Class 9 Science Chapter 10, “Sound Waves: Characteristics and Applications,” is part of the new NCERT “Exploration” textbook (2026-27 session) and covers how sound is produced by vibrating objects, how it travels as a longitudinal mechanical wave through compressions and rarefactions, its characteristics (wavelength, frequency, time period, amplitude, speed), reflection of sound (echo and reverberation), the human audible range, and applications of ultrasonic and infrasonic waves such as SONAR and medical imaging. These solutions were cross-checked against LearnCBSE, TiwariAcademy, and other current “Exploration”-edition resources, and every numerical answer was independently recomputed to verify accuracy before being presented here.
Last Updated: September 23, 2026
NCERT Solutions for Class 9 Science Chapter 10: Sound Waves: Characteristics and Applications
Revise, Reflect, Refine (NCERT Textbook, Page No. 204-206)
1. Which observation best supports the idea that sound is a mechanical wave?
(i) Sound shows reflection
(ii) Sound needs a medium to propagate
(iii) Sound has frequency
(iv) Sound carries energy
(ii) Sound needs a medium to propagate. A mechanical wave, by definition, requires a material medium (solid, liquid or gas) to travel. The fact that sound cannot travel through vacuum (as shown by the ringing-bell-in-a-jar experiment) is the observation that specifically proves sound is mechanical; the other options are properties shared by many kinds of waves, including non-mechanical ones.
2. For a sound wave propagating in a medium, increasing its frequency will increase its
(i) wavelength
(ii) speed
(iii) number of compressions per second
(iv) time period
(iii) number of compressions per second. Frequency is defined as the number of compressions (or complete oscillations) produced per second, so raising the frequency directly raises this count. The speed of sound depends only on the medium and stays constant, while wavelength (λ = v/ν) and time period (T = 1/ν) both decrease as frequency increases.
3. If 20 compressions pass a point in 4 seconds, the frequency is
(i) 80 Hz
(ii) 5 Hz
(iii) 10 Hz
(iv) 0.2 Hz
(ii) 5 Hz. Frequency = number of compressions ÷ time = 20 ÷ 4 = 5 Hz.
4. In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify your answer.
It will produce reverberation, not an echo. The human ear retains the sensation of a sound for about 0.1 s, so a reflected sound must arrive at least 0.1 s after the original sound to be perceived as a distinct, separate echo. Since the reflected sound here returns in only 0.05 s (less than 0.1 s), it overlaps with the original sound and is heard as a prolonged, blended sound — reverberation.
5. Graphs representing two sound waves are given in Fig. 10.30. If the scales on the X and Y axes of the two graphs are the same, which of the two sound waves has (i) greater wavelength, and (ii) smaller amplitude?
(i) Wave (a) has the greater wavelength, because fewer complete waves fit into the same horizontal distance compared to wave (b), meaning each cycle in (a) spans a larger distance. (ii) Wave (a) also has the smaller amplitude, because its crests and troughs show a smaller maximum displacement (density variation) from the mean line than those of wave (b).
6. The sound waves emitted by three sources A, B and C are represented in Fig. 10.31. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.
Frequency depends on how many complete oscillations occur within a given distance/time — more crowded waves mean higher frequency, and more spread-out waves mean lower frequency. So on the three curves in Fig. 10.31: the curve with the greatest number of oscillations (most tightly packed waves) should be labelled A (maximum frequency); the curve with the fewest oscillations (most widely spaced waves) should be labelled C (minimum frequency); and the remaining curve, with an in-between number of oscillations, should be labelled B.
7. Draw a graph to represent a sound wave for which the density amplitude is 3 units, and wavelength is 4 cm.
Plot distance (cm) on the horizontal axis and density on the vertical axis, with a horizontal mean/average-density line. Draw a smooth sinusoidal curve that rises to a maximum of +3 units above the mean line (a compression peak) and falls to a minimum of −3 units below it (a rarefaction trough), with the distance from one compression peak to the next compression peak (one full cycle) equal to 4 cm.
8. In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?
There are two physics errors. First, outer space is essentially a vacuum with no material medium, and since sound needs a medium to travel, no sound at all should be heard from an explosion in space — only the flash of light would be visible. Second, even in a situation where both light and sound could be produced together, light (3 × 10⁸ m/s) travels enormously faster than sound (a few hundred m/s), so a distant observer would never see the flash and hear the “bang” simultaneously.
9. A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 ms⁻¹, find its time period.
Given: λ = 3.44 m, v = 344 m/s.
Using v = λν, frequency ν = v/λ = 344/3.44 = 100 Hz.
Time period, T = 1/ν = 1/100 = 0.01 s.
(This is the same as directly using T = λ/v = 3.44/344 = 0.01 s.)
10. A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 ms⁻¹ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
Given: speed of sound in seawater = 1525 m/s; total (to-and-fro) time for the echo = 5 s.
Total distance travelled by the signal = speed × time = 1525 × 5 = 7625 m (this is the round-trip distance, down to the wreck and back up).
Depth of the wreckage = 7625 ÷ 2 = 3812.5 m.
11. A vehicle is fitted with an ultrasonic distance sensor as part of a parking assistance system, which provides echolocation, while the driver is reversing the vehicle. It emits an ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by the ultrasonic wave to travel to the obstacle and come back? Assume the speed of the ultrasonic wave in air to be 345 ms⁻¹.
Given: distance to obstacle = 1.2 m; speed = 345 m/s.
Total distance travelled to-and-fro = 2 × 1.2 = 2.4 m.
Time taken, t = distance/speed = 2.4/345 ≈ 0.00696 s ≈ 0.007 s (about 7 milliseconds).
12. The speed of sound in air is about 331 ms⁻¹ at 0°C and nearly 344 ms⁻¹ 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? Assume that all other conditions remain unchanged.
Given: distance = 1720 m; v at 22°C = 344 m/s; v at 0°C = 331 m/s.
Time at 22°C, t₁ = 1720/344 = 5 s.
Time at 0°C, t₂ = 1720/331 ≈ 5.196 s ≈ 5.20 s.
Extra time taken, Δt = t₂ − t₁ ≈ 5.20 − 5.00 = 0.20 s. So the thunder takes roughly 0.2 s longer to cover the same distance in the colder air.
13. The variation of density of medium for a sound wave propagating with a speed of 340 ms⁻¹ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.
From the figure, the marked distance of 8 cm spans two complete wave cycles, so wavelength λ = 8 cm ÷ 2 = 4 cm = 0.04 m.
Using v = λν, frequency ν = v/λ = 340/0.04 = 8500 Hz.
So the wavelength is 4 cm (0.04 m) and the frequency is 8500 Hz.
14. The graphical representation of two sound waves A and B propagating at the same speed of 345 ms⁻¹ is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.
From the graph, wavelength of wave A ≈ 2.5 cm = 0.025 m, and wavelength of wave B ≈ 5 cm = 0.05 m.
Using ν = v/λ:
Frequency of A = 345/0.025 = 13,800 Hz.
Frequency of B = 345/0.05 = 6,900 Hz.
Wave A, with the shorter wavelength, has the higher frequency, as expected since both waves travel at the same speed.
15. Two identical sound sources are placed at A and B, one in air and one submerged in water (Fig. 10.34). Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times that of B, what is the ratio between the speeds of sound in air and water?
Let the distance from each source to the cliff be d, so each sound travels a round-trip distance of 2d.
Time for A (in air): t_A = 2d/v_air. Time for B (in water): t_B = 2d/v_water.
Given t_A = 4.5 × t_B, so 2d/v_air = 4.5 × (2d/v_water), which gives 1/v_air = 4.5/v_water, i.e., v_water = 4.5 × v_air.
Therefore v_air : v_water = 1 : 4.5 = 2 : 9. This makes physical sense because sound genuinely travels faster in water than in air, so it takes less time to cover the same distance in water, matching the given ratio.
In-Text Questions (Think It Over, Pause and Ponder, What If, and Chapter Activities)
Think It Over (Page No. 184) — Two astronauts are repairing the arm of a space station together during a spacewalk. Can they talk to each other and hear the sounds of metal clanking as they do on Earth?
No. Outer space is essentially a vacuum, and since sound is a mechanical wave that needs a material medium (solid, liquid or gas) to travel, it cannot propagate between the astronauts. This is why astronauts use radio communication (electromagnetic waves, which do not need a medium) instead of speaking directly to each other.
Think It Over (Page No. 184) — How do most bats use sound to locate their prey in the dark at night?
Bats emit high-frequency ultrasonic sound waves (above the human hearing range of 20 kHz). These waves reflect off nearby objects and prey as echoes, and the bat’s ears detect the returning echoes to judge the direction, distance, and size of the prey. This technique is called echolocation.
Think It Over (Page No. 184) — Which form of energy gets converted to sound energy? How is sound produced, and how does it reach our ears?
Mechanical (vibrational) energy is converted into sound energy. Sound is produced when an object vibrates; these vibrations disturb the surrounding medium, creating alternating compressions and rarefactions that travel outward as a wave. When this wave reaches the ear, it makes the eardrum vibrate, which we perceive as sound.
Pause and Ponder (Page No. 186) — Explore various ways of producing sound.
Sound can be produced by plucking a stretched string (guitar, sitar), blowing air into a tube (flute, trumpet), striking an object (drum, bell, tabla), the vibration of vocal cords (human speech), and rubbing or shaking objects (a violin bow on strings, rattles). In every case, sound arises from some part of the object vibrating.
Pause and Ponder (Page No. 186) — Make a list of different types of musical instruments and identify their vibrating parts which produce sound.
Guitar, violin and sitar — vibrating stretched strings; flute — vibrating column of air inside the tube; trumpet — vibrating air column set in motion by the player’s vibrating lips; drum and tabla — vibrating stretched membrane (skin); bells — vibrating metal body.
Pause and Ponder (Page No. 187) — Would you hear sound in a vacuum?
No. Sound is a mechanical wave and needs particles of a medium to carry the vibrations forward. A vacuum has no particles, so there is nothing to compress and rarefy, and hence no sound can travel or be heard.
Pause and Ponder (Page No. 188) — Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out. Reason (R): Sound requires a medium to travel. Choose the correct statement: (i) Both A and R are true, but R is not the correct explanation of A. (ii) Both A and R are true, and R is the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.
(ii) Both A and R are true, and R correctly explains A. As air is pumped out of the jar, a near-vacuum is created; since sound needs a material medium to propagate and there is (almost) no medium left inside the jar, the sound of the ringing bell cannot reach an observer outside, even though the bell continues to vibrate.
Pause and Ponder (Page No. 191) — Assertion (A): Compressions and rarefactions move through the medium. Reason (R): Individual particles of the medium continuously move forward with the wave. Choose the correct statement: (i) Both A and R are true, but R is not the correct explanation of A. (ii) Both A and R are true, and R is the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.
(iii) A is true, but R is false. Compressions and rarefactions (the disturbance/energy) do move through the medium from source to listener, but the individual particles of the medium do not travel along with the wave — they simply oscillate back and forth about their own mean position. Since R incorrectly claims the particles move forward with the wave, it is a false statement and cannot be the correct explanation of A.
Pause and Ponder (Page No. 192) — When sound travels from a tuning fork to your ear, which of the following actually reaches your ear? (i) Air particles near the tuning fork (ii) Energy carried by sound waves (iii) The tuning fork material (iv) A continuous stream of compressed air
(ii) Energy carried by sound waves. The vibrating tuning fork disturbs the air particles next to it, and this disturbance (as alternating compressions and rarefactions) is passed from particle to particle through the medium. The particles themselves stay near their original positions; only the energy of the disturbance travels all the way to the ear.
Pause and Ponder (Page No. 193) — The variation of density of the medium for two sound waves is shown in Fig. 10.17(a) and (b). Label compression and rarefaction by C and R on it. In the graphs in Fig. 10.17(c) and (d), label the axes and draw the curves corresponding to Fig. 10.17(a) and (b).
In the density-versus-distance pictures, the regions where the medium’s particles are crowded closer together (density above average) should be labelled C (compression), and the regions where particles are spread further apart (density below average) should be labelled R (rarefaction). On the corresponding graphs (c) and (d), the horizontal axis should be labelled “distance” and the vertical axis “density,” with the curve rising to a peak at each compression and dipping to a trough at each rarefaction — a smooth, continuous sinusoidal curve for each wave.
Pause and Ponder (Page No. 195) — Conduct Activity 10.1 once again with a thick rubber band and then with a thin rubber band. Does the thin rubber band vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber band?
Yes, the thin rubber band vibrates faster than the thick one. A faster vibration means the thin band has a higher frequency (and therefore produces a higher-pitched sound) and, since frequency and time period are inversely related (T = 1/ν), a correspondingly smaller time period. The thick rubber band, vibrating more slowly, has a lower frequency, a larger time period, and produces a lower-pitched sound.
Pause and Ponder (Page No. 195) — If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?
Frequency = 20 Hz means 20 oscillations every second. In one minute (60 s), total oscillations = 20 × 60 = 1200. So the piston completes 1200 oscillations per minute.
Pause and Ponder (Page No. 195) — For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?
Reading the graph, one complete wavelength (crest to crest) is about 3 cm, so half the wavelength is λ/2 = 3/2 = 1.5 cm.
Pause and Ponder (Page No. 197) — Table 10.1 shows the speed of sound in a few media at atmospheric pressure (air ≈ 340 m/s, water ≈ 1500 m/s, steel ≈ 5000 m/s at 15°C). Compare the speeds in different media by finding the ratio of (i) the speed of sound in water with respect to the speed in air, and (ii) the speed of sound in steel with respect to the speed in water.
(i) Ratio of speed in water to speed in air = 1500/340 = 75:17 (≈ 4.4 times faster in water than in air).
(ii) Ratio of speed in steel to speed in water = 5000/1500 = 10:3 (≈ 3.3 times faster in steel than in water). Note: some solution sets carelessly print this second ratio as the decimal “10.3,” which is incorrect — the correctly simplified whole-number ratio is 10:3, equal to approximately 3.33.
Pause and Ponder (Page No. 197) — Two friends are standing along a steel fence at a distance of 340 m from each other (Fig. 10.23). Gunjan places her ear over the fence, and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air given in Table 10.1, calculate the time difference between the sound that reached Gunjan through the air and through the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least 0.1 s to be heard separately.)
Distance = 340 m; speed in air = 340 m/s; speed in steel = 5000 m/s.
Time through air, t_air = 340/340 = 1 s. Time through steel, t_steel = 340/5000 = 0.068 s.
Time difference, Δt = 1 − 0.068 = 0.932 s. Since 0.932 s is much greater than the 0.1 s needed to hear two sounds separately, yes, Gunjan would clearly be able to distinguish the two sounds — she would hear the sound travelling through the steel fence first (arriving almost instantly), followed nearly a second later by the same sound arriving through the air.
Pause and Ponder (Page No. 201) — An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 ms⁻¹.
Total (round-trip) distance travelled by sound in 0.2 s = speed × time = 343 × 0.2 = 68.6 m.
Since this is the to-and-fro distance, the minimum distance to the reflecting surface = 68.6/2 = 34.3 m.
Pause and Ponder (Page No. 203) — Sound travels much farther in water than light and is thus used for various underwater applications. A sonar signal sent to find the depth of the ocean takes 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500 ms⁻¹?
Total distance travelled by the signal (down and back up) = speed × time = 1500 × 4 = 6000 m.
Depth of the ocean = 6000/2 = 3000 m.
What If… (Page No. 202) — What if humans could detect ultrasonic waves like dogs can? What would be the advantages and disadvantages?
Advantages: humans could detect very faint or high-frequency vibrations, which could aid in sensing machine faults, medical diagnosis, or detecting hidden/distant objects without instruments. Disadvantages: our surroundings are full of natural and man-made ultrasonic vibrations, so we would be constantly bombarded with additional “noise,” making it harder to focus, causing discomfort or fatigue, and disturbing sleep and concentration.
Activity 10.1: Let Us Explore (Page No. 185) — Aim: To show that sound is produced due to the vibration of an object (using a stretched rubber band).
When the rubber band stretched over a box is plucked, it is seen vibrating rapidly and a sound is heard; the sound continues only as long as the vibration continues and stops the instant the vibration is stopped by hand. Changing the tension in the band changes the pitch of the sound produced. Conclusion: sound is produced by vibrating objects, and it stops as soon as the vibration stops.
Activity 10.2: Let Us Explore (Page No. 186) — Aim: To show that sound is produced due to vibrations, using a tuning fork.
When a tuning fork is struck on a rubber pad, its prongs are seen to vibrate and a sound is heard; bringing the vibrating prongs close to the ear makes the sound clearer, and touching the prongs to the surface of water produces ripples, visually confirming the vibration. Conclusion: the tuning fork produces sound only while its prongs are vibrating, confirming that vibration is essential for producing sound.
Activity 10.3: Let Us Investigate (Page No. 186-187) — Aim: To show that sound can travel through solids.
Tapping a wooden desk while listening through the air gives a faint sound, but placing the ear directly against the desk while it is tapped makes the sound noticeably louder and clearer. Conclusion: sound travels through solids, and it travels more effectively through a solid (where particles are closely packed) than through air.
Activity 10.4: Let Us Investigate (Page No. 187) — Aim: To show that sound can travel through liquids and requires a material medium.
When two spoons are tapped together underwater, the resulting sound can still be clearly heard by an observer with their ear near the water’s surface, showing that sound travels through the water and then through air to reach the ear. Conclusion: sound can travel through solids, liquids, and gases, but it always requires some material medium — it cannot travel through a vacuum, where there are no particles to carry the vibration.
Activity 10.5: Let Us Observe (Page No. 188) — Aim: To study how a disturbance travels through a medium using a slinky.
Pushing and pulling one end of a stretched slinky creates a disturbance that travels along its length, forming regions where the coils bunch closely together (compression) and regions where they spread apart (rarefaction). A marked point on the slinky is seen only to oscillate back and forth about its original position — it does not travel forward with the disturbance. Conclusion: a wave transfers energy through a medium without transferring the matter of the medium itself; this models exactly how sound propagates through air.
Activity 10.6: Let Us Experiment (Page No. 191) — Aim: To show that sound is a form of energy that can cause vibrations in objects, using grains on a stretched sheet/membrane.
When a loud sound is produced near a membrane with light grains (like semolina) scattered on it, the grains are seen jumping and moving even though nothing physically touches the sheet; louder sounds cause more vigorous movement of the grains, and the grains stop moving as soon as the sound stops. Conclusion: sound carries energy that can be transferred through the air to make a distant surface vibrate, proving sound is a genuine form of energy.
Activity 10.7: Let Us Experiment (Demonstration Activity, Page No. 194) — Aim: To study how the frequency of sound changes across different musical notes using a sound-frequency mobile app.
Recording the notes Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa (a musical octave) with a frequency-analysis app shows that “Sa” has the lowest frequency of the set and that frequency increases steadily as the notes rise toward the higher “Sa.” Conclusion: different musical notes correspond to different, systematically related frequencies of vibration.
Activity 10.8: Let Us Experiment (Demonstration Activity, Page No. 198-199) — Aim: To study how frequency affects hearing and to determine the approximate range of human hearing.
Playing tones of increasing frequency (starting around 100 Hz and going up toward 1000 Hz and beyond) makes the pitch sound progressively higher and sharper, while reducing the frequency below about 50 Hz makes the sound grow fainter until, around 20 Hz or below, it becomes inaudible. Conclusion: the human ear can hear sounds only within a limited frequency range of roughly 20 Hz to 20,000 Hz (20 kHz); sounds below this range are called infrasonic, and sounds above it are called ultrasonic.
Why This Chapter Matters
Sound Waves: Characteristics and Applications extends the general idea of wave motion that students meet elsewhere in Class 9 Science, applying concepts like wavelength, frequency, time period, amplitude and speed to a wave students experience constantly in daily life. The activity-based approach — using rubber bands, tuning forks, slinkies and smartphone apps — builds the same scientific-inquiry skills needed across the “Exploration” textbook, while the chapter’s applications (SONAR, echolocation, ultrasonography, industrial flaw detection) connect physics directly to technology, medicine and everyday problem solving, and lay the groundwork for the more advanced treatment of sound and wave optics that students will encounter in Class 10 and beyond.
Extra Questions | Revision Notes | Formulas Handbook
Chapter Quiz — Test Your Understanding
Class 9 Science Chapter 10 – Notes and Extra Questions
Along with these NCERT Solutions, students can also use the Class 9 Science Chapter 10 Extra Questions and Class 9 Science Chapter 10 Revision Notes for quick revision and extra practice.
- Chapter 1: Entering the World of Secondary Science – Free PDF Download
- Chapter 2: Cell: The Building Block of Life – Free PDF Download
- Chapter 3: Tissues in Action – Free PDF Download
- Chapter 4: Describing Motion Around Us – Free PDF Download
- Chapter 5: Exploring Mixtures and their Separation – Free PDF Download
- Chapter 6: How Forces Affect Motion – Free PDF Download
- Chapter 7: Work, Energy and Simple Machines – Free PDF Download
- Chapter 8: Journey Inside the Atom – Free PDF Download
- Chapter 9: Atomic Foundations of Matter – Free PDF Download
- Chapter 11: Reproduction: How Life Continues – Free PDF Download
- Chapter 12: Patterns in Life: Diversity and Classification – Free PDF Download
- Chapter 13: Earth as a System: Energy, Matter and Life – Free PDF Download
Frequently Asked Questions
What is the difference between an echo and reverberation, and what minimum distance is needed to hear an echo?
An echo is a distinct repetition of a sound heard after it reflects off a surface, and it is heard separately only if it reaches the ear at least 0.1 s after the original sound (because the human ear retains a sound sensation for about 0.1 s). Reverberation is the persistence or blending of sound caused by multiple reflections arriving too close together in time (under 0.1 s) to be told apart. Taking the speed of sound in air as about 340 m/s, the minimum distance from a reflecting surface needed to hear a clear echo works out to roughly 17 m (since sound must cover twice that distance, to the surface and back, in at least 0.1 s).
Why can’t astronauts hear sound directly in space, even though they can see each other clearly?
Sound is a mechanical wave, meaning it needs a material medium — solid, liquid or gas — made of particles that can be compressed and rarefied to carry the vibration forward. Space is essentially a vacuum with no such medium, so sound cannot propagate through it at all, even though light (an electromagnetic wave) travels through vacuum without any issue and lets astronauts see each other perfectly well.
What is the audible range of the human ear, and what do “infrasonic” and “ultrasonic” mean?
Humans can typically hear sound frequencies between about 20 Hz and 20,000 Hz (20 kHz). Sound waves with frequency below 20 Hz are called infrasonic waves (detectable by animals like elephants, and produced by events such as earthquakes), while sound waves with frequency above 20 kHz are called ultrasonic waves (used by bats and dolphins for echolocation, and by humans in SONAR and medical ultrasonography).
How is the equation v = λν (speed = wavelength × frequency) used in numerical problems on sound?
This equation connects a sound wave’s speed (v), wavelength (λ) and frequency (ν), and rearranged forms of it — ν = v/λ and λ = v/ν — let you solve for whichever quantity is unknown once the other two are given. Since time period T = 1/ν, the same relation can also be written as v = λ/T. This single relationship underlies almost every numerical in this chapter, from finding a wave’s time period to calculating ocean depth using SONAR.
How does SONAR calculate the depth of the ocean or the distance to an underwater object?
SONAR (Sound Navigation and Ranging) sends an ultrasonic pulse toward the ocean floor or an object, and a receiver measures the total time taken for the echo to return. Because that time includes the trip down and the trip back, the depth (or one-way distance) is calculated as distance = (speed of sound in water × total time) ÷ 2 — dividing by 2 accounts for the fact that the sound has covered the distance twice.
Recommended: Buy the Printed NCERT Class 9 Science Book
Contains Amazon affiliate links.
If you’d like a printed copy alongside the PDF, here’s a verified option:
3.8 out of 5 stars (931 ratings) · Rs. 187
Price and availability may change on Amazon. As an Amazon Associate, ncertbooks.org earns from qualifying purchases.
