1 s at ~340 m s⁻¹.
14. Name the instrument commonly used to produce nearly single-frequency sound in experiments.
    Answer:
    A tuning fork — a U-shaped metal bar (steel or aluminium) with prongs (tines) — produces nearly single-frequency sound when struck.
15. What is reverberation?
    Answer:
    Reverberation is the persistence of sound in a large hall due to multiple reflections from surfaces, where reflected sounds arrive within 0.05 s of each other.

NCERT Class 9 Science Exploration Chapter 10 Short Answer Type Questions with Explanation.

1. Explain with an example how sound is produced.
   Answer:
   Sound is produced by vibrations. When a tuning fork is struck on a rubber pad, its prongs vibrate, disturbing surrounding air and creating alternating compressions and rarefactions that propagate as sound and reach our ears.
2. Why does sound need a medium to propagate? Give experimental evidence.
   Answer:
   Sound is a mechanical wave that requires particles to transmit energy. In the vacuum bell jar experiment, as air is pumped out, the ringing bell becomes inaudible, proving sound cannot propagate in vacuum without a medium.
3. How are compressions and rarefactions formed by an oscillating piston in a tube?
   Answer:
   When the piston moves forward, it compresses nearby air, forming a compression. When it moves back, the air expands, forming a rarefaction. Continuous oscillation produces alternating compressions and rarefactions that travel as a sound wave.
4. Differentiate between longitudinal and transverse waves with examples.
   Answer:
   In longitudinal waves, particles vibrate parallel to the direction of propagation (e.g., sound waves). In transverse waves, particles vibrate perpendicular to propagation (e.g., light waves). Sound is a longitudinal mechanical wave; light is a transverse, non-mechanical wave.
5. Why does the intensity of sound decrease with distance from the source?
   Answer:
   As sound travels outward, it spreads over an increasingly larger area. Since energy must be conserved but is distributed over a larger surface, the energy per unit area (intensity) decreases with distance from the source.
6. What is the difference between loudness and intensity of sound?
   Answer:
   Intensity is a measurable physical quantity — sound energy passing through unit area per unit time. Loudness is the subjective human perception of that intensity and depends on the listener’s hearing ability, not just on the wave’s amplitude.
7. Explain how SONAR works to locate underwater objects.
   Answer:
   SONAR sends ultrasonic pulses into water. These reflect off underwater objects and return as echoes. By measuring the time taken for the echo to return and knowing the speed of sound in water, the distance of the object is calculated as: distance = (v × t)/2.
8. What is pitch and how is it related to frequency?
   Answer:
   Pitch is the human perception of frequency. High-frequency sounds (e.g., whistle) have high pitch; low-frequency sounds (e.g., thunder) have low pitch. In general, pitch increases with frequency, although the exact mathematical relationship between them is complex.
9. Why does sound travel faster in solids than in liquids or gases?
   Answer:
   Solids have particles closely packed with strong intermolecular forces, enabling faster and more efficient transfer of vibrational energy. Sound travels about 4–5 times faster in water and 15–20 times faster in solids than in air.
10. What do we mean when we say sound is a longitudinal wave?
    Answer:
    In a longitudinal wave, the particles of the medium vibrate parallel to the direction in which the wave travels. In sound, air particles move back and forth in the same direction as the compressions and rarefactions propagate through the medium.

NCERT Class 9 Science Exploration Chapter 10 Long Answer Type Questions with Explanation.

Answer:
Sound propagation through air is effectively modelled using a long tube filled with air, with an oscillating piston at one end and the other end open.
When the piston is stationary: The air inside has a uniform average density throughout.
Forward movement of piston: The piston pushes nearby air particles forward, compressing them into a smaller volume. This creates a region of higher-than-average density called a compression (C). The compressed particles collide with their neighbours ahead, passing the compression forward. Those particles then collide further ahead, and so on – the compression moves forward through the tube without the particles themselves travelling along.
Backward movement of piston: The piston pulls back, and the nearby air particles spread out into the space created. This forms a region of lower-than-average density called a rarefaction (R). This too travels forward through the tube by the particle collision mechanism.
Continuous oscillation: As the piston moves back and forth repeatedly, it generates alternating compressions and rarefactions. These travel away from the source as a sound wave.
Key principle: The particles of the medium do NOT travel with the wave. Each particle only oscillates about its own mean position parallel to the direction of wave propagation. This is why sound is called a longitudinal wave. What actually travels is the disturbance in density — the energy — not the matter itself.
When the medium is unconfined (open air rather than a tube), these compressions and rarefactions spread outward in all directions from the source as spherical waves, reaching a listener and being perceived as sound.

Answer:
Sound waves are completely described by five interrelated characteristics:

1. Wavelength (λ):
   The distance between two consecutive compressions (crests) or two consecutive rarefactions (troughs) is the wavelength. It represents one complete cycle of the density oscillation in space. SI unit: metre (m).
2. Frequency (ν):
   The number of complete density oscillations that occur at a fixed point in the medium per unit time is called frequency. High frequency sounds are perceived as high pitch. SI unit: hertz (Hz) = s⁻¹.
3. Time Period (T):
   The time taken for one complete density oscillation at a fixed point is the time period. Frequency and time period are inversely related: ν = 1/T (Equation 10.1)
4. Amplitude:
   The maximum change in density of the medium from average density (in a compression or rarefaction) is the amplitude. Greater amplitude means more energy and greater loudness. Amplitude does not affect the speed or frequency of sound.
5. Speed (v):
   The speed of sound is the distance travelled by a point on the wave (crest or trough) per unit time. Speed depends on the medium: fastest in solids (~5000 m s⁻¹ in steel), intermediate in liquids (~1500 m s⁻¹ in water), and slowest in gases (~340 m s⁻¹ in air at room temperature).
   Derivation of v = λν:
   In one time period (T), the wave travels a distance of exactly one wavelength (λ). Using the basic relation speed = distance/time: v = λ/T
   Since ν = 1/T, substituting:
   v = λ × ν (Equation 10.2)
   Numerical Example:
   A sound wave in air has frequency 440 Hz and speed 344 m s⁻¹. Find its wavelength.
   λ = v/ν = 344/440 = 0.782 m.