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

Question 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
Answer: (ii) Sound needs a medium to propagate.
In simple words: A mechanical wave requires a material medium to move. Since sound cannot travel through empty space and must have air, water, or another substance to move through, this fact proves sound is mechanical.

Exam Tip: Remember that all mechanical waves need a medium - this is their defining feature. The vacuum bell jar experiment is the classic proof that sound cannot travel without one.

Question 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
Answer: (iii) number of compressions per second.
In simple words: Frequency counts how many vibrations or compressions happen each second. When frequency goes up, more compressions pass a point in each second.

Exam Tip: Use the formula v = fλ to remember why wavelength and speed don't change - speed depends only on the medium, not on frequency.

Question 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
Answer: (ii) 5 Hz
In simple words: Frequency means how many compressions go past one spot in one second. If 20 compressions take 4 seconds, then in 1 second there are 20 divided by 4, which is 5 compressions per second - that is 5 Hz.

Exam Tip: Always divide the total number of compressions by the total time to find frequency. The units will be Hz when time is in seconds.

Question 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.
Answer: The sound will create reverberation, not an echo. For an echo to be heard as a separate sound, the reflected sound must arrive at least 0.1 second after the original sound is made. Since 0.05 s is less than 0.1 s, the reflected sound mixes with the original sound and cannot be heard as a distinct echo. When reflected sounds blend together like this, the effect is called reverberation.
In simple words: If the bounce-back comes back too fast - less than 0.1 seconds - your ear hears one long sound instead of two separate sounds. That blended sound is reverberation.

Exam Tip: The 0.1 second rule is critical - memorise it. Any time gap shorter than this produces reverberation; 0.1 seconds or longer produces a distinct echo.

Question 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?
Answer:
(i) Greater wavelength: Wave (a) has the greater wavelength. Since graph (a) shows fewer complete cycles over the same horizontal distance, each cycle takes up more space. This means the distance between one crest and the next crest (or one trough and the next trough) is larger in wave (a).

(ii) Smaller amplitude: Wave (a) has smaller amplitude. The height of the peaks and the depth of the valleys from the centre line are both less for wave (a) compared to wave (b).
In simple words: Wave (a) has longer, stretched-out waves and shorter, quieter bumps. Wave (b) has short, squeezed waves and taller, louder bumps.

Exam Tip: Count the number of complete cycles (crests or troughs) visible in each graph over the same distance - fewer cycles means longer wavelength. Measure the vertical distance from the centre line to a peak for amplitude.

Question 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.
Answer: Frequency depends on how many wave cycles fit into the same horizontal distance. The curve that shows the most oscillations in that space has the highest frequency; the curve with the fewest oscillations has the lowest frequency.

From the figure:
- The green curve displays the most oscillations - this means maximum frequency, so A = Green curve
- The blue curve shows the fewest oscillations - this means minimum frequency, so C = Blue curve
- The red curve is in between - so B = Red curve
In simple words: Count how many complete up-and-down wiggles each curve makes in the same length. Green wiggles most (highest frequency), blue wiggles least (lowest frequency), red is in the middle.

Exam Tip: Frequency is not about how tall or short the wave is - it is about how many cycles are packed into a given space. Taller and shorter only affect amplitude.

Question 7. Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
Answer: To draw this graph, set up a coordinate system with distance on the horizontal axis and density on the vertical axis. Mark a centre line to show the average density. The highest point (crest) should be 3 units above this line, and the lowest point (trough) should be 3 units below. The distance from one crest to the next crest (or one trough to the next trough) must be exactly 4 cm. A complete wave cycle showing these measurements would have its crest at +3 units, its trough at -3 units, and span 4 cm horizontally from crest to crest.
In simple words: Draw a wavy line that bounces up 3 units and down 3 units from the middle, and make sure each complete wave (crest to crest) takes up 4 cm of space.

Exam Tip: Mark the amplitude clearly with vertical arrows showing the 3-unit distance. Mark the wavelength with a horizontal line or arrow showing the 4 cm distance between matching points on successive cycles.

Question 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?
Answer: Two major errors appear in this scene. First, sound cannot travel through space because space is almost completely empty - a near vacuum - and sound is a mechanical wave that always requires a material medium such as air, water, or solid material to move through. No medium means no sound wave can form or spread. Second, even if sound could somehow travel in space, light travels vastly faster than sound. Light moves at about 300,000 km per second, while sound travels at roughly 340 m per second in air. This enormous speed difference means that if both light and sound were produced by the explosion, we would see the flash almost instantly but would hear the sound much later, or possibly not at all if sound could not travel anyway. In reality, only the bright flash would be visible in space - no sound should be heard.
In simple words: Sound needs air or some material to travel, and space has neither. Light is far faster than sound, so they could never reach us at the same time anyway.

Exam Tip: Remember the two conditions: sound requires a medium (ruled out in space), and light travels much faster than sound. Both must be mentioned for full marks.

Question 9. A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s⁻¹, find its time period.
Answer: Given: wavelength λ = 3.44 m and speed v = 344 m s⁻¹. Using the wave relation v = fλ, rearrange to find frequency: f = v / λ = 344 / 3.44 = 100 Hz. Now, since time period T is the reciprocal of frequency, T = 1 / f = 1 / 100 = 0.01 s. Therefore, the time period of the sound wave is 0.01 s.
In simple words: First find how many vibrations happen per second (frequency = 100). Then find how long one vibration takes (time period = 1/100 = 0.01 seconds).

Exam Tip: Always use v = fλ to find frequency first, then convert to time period using T = 1/f. Show both steps for full marks.

Question 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 m s⁻¹ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
Answer: The sonar pulse travels down to the wreck and then back up to the ship. The total time for this round trip is 5 seconds. The one-way travel time to reach the wreck is therefore 5 / 2 = 2.5 seconds. Using the relation distance = speed × time, the distance to the wreck = 1525 × 2.5 = 3812.5 m. The wreckage is located approximately 3812.5 m (or about 3.8 km) below the surface of the ocean.
In simple words: The sound goes down 2.5 seconds and comes back up 2.5 seconds. Multiply the speed by the one-way time to find how deep the wreck is.

Exam Tip: Always divide the total echo time by 2 - the signal goes down and comes back, so you only want half the total time for the one-way distance.

Question 11. A vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits 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 ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be 345 m s⁻¹.
Answer: The ultrasonic pulse travels from the vehicle to the obstacle and back. The distance to the obstacle is 1.2 m, so the total distance travelled is 2 × 1.2 = 2.4 m. Using time = distance / speed, the time taken = 2.4 / 345 = 0.00696 s, which rounds to approximately 0.007 s or 7 milliseconds. This is the time for the ultrasonic wave to go to the obstacle and return to the sensor.
In simple words: The sound goes out 1.2 metres and comes back 1.2 metres, so it travels a total of 2.4 metres. Divide by the speed to find how long this takes.

Exam Tip: Never forget to multiply the one-way distance by 2 - the sound must travel out and back. Always show your calculation clearly.

Question 12. The speed of sound in air is about 331 m s⁻¹ at 0°C and nearly 344 m s⁻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.
Answer: At 22°C, the speed is 344 m s⁻¹, so the time taken = 1720 / 344 = 5 s. At 0°C, the speed is 331 m s⁻¹, so the time taken = 1720 / 331 ≈ 5.20 s. The difference in time = 5.20 - 5.00 = 0.20 s. Therefore, when the temperature drops from 22°C to 0°C, sound takes about 0.2 seconds extra to travel the same 1720 m distance.
In simple words: Sound moves slower in cold air, so it takes longer to travel the same distance. The colder the air, the longer the journey takes.

Exam Tip: Calculate the time at each temperature separately, then subtract to find the difference. Pay careful attention to the direction of the temperature change.

Question 13. The variation of density of medium for a sound wave propagating with a speed of 340 m s⁻¹ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.
Answer: From the figure, the marked distance of 8 cm represents two complete wavelengths. Therefore, 2λ = 8 cm, which gives λ = 4 cm. Converting to metres: λ = 0.04 m. Using the wave equation v = fλ, rearrange to find frequency: f = v / λ = 340 / 0.04 = 8500 Hz. The wavelength of the sound wave is 4 cm (or 0.04 m) and the frequency is 8500 Hz.
In simple words: The distance shown contains two complete waves, so one wave is half that distance. Use the speed and wavelength to find how many waves pass per second.

Exam Tip: Always count complete cycles in the figure carefully. Convert cm to metres before using the wave equation. Show both wavelength and frequency in your final answer.

Question 14. The graphical representation of two sound waves A and B propagating at the same speed of 345 m s⁻¹ is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.
Answer: Wave A completes more cycles across the same distance, indicating a shorter wavelength. Wave B completes fewer cycles across the same distance, indicating a longer wavelength. From the graph: wavelength of wave A ≈ 2.5 cm and wavelength of wave B ≈ 5.0 cm. Converting to metres: λ₁ = 0.025 m and λ₂ = 0.05 m. Since both waves travel at the same speed of 345 m s⁻¹, using f = v / λ: for wave A, f₁ = 345 / 0.025 = 13,800 Hz; for wave B, f₂ = 345 / 0.05 = 6,900 Hz. Wave A has wavelength 2.5 cm and frequency 13,800 Hz. Wave B has wavelength 5.0 cm and frequency 6,900 Hz.
In simple words: Wave A has shorter, tightly-packed waves and vibrates faster. Wave B has longer, stretched-out waves and vibrates slower. But they travel at the same speed.

Exam Tip: Remember that speed is constant for both waves (same medium), so a shorter wavelength must mean higher frequency. Count the cycles carefully on the graph.

Question 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 than that of B, what is the ratio between the speeds of sound in air and water?
Answer: Let the distance from each source to the cliff be d. Since the sound travels to the cliff and returns, the total distance in each case is 2d. For source A in air: time taken = 2d / v_air. For source B in water: time taken = 2d / v_water. According to the problem: 2d / v_air = 4.5 × (2d / v_water). Cancelling 2d from both sides: 1 / v_air = 4.5 / v_water. Rearranging: v_water = 4.5 × v_air. This means v_air : v_water = 1 : 4.5. To express in whole numbers: v_air : v_water = 2 : 9. Since sound takes longer in air to travel the same distance, it must be moving slower in air than in water. The ratio of the speed of sound in air to water is 2 : 9.
In simple words: Sound takes more than four times as long in air as in water, so sound travels much faster in water. The speed ratio is 2 to 9.

Exam Tip: The time ratio is the inverse of the speed ratio - if A takes 4.5 times longer than B, then A is 1/4.5 times as fast, or equivalently, B is 4.5 times faster. Use this inverse relationship to convert between time ratios and speed ratios.

Class 9 Science Exploration Chapter 10 Important Formulae

Class 9 Science Exploration Chapter 10 Extra Question Answers

Very Short Answer Type Questions

Question 1. What is the source of sound?
Answer: Any object that moves back and forth creates sound. When something vibrates, it disturbs the air (or another material) around it and produces sound waves that we can hear.
In simple words: Sound comes from things that shake or move. Without movement, there is no sound.

Exam Tip: Remember that vibration and sound are connected - all sound begins with a vibrating object.

Question 2. What is vibration?
Answer: Vibration is the repeating back-and-forth motion (called oscillation) that an object makes around its rest position. This motion happens over and over in a regular pattern.
In simple words: Vibration is when something shakes back and forth really fast, always returning to where it started before shaking again.

Exam Tip: The key word is "periodic" - the motion repeats at regular time intervals. This is how vibrations differ from random movements.

Question 3. What is a compression in a sound wave?
Answer: A compression is an area in the medium where particles are packed closer together than they normally are. The particle density in this region is higher than average. Compressions form when the vibrating source moves in one direction, pushing particles forward.
In simple words: A compression is a crowded zone where many particles bunch up together in a small space.

Exam Tip: Compressions and rarefactions alternate as the sound wave travels. Knowing where each occurs helps you understand how sound moves.

Question 4. What is rarefaction in a sound wave?
Answer: A rarefaction is an area in the medium where particles are spread further apart than they normally are. The particle density in this region is lower than average. Rarefactions form when the vibrating source moves in the opposite direction, pulling particles back.
In simple words: A rarefaction is a thin zone where particles are spread out and farther apart.

Exam Tip: Think of compressions as "push" and rarefactions as "pull" - the back-and-forth vibration creates both.

Question 5. State the SI unit of frequency.
Answer: The SI unit of frequency is hertz (Hz). One hertz equals one oscillation per second, or s⁻¹. It measures how many vibrations or cycles occur in a single second.
In simple words: Hertz is the count of how many times something wiggles back and forth in one second.

Exam Tip: Always use Hz for frequency when answering exam questions - it is the standard unit. An older unit called "cycles per second" means the same thing.

Question 6. Write the relation between frequency (ν) and time period (T).
Answer: Frequency and time period are inversely related through the equation: \( \nu = \frac{1}{T} \). This means that as frequency increases, time period decreases, and vice versa. If something vibrates very fast (high frequency), each vibration takes very little time (short period).
In simple words: Fast vibrations (high frequency) take less time each. Slow vibrations (low frequency) take more time each.

Exam Tip: This inverse relationship is fundamental - use it to convert between frequency and time period in any problem.

Question 7. What is the human audible range of sound?
Answer: The human ear can detect sounds between 20 Hz and 20,000 Hz (also written as 20 kHz). This range varies somewhat depending on the person's age and individual hearing ability. Younger people can usually hear higher frequencies, while this upper limit decreases with age.
In simple words: Humans can hear sounds that vibrate between 20 and 20,000 times per second. Sounds faster or slower than this are not heard by our ears.

Exam Tip: Memorise the range 20 Hz to 20 kHz - it is important for understanding ultrasonic and infrasonic waves.

Question 8. Why can astronauts not hear each other directly during a spacewalk?
Answer: Space is almost completely empty - it contains almost no air or any other material substance. Sound is a mechanical wave that must travel through a physical medium such as air, water, or solid material. Since no medium exists in space, sound cannot form or spread there, even though the astronauts are close to each other.
In simple words: Sound needs air or something solid to travel through. Space is mostly nothing, so sound cannot move there at all.

Exam Tip: This is a key fact proving that sound is a mechanical wave - without a medium, sound cannot exist or travel.

Question 9. Define the wavelength of a sound wave.
Answer: Wavelength is the distance between two adjacent crests (peaks) or between two adjacent troughs (valleys) of a wave. It also equals the distance over which one complete cycle of the wave pattern occurs. The standard unit is the metre (m).
In simple words: Wavelength is how far apart two matching points on the wave are - for example, from one bump up to the next bump up.

Exam Tip: You can measure wavelength between any two matching parts - crest to crest, trough to trough, or compression to compression.

Question 10. What is the speed of sound in dry air at 0°C and 22°C?
Answer: The speed of sound in dry air is approximately 331 m s⁻¹ at 0°C (freezing point) and about 344 m s⁻¹ at 22°C (room temperature). Sound moves faster in warmer air because the particles have more energy and transmit vibrations more quickly.
In simple words: Sound travels slower in cold air and faster in warm air. The warmer the air, the quicker the sound moves.

Exam Tip: Know these two reference speeds - 331 at 0°C and 344 at 22°C - they appear frequently in exam problems.

Question 11. What are ultrasonic waves? Give one example of an animal that detects them.
Answer: Ultrasonic waves are sound waves with frequencies above 20,000 Hz (20 kHz). These frequencies are too high for the human ear to detect, making them inaudible to us. Many animals can hear ultrasonic waves. Bats produce and detect ultrasonic waves to navigate and locate prey in the dark through echolocation.
In simple words: Ultrasonic waves are very high-pitched sounds that humans cannot hear but many animals can. Bats use them to find food at night.

Exam Tip: Be ready to explain echolocation - it is the practical application of ultrasonic waves that appears in many exam questions.

Question 12. What is echolocation?
Answer: Echolocation is a biological ability by which animals send out sound waves and then listen to the echoes that bounce back from objects. By analysing these echoes, the animal determines the location, distance, and sometimes the shape of objects around it. Bats, dolphins, and whales all use this method to navigate and hunt in darkness or murky water.
In simple words: An animal makes a sound, listens to how it bounces back, and uses that information to find where things are.

Exam Tip: Echolocation is the animal version of SONAR technology used by ships - the principle is identical.

Question 13. What is the minimum distance from a reflecting wall to hear an echo?
Answer: The minimum distance is 17 m from the reflecting wall. Sound must travel from the source to the wall and back to the listener, covering a total of 34 m. At a speed of approximately 340 m s⁻¹, this round-trip journey takes at least 0.1 second. For the ear to hear the reflected sound as a separate echo rather than a blur mixed with the original sound, this 0.1 second delay is essential.
In simple words: If you are less than 17 metres away from a wall, the sound bounces back so fast it mixes with the original sound and you hear a blur, not a clear echo.

Exam Tip: The 0.1 second is the key threshold - memorise it. At exactly 17 m distance, the echo returns in exactly 0.1 s.

Question 14. Name the instrument commonly used to produce nearly single-frequency sound in experiments.
Answer: A tuning fork is the standard instrument. It is a U-shaped metal bar, usually made of steel or aluminium, with two parallel prongs called tines. When struck against a rubber pad or similar object, the prongs vibrate and produce a nearly pure sound of a single, fixed frequency. Different tuning forks produce different frequencies depending on their size and thickness.
In simple words: A tuning fork is a metal fork that rings with one clear, pure note when you hit it. Scientists use it because it makes a sound of exactly one frequency.

Exam Tip: Be able to describe the tuning fork and explain how it creates a pure, single frequency - this is important for understanding sound in experiments.

Question 15. What is reverberation?
Answer: Reverberation is the persistence (lingering) of sound in a large enclosed space due to multiple reflections from the walls, ceiling, and floor. When reflected sound waves arrive back at the listener within 0.05 seconds of each other, they blend together and are heard as one drawn-out sound rather than separate echoes. This creates the characteristic echo-like prolonging of sound in concert halls, auditoriums, and large rooms.
In simple words: Reverberation is when a sound bounces around a big room so many times so quickly that the bounces blend together, making the sound seem to linger.

Exam Tip: Remember the 0.05 second time limit - reflections closer together than this create reverberation, not distinct echoes.

Short Answer Type Questions

Question 1. Explain with an example how sound is produced.
Answer: Sound is created by vibrating objects. When a tuning fork is struck on a rubber pad, the two prongs move rapidly back and forth. As they move forward, they push the surrounding air particles together, creating a high-density region called a compression. As they move backward, they leave a gap, causing air particles to spread out into a low-density region called a rarefaction. This alternating pattern of compressions and rarefactions travels outward through the air as a sound wave. When these waves reach our ears, we perceive them as sound.
In simple words: The tuning fork shakes back and forth very fast. This shaking pushes and pulls the air around it, creating waves that travel to your ears and become sound.

Exam Tip: Always include the role of vibration in sound production and mention the formation of compressions and rarefactions for a complete answer.

Question 2. Why does sound need a medium to propagate? Give experimental evidence.
Answer: Sound is a mechanical wave that travels by transmitting vibrational energy through the particles of a medium. It cannot travel through empty space because there are no particles to pass the vibrations along. This is proven by the classic vacuum bell jar experiment: a ringing bell is placed inside a glass jar, and the air is gradually pumped out to create a vacuum. As the air pressure inside drops, the sound of the ringing bell becomes progressively fainter and eventually becomes completely inaudible. Yet the bell continues to ring silently inside the vacuum. This demonstrates conclusively that sound cannot propagate without a medium.
In simple words: In the vacuum experiment, as you remove the air, the bell's sound fades until you cannot hear it at all, even though the bell is still ringing. This proves sound needs air to travel.

Exam Tip: Be ready to describe the vacuum bell jar experiment in detail - it is the key experimental evidence that sound requires a medium.

Question 3. How are compressions and rarefactions formed by an oscillating piston in a tube?
Answer: Picture a tube filled with air with a piston at one end that moves back and forth. When the piston pushes inward (forward), it compresses the air particles in front of it, squeezing them into a smaller space. This crowding of particles creates a compression - a zone of higher-than-normal density. As the piston then pulls outward (backward), it creates empty space that the air particles spread out to fill. This spreading of particles into a larger volume creates a rarefaction - a zone of lower-than-normal density. As the piston continues its oscillating motion, it repeatedly generates alternating compressions and rarefactions that propagate down the tube as a sound wave.
In simple words: The piston pushes forward to squeeze air (compression) then pulls back to spread air out (rarefaction). This push-pull motion repeating over and over creates the sound wave.

Exam Tip: Use the terms "compression" and "rarefaction" correctly in your explanation - compression is crowded, rarefaction is spread out.

Question 4. Differentiate between longitudinal and transverse waves with examples.
Answer: In longitudinal waves, the particles of the medium oscillate parallel to (in the same direction as) the wave's direction of travel. The wave consists of alternating compressions and rarefactions. Sound waves are longitudinal - air particles vibrate back and forth in the same direction the sound travels. In transverse waves, particles oscillate perpendicular to (at right angles to) the wave's direction of travel. The wave appears as a series of crests and troughs. Light waves are transverse - the electric and magnetic fields oscillate perpendicular to the direction light travels. Sound is a longitudinal mechanical wave because it requires a medium and travels via particle oscillations along its path. Light is a transverse non-mechanical (or electromagnetic) wave that does not require a medium.
In simple words: Longitudinal waves shake forward and backward like a spring being pushed and pulled. Transverse waves shake up and down like a rope flicked at one end. Sound is longitudinal; light is transverse.

Exam Tip: Remember the key difference: longitudinal particles move along the wave direction; transverse particles move across the wave direction.

Question 5. Why does the intensity of sound decrease with distance from the source?
Answer: Sound energy spreads outward from its source in all directions, forming expanding spheres. The total energy remains constant (energy is conserved), but as the distance increases, this fixed amount of energy is distributed over an increasingly larger surface area. Since intensity is the amount of energy passing through a unit area per unit time, the same energy spread over a larger area results in lower intensity per unit area. This is why sound becomes quieter as you move farther from the source - the energy density decreases, even though the total energy is conserved.
In simple words: Imagine spreading a fixed amount of paint over a larger and larger area - the same paint makes a thinner layer on the bigger area, just like sound energy makes a weaker effect over larger distances.

Exam Tip: This concept relates to the inverse square law in physics - intensity drops as the square of the distance increases.

Question 6. What is the difference between loudness and intensity of sound?
Answer: Intensity is a measurable physical quantity - the amount of sound energy passing through a unit area per unit time, measured in watts per square metre (W m⁻²). It depends solely on the properties of the wave itself (such as amplitude and frequency) and the medium. Loudness, by contrast, is how loud the sound seems to a person listening to it - it is a subjective human perception that depends on intensity but also on many other factors including the listener's age, hearing ability, and psychological state. Two people hearing the same sound (same intensity) may perceive different loudness. Additionally, the human ear is not equally sensitive to all frequencies, so identical intensities at different frequencies may seem like different loudness levels.
In simple words: Intensity is a fact you can measure with an instrument. Loudness is how loud it sounds to your ear, which depends on your hearing and what you are paying attention to.

Exam Tip: Intensity is physical and measurable; loudness is biological and subjective - this distinction is crucial for full marks.

Question 7. Explain how SONAR works to locate underwater objects.
Answer: SONAR stands for Sound Navigation and Ranging. The SONAR equipment on a ship sends out a burst of ultrasonic sound waves (frequency above 20 kHz) downward into the water. When these waves strike an underwater object such as a sunken ship or the ocean floor, they bounce back as an echo. The equipment detects when this echo returns and measures the time interval between transmission and echo reception. Using the formula distance = (speed × time) / 2, and knowing the speed of sound in seawater (approximately 1500 m s⁻¹), the distance to the object is calculated. Dividing by 2 accounts for the round-trip journey. The greater the time delay, the farther away the object is.
In simple words: SONAR sends a sound down to the bottom and listens for the bounce-back. The longer you wait for the echo, the deeper the object is.

Exam Tip: Always remember to divide the distance by 2 - the sound goes down and comes back, so the object is only halfway as far as the sound actually travelled.

Question 8. What is pitch and how is it related to frequency?
Answer: Pitch is the subjective sensation of how "high" or "low" a sound seems to a listener - it is a property of human perception. Frequency is the objective physical property of a wave - the number of vibrations per second. In general, pitch increases with frequency: sounds with higher frequencies (like a whistle or a small bell) are perceived as having higher pitch, while sounds with lower frequencies (like thunder or a big drum) are perceived as having lower pitch. However, the relationship between pitch and frequency is not perfectly linear across all ranges - the human ear's sensitivity to frequency changes varies in a complex way, making the pitch-frequency relationship somewhat non-linear, especially at very high or very low frequencies.
In simple words: Frequency is the real vibration rate you can measure. Pitch is what your ear hears. Usually higher frequency sounds like higher pitch, but not always in exactly the same way.

Exam Tip: Pitch is subjective and psychological; frequency is objective and physical. Higher frequency generally means higher pitch, but they are not identical concepts.

Question 9. Why does sound travel faster in solids than in liquids or gases?
Answer: Sound travels by the vibrations of particles passing energy to their neighbours. In solids, the particles are tightly packed together with strong intermolecular forces holding them in place. This tight packing and strong bonding allows vibrational energy to be transmitted very efficiently and rapidly from particle to particle. In liquids, particles have more freedom of movement and weaker intermolecular bonds, allowing slower energy transmission. In gases like air, particles are far apart and interact weakly, resulting in the slowest sound transmission. Measured speeds illustrate this progression: sound travels at approximately 5000 m s⁻¹ in steel (solid), about 1500 m s⁻¹ in water (liquid), and roughly 340 m s⁻¹ in air (gas). Sound travels about 4-5 times faster in water than in air, and about 15-20 times faster in solids than in air.
In simple words: In solids, particles are packed tightly and bonded strongly, so vibrations pass along very quickly. In gases, particles are far apart and bonded weakly, so vibrations travel slowly.

Exam Tip: Remember the approximate speeds in each medium - this helps you understand why solids are best for transmitting sound.

Question 10. What do we mean when we say sound is a longitudinal wave?
Answer: A longitudinal wave is one in which the particles of the medium vibrate parallel to the direction the wave itself is travelling. In sound waves, as the wave moves forward, the individual air (or water, or solid) particles also move back and forth in the forward-backward direction - the same direction as the wave's motion. The particles do not move sideways across the wave; instead, they oscillate along the path that the wave is taking. This is why sound is a longitudinal mechanical wave - it relies on particles moving along its direction of propagation, alternately compressing (pushing forward) and rarefacting (moving back) to carry the energy forward.
In simple words: Sound travels forward, and the air shakes forward and backward in the same direction the sound is going. That forward-and-backward shaking is what makes sound a longitudinal wave.

Exam Tip: Contrast this with transverse waves where particles shake perpendicular to the wave direction - sound never does this.

Long Answer Type Questions

Question 1. Describe the propagation of sound through air using the piston-in-tube model. Explain the role of compressions and rarefactions and state what actually travels in a sound wave.
Answer: Sound propagation through air can be effectively modelled using a long tube filled with air, with an oscillating piston at one end and an open end on the other side. Initially when the piston is not moving, the air inside has a uniform average density throughout. When the piston pushes forward (inward), it forces the nearby air particles forward into a smaller volume, compressing them together. This creates a region of higher-than-average density called a compression. The compressed particles collide with their neighbours ahead, transferring the compression forward. Those neighbouring particles then collide with particles further ahead, passing the compression onward through the tube. Crucially, the individual air particles themselves do not travel along with the compression - each particle only oscillates about its own fixed mean position, while the disturbance (compression) moves forward through space. When the piston pulls backward (outward), it leaves a partial vacuum that the air particles expand to fill, creating a region of lower-than-average density called a rarefaction. This rarefaction also propagates forward by the same particle-collision mechanism. As the piston continues to oscillate back and forth repeatedly, it generates a continuous stream of alternating compressions and rarefactions that travel away from the source as a sound wave. The key principle is that the particles of the medium do NOT travel with the wave - each particle simply oscillates back and forth parallel to the direction the wave is travelling. What actually travels is the disturbance in density (the pattern of compressions and rarefactions) and the energy carried within it. When the medium is not confined in a tube but is open air, these compressions and rarefactions spread outward in all directions from the source as spherical (three-dimensional) waves, eventually reaching a listener's ear where they are perceived as sound.
In simple words: The piston pushes and pulls, creating squeezes (compressions) and spreads (rarefactions) that ripple outward. The air particles just wiggle back and forth in place - they do not travel with the wave. What travels is the pattern of squeezes and spreads, carrying energy to your ear.

Exam Tip: Emphasise that particles vibrate in place and do not travel - only the disturbance travels. This distinction is critical for understanding why sound is a mechanical wave.

Question 2. Explain the characteristics of a sound wave - wavelength, frequency, time period, amplitude, and speed. Derive the relation connecting speed, wavelength and frequency. Give one worked numerical example.
Answer: Sound waves are fully described by five interconnected characteristics:

1. Wavelength (λ): The distance between two successive crests (compressions) or two successive troughs (rarefactions) is the wavelength. It represents the spatial extent of one complete cycle of density oscillation. SI unit: metre (m).

2. Frequency (ν): The number of complete density oscillations occurring at any fixed point in the medium in one second is the frequency. High-frequency sounds are perceived as high-pitched, while low-frequency sounds seem low-pitched. SI unit: hertz (Hz) or s⁻¹.

3. Time Period (T): The duration required for one complete density oscillation at a fixed point is the time period. Frequency and time period are inversely proportional to each other, given by the relation: \( \nu = \frac{1}{T} \)

4. Amplitude (A): The maximum change in density (or maximum displacement) of the medium from its average (equilibrium) value during a compression or rarefaction is the amplitude. Greater amplitude corresponds to more energy and louder sound. Amplitude does not affect the speed or frequency of sound in a given medium.

5. Speed (v): The speed of sound is the distance travelled by any point on the wave (such as a crest or compression) per unit time. Speed depends on the properties of the medium: sound travels fastest in solids (approximately 5000 m s⁻¹ in steel), at intermediate speeds in liquids (approximately 1500 m s⁻¹ in water), and slowest in gases (approximately 340 m s⁻¹ in air at room temperature).

Derivation of v = λν: Consider one complete time period (T). During this time, the wave travels a distance equal to exactly one wavelength (λ). Using the basic definition speed = distance / time: \( v = \frac{\lambda}{T} \) Since frequency ν = 1/T, we can substitute: \( v = \lambda \times \nu \) This is the fundamental wave equation relating speed, wavelength, and frequency.

Numerical Example: A sound wave in air has frequency 440 Hz and travels at a speed of 344 m s⁻¹. Find its wavelength. Using the equation: \( \lambda = \frac{v}{\nu} = \frac{344}{440} = 0.782 \text{ m} \) Therefore, the wavelength is approximately 0.78 metres or 78 centimetres.
In simple words: Frequency tells you how many times per second the wave wiggles. Wavelength tells you how far apart the wiggles are. Speed tells you how fast the wiggles travel. When you multiply wavelength times frequency, you get the speed - they are all connected.

Exam Tip: Be able to define all five characteristics and show the derivation of v = λν clearly. Use the example formula to solve similar problems - this relationship appears in almost every sound-wave problem.

Comparison: Echo vs. Reverberation

Comparison: Infrasonic vs. Ultrasonic Waves