GSEB Class 9 Science Solutions Chapter 12 Sound

Question 1. How does the sound produced by a vibrating object in a medium reach your ear?
Answer: When an object vibrates, it pushes the nearby particles of the medium, making them vibrate too. In this way, the vibrations created by an object are passed from one particle to another until they finally reach your ear.
In simple words: A vibrating object makes nearby air particles vibrate, and these vibrations pass from one particle to the next until they get to your ear.

Exam Tip: Remember that sound travels through vibrations passed from particle to particle in a medium, not by the movement of the particles themselves over long distances.

Question 1. Explain how sound is produced by your school bell.
Answer: When the school bell is hit with a hammer, it starts to vibrate. This vibration causes sound waves to be created and travel through the air.
In simple words: Hitting the school bell makes it shake, which then creates sound waves.

Exam Tip: Always link sound production to vibration. Any object that makes a sound is vibrating.

Question 2. Why are sound waves called mechanical waves?
Answer: Waves that need a material medium to travel are called mechanical waves. Since sound waves also need a material medium to move, they are known as mechanical waves.
In simple words: Sound waves are mechanical because they need something (like air or water) to travel through.

Exam Tip: The key characteristic of a mechanical wave is its reliance on a medium; without one, it cannot propagate.

Question 3. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer: Sound needs a material medium for its travel. Since there is no atmosphere on the moon, a person cannot hear any sound produced by another person there.
In simple words: No, you cannot hear sound on the moon because it has no air, which sound needs to travel.

Exam Tip: This question tests your understanding of the essential requirement of a medium for sound propagation. Mentioning the absence of an atmosphere on the moon is crucial.

Question 1. Which property determines
(a) loudness
(b) pitch?
Answer:
(a) The amplitude of the wave determines loudness.
(b) The frequency of the wave determines pitch.
In simple words: How loud a sound is depends on its amplitude, while how high or low it sounds (pitch) depends on its frequency.

Exam Tip: Clearly differentiate between amplitude and frequency in relation to sound characteristics; amplitude affects intensity (loudness) and frequency affects pitch.

Question 2. Guess which sound has a higher pitch: guitar or car horn?
Answer: A guitar generally has a higher pitch than a car horn, assuming the guitar is tuned correctly.
In simple words: A guitar usually makes a higher-pitched sound than a car horn.

Exam Tip: Pitch refers to how high or low a sound is perceived; higher frequency means higher pitch.

Question 1. Define the following terms for a longitudinal wave: wavelength, frequency, time period and amplitude of a sound wave.
Answer: For a longitudinal wave:

• Wavelength: This is the distance between two successive compressions or two successive rarefactions.
• Frequency: This refers to the number of waves produced per second.
• Time period: This is the time taken by two successive compressions or rarefactions to cross a fixed point.
• The amplitude of wave: This is the magnitude of the maximum disturbance on either side of the average value.

Exam Tip: Be precise with definitions, especially distinguishing between compressions/rarefactions for longitudinal waves when defining wavelength and time period.

Question 2. How are the wavelength and frequency of sound wave related to its speed?
Answer: The speed of sound is calculated by multiplying its frequency by its wavelength.
In simple words: The speed of a sound wave is found by taking its frequency and multiplying it by its wavelength.

Exam Tip: Remember the fundamental wave equation: Speed = Frequency × Wavelength (\( v = f\lambda \)).

Question 3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer: Given,
Frequency \( (f) = 220 \, \text{Hz} \)
Speed \( (v) = 440 \, \text{m/s} \)
Using the formula: Wavelength \( (\lambda) = \frac{\text{Speed}}{\text{Frequency}} \)
\( \lambda = \frac{v}{f} \)
\( \lambda = \frac{440 \, \text{m/s}}{220 \, \text{Hz}} \)
\( \lambda = 2 \, \text{m} \)
In simple words: To find the wavelength, you divide the speed of the sound by how many waves it makes per second (its frequency). In this problem, 440 divided by 220 gives 2 meters.

Exam Tip: Always ensure that units are consistent (e.g., m/s for speed, Hz for frequency, m for wavelength) before performing calculations.

Question 4. A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compression from the source?
Answer: Given,
Frequency \( (f) = 500 \, \text{Hz} \)
The time interval between successive compressions is the time period \( (T) \).
\( T = \frac{1}{f} \)
\( T = \frac{1}{500} \)
\( T = 0.002 \, \text{s} \)
In simple words: The time between two peaks of a wave is its time period. To find it, you divide one by the frequency. So, for a 500 Hz sound, it's 1 divided by 500, which is 0.002 seconds.

Exam Tip: Understand that the time interval between successive compressions (or rarefactions) is precisely the time period of the wave, and it is the reciprocal of the frequency.

Question 1. Distinguish between loudness and intensity of sound.
Answer:
Loudness: The sensation produced in the ear that helps us tell the difference between a loud and a quiet sound is called loudness.
Intensity: The intensity of sound at any point in space is defined as the amount of energy passing per unit of time per unit of area in the direction perpendicular to that area.
In simple words: Loudness is how strong a sound feels to your ear. Intensity is a precise measurement of the sound energy flowing through a specific area in a certain amount of time.

Exam Tip: Loudness is a subjective perception (how the ear interprets sound), while intensity is an objective, measurable physical quantity related to the energy of the sound wave.

Question 1. In which of the three media air, water or iron, does sound travel the fastest at a particular temperature?
Answer: Sound travels fastest in iron because it is a solid medium. Generally, sound moves quicker through solids, then liquids, and slowest through gases.
In simple words: Sound moves fastest through iron because it's a solid.

Exam Tip: Remember the general rule that sound speed increases with the density and elasticity of the medium, making it fastest in solids and slowest in gases.

Question 1. An echo was heard after 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Answer: Given,
Time \( (t) = 3 \, \text{s} \)
Speed \( (v) = 342 \, \text{m/s} \)
The total distance traveled by the sound for an echo is \( 2 \times \text{distance to reflector} \).
So, \( 2d = v \times t \)
Distance \( d = \frac{v \times t}{2} \)
\( d = \frac{342 \times 3}{2} \)
\( d = \frac{1026}{2} \)
\( d = 513 \, \text{m} \)
In simple words: To find the distance to the reflecting surface, you multiply the speed of sound by the time it took for the echo, then divide by two because the sound traveled there and back. Here, 342 times 3, divided by 2, equals 513 meters.

Exam Tip: For echo problems, always remember to divide the total distance traveled by 2, as the sound covers the distance to the reflecting surface and back.

Question 1. Why are the ceilings of concert halls curved?
Answer: The ceilings of concert halls are curved so that sound, after reflecting off them, can reach all the corners of the hall evenly. This design helps to distribute the sound throughout the audience.
In simple words: Concert hall ceilings are curved to bounce sound all over the room, making sure everyone hears well.

Exam Tip: Curved surfaces are used in auditoriums to evenly disperse sound waves, improving acoustics by preventing dead spots and focusing sound.

Question 1. What is the audible range of the average human ear?
Answer: The average human ear can hear sounds with frequencies ranging from 20 Hz to 20,000 Hz.
In simple words: Most people can hear sounds that are between 20 Hz and 20,000 Hz.

Exam Tip: Know the specific frequency range for human hearing, as questions often involve identifying sounds as infrasound, audible, or ultrasound.

Question 2. What is the range of frequencies associated with
(a) Infrasound
(b) Ultrasound?
Answer:
(a) Infrasound: The frequency is less than 20 Hz.
(b) Ultrasound: The frequency is between 20 Hz and 20 KHz (20,000 Hz).
In simple words: Infrasound has a frequency lower than 20 Hz, while ultrasound has a frequency higher than 20,000 Hz.

Exam Tip: Be precise with the frequency limits for infrasound (below human hearing) and ultrasound (above human hearing).

Question 1. A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Answer: Given,
Time \( (t) = 1.02 \, \text{s} \)
Speed \( (v) = 1531 \, \text{m/s} \)
The total distance traveled by the sonar pulse is \( 2 \times \text{distance to cliff} \).
So, \( 2d = v \times t \)
Distance \( d = \frac{v \times t}{2} \)
\( d = \frac{1531 \times 1.02}{2} \)
\( d = \frac{1561.62}{2} \)
\( d = 780.81 \, \text{m} \)
In simple words: To find the distance to the cliff, multiply the speed of sound by the total time the pulse traveled and then divide by two. The calculation is 1531 times 1.02, then divided by 2, which gives 780.81 meters.

Exam Tip: This is another echo-ranging problem. Remember to divide by 2 for the one-way distance to the object.

In-Text Activities Solved

(Textbook Page 160)

Activity 12.1

Discussion and conclusion
Answer:

• When a vibrating object is brought near the ear, a sound is heard.
• Touching one of the prongs of a tuning fork makes you feel a vibrating sensation.
• When a table tennis ball is gently touched with the prong of a tuning fork, the ball starts to vibrate as some energy from the tuning fork is transferred to it.

• After touching the water surface with one of the prongs, the water starts vibrating, and waves are created.
• When both prongs of the tuning fork are lowered into the water in the glass, the water splashes out. The energy from the vibrating tuning fork is changed into kinetic energy of the water, which causes it to splash out of the glass.

Exam Tip: This activity demonstrates the transfer of energy from a vibrating object to another medium (ball or water), showing that vibrations cause sound and energy propagation.

Activity 12.5

Reflection of sound

Discussion and conclusion
Answer: Here, the angle of incidence equals the angle of reflection. The incident sound, reflected sound, and the normal all lie in the same plane. This shows that both the first and second laws of reflection of sound are satisfied.

In simple words: When sound reflects, the incoming angle is the same as the outgoing angle. The original sound, the bounced sound, and an imaginary line (normal) all line up. This matches the rules of reflection.

Exam Tip: The laws of reflection of sound are analogous to the laws of reflection of light. Clearly state both laws and how the diagram demonstrates them.

Gujarat Board Class 9 Science Sound Textbook Questions and Answers

Question 1. What is sound and how is it produced?
Answer: Sound is a form of energy. It is created because of the vibrations of different types of objects. These vibrations cause disturbances that travel through a medium, which we perceive as sound.
In simple words: Sound is energy made when things vibrate.

Exam Tip: Emphasize that sound is a form of energy and that vibration is the fundamental cause of its production.

Question 2. Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of the sound.
Answer: When a vibrating object moves forward in the air, it pushes and squeezes the air directly in front of it. This creates a region of high pressure, which is called a compression (C). This compression then begins to move away from the vibrating object. When the vibrating object moves backward, it forms a region of low pressure known as a rarefaction (R). As the object quickly moves back and forth, it creates a series of compressions and rarefactions in the air. These areas of high and low pressure then travel through the medium as a sound wave.

In simple words: When something vibrates, it pushes air forward to make a crowded spot (compression) and pulls back to make an empty spot (rarefaction). These crowded and empty spots then travel as a sound wave.

Exam Tip: Clearly explain how the forward and backward motion of the vibrating object leads to regions of high and low pressure, respectively. The diagram should illustrate these regions accurately.

Question 3. Cite an experiment to show that sound needs a material medium for its propagation.
Answer: An electric bell can be placed inside an empty bell-jar that is connected to a vacuum pump. Initially, you can hear the sound of the bell ringing clearly. As air is pumped out of the bell-jar using the vacuum pump, you will notice that the sound of the ringing bell decreases. If all the air is removed, no sound can be heard from the ringing bell, even though you can still see it vibrating. This shows that sound cannot travel through a vacuum, proving it needs a material medium to propagate.

In simple words: When you pump air out of a bell jar with a ringing bell inside, the sound gets quieter until you can't hear it at all. This shows that sound needs air to travel.

Exam Tip: Clearly describe the setup and the observations. The key inference is that the decreasing sound volume as air is removed demonstrates sound's reliance on a medium.

Question 4. Why is a sound wave called a longitudinal wave?
Answer: A sound wave is called a longitudinal wave because the particles of the medium vibrate back and forth in the same direction that the wave energy is traveling.
In simple words: Sound waves are longitudinal because the particles move in the same direction as the wave itself.

Exam Tip: Distinguish longitudinal waves (particle motion parallel to wave direction) from transverse waves (particle motion perpendicular to wave direction).

Question 5. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer: The quality (or timber) of sound is the characteristic that helps you tell one sound from another, even when they have the same pitch and loudness. This unique quality allows you to recognize your friend's voice.
In simple words: The unique "quality" or "timber" of sound helps you recognize your friend's voice, even if other sounds have the same loudness and pitch.

Exam Tip: Quality or timber is related to the waveform and overtones present, allowing distinction between different sound sources even if pitch and loudness are identical.

Question 6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after flash is seen, why?
Answer: Flash and thunder happen at the same time. However, the speed of light is much greater than the speed of sound. Because of this, you see the flash of light earlier than you hear the thunder, even though both events occur at the same moment.
In simple words: You see lightning before you hear thunder because light travels much faster than sound.

Exam Tip: This question highlights the significant difference in the speeds of light and sound in air. Quantifying the speeds can further strengthen your answer.

Question 12. Give two practical applications of reflection of sound waves.
Answer:

• Reflection of sound is utilized to measure the distance and speed of underwater objects. This technique is known as Sonar.
• The functioning of a stethoscope also relies on the reflection of sound. In a stethoscope, the sound of the patient's heartbeat reaches the doctor's ear through multiple reflections of sound inside the tubes.

Exam Tip: Focus on applications where sound reflection is purposefully engineered for a specific function, such as measurement or amplification.

Question 13. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top?
Answer: Given,
Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)
Speed of sound \( u_s = 340 \, \text{m/s} \)
Height of tower \( s = 500 \, \text{m} \)

Initial velocity of the stone, \( u_i = 0 \).
Let \( t_1 \) be the time taken for the stone to fall to the base of the tower.
Using the equation of motion: \( s = u_i t_1 + \frac{1}{2} g t_1^2 \)
\( 500 = (0)t_1 + \frac{1}{2} \times 10 \times t_1^2 \)
\( 500 = 5 t_1^2 \)
\( t_1^2 = \frac{500}{5} \)
\( t_1^2 = 100 \)
\( t_1 = \sqrt{100} \)
\( t_1 = 10 \, \text{s} \)

Now, let \( t_2 \) be the time taken for the sound of the splash to reach the top from the base of the tower.
\( \text{Distance} = \text{Speed} \times \text{Time} \)
\( s = u_s t_2 \)
\( t_2 = \frac{s}{u_s} \)
\( t_2 = \frac{500 \, \text{m}}{340 \, \text{m/s}} \)
\( t_2 \approx 1.47 \, \text{s} \)

Therefore, the splash is heard at the top after a total time \( t \), where:
\( t = t_1 + t_2 \)
\( t = 10 \, \text{s} + 1.47 \, \text{s} \)
\( t = 11.47 \, \text{s} \)
In simple words: First, figure out how long the stone takes to fall (10 seconds). Then, figure out how long the splash sound takes to travel up (about 1.47 seconds). Add these two times together to get the total time until the splash is heard at the top, which is 11.47 seconds.

Exam Tip: This problem involves two distinct phases: the fall of the stone (kinematics under gravity) and the upward travel of sound (constant speed). Calculate each time separately and then sum them up.

Question 14. A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer: Given,
Speed of sound \( v = 339 \, \text{m/s} \)
Wavelength of sound wave \( \lambda = 1.5 \, \text{cm} = 1.5 \times 10^{-2} \, \text{m} \)
Using the formula: Speed \( v = \text{Frequency} \times \text{Wavelength} \)
Frequency \( f = \frac{v}{\lambda} \)
\( f = \frac{339 \, \text{m/s}}{1.5 \times 10^{-2} \, \text{m}} \)
\( f = \frac{339}{0.015} \)
\( f = 22600 \, \text{Hz} \)

The frequency range for audible sound for humans is between 20 Hz and 20,000 Hz. Since the calculated frequency of the given sound is 22,600 Hz, which is more than 20,000 Hz, it is not audible to humans.
In simple words: First, calculate the wave's frequency by dividing its speed by its wavelength. The frequency is 22,600 Hz. Because this number is higher than 20,000 Hz, humans cannot hear this sound.

Exam Tip: Convert all units to SI (meters, seconds, Hz) before calculation. Also, compare the final frequency with the human audible range (20 Hz to 20,000 Hz) to determine audibility.

Question 15. What is reverberation? How can it be reduced?
Answer: Reverberation occurs when a sound made in a large hall continues to echo due to repeated reflections from the walls. This persistence of sound lasts until its intensity decreases to a point where it is no longer audible. To reduce reverberation, materials that absorb sound (like curtains, carpets, and acoustic panels) can be used on walls and ceilings, or the room's geometry can be adjusted to disperse sound more effectively.
In simple words: Reverberation is when sound keeps bouncing around in a big room, making it last too long. You can stop it by adding soft materials that soak up sound.

Exam Tip: Clearly define reverberation and provide practical methods for its reduction, focusing on sound-absorbing materials and architectural design principles.

Question 16. What is the loudness of sound? What factor does it depend on?
Answer: Loudness of sound refers to how strong or weak a sound is perceived by the human ear. A loud sound has high energy. Loudness mainly depends on the amplitude of vibrations. In fact, loudness is directly proportional to the square of the amplitude of vibrations.
In simple words: Loudness is how strong a sound feels. It depends on how big the sound wave's vibrations are (its amplitude); bigger vibrations mean louder sound.

Exam Tip: Remember that loudness is a subjective perception linked to the objective physical quantity of amplitude, specifically proportional to the square of the amplitude.

Question 17. Explain how bats use ultrasound to catch prey.
Answer: Bats use ultrasound to catch prey through a process called echolocation. They emit high-frequency ultrasonic waves. These waves travel through the air, hit objects (like prey), and reflect back. The bat's ears detect these reflected waves. By analyzing the nature of these echoes, the bat can determine the precise location, size, and movement of its prey.
In simple words: Bats send out high-pitched sounds (ultrasound). These sounds bounce off prey and come back, telling the bat exactly where the prey is and what it's doing.

Exam Tip: The key terms here are 'ultrasound' and 'echolocation'. Explain the emission, reflection, and detection process, and how the bat interprets the echoes.

Question 18. How is ultrasound used for cleaning?
Answer: Ultrasound is used for cleaning by placing objects in a special cleaning solution. High-frequency ultrasound waves are then passed through this solution. These intense waves create cavitation (the formation and collapse of tiny bubbles) that detaches and removes dirt, grease, and other contaminants from the surfaces of the objects.
In simple words: Ultrasound cleans things by sending strong sound waves through a cleaning liquid. These waves shake off dirt and grime from the objects.

Exam Tip: Highlight the use of high-frequency waves and the mechanism of dirt detachment (cavitation or intense vibrations) in the cleaning process.

Question 19. Explain the working and application of a sonar.
Answer: The acronym SONAR stands for Sound Navigation and Ranging. Sonar is a device that uses ultrasonic waves to measure the distance, direction, and speed of underwater objects.

1. A sonar system consists of a transmitter and a detector, typically installed in a boat or a ship.
2. The transmitter generates and sends out ultrasonic waves into the water. These waves travel through the water until they strike an object on the seabed or an underwater feature.
3. After hitting the object, the ultrasonic waves are reflected back towards the sonar device and are sensed by the detector.
4. The detector converts these reflected ultrasonic waves into electrical signals, which are then processed and interpreted.
5. By knowing the speed of sound in water and measuring the time interval between sending out the pulse and receiving its echo, the distance to the reflecting object can be calculated.
6. If \( t \) is the time interval between transmission and reception of the ultrasound signal, and \( v \) is the speed of sound through seawater, then the total distance \( 2d \) traveled by the ultrasound is given by \( 2d = v \times t \).
7. This method is called echo-ranging. The sonar technique is widely used to determine the depth of the sea and to locate underwater hills, valleys, submarines, icebergs, and sunken ships.

In simple words: Sonar uses sound waves to find things underwater. A transmitter sends out waves, which bounce off objects and return to a detector. By measuring the time it takes, it can tell how far away objects are and what the seabed looks like.

Exam Tip: Clearly define SONAR, explain its two main components (transmitter and detector), describe the step-by-step process of transmitting and receiving ultrasonic waves, and list its common applications.

Question 20. A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Answer: Given,
Time taken to hear the echo \( t = 5 \, \text{s} \)
Distance of the object from the submarine \( d = 3625 \, \text{m} \)

The total distance traveled by the sonar waves during transmission and reception in water is \( 2d \).
\( \text{Total distance} = 2 \times 3625 \, \text{m} = 7250 \, \text{m} \)

Velocity of sound in water \( v = \frac{\text{Total distance}}{\text{Time taken}} \)
\( v = \frac{2d}{t} \)
\( v = \frac{2 \times 3625}{5} \)
\( v = \frac{7250}{5} \)
\( v = 1450 \, \text{m/s} \)
In simple words: The sound travels twice the distance to the object (there and back). So, the total distance is 7250 meters. Divide this by the time (5 seconds) to get the speed of sound, which is 1450 meters per second.

Exam Tip: Remember to calculate the total distance (2d) for the sound's round trip. Use the formula speed = distance/time and ensure correct units for the final answer.

Question 21. Explain how defects in a metal block can be detected using ultrasound.
Answer: Defects in a metal block can be detected using ultrasound because flaws or cracks do not allow ultrasound to pass through them and instead reflect the waves. This property is used to identify defects in metal blocks. Ultrasound waves are passed through one end of a metal block, and detectors are placed on the other end. If there is a defective part within the metal block, it will not permit the ultrasound to pass through it, causing the waves to be reflected. As a result, the ultrasound will not reach the detector on the other side. This absence of detection signals indicates the presence and location of defects in the metal block.

In simple words: Ultrasound waves are sent through a metal block. If there's a flaw, the waves bounce back instead of passing through to detectors on the other side. This tells you where the defect is.

Exam Tip: Explain the principle of reflection/absorption by defects and how the absence of signals at specific detectors reveals the location of flaws. The diagram should clearly show waves passing through non-defective areas and being reflected by defects.

Question 22. Explain how the human ear works.
Answer: The human ear is an extremely sensitive organ that allows us to hear. It works by converting pressure changes in the air (sound waves) into electrical signals that travel to the brain via the auditory nerve. Here is how it functions:

1. The outer ear, called the pinna, collects sound waves from the surroundings.
2. These collected sound waves travel through the auditory canal.
3. At the end of the auditory canal is a thin membrane known as the eardrum or tympanic membrane.
4. When a compression (high-pressure region) in the sound wave reaches the eardrum, the pressure on its outside surface increases, pushing the eardrum inward.
5. Conversely, when a rarefaction (low-pressure region) reaches the eardrum, the pressure outside decreases, causing the eardrum to move outward. This back-and-forth movement makes the eardrum vibrate.
6. These vibrations are then greatly amplified by three tiny bones in the middle ear: the hammer (malleus), anvil (incus), and stirrup (stapes).
7. The middle ear transmits these amplified pressure variations to the inner ear.
8. In the inner ear, specifically within the cochlea, these pressure variations are converted into electrical signals.
9. Finally, these electrical signals are sent to the brain through the auditory nerve, and the brain interprets them as sound.

In simple words: The outer ear gathers sound, which travels to the eardrum and makes it vibrate. These vibrations get bigger in the middle ear bones and then turn into electrical signals in the inner ear. These signals go to the brain, and that's how we hear.

Exam Tip: Describe the function of each main part of the ear (outer, middle, inner ear) in a sequential manner, explaining how sound energy is transferred and transformed at each stage.