Answer: (ii) Sound needs a medium to propagate.

Answer: (iii) number of compressions per second.

Answer: (ii) 5 Hz.

Answer: It will produce reverberation, not an echo.

Answer:
(i) Greater wavelength: Wave (a) has greater wavelength.

(ii) Smaller amplitude: 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 and 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 = 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. ⇒ 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.

Answer: Any vibrating object is the source of sound. The object that produces sound by its vibrations is called the source.

Answer: Vibration is the periodic to-and-fro motion (oscillation) of an object about its mean position of rest.

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.

Answer: Rarefaction is a region where the density of the medium is lower than the average density, formed when the source moves backward.

Answer: The SI unit of frequency is hertz (Hz), equivalent to one oscillation per second (s⁻¹).

Answer: ν = 1/T. Frequency and time period are inversely related; a higher frequency corresponds to a shorter time period.

Answer: The human audible range is from 20 Hz to 20,000 Hz (20 kHz). This range varies with age and person.

Answer: Outer space is a near vacuum — there is no medium. Since sound requires a material medium to propagate, it cannot travel in space.

Answer: Wavelength (λ) is the distance between two consecutive crests or two consecutive troughs of a wave. Its SI unit is metre (m).

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.

Answer: Ultrasonic waves have frequency above 20 kHz, inaudible to humans. Bats detect and use ultrasonic waves for echolocation.

Answer: Echolocation is the ability to locate objects by emitting sound waves and detecting the reflected echoes. Bats, dolphins, and whales use it.

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.

Answer: A tuning fork — a U-shaped metal bar (steel or aluminium) with prongs (tines) — produces nearly single-frequency sound when struck.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.
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.

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.