Answer:

1. No, astronauts on a spacewalk cannot talk to each other or hear sounds like metal clanking as they would on Earth. This is because there is a near-vacuum in outer space, and sound cannot propagate without a medium. Sound needs a material medium (like air, water, or solid) to travel.
2. Most bats use echolocation to locate their prey in the dark. They emit short bursts of ultrasonic waves that reflect off objects and prey. By sensing the returning echoes, the bat can determine the position, distance, and even the size and shape of its prey and obstacles. This ability to locate objects using reflected sound waves is called echolocation.

Answer:

Sound can be produced by various vibrating objects, including:

• Striking or hitting objects: e.g., tabla, drum, gong, doorbell, cymbals. The vibrating part is the surface that is struck.
• Plucking or strumming: e.g., guitar, sitar, veena, violin. The vibrating part is the string.
• Blowing air: e.g., flute, shehnai, trumpet. The vibrating part is the air column inside the instrument.
• Shaking: e.g., tambourine, maracas. The vibrating part is the surface or the beads inside.
• Rubbing: e.g., a violin bow on strings. The vibrating part is the string.

Other examples include human vocal cords (vibrating flaps in the larynx) and a tuning fork (vibrating prongs). The object that produces sound is called its 'source'.

Answer:

The correct statement is (ii) Both A and R are true, and R is the correct explanation of A.

In the vacuum bell jar experiment, as air is removed, the sound of the bell becomes faint and eventually cannot be heard, even though the bell is still ringing. This is because the air (medium) that carries the sound waves to our ears is removed. Sound cannot travel through a vacuum. Therefore, the reason (R) correctly explains the assertion (A).

Answer:

The correct statement is (iii) A is true, but R is false.

Compressions and rarefactions are disturbances that move (propagate) through a medium, carrying energy. However, the individual particles of the medium do not move forward with the wave. They only oscillate or vibrate about their mean positions. The disturbance (compression and rarefaction) is passed on through collisions from one particle to the next, but the particles themselves do not travel with the wave. Therefore, the assertion is true, but the reason is false.

Answer:

The correct answer is (ii) Energy carried by sound waves.

When a tuning fork vibrates, it transfers its energy to the particles of the surrounding medium (air). These particles vibrate and pass the energy to their neighbours in the form of a sound wave (compressions and rarefactions). The air particles themselves do not travel from the tuning fork to your ear; they only oscillate about their mean positions. It is the energy carried by the wave that travels and reaches your ear, causing your eardrum to vibrate and enabling you to hear the sound.

Answer:

The task in this question requires labelling diagrams, which cannot be fully represented in text. However, the solution can be described as follows:

For Fig. 10.17 (a) and (b), the regions where the dots are closer together represent compressions (C), and the regions where they are farther apart represent rarefactions (R).

For the graph in Fig. 10.17 (c) and (d), the X-axis should be labelled 'Distance' and the Y-axis should be labelled 'Density'. The curves will represent the variation of density with distance. The crests (high points) on the graph correspond to compressions (C), and the troughs (low points) correspond to rarefactions (R). The curve for Fig. 10.17 (c) will have a larger amplitude (greater density changes) compared to the curve for Fig. 10.17 (d).

C = compressions; density increases   |   R = Rarefactions; density decreases

Answer:

Yes, a thin rubber band generally vibrates faster than a thick rubber band.

A faster vibration means a higher frequency. Therefore, the sound produced by the thin rubber band has a higher frequency than the sound produced by the thick rubber band. Since frequency and time period are inversely related (T = 1/ν), the sound produced by the thin rubber band has a shorter time period compared to the sound produced by the thick rubber band.

In simpler terms, the thin rubber band produces a sound with a higher pitch (shrillness) due to its higher frequency, while the thick rubber band produces a sound with a lower pitch.

Answer:

Frequency is defined as the number of oscillations per second. A frequency of 20 Hz means the piston completes 20 oscillations in 1 second.

To find the number of oscillations in 1 minute (60 seconds), we multiply the frequency by the time:

Oscillations per minute = Frequency × Time (in seconds) = 20 Hz × 60 s = 1200 oscillations.

Therefore, the piston completes 1200 oscillations per minute.

Answer:

The solution depends on the graphical representation of the sound wave (Fig. 10.19). Wavelength (λ) is the distance between two consecutive crests (or troughs). By measuring the distance between two consecutive crests on the given graph, you can find the wavelength. Half of the wavelength is λ/2.

Answer:

From Table 10.1, the speed of sound in air is 340 m/s, in water is 1500 m/s, and in steel is 5000 m/s.

(i) Ratio of speed in water to speed in air = Speed in water / Speed in air = 1500 m/s / 340 m/s ≈ 4.4. Sound travels about 4.4 times faster in water than in air.

(ii) Ratio of speed in steel to speed in water = Speed in steel / Speed in water = 5000 m/s / 1500 m/s ≈ 3.3. Sound travels about 3.3 times faster in steel than in water.

Answer:

Given:

• Distance between the two friends, d = 340 m
• Speed of sound in air, v_air = 340 m/s (from Table 10.1)
• Speed of sound in steel, v_steel = 5000 m/s (from Table 10.1)

Step 1: Calculate the time taken for sound to travel through air.

Time_air = Distance / Speed_air = 340 m / 340 m/s = 1.0 s

Step 2: Calculate the time taken for sound to travel through steel.

Time_steel = Distance / Speed_steel = 340 m / 5000 m/s = 0.068 s

Step 3: Calculate the time difference.

Time difference = Time_air - Time_steel = 1.0 s - 0.068 s = 0.932 s

Step 4: Determine if the sounds can be distinguished.

Since the time difference (0.932 s) is greater than the minimum required time difference of 0.1 s, Gunjan would be able to hear the two sounds separately.

Answer:

Given:

• Time interval between the emission of sound and the arrival of echo, t = 0.2 s
• Speed of sound in air, v = 343 m/s

The sound must travel from the source to the reflecting surface and back to the listener, so the total distance travelled by the sound is twice the distance from the source to the reflecting surface (d).

Total distance = v × t = 343 m/s × 0.2 s = 68.6 m

Therefore, the minimum distance of the reflecting surface (d) = Total distance / 2 = 68.6 m / 2 = 34.3 m

The reflecting surface should be placed at a minimum distance of 34.3 m from the source.

Answer:

Given:

• Time for the sonar signal to go to the bottom and return, t = 4 s
• Speed of sound in seawater, v = 1500 m/s

The sonar signal travels down to the ocean floor and back up to the ship. So, the total distance travelled by the signal is twice the depth (d) of the ocean at that point.

Total distance = v × t = 1500 m/s × 4 s = 6000 m

Depth of the ocean, d = Total distance / 2 = 6000 m / 2 = 3000 m

The depth of the ocean at that location is 3000 m (or 3 km).

Answer:

In humans, sound is produced by the vibration of vocal cords, which are tightly stretched muscular flaps located inside the voice box or larynx in the throat. The tongue, lips, mouth, and nasal cavity help convert the basic sound into speech or music.

Some animals produce sound by striking or rubbing certain body parts. For example, grasshoppers and crickets rub their wings or legs to produce sound. Bats use their larynx to produce ultrasonic waves. Many other animals, like birds, use a syrinx (a vocal organ) to produce sound.

Answer:

These sounds are produced when air or gases are heated rapidly, causing them to expand very quickly in a short time. This rapid expansion creates a sudden disturbance in the air density, which travels outward as a loud sound pulse. When this disturbance reaches our ears, we perceive it as a sudden loud sound.

A similar but more complex disturbance is produced when a supersonic aircraft flies at a speed higher than the speed of sound, creating a loud sound known as a sonic boom.

Answer:

This is a practical activity. The solution involves:

1. Downloading and using the Phyphox app (or similar sound analysis apps).
2. Using the 'Audio Scope' or 'Audio Spectrum' feature to produce and observe sound waves in real-time.
3. Whistling carefully to produce a sound close to a single frequency, which will appear as a clean sine wave pattern on the graph.
4. Observing how different sounds produce different wave patterns and frequencies.

This demonstrates that a pure tone (like a whistle or tuning fork) is a sound of a single frequency, which can be graphically represented.

Answer:

Everyone's voice sounds unique because it is shaped by the throat, mouth, and nasal cavities in addition to the vocal cords. The frequency of the sound produced by the vocal cords is the fundamental pitch, but the overtones and the way the sound is shaped by the vocal tract give each voice its unique character or 'timbre'.

During adolescence, boys' vocal cords lengthen and thicken, vibrating less frequently, which 'deepens' their voice. Similarly, women and children typically have shorter and thinner vocal cords, which vibrate at higher frequencies, resulting in higher-pitched voices. The unique combination of these factors makes each person's voice distinct.

Answer:

This is an informative 'Threads of Curiosity' text explaining the concepts of tone, musical notes, fundamental frequency, overtones, timbre, and octave.

• Tone: A sound of a single frequency (e.g., tuning fork).
• Musical Note: A combination of a fundamental frequency (the lowest) and its overtones (higher frequencies). This combination creates a rich and pleasant sound.
• Timbre: The quality that makes different instruments sound unique even when playing the same note at the same loudness. It depends on the pattern and intensity of overtones, which is determined by the instrument's shape, material, and construction.
• Octave: The interval between two notes where one has double the frequency of the other (e.g., 200 Hz and 400 Hz).

Answer:

This is an informative 'Threads of Curiosity' text explaining the role of the 'syaahi' in Indian percussion instruments like the tabla and mridangam.

The syaahi is a black patch at the centre of the drum head. It alters the vibration of the membrane, allowing the instrument to produce a rich variety of sounds and a level of tonal control rarely found in other drums. This contributes to the unique and complex sound of these instruments.

Answer:

This is an interesting cultural fact mentioned in the chapter. It highlights how sound and music are deeply integrated into human societies in unique ways. The tradition of Kongthong village shows that sound can be used not just for communication, but also as a form of identity and cultural expression. Each person having a unique musical 'tune name' is a beautiful example of how sound can be personalized and meaningful beyond words.

Answer:

The correct answer is (ii) Sound needs a medium to propagate.

A mechanical wave is a wave that requires a material medium to propagate. Sound cannot travel through a vacuum, as demonstrated by the vacuum bell jar experiment. This is the defining characteristic of a mechanical wave. While reflection, frequency, and energy are also properties of sound, they are not unique to mechanical waves (e.g., light is a transverse wave that reflects and carries energy but does not need a medium).

Answer:

The correct answer is (iii) number of compressions per second.

Frequency is defined as the number of complete oscillations (or density variations) per unit time. An increase in frequency means more compressions (and rarefactions) pass a point in the medium per second. The speed of sound depends on the medium, not the frequency. Wavelength is inversely proportional to frequency (v = fλ), so increasing frequency decreases wavelength. Time period (T = 1/f) is inversely proportional to frequency, so it decreases when frequency increases.

Answer:

The correct answer is (ii) 5 Hz.

Frequency is the number of waves (or compressions) passing a point per second. In 4 seconds, 20 compressions pass.

Frequency (ν) = Number of compressions / Time taken = 20 / 4 s = 5 s^-1 = 5 Hz

Answer:

It will produce a reverberation.

An echo is heard when the reflected sound reaches the ear after a time delay of at least 0.1 s, allowing the brain to distinguish it from the original sound. If the reflected sound reaches the ear within less than 0.1 s, it blends with the original sound, causing a prolongation of the sound known as reverberation.

In this case, the reflected sound arrives after only 0.05 s, which is less than 0.1 s. Therefore, it will not produce a distinct echo; instead, it will cause reverberation.

Answer:

The solution depends on the graphical representation of the two sound waves (Fig. 10.30).

(i) Greater wavelength: The wave that has a larger distance between two consecutive crests (or troughs) has the greater wavelength. Wave (a) has the greater wavelength.

(ii) Smaller amplitude: The wave that has a smaller vertical distance from the mean position to the crest (or trough) has the smaller amplitude. Wave (a) has the smaller amplitude.

Answer:

Frequency corresponds to how closely packed the wave cycles are within a given distance. The wave with the shortest wavelength (most number of crests in a given distance) has the highest frequency, and the wave with the longest wavelength has the lowest frequency.

• Source A (Maximum frequency): The curve with the most wave cycles packed into the given distance.
• Source C (Minimum frequency): The curve with the fewest wave cycles across the distance.
• Source B: The curve with a wavelength between A and C.

Answer:

The graph should plot density on the Y-axis and distance on the X-axis. The density amplitude is 3 units, meaning the maximum change in density from the mean is 3 units. The wavelength is 4 cm, meaning the distance between two consecutive crests (or troughs) is 4 cm.

The graph will show a sine wave pattern oscillating between +3 and -3 units relative to the mean density, with a period of 4 cm along the X-axis. The crests represent compressions and the troughs represent rarefactions.

Answer:

There are two main errors in this movie depiction:

1. Sound cannot travel in space: Space is a near-vacuum, meaning there is no material medium (like air) for sound waves to propagate. Sound is a mechanical wave and requires a medium to travel. Therefore, the explosion of a spacecraft in space would not produce any sound that could be heard by anyone in the vicinity. The movie incorrectly shows a sound accompanying the explosion.
2. Sound travels slower than light: Even in a medium, sound travels much slower than light. If there were air to carry sound, the flash of light would be seen before the sound is heard (like during a thunderstorm on Earth). The movie shows both light and sound reaching the viewer at the same time, which is physically inaccurate.

The correct depiction would be to show only a flash of light with no accompanying sound, as the explosion would be silent in space.

Answer:

Given:

• Wavelength, λ = 3.44 m
• Speed of sound, v = 344 m/s

Step 1: Calculate the frequency.

Using the relation v = λ × ν, where ν is the frequency, we have:

ν = v / λ = 344 m/s / 3.44 m = 100 Hz

Step 2: Calculate the time period.

The time period (T) is the inverse of frequency: T = 1/ν

T = 1 / 100 Hz = 0.01 s

The time period of the sound wave is 0.01 s.

Answer:

Given:

• Time for the sonar signal to travel to the wreckage and back, t = 5 s
• Speed of sound in seawater, v = 1525 m/s

The sonar signal travels from the ship down to the wreckage and back. So, the total distance travelled by the signal is twice the depth (d) of the wreckage below the ship.

Total distance = v × t = 1525 m/s × 5 s = 7625 m

Depth of the wreckage, d = Total distance / 2 = 7625 m / 2 = 3812.5 m

The wreckage is located approximately 3812.5 m (or about 3.8 km) below the ship.

Answer:

Given:

• Distance from the obstacle, d = 1.2 m
• Speed of ultrasonic wave in air, v = 345 m/s

The ultrasonic wave travels from the sensor to the obstacle and back. The total distance travelled by the wave is twice the distance to the obstacle: 2d.

Total Distance = 2d = 2 × 1.2 m = 2.4 m

Time taken for the wave to travel to the obstacle and come back:

t = Total distance / Speed = 2.4 m / 345 m/s ≈ 0.00695 s

The time taken by the ultrasonic wave to travel to the obstacle and come back is approximately 0.00695 s.

Answer:

Distance: 1720 m

Speed of Sound at 22°C: 344 m/s

Speed of sound at 0°C: 331 m/s

1. Time taken at 22 °C

Speed = 344 m/s | Distance = 1720 m

Time = Distance ÷ Speed

t₁ = 1720 ÷ 344 = 5 s

2. Time taken at 0 °C

Speed = 331 m/s | Distance = 1720 m

t₂ = 1720 ÷ 331 = 5 + (65 ÷ 331) ≈ 5.196 s

(331 × 5 = 1655 ; 1720 − 1655 = 65 ; 65 ÷ 331 ≈ 0.196)

3. Extra time taken

Extra time = Time at 0 °C − Time at 22 °C

= 5.196 − 5 = 0.196 s

Roughly → 0.2 seconds (about 1/5 of a second)

Final answer: The thunder will take roughly 0.2 seconds extra when temperature drops from 22 °C to 0 °C.

Answer:

From Fig. 10.32: The distance between two consecutive compressions (or rarefactions) is given as 8 cm. That distance represents the wavelength (λ). Speed of sound, v = 340 m/s.

1. Wavelength (λ)

In the figure, the distance between two successive compressions (or rarefactions) is 8 cm.

Wavelength = distance between two consecutive compressions

2λ = 8 cm

λ = 0.08/2 m (since 1 m = 100 cm)

✅ λ = 0.04 m

2. Frequency (f)

Use the wave formula: v = f × λ

f = v ÷ λ

f = 340 ÷ 0.04

340 ÷ 0.04 = 340 ÷ (4/100) = 340 × 100 ÷ 4 = 34000 ÷ 4 = 8500 Hz

✅ f = 8500 Hz (or 8.5 kHz)

Final Answer: The wavelength is 0.04 m (or 4 cm) and the frequency is 8500 Hz.

Answer:

Given:

• Speed of sound, v = 345 m s⁻¹

Step 1: Find the wavelength of wave A

Wavelength is the distance between two successive points in the same phase, such as two consecutive crests or two consecutive troughs.

From the graph, wave A completes one complete cycle in a distance of 2.5 cm.

Therefore, λ_A = 2.5 cm = 0.025 m

Step 2: Find the frequency of wave A

We know, v = fλ

Therefore, f = v / λ

For wave A:

f_A = 345 / 0.025

f_A = 13,800 Hz

Step 3: Find the wavelength of wave B

From the graph, wave B completes one complete cycle in a distance of 5.0 cm.

Therefore, λ_B = 5.0 cm = 0.05 m

Step 4: Find the frequency of wave B

Using, f = v / λ

For wave B:

f_B = 345 / 0.05

f_B = 6,900 Hz

Final Answer:

• Wavelength of wave A = 2.5 cm
• Frequency of wave A = 13,800 Hz
• Wavelength of wave B = 5.0 cm
• Frequency of wave B = 6,900 Hz

Answer:

Given:

• Two identical sound sources are placed at A and B.
• Sound from A travels through air.
• Sound from B travels through water.
• Time taken by sound to return to A is 4.5 times the time taken by sound to return to B.

Step 1: Use the relation between speed, distance and time

We know that: Speed = Distance / Time

The sound waves travel to the cliff and then return, so both sounds cover the same total distance. Let the total distance travelled by each sound be d.

Step 2: Write the times for the two sounds

Let:

• Speed of sound in air = v_a
• Speed of sound in water = v_w
• Time taken by sound in air = t_a
• Time taken by sound in water = t_w

Therefore, t_a = d / v_a and t_w = d / v_w

According to the question: t_a = 4.5t_w

Step 3: Find the ratio of speeds

Substituting the expressions for time:

d / v_a = 4.5(d / v_w)

Cancel d from both sides:

1 / v_a = 4.5 / v_w

v_w = 4.5v_a

Hence, v_a : v_w = 1 : 4.5 = 2 : 9

Final Answer:

The ratio of the speed of sound in air to the speed of sound in water is: Speed in air : Speed in water = 2 : 9

Key idea: For the same distance, a sound that takes more time has a lower speed. Since the sound in air takes 4.5 times longer, sound travels 4.5 times faster in water.