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Sound Waves: Characteristics and Applications Class 9 Science Notes

Sound Waves: Characteristics and Applications

Sound is a form of energy that we experience through our sense of hearing. It allows us to be aware of our surroundings, from the chirping of birds to the music on our phones.

Production of Sound

The fundamental principle behind sound production is vibration. A vibration is a rapid, periodic to-and-fro motion, also known as an oscillation. When an object vibrates, it can produce sound.

Plucking a stretched rubber band or a guitar string causes it to vibrate and create sound.
Striking a bell or a metal object makes it vibrate, producing a ringing sound. This property in metals is called sonority.
In a flute, the sound is produced by the vibration of the air column inside the hollow pipe.

The object that vibrates to produce sound is called the source of the sound.

How Humans and Animals Produce Sound

In humans and many animals, sound originates from the vocal cords, which are muscular flaps located in the larynx (or voice box) in the throat. When air passes through them, they vibrate. The tongue, lips, mouth, and nasal cavity then shape this sound into speech or music.

Some animals, like grasshoppers and crickets, produce sound by rubbing or striking body parts together, such as their wings or legs.

Tuning fork

A tuning fork is a standard U-shaped metal instrument used in sound experiments. The two sides of the 'U' are called prongs or tines. When a prong is struck on a soft pad, the fork vibrates and produces a sound of a nearly single frequency. The vibrations are so fast they are often invisible, but touching a vibrating prong to the surface of water will create visible waves, confirming the motion.

Propagation of Sound

For sound to travel from a source to our ears, it needs a path. Sound can travel through solids, liquids, and gases. The material through which sound travels is called a medium.

Example

If you place your ear against a long wooden desk and a friend scratches the other end, you'll hear the sound clearly through the solid wood. Similarly, striking two spoons together underwater demonstrates that sound can travel through liquids.

Sound needs a medium to propagate

Sound cannot travel through a vacuum, which is a space with no matter. This is a key characteristic of sound.

Note

Sound is a mechanical wave, which means it requires a material medium (solid, liquid, or gas) for its propagation.

The classic bell jar experiment proves this. An electric bell is placed inside a sealed glass jar.

With air inside the jar, the ringing bell is clearly heard.
As air is pumped out using a vacuum pump, the sound becomes fainter and fainter.
In a near-vacuum, the bell can be seen ringing, but almost no sound is heard.
When air is allowed back in, the sound returns to its original loudness.

This is why astronauts on a spacewalk cannot talk to each other directly. Since outer space is a near-vacuum, there is no medium to carry the sound waves. They must use radios built into their helmets to communicate.

Sound Waves

Sound travels through a medium not by moving the particles from the source to the listener, but by creating a disturbance that passes from particle to particle.

Imagine a vibrating object, like a tuning fork, in the air.

As it moves forward, it pushes and compresses the air particles in front of it, creating a region of high density and pressure. This is called a compression (C).
As it moves backward, it creates a region of low density and pressure. This is called a rarefaction (R).

As the object continues to vibrate, it produces a series of alternating compressions and rarefactions that travel away from the source. This propagating disturbance is what we call a sound wave.

The individual particles of the medium (like air molecules) only oscillate back and forth around their fixed positions. It is the disturbance, or energy, that travels, not the particles themselves.

Longitudinal Waves

In a sound wave, the particles of the medium vibrate back and forth in a direction that is parallel to the direction the wave is traveling. Waves with this characteristic are known as longitudinal waves. Sound is a longitudinal wave.

Note

There are also transverse waves, where particles vibrate perpendicular to the direction of wave propagation. Light is an example of a transverse wave. Unlike sound, light does not need a medium and can travel through a vacuum.

Energy of Sound Waves

The fact that sound can cause objects to move demonstrates that it carries energy. When a sound source vibrates, it transfers energy to the particles of the surrounding medium. This energy is then passed along from particle to particle as the wave propagates.

Example

If you stretch a thin sheet over a bowl and sprinkle some grains on it, making a loud sound near the bowl will cause the grains to jump. The energy from the sound wave makes the sheet vibrate, which in turn moves the grains.

Graphical Representation of a Sound Wave

A sound wave can be represented graphically by plotting the density of the medium against distance from the source at a specific moment in time.

The horizontal axis represents distance.
The vertical axis represents density.
A horizontal line shows the average, undisturbed density of the medium.
The wave-like curve shows how density varies.
- The highest points of the graph are called crests and correspond to the centers of compressions (maximum density).
- The lowest points are called troughs and correspond to the centers of rarefactions (minimum density).

Characteristics of a Sound Wave

Several key properties are used to describe a sound wave.

Wavelength, frequency and time period

Wavelength ( $\lambda$ ): The distance between two consecutive compressions (crests) or two consecutive rarefactions (troughs). Its SI unit is the metre (m).
Frequency ( $\nu$ ): The number of complete oscillations (or the number of compressions/rarefactions) that pass a fixed point per unit time. Its SI unit is the hertz (Hz), where $1 \text{ Hz} = 1 \text{ oscillation per second } (s^{-1})$ .
Time Period (T): The time it takes for one complete oscillation to pass a fixed point. Its SI unit is the second (s).

Frequency and time period are inversely related. A wave with a high frequency has a short time period. $v=\frac{1}{T}$

Example

Example If there are 10 density oscillations in 2 seconds at a given position, then calculate the (i) frequency of sound wave, and (ii) its time period.

To Find

(i) Frequency of the sound wave, $\nu$ (ii) Time period, T

Formula

$\text{Frequency} = \frac{\text{number of oscillations}}{\text{time taken}}$

$\text{Time Period} = \frac{1}{\text{Frequency}}$

Solution

(i) Calculate the frequency

$\nu = \frac{10}{2 \mathrm{~s}} = 5 \mathrm{~Hz}$

Answer for part (i) = $5 \text{ Hz}$

(ii) Calculate the time period

The time for one oscillation is: $T = \frac{2 \mathrm{~s}}{10} = 0.2 \mathrm{~s}$ Alternatively, using the frequency: $T = \frac{1}{\nu} = \frac{1}{5 \text{ Hz}} = 0.2 \text{ s}$

Answer for part (ii) = $0.2 \text{ s}$

Amplitude and intensity of the sound waves

Amplitude: The maximum change in the density (or pressure) of the medium from its average value. It is represented by the height of a crest or depth of a trough on the wave graph. A wave with a larger amplitude carries more energy.
Intensity: The amount of sound energy that passes through a unit area in a unit of time. A higher amplitude wave has a higher intensity.

Speed of Sound

The speed of sound (v) is the speed at which the disturbance (a compression or rarefaction) travels through the medium. It is defined as the distance a point on the wave, such as a crest, travels per unit time.

The speed, wavelength, and frequency of a wave are related by the following equation: $\text{speed} = \frac{\text{distance}}{\text{time}} \quad \Rightarrow v=\frac{\lambda}{T}$ Since $\nu = 1/T$ , we can also write: $v = \lambda \times v$ $\text{speed} = \text{wavelength} \times \text{frequency}$

The speed of sound depends on the properties of the medium.

It is fastest in solids, slower in liquids, and slowest in gases.
It also depends on temperature and humidity. For instance, the speed of sound in air increases as the temperature increases. At $0 \text{ °C}$ , it is about $331 \text{ m s}^{-1}$ , and at $22 \text{ °C}$ , it is about $344 \text{ m s}^{-1}$ .

Example

Example Human hearing roughly spans 20 Hz to 20 kHz. What are the corresponding wavelengths in air for these two frequencies? Use the speed of sound in air as

344 \mathrm{~m} \mathrm{~s}^{-1}

.

Given

Speed of sound in air, $v = 344 \text{ m s}^{-1}$
(i) Frequency, $\nu = 20 \text{ Hz}$
(ii) Frequency, $\nu = 20 \text{ kHz} = 20000 \text{ Hz}$

To Find

The corresponding wavelengths ( $\lambda$ ) for each frequency.

Formula

$\lambda = \frac{\text{speed of the wave}}{\text{frequency}} = \frac{v}{\nu}$

Solution

(i) Wavelength for 20 Hz

$\lambda = \frac{344 \mathrm{~m} \mathrm{~s}^{-1}}{20 \mathrm{~s}^{-1}} = 17.2 \mathrm{~m}$

Answer for part (i) = $17.2 \text{ m}$

(ii) Wavelength for 20 kHz

$\lambda = \frac{344 \mathrm{~m} \mathrm{~s}^{-1}}{20000 \mathrm{~s}^{-1}} = 0.0172 \mathrm{~m} = 1.72 \mathrm{~cm}$

Answer for part (ii) = $1.72 \text{ cm}$

Example

Example During a thunderstorm, lightning is seen before thunder is heard because sound travels much slower than light. If the time delay between seeing the lightning flash and hearing the thunder is measured to be 5 s, estimate the distance to the lightning strike. Use the speed of sound in air as

340 \mathrm{~m} \mathrm{~s}^{-1}

. Assume that light reaches you almost instantaneously.

Given

Time delay, $t = 5 \text{ s}$
Speed of sound, $v = 340 \text{ m s}^{-1}$

To Find

Distance to the lightning strike.

Formula

$\text{Distance} = \text{speed} \times \text{time}$

Solution

Substitute the given values into the formula:

$\text{Distance} = 340 \mathrm{~m} \mathrm{~s}^{-1} \times 5 \mathrm{~s} = 1700 \mathrm{~m}$ This is equal to $1.7$ km.

Final Answer The lightning struck about $1.7 \text{ km}$ away.

Example

Example From the graphical representation of a sound wave propagating in steel (Fig. 10.22), find its wavelength. Calculate its frequency and time period if the speed of sound in steel is

5000 \mathrm{~m} \mathrm{~s}^{-1}

.

Given

From the graph, wavelength $\lambda = 50 \text{ m}$
Speed of sound in steel, $v = 5000 \text{ m s}^{-1}$

To Find

(i) Frequency, $\nu$ (ii) Time period, T

Formula

$v = \frac{v}{\lambda}$ $T = \frac{1}{v}$

Solution

(i) Calculate the frequency

$v = \frac{5000 \mathrm{~m} \mathrm{~s}^{-1}}{50 \mathrm{~m}} = 100 \mathrm{~Hz}$

Answer for part (i) = $100 \text{ Hz}$

(ii) Calculate the time period

$T = \frac{1}{100 \mathrm{~Hz}} = 0.01 \mathrm{~s}$

Answer for part (ii) = $0.01 \text{ s}$

Human perception of sound

While physical properties like frequency and amplitude can be measured precisely, our experience of sound is subjective.

Pitch: Our perception of frequency. A high-frequency sound (like a whistle) is perceived as having a high pitch (shrill). A low-frequency sound (like thunder) is perceived as having a low pitch (deep).
Loudness: Our perception of amplitude. A sound with a larger amplitude is perceived as being louder. Loudness is commonly measured in decibels (dB).

Unwanted or harmful sound is called noise. Prolonged exposure to loud sounds can cause noise pollution and lead to health issues, including hearing loss.

Audible Range of Hearing

Humans can only hear sounds within a specific range of frequencies, known as the audible range, which is typically from 20 Hz to 20,000 Hz (or 20 kHz).

Sounds with frequencies below 20 Hz are called infrasonic waves (or infrasound). Elephants can detect infrasound.
Sounds with frequencies above 20 kHz are called ultrasonic waves (or ultrasound). Dogs, cats, bats, and dolphins can detect ultrasound.

Reflection of Sound

Just like light, sound waves can bounce off surfaces. This is called the reflection of sound. Sound follows the same laws of reflection: the angle of incidence is equal to the angle of reflection, and the incident wave, reflected wave, and the normal to the surface all lie in the same plane.

Echo

An echo is a reflected sound that we hear distinctly after the original sound. For our brain to distinguish between the original sound and its reflection, there must be a time gap of at least 0.1 seconds between them.

We can calculate the minimum distance to a reflecting surface to hear an echo. If the speed of sound is about $340 \text{ m s}^{-1}$ :

Distance traveled by sound in 0.1 s = $340 \text{ m s}^{-1} \times 0.1 \text{ s} = 34 \text{ m}$ .
This is the total distance to the surface and back.
Therefore, the minimum distance to the reflecting surface must be half of this, which is 17 m.

Echoes are clearer from hard, smooth surfaces. Soft surfaces like curtains absorb sound, reducing reflection.

Example

Example You clap in an empty corridor and hear an echo after 0.5 s. If the speed of sound in air is

340 \mathrm{~m} \mathrm{~s}^{-1}

, calculate your distance from the wall.

Given

Total time for sound to travel to the wall and back, $t = 0.5 \text{ s}$
Speed of sound, $v = 340 \text{ m s}^{-1}$

To Find

Distance from the wall.

Formula

The total distance traveled is $d = v \times t$ . The distance to the wall is half of this total distance. $\text{distance from wall} = \frac{v \times t}{2}$

Solution

Substitute the given values into the formula:

$\text{distance from wall} = \frac{340 \mathrm{~m} \mathrm{~s}^{-1} \times 0.5 \mathrm{~s}}{2} = \frac{170 \text{ m}}{2} = 85 \text{ m}$

Final Answer Your distance from the wall is $85 \text{ m}$ .

Reverberation

In a large hall or auditorium, sound can undergo multiple reflections from the walls, ceiling, and floor. The persistence of sound due to these repeated reflections is called reverberation. If the reflections arrive very quickly (with a time difference of less than 0.05 s), they merge with the original sound, which can make it sound garbled or unclear.

Concert halls are designed to control reverberation. Curved ceilings help to distribute sound evenly, while sound-absorbing materials like carpets, curtains, and upholstered seats reduce unwanted reflections.

Ultrasonic and Infrasonic Waves, and their Applications

Sounds outside the human audible range have many important uses.

Echolocation

Echolocation is the process of using reflected sound waves to locate objects.

Bats are a prime example. They emit short bursts of high-frequency ultrasonic waves. By listening to the echoes that bounce back from objects and prey, they can navigate and hunt in complete darkness. Dolphins and whales also use echolocation.

Sonar

Humans have adopted this principle in a technology called SONAR (Sound Navigation and Ranging).

Sonar systems are used on ships to map the seabed or locate underwater objects like submarines, shipwrecks, or schools of fish.
The system sends out ultrasonic waves and detects the reflected waves (echoes). By measuring the time it takes for the echo to return, the distance to the object can be calculated.

Example

Example A naval sonar signal sent into seawater returns after 0.90 s. The speed of sound in seawater is

1530 \mathrm{~m} \mathrm{~s}^{-1}

. How far is the object?

Given

Total time for the signal to travel to the object and back, $t_{total} = 0.90 \text{ s}$
Speed of sound in seawater, $v = 1530 \text{ m s}^{-1}$

To Find

Distance to the object.

Formula

$\text{Distance} = \text{speed} \times \text{time}$

Solution

First, find the time taken for the signal to travel one way to the object. This is half the total time.

$t_{one-way} = \frac{0.90 \mathrm{~s}}{2} = 0.45 \mathrm{~s}$

Now, calculate the distance using this one-way travel time.

$\text{Distance} = 1530 \mathrm{~m} \mathrm{~s}^{-1} \times 0.45 \mathrm{~s} = 688.5 \mathrm{~m}$

Final Answer The object is $688.5 \text{ m}$ away.

Sound Waves: Characteristics and ApplicationsClass 9 Science Notes

Sound Waves: Characteristics and Applications

Production of Sound

How Humans and Animals Produce Sound

Tuning fork

Propagation of Sound

Sound needs a medium to propagate

Sound Waves

Longitudinal Waves

Energy of Sound Waves

Graphical Representation of a Sound Wave

Characteristics of a Sound Wave

Wavelength, frequency and time period

To Find

Formula

Solution

Amplitude and intensity of the sound waves

Speed of Sound

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Given

To Find

Formula

Solution

Human perception of sound

Audible Range of Hearing

Reflection of Sound

Echo

Given

To Find

Formula

Solution

Reverberation

Ultrasonic and Infrasonic Waves, and their Applications

Echolocation

Sonar

Given

To Find

Formula

Solution