Sound Class 9 Notes

Sound Class 9 Notes

Introduction:

• Sound is a form of energy that produces the sensation of hearing. It is generated by the vibration of objects and travels through a medium such as air, water, or solids in the form of waves.

Sources of Sound:

Any vibrating body is a source of sound, such as:

• Tuning fork
• Drum
• Loudspeaker
• Vocal cords

Importance of Sound:

• Helps in communication
• Used in music and entertainment
• Useful in medical diagnosis (ultrasound)
• Used in navigation (SONAR)

        Thus, sound plays an important role in our daily life and is an essential topic in Class 9 Physics.

Definition:

• Sound is a form of energy produced by vibrating objects that travels through a material medium in the form of waves and produces the sensation of hearing.
• Scientifically/Technically Sound is a kind of wave.

Wave:

• A wave is a disturbance that travels through a medium or space and transfers energy from one point to another without the actual transfer of matter.

Key Points:

• A wave transfers energy, not particles
• It can travel through a medium (like sound waves)
• Some waves can travel in vacuum (like light waves)

Example:

• When a stone is dropped into water, ripples move outward. The water particles move up and down, but the wave travels forward.

Types of Wave:

- Waves are mainly classified into two types based on whether they require a medium for propagation.

1. Mechanical Waves

Definition:
Mechanical waves are waves that require a material medium (solid, liquid, or gas) to travel.

Key Features:

• Cannot travel in vacuum
• Energy is transferred through vibrations of particles
• Can be transverse or longitudinal

Examples:

• Sound waves
• Water waves
• Waves on a stretched string

2. Electromagnetic Waves

Definition:
Electromagnetic waves are waves that do not require any material medium and can travel through vacuum.

Key Features:

• Produced by oscillating electric and magnetic fields
• Always transverse waves
• Travel at the speed of light in vacuum

Examples:

• Light waves
• Radio waves
• X-rays
• Microwaves

- Again waves are classified into two types based on the direction of vibration of particles with respect to the direction of wave propagation:

1. Transverse Waves

In a transverse wave, the particles of the medium vibrate perpendicular to the direction of propagation of the wave.

Examples:

• Waves on a stretched string
• Ripples on the surface of water
• Light waves (electromagnetic)

Characteristics:

• Have crests and troughs
• Can travel through solids and on liquid surfaces

2. Longitudinal Waves

In a longitudinal wave, the particles of the medium vibrate parallel to the direction of propagation of the wave.

Examples:

• Sound waves in air
• Compression waves in a spring

Characteristics:

• Have compressions and rarefactions
• Can travel through solids, liquids, and gases

Scientific/Technical Definition of Sound:

• Sound is a mechanical longitudinal wave produced by vibrating objects that propagates through a material medium in the form of compressions and rarefactions, transferring energy without the transfer of matter.

OR

• In simple words, Sound is a kind of mechanical and longitudinal wave which propagates in the form of compressions and rarefactions

Production of Sound

How is Sound Produced?

Sound is produced when an object vibrates.

• Vibration means a rapid to-and-fro (back and forth) motion of an object.

• When an object vibrates, it causes the surrounding air particles to vibrate.

• These vibrations travel as sound waves and reach our ears.

Important Point:

• If there are no vibrations, no sound is produced.

Propagation of Sound

What is Propagation of Sound?

• Propagation of sound means the movement or transmission of sound waves from the source to the listener through a material medium.
• Sound propagates by transferring energy, not matter.

Medium Required for Sound Propagation

• Sound needs a material medium to travel.

Medium	Speed of Sound (Approx.)
Solids	Fastest
Liquids	Slower than solids
Gases (air)	Slowest
Vacuum	Cannot propagate

Note:- Sound cannot travel in vacuum because there are no particles to vibrate.

How Sound Propagates in Air

Sound travels in air as longitudinal waves.

Step-by-Step Explanation:

1. A vibrating object produces sound.

2. The vibration compresses nearby air particles.

3. These particles transfer energy to adjacent particles.

4. This creates regions of:

   • Compression (high pressure, particles close together)

   • Rarefaction (low pressure, particles far apart)

5. These compressions and rarefactions travel as sound waves.

Important Note:- Air particles only vibrate about their mean position; they do not travel with the wave.

Longitudinal Waves

• Sound waves are longitudinal waves.
• In these waves:
• Particle vibration is parallel to the direction of wave travel.

Key Terms Related to Propagation of Sound

1. Compression

• Region of high pressure

• Particles are close together

2. Rarefaction

• Region of low pressure

• Particles are spread apart

Propagation of Sound in Different Media

1. In Solids

• Particles are tightly packed.
• Sound travels fastest.

2. In Liquids

• Particle spacing is more than solids.
• Sound travels faster than gases but slower than solids.

3. In Gases

• Particles are far apart.
• Sound travels slowest.

Speed of Sound

• Speed depends on:

  1. Nature of the medium

  2. Temperature (in gases)

  3. Humidity (in air)

Speed of sound in air at 25°C ≈ 346 m/s

Why Sound Cannot Propagate in Vacuum

• No particles to vibrate
• No compressions or rarefactions can form

Sound Needs a medium to travel

Why Does Sound Need a Medium (Solid/Liquid/Gas)?

Sound travels by causing vibrations of particles of the medium.

• When an object vibrates, it transfers energy to nearby particles.
• These particles vibrate and pass the energy to neighboring particles.
• In this way, sound energy travels through the medium.

- Without particles, sound cannot travel.

Experiment: Bell Jar Experiment

• This experiment proves that sound needs a medium.

Procedure:

1. Place an electric bell inside a glass bell jar.

2. Connect it to a power supply.

3. Switch on the bell – sound is heard.

4. Gradually remove air from the jar using a vacuum pump.

Observation:

• Sound becomes faint and finally inaudible.
• The hammer of the bell can still be seen vibrating.

Conclusion:

• Sound needs a material medium to travel.

Sound is a Longitudinal Wave

• Sound waves can be easily understood using a slinky spring. This experiment clearly shows that sound is a longitudinal wave.

Slinky Experiment to Explain Sound Wave

Apparatus

• A long slinky spring
• Two students

Procedure

1. Stretch the slinky on a smooth floor between two students.

2. One student holds one end of the slinky fixed.

3. The other student gives a push and pull along the length of the slinky (not sideways).

Observation

• When the slinky is pushed forward, the coils come closer together → Compression
• When the slinky is pulled backward, the coils move far apart → Rarefaction
• These compressions and rarefactions move along the slinky.

Explanation

• Each coil of the slinky vibrates to and fro in the same direction as the wave travels.
• The coils do not move forward permanently; they only oscillate about their mean positions.
• The moving pattern of compressions and rarefactions represents a sound wave.

Conclusion

• Since the particles of the medium (slinky coils) vibrate parallel to the direction of wave propagation, sound is a longitudinal wave.

Relation of Motion of Slinky with Sound

Slinky Motion	Sound Wave
Closely packed coils	Compression
Widely spaced coils	Rarefaction
Push–pull motion	Vibration of particles
Direction of coil vibration	Direction of sound propagation

Characteristics of a Sound wave

• Sound waves have certain important characteristics that help us describe and understand sound. These characteristics are based on the vibrations of particles in the medium.

1. Amplitude

Definition:

• Amplitude is the maximum displacement of the particles of the medium from their mean position.

Effect on Sound:

• Determines the loudness of sound
• Greater amplitude → Louder sound
• Smaller amplitude → Faint sound

Unit: metre (m)

- Loud sound has large compressions and rarefactions.

2. Wavelength (λ)

Definition:

• Wavelength is the distance between two successive compressions or two successive rarefactions.

Effect on Sound:

• Affects the pitch of sound
• Shorter wavelength → higher pitch
• Longer wavelength → lower pitch

Unit: metre (m)

3. Frequency (f)

Definition:

• Frequency is the number of vibrations per second produced by a sound source.

Effect on Sound:

• Determines the pitch (shrillness) of sound
• Higher frequency → High-pitched (shrill) sound
• Lower frequency → Low-pitched (deep) sound

Unit: hertz (Hz)

4. Time Period (T)

Definition:

• Time period is the time taken to complete one vibration.

Relation with Frequency:

$$T=\frac{1}{\mathrm{f}}$$

Unit: second (s)

5. Speed (Velocity) of Sound

(Relation between speed/velocity, frequency and wavelength of Sound)

Definition:

• Speed of sound is the distance travelled by sound per unit time.

Formula:

$$v=f\times \lambda$$

Depends on:

• Nature of medium (solid > liquid > gas)
• Temperature of the medium

Does NOT depend on:

• Amplitude
• Loudness

6. Loudness or Softness

• Sensation of sound perceived by human ear
• Depends on amplitude
• Unit: decibel (dB)

7. Intensity

• Amount of sound energy passing through a unit area
• Depends on amplitude

8. Pitch

• Pitch is the sensation of the human ear that enables us to distinguish between shrill (high-pitched) and deep (low-pitched) sounds.

Relation with Frequency

Pitch depends on the frequency of the sound wave.

• High frequency → High pitch (shrill sound)
• Low frequency → Low pitch (deep sound)

Example:

• Sound of a mosquito → high pitch
• Sound of a drum → low pitch

9. Quality or Timbre

• Quality (Timbre) is the characteristic of sound that enables us to differentiate between two sounds of same pitch and same loudness, coming from different sources.

Explanation

• Two instruments may produce sounds with the same frequency (pitch) and same amplitude (loudness).
• Still, they sound different because of their waveform.
• This difference in waveform gives rise to different timbre.

Cause of Timbre

Timbre depends on:

• Shape of the sound wave
• Number and relative strength of overtones (harmonics) present along with the fundamental note

Example

• Sound produced by a flute and a violin may have the same pitch and loudness.
• Still, we can easily identify them because their quality (timbre) is different.
• The Sound which is more pleasant is said to be a rich quality Sound.

10. Tone

• A tone is a sound that is produced by regular and periodic vibrations and has a single frequency.

Explanation

• When a sound contains only one frequency, it is called a tone.
• Pure tones are rare in nature.
• Sounds produced by tuning forks are nearly pure tones.

Example

• Sound of a tuning fork
• Sound produced by an electronic signal generator

11. Note

• A note is a sound produced by regular periodic vibrations and consists of a mixture of different frequencies (fundamental frequency and overtones).

Explanation

• Most musical instruments do not produce a single frequency.
• Along with the fundamental frequency, several overtones (harmonics) are produced.
• The combination of these frequencies gives rise to a note.

Example

• Sound produced by flute, harmonium, guitar, violin
• Human voice while singing

13. Noise

• Noise is a sound produced by irregular and non-periodic vibrations, which is unpleasant to the human ear.

Explanation

• In noise, vibrations occur randomly.
• Noise does not have a definite pitch.
• It lacks a regular pattern of frequencies.

Examples

• Traffic sound
• Horns and loudspeakers
• Machinery in factories
• Construction work
• Firecrackers

Numerical

Numerical 1

The frequency of a sound wave is 500 Hz and its wavelength is 0.68 m.
Find the speed of sound.

Given:
Frequency $500$ Hz
Wavelength $0.68$ m

Formula:
$v=$f×λ

Solution:

v = 500 × 0.68

v = 340 m/s

Numerical 2

The speed of sound in air is 330 m/s.
If the frequency of a sound wave is 660 Hz, find its wavelength.

Given:
Speed $330$m/s
Frequency $660$ Hz

Formula:

$$\mathrm{\lambda }= \frac{v}{\mathrm{f}}$$

Solution:

$$\mathrm{\lambda }= \frac{330}{660}$$

$$\lambda \; = 0.5\text{ m}$$

Numerical 3

A sound wave has a wavelength of 1.5 m.
If the speed of sound is 300 m/s, calculate the frequency.

Given:
Wavelength $1.5$ m
Speed $300$ m/s

Formula:

$$\mathrm{f }= \frac{v}{\mathrm{\lambda }}$$

Solution:

$$\mathrm{f }= \frac{300}{1.5}$$

$$\frac{\mathrm{<br>}}{}$$

$$\frac{\mathrm{f\; =\; 200\; Hz}}{}$$

Numerical 4

The time period of a sound wave is 0.002 s and its wavelength is 0.68 m.
Find the speed of sound.

Given:
Time period $0.002$ s
Wavelength $0.68$ m

Formula:

$$\mathrm{v }= \frac{\lambda }{\mathrm{T}}$$

Solution:

$$\mathrm{v }= \frac{0.68}{0.002}$$

$$\frac{\mathrm{<br>}}{}$$

$$v\; =340\text{ m/s}$$

Numerical 5

A sound wave has wavelength 0.5 m and time period 0.0015 s.
Find:

1. Frequency                            2. Speed of sound

Given:
$0.5$ m
$0.0015$ s

Step 1: Frequency

$$\mathrm{f }= \frac{1}{T}$$

$$\frac{\mathrm{<br>}}{}$$

$$f\; =\frac{1}{0.0015}$$

$$\frac{\mathrm{<br>}}{}$$

$$f\; =666.7\text{ Hz}$$

Step 2: Speed

$$\mathrm{v }= \lambda \times f$$

$$\mathrm{<br>}$$

$$v\; = 0.5\times 666.7$$

$$\mathrm{<br>}$$

$$v\; = 333.3\text{ m/s}$$

$$\text{<b>Practice Numerical:</b>}$$

1) Find the frequency of a wave whose time period is 0.002 second. (Click here for Answer)

2) A source of wave produces 40 crests and 40 troughs in 0.4 second. Find the frequency of the wave. (Click here for Answer)

3) Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium. (Click here for Answer)

4) A sound has 13 crests and 15 troughs in 3 seconds. When the second crest is produced the first is 2cm away from the source? Calculate a) The wavelength, b) The frequency, c) The wave speed (Click here for Answer)

5) A wave moves a distance of 8 m in 0.05 s. Find the velocity of the wave. (Click here for Answer)

Reflection of Sound

• Reflection of sound is the phenomenon in which sound waves bounce back after striking a hard and smooth surface.

Experiment: To Demonstrate Reflection of Sound

Aim

To show that sound waves get reflected from a hard surface.

Materials Required

• Two identical cardboard tubes (or long pipes)
• A hard reflecting surface (smooth wall or large wooden board)
• A source of sound (ticking clock / mobile phone / tuning fork)

Procedure

1. Place the reflecting surface (wall or wooden board) vertically.

2. Hold one cardboard tube close to the sound source (clock or tuning fork).

3. Place the second tube near your ear.

4. Arrange both tubes such that they make equal angles with the normal to the reflecting surface.

5. Slowly change the angle of the second tube until the sound is heard clearly in your ear.

Observation

• The sound is heard clearly only when the angles of the two tubes are equal.
• This shows that sound waves are reflected from the hard surface.

Conclusion

• Sound waves follow the laws of reflection.
• The angle of incidence equals the angle of reflection.

Echo

• An echo is the repetition of a sound caused by the reflection of sound waves from a distant surface.

Conditions for Hearing an Echo

• The distance between the source and reflecting surface should be at least 17.2 m.
• The time gap between original sound and reflected sound should be ≥ 0.1 second.
• The reflecting surface must be hard (wall, hill, building).

Solved Numerical on Echo

Question: 1
A boy shouts in front of a cliff and hears an echo after 2 seconds. Find the distance of the cliff.
(Speed of sound = 340 m/s)

Solution:

$$d = \frac{340 \times 2}{2} = 340 \text{ m}$$

Question: 2
A person hears the echo after 3 s from a wall at a distance of 510 m. Find the speed of sound.

Solution:

$$v = \frac{2d}{t} = \frac{2 \times 510}{3} = 340 \text{ m/s}$$

Question: 3
A man is standing 10 m away from a wall. Will he hear an echo? (v = 340 m/s)

Solution:

$$t=\frac{2d}{v}=\frac{20}{340}=0.058s$$

Since 0.058 s < 0.1 s, echo cannot be heard.

Reverberation

• Reverberation is the persistence of sound in an enclosed space due to multiple reflections of sound.

Examples

• Sound heard in a large hall or auditorium.
• Sound heard inside empty rooms.

Reduction of Reverberation

• Using sound-absorbing materials like curtains, carpets, acoustic panels, and false ceilings.

Applications of Reflection of Sound

1. Megaphone and Loudspeaker

• Sound waves are reflected forward, increasing loudness.

2. Stethoscope

• Multiple reflections of sound inside the tube help doctors hear heartbeat clearly.

3. Soundboards

• Used in auditoriums to reflect sound towards the audience.

4. Hearing Aids

• Use reflection and amplification of sound.

Range of hearing

• The range of hearing is the frequency range of sound waves that a human ear can hear.

Range of Human Hearing:-

• A normal human ear can hear sounds having frequencies from:

$$\boxed{{\displaystyle 20\text{ Hz to }20,000\text{ Hz (20 kHz)}}}$$

• Below 20 Hz → cannot be heard

• Above 20,000 Hz → cannot be heard

Types of Sound Based on Frequency

1) Infrasonic Sound

• Frequency: Less than 20 Hz

• Cannot be heard by humans

• Produced by:

  • Earthquakes

  • Volcanoes

  • Large animals like elephants, whales

• Used by animals for long-distance communication

2) Audible Sound

• Frequency: 20 Hz – 20,000 Hz

• Can be heard by humans

• Includes:

• Human speech

• Music

• School bell

3) Ultrasonic Sound

• Frequency: Greater than 20,000 Hz

• Cannot be heard by humans

• Used in:

• SONAR (finding depth of sea)

• Medical ultrasound (sonography)

• Bats and dolphins for navigation

Range of Hearing of Different Animals

Living Being	Range of Hearing
Human	20 Hz – 20,000 Hz
Dog	Up to 45,000 Hz
Cat	Up to 64,000 Hz
Bat	20,000 – 120,000 Hz
Dolphin	Up to 150,000 Hz
Elephant	Below 20 Hz

$$\text{Applications of Ultrasound}$$

• Ultrasound refers to sound waves having frequency greater than 20,000 Hz (20 kHz). These sounds cannot be heard by humans, but they have many useful applications.

1) Medical Diagnosis (Ultrasonography)

• Used to view internal organs like:

  • Heart

  • Liver

  • Kidney

• Used to monitor growth of fetus during pregnancy

• Detects:

  • Kidney stones

  • Tumours

• Safe because it does not use harmful X-rays

2) Industrial Testing

• Used to detect cracks and flaws in metal blocks

• Very important in:

• Aircraft industry

• Bridge construction

• Railways

3) Cleaning of Delicate Parts

• Used to clean:

  • Jewellery

  • Watch parts

  • Electronic components

• Ultrasound removes dirt from tiny gaps that cannot be cleaned by hand

4) Breaking Kidney Stones (Lithotripsy)

• High-intensity ultrasound is used to break kidney stones into small pieces

• Stones are later removed naturally through urine

5) Echocardiography

• Special ultrasound technique used to examine:

  • Structure of the heart

  • Blood flow

• Helps in detecting heart diseases

6) Welding and Drilling

• Ultrasound is used for:

  • Welding plastics

  • Drilling hard materials like glass and ceramics

$$\text{7) SONAR}$$

• SONAR stands for Sound Navigation and Ranging.
• $$\text{It is a technique that uses sound waves to detect objects underwater, measure distances, and even map the ocean floor. }$$
• $$\text{SONAR technology is widely used in submarines, ships, and oceanographic research to navigate, communicate, and find underwater objects.}$$

How Does SONAR Work?

• The basic working principle of SONAR is similar to echolocation used by bats. 
• It involves sending out a sound wave and measuring how long it takes for the sound to return after bouncing off an object. 

Formula:

• The distance between the SONAR device and the object can be calculated using the formula:

$$\text{Distance}=\frac{\text{Speed of sound}\times \text{Time taken for echo to return}}{\mathrm{2}}$$

• The distance is divided by 2 because the sound wave has to travel to the object and then return back.

Applications of SONAR:

1. Navigation: SONAR helps ships and submarines navigate through the ocean by detecting obstacles and mapping underwater terrain.

2. Fisheries: Fishermen use SONAR to locate schools of fish in the ocean.

3. Oceanography: SONAR is used to study and map the ocean floor, its depth, and underwater features like mountains, valleys, and ridges.

4. Submarine Detection: Military forces use SONAR to detect enemy submarines and other underwater vessels.

5. Search and Rescue: SONAR can help locate objects lost underwater, such as sunken ships or aircraft.

6. Underwater Communication: Some SONAR systems are used for communication between underwater vehicles or between submarines and ships.

Numericals on SONAR:

1) A SONAR system emits a sound wave in water. The sound wave takes 2 seconds to travel to the object and back. Calculate the distance of the object from the SONAR system. (Speed of sound in water = 1500 m/s)

Solution:

We can use the formula for distance:

$$\text{Distance}=\frac{\text{Speed of sound }\times \text{Time taken for echo to return / }}{\mathrm{2}}$$

Given:

• Speed of sound in water = 1500 m/s

• Time for sound to return = 2 seconds

Now, substituting the values into the formula:

$$\text{Distance}=\frac{\mathrm{1500 }\times \mathrm{2\; / }}{2}=1500\text{m}$$

Answer: The distance of the object from the SONAR system is 1500 meters.

2) A SONAR system detects an object at a distance of 3000 meters. If the speed of sound in water is 1500 m/s, how much time does it take for the sound wave to travel to the object and back?

Solution:

We can use the formula for time:

$$\text{Time}=\frac{\text{Distance}\times \mathrm{2\; / }}{\text{Speed of sound}}$$

Given:

• Distance to the object = 3000 m

• Speed of sound in water = 1500 m/s

Now, substituting the values into the formula:

$$\text{Time}=\frac{3000\times \mathrm{2\; / }}{1500}=4\text{seconds}$$

Answer: The time taken for the sound wave to travel to the object and back is 4 seconds.

Structure of Human Ear

• The ear is not just an organ for hearing, but it also helps in maintaining balance. 

Parts of the Human Ear

The human ear is divided into three main sections:

1. Outer Ear

2. Middle Ear

3. Inner Ear

Each part plays a crucial role in the process of hearing and balance.

1. Outer Ear

The outer ear consists of two main parts:

1) Pinna (Auricle):

• It is the visible part of the ear 
• The main function of the pinna is to collect sound waves and direct them into the ear canal.

2) Ear Canal (External Auditory Canal):

• The ear canal is a tube-like structure that leads sound waves from the pinna towards the eardrum.
• It is about 2.5 cm long in adults.

3) Eardrum (Tympanic Membrane):

• The eardrum is a thin, flexible membrane at the end of the ear canal.
• It vibrates when sound waves hit it, converting sound energy into mechanical vibrations. These vibrations are then passed on to the middle ear.

2. Middle Ear

• The middle ear contains three tiny bones called ossicles. These bones amplify and transmit sound vibrations from the eardrum to the inner ear.

1) Ossicles:

• There are three small bones in the middle ear:

1) Malleus (Hammer): 

• The first bone that is attached to the eardrum. It receives vibrations from the eardrum.

2) Incus (Anvil): 

• The second bone that is connected to the malleus. It transfers the vibrations from the malleus to the stapes.

3) Stapes (Stirrup): 

• The third and smallest bone. It transmits vibrations to the oval window, which is the entrance to the inner ear.

3. Inner Ear

• The inner ear is where the actual process of hearing occurs. It contains specialized structures that convert sound vibrations into electrical signals that the brain can understand.

1) Cochlea:

• The cochlea is a spiral-shaped, fluid-filled structure. 
• It convert mechanical vibrations into electrical signals, which are sent to the brain via the auditory nerve.

2) Auditory Nerve:

• The auditory nerve carries the electrical signals from the cochlea to the brain, which interprets them as sound. 

3) Semicircular Canals:

• The semicircular canals are part of the inner ear, but their main function is related to balance, not hearing.

Process of Hearing:

1) Sound Wave Collection: The pinna collects sound waves from the environment and directs them into the ear canal.

2) Vibration of Eardrum: The sound waves travel through the ear canal and hit the eardrum, causing it to vibrate.

3) Amplification by Ossicles: The vibrations are transmitted to the ossicles (malleus, incus, stapes) in the middle ear, where they are amplified.

4) Transmission to Cochlea: The stapes transmits the vibrations to the oval window of the cochlea in the inner ear.

5) Conversion to Electrical Signals: The movement of fluid inside the cochlea stimulates hair cells, which convert the mechanical vibrations into electrical signals.

6) Signal Transmission to Brain: The electrical signals are sent through the auditory nerve to the brain, which interprets them as sound.

Practice Numerical:

1) A person standing near a cliff makes a sound. The sound wave travels to the cliff and echoes back in 3 seconds. If the speed of sound in air is 340 m/s, calculate the distance between the person and the cliff. (Answer = 510 m)

2) A sound takes 5 seconds to travel to a building and echo back. If the speed of sound in air is 330 m/s, calculate the distance from the source of the sound to the building. (Answer = 825 m)

3) A SONAR device sends a sound wave to the seabed, and the echo returns in 10 seconds. If the speed of sound in water is 1500 m/s, calculate the depth of the water at that location. (Answer = 7500 m)

4) A boy shouts in front of a cliff and hears an echo after 2.5 seconds. Find the distance of the cliff. (Speed of sound = 342 m/s) (Answer = 427.5 m)

5) A SONAR device detects an object underwater. The time taken for the sound wave to travel to the object and return is 8 seconds. If the speed of sound in water is 1400 m/s, calculate the distance of the object from the SONAR device. (Answer = 5600 m)

$$$$

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