Class 9Science · PhysicsFull chapter

Sound

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Production and Propagation of Sound

Quick answer Sound is produced when an object vibrates, and it travels outward from the source only through a material medium as a longitudinal wave.

Every sound we hear starts with something vibrating — moving rapidly to and fro about a fixed position. When you pluck a guitar string, strike a tabla, or speak, the string, the drum membrane, or your vocal cords vibrate rapidly. These vibrations disturb the particles of the medium (usually air) touching the vibrating body.

A vibrating particle does not travel all the way to your ear by itself. Instead, it pushes the neighbouring particles, which push the next ones, and so on. Each particle vibrates about its own mean position and passes the disturbance forward, carrying energy but not matter. This travelling disturbance is called a sound wave.

In air, the particles vibrate parallel to the direction in which the sound travels. Such a wave is called a longitudinal wave. As the vibrating source moves forward, it pushes air particles closer together, creating a region of high pressure and high density called a compression (C). As it moves back, it creates a region of low pressure and low density called a rarefaction (R). A sound wave is really a series of compressions and rarefactions travelling one after another through the medium.

Worked example: A bell is rung inside a glass jar connected to a vacuum pump. Initially, with air inside the jar, the ringing is heard clearly. As air is pumped out, the sound becomes fainter and fainter, and once almost all the air is removed, no sound is heard at all — even though the hammer can still be seen striking the bell through the glass. This shows that sound needs a material medium (solid, liquid or gas) to travel, and it cannot travel through vacuum. This is also why astronauts on a spacewalk cannot talk to each other directly through the vacuum of space and must use radio signals instead.

Nature of a sound wave Sound wave = longitudinal mechanical wave Needs a material medium; cannot pass through vacuum
Remember
  • Sound is produced by a vibrating object and stops when the vibration stops.
  • Sound travels as a longitudinal wave: particle vibration is parallel to the direction of wave travel.
  • A sound wave consists of alternate compressions (high pressure/density) and rarefactions (low pressure/density).
  • Sound needs a material medium (solid, liquid or gas); it cannot travel through vacuum.
  • The wave carries energy from the source to the listener without any net transfer of matter.

Characteristics of a Sound Wave

Quick answer A sound wave is described by its wavelength, frequency, time period, amplitude, and speed, connected by the key relation v = f × λ.

Just like any wave, a sound wave can be described completely using a few measurable quantities.

Wavelength (λ): The distance between the centres of two consecutive compressions or two consecutive rarefactions is called the wavelength. It is measured in metres (m).

Frequency (f or ν): The number of complete oscillations (or the number of compressions, or the number of rarefactions) produced per second is the frequency of the sound wave. Its SI unit is the hertz (Hz); 1 Hz means one complete vibration per second. Frequency decides the pitch of a sound — a high-frequency sound has a high (shrill) pitch, and a low-frequency sound has a low (grave) pitch.

Time period (T): The time taken to complete one full oscillation (one compression plus one rarefaction) is the time period. It is the reciprocal of frequency.

Amplitude (A): The maximum displacement of the particles of the medium from their mean (rest) position, on either side, is called the amplitude of the wave. A larger amplitude makes the sound louder; a smaller amplitude makes it softer. Amplitude does not change the pitch.

Speed (v): The distance travelled by the wave in one second is its speed, measured in m/s. Speed, frequency and wavelength are connected by one of the most useful relations in this chapter:

v = f × λ

Worked example: A sound wave has a frequency of 2 kHz (2000 Hz) and a wavelength of 35 cm (0.35 m) in a certain medium. Find its speed, and the time it will take to travel a distance of 1.5 km.

Speed: v = f × λ = 2000 Hz × 0.35 m = 700 m/s.

Time to travel 1.5 km (1500 m): time = distance ÷ speed = 1500 m ÷ 700 m/s ≈ 2.14 s.

Speed of a wave v = f × λ v in m/s, f in Hz, λ in m
Time period T = 1/f T in seconds
Frequency f = 1/T Reciprocal of time period
Remember
  • Wavelength (λ) is the distance between two consecutive compressions or two consecutive rarefactions; SI unit metre (m).
  • Frequency (f) is the number of oscillations per second, measured in hertz (Hz); it decides the pitch of the sound.
  • Time period (T) is the time for one complete oscillation; T = 1/f.
  • Amplitude decides the loudness of the sound — larger amplitude means louder sound.
  • Speed of a sound wave relates to frequency and wavelength by v = f × λ.

Speed of Sound and the Audible Range of Hearing

Quick answer Sound travels fastest through solids and slowest through gases, and the human ear can normally detect only frequencies between 20 Hz and 20,000 Hz.

The speed of sound depends on the medium through which it travels. Since sound propagates by particles colliding with their neighbours, it travels fastest where particles are packed closely together and are strongly linked — that is, in solids. It travels slower in liquids, and slowest in gases, where particles are far apart.

In general: speed of sound in solids > speed of sound in liquids > speed of sound in gases.

For example, sound travels at about 346 m/s in air at 25°C (often approximated as 340–344 m/s in numerical problems), about 1493 m/s in water at 25°C, and about 5960 m/s in steel — these are the standard reference values used in the NCERT speed-of-sound table. The speed of sound in a gas such as air also increases as the temperature of the gas increases, because warmer particles vibrate faster and transfer the disturbance more quickly.

The human ear is not able to detect every frequency of sound. A healthy young person can normally hear sound with frequencies from about 20 Hz to 20,000 Hz (20 kHz). This range is called the audible range of hearing.

Sound waves with a frequency below 20 Hz are called infrasonic sound (or infrasound). Humans cannot hear these, but animals like elephants, rhinoceroses and whales produce and use infrasound to communicate over long distances, and such very low-frequency vibrations are also associated with events like earthquakes.

Sound waves with a frequency above 20,000 Hz are called ultrasonic sound (or ultrasound). Bats and dolphins can produce and detect ultrasound; bats use it to locate obstacles and prey in the dark, a technique called echolocation.

Worked example: A dog can hear sound of frequency 25,000 Hz, but a human standing next to it cannot. Since the human audible range is only up to about 20,000 Hz, a frequency of 25,000 Hz lies in the ultrasonic range, above the upper limit of human hearing. A dog's audible range extends higher than a human's, so the dog can detect it while the human cannot.

Order of speed of sound v(solid) > v(liquid) > v(gas) Depends on how tightly the particles of the medium are packed
Human audible range 20 Hz ≤ f ≤ 20,000 Hz Below 20 Hz: infrasonic; above 20,000 Hz: ultrasonic
Remember
  • Speed of sound: solids > liquids > gases, because particles are more closely packed and transfer vibrations faster.
  • Speed of sound in a gas increases as temperature increases.
  • The audible range for a normal human ear is about 20 Hz to 20,000 Hz.
  • Infrasonic sound has frequency below 20 Hz (used by elephants, whales; linked to earthquakes).
  • Ultrasonic sound has frequency above 20,000 Hz (used by bats and dolphins for echolocation).

Reflection of Sound: Echo and Reverberation

Quick answer Sound reflects off hard, large surfaces obeying the same laws as light, and a reflected sound heard as a distinct repetition is called an echo.

Like light, sound also bounces back when it strikes a large, hard surface, such as a wall, a cliff, or a hillside. This is called reflection of sound, and it obeys the same laws as the reflection of light:

  • The angle of incidence is equal to the angle of reflection.
  • The incident sound wave, the reflected sound wave, and the normal at the point of incidence all lie in the same plane.

When a reflected sound reaches a listener late enough to be heard as a separate, repeated sound (rather than blending with the original), it is called an echo. This happens because the human ear keeps sensing a sound for about 0.1 second after the source stops (this is called the persistence of hearing). So, for an echo to be heard distinctly, the reflected sound must reach the ear at least 0.1 s after the original sound.

This sets a minimum distance between the source/listener and the reflecting surface. If v is the speed of sound, the sound must travel to the surface and back (a distance of 2d, where d is the distance to the surface) in at least 0.1 s.

Worked example: Taking the speed of sound in air as 344 m/s, find the minimum distance of a reflecting surface from a source so that an echo can be heard distinctly.

Total distance travelled by sound (to the surface and back) = v × t = 344 m/s × 0.1 s = 34.4 m.

Minimum distance of the surface, d = 34.4 m ÷ 2 = 17.2 m.

So the reflecting surface must be at least about 17.2 m away for a distinct echo to be heard in air.

In large halls or auditoriums, sound reflects again and again off the walls, ceiling and floor before dying out, causing the original sound to persist and become blurred — this is called reverberation. Excessive reverberation is reduced by covering the walls and ceiling with sound-absorbent materials such as compressed fibreboard, rough plaster, or heavy curtains.

Law of reflection of sound ∠(incidence) = ∠(reflection) Incident wave, reflected wave and normal lie in the same plane
Minimum distance for a distinct echo d(min) = (v × t)/2 t = 0.1 s (persistence of hearing); v = speed of sound in the medium
Remember
  • Reflection of sound follows the same laws as reflection of light (angle of incidence = angle of reflection).
  • An echo is a distinctly heard repetition of sound caused by reflection off a large, hard surface.
  • The human ear retains a sound sensation for about 0.1 s (persistence of hearing), so an echo needs at least this much delay to be heard separately.
  • In air (v ≈ 344 m/s), the reflecting surface must be at least about 17.2 m away for an echo to be heard.
  • Reverberation is the persistence of sound in a large enclosed space due to repeated reflections; it is controlled using sound-absorbent materials.

Ultrasound and the Human Ear

Quick answer Ultrasound (frequency above 20 kHz) has many practical uses such as SONAR and medical imaging, and the human ear converts sound vibrations into nerve signals the brain can understand.

Ultrasound is sound of frequency greater than 20,000 Hz — too high for humans to hear, but extremely useful because it travels in narrow, well-defined beams and can carry a lot of energy.

Some common uses of ultrasound:

  • Cleaning hard-to-reach parts of objects such as spiral tubes and odd-shaped machine parts: objects are placed in a cleaning solution, and ultrasonic waves are passed through the solution to dislodge dirt and grease particles without damaging the object.
  • Detecting cracks and flaws inside metal blocks used in structures and machines, which are otherwise invisible from the outside; ultrasonic waves pass through the metal and are reflected back by any flaw, revealing its presence and position.
  • In medicine, ultrasound is used to obtain images of internal organs (ultrasonography), to examine the heart (echocardiography), and to break down kidney stones into fine grains without surgery.
  • In SONAR (SOund Navigation And Ranging), a ship or submarine sends out ultrasonic waves and detects the echo reflected from an object underwater, such as the sea bed, a shoal of fish, or another vessel. Measuring the time taken for the echo to return allows the distance of the object to be calculated, since the reflected wave travels the distance to the object and back.

Worked example: A SONAR device on a ship sends an ultrasonic signal that returns as an echo from the sea bed after 2 s. If the speed of sound in sea water is 1500 m/s, find the depth of the sea at that point.

Total distance travelled by the signal (down to the sea bed and back up) = v × t = 1500 m/s × 2 s = 3000 m.

Depth of the sea, d = 3000 m ÷ 2 = 1500 m.

The human ear receives sound waves and converts them into electrical signals the brain can interpret. It has three main parts:

  • Outer ear: the visible part (pinna) collects sound waves from the surroundings and funnels them through the auditory canal to the eardrum (tympanic membrane), which vibrates back and forth.
  • Middle ear: contains three small, delicate bones — the hammer, anvil and stirrup — joined end to end. These bones amplify the vibrations of the eardrum and pass them on to the inner ear.
  • Inner ear: the vibrations enter a coiled, fluid-filled structure called the cochlea, which converts the pressure vibrations into electrical signals. These signals travel through the auditory nerve to the brain, which interprets them as sound.
SONAR distance/depth d = (v × t)/2 v = speed of sound in water (≈1500 m/s); t = total time for the echo to return
Remember
  • Ultrasound is sound with frequency above 20,000 Hz, used because it can travel in a directed beam.
  • Applications of ultrasound include cleaning, detecting internal flaws in metals, medical imaging (ultrasonography, echocardiography), and breaking kidney stones.
  • SONAR uses reflected ultrasonic waves to find the distance/depth of an underwater object: d = (v × t)/2.
  • The human ear has three main parts: outer ear (pinna, ear canal, eardrum), middle ear (hammer, anvil, stirrup), and inner ear (cochlea).
  • The cochlea converts vibrations into electrical signals, carried by the auditory nerve to the brain.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

Sound wave = longitudinal mechanical wave
Nature of a sound wave
v = f × λ
Speed of a wave
T = 1/f
Time period
f = 1/T
Frequency
v(solid) > v(liquid) > v(gas)
Order of speed of sound
20 Hz ≤ f ≤ 20,000 Hz
Human audible range
∠(incidence) = ∠(reflection)
Law of reflection of sound
d(min) = (v × t)/2
Minimum distance for a distinct echo
d = (v × t)/2
SONAR distance/depth

Test yourself

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0 correct · 0/12 answered
Q1 Production of sound easy

Sound is produced in an object mainly due to its:

Q2 Propagation of sound easy

Sound cannot travel through:

Q3 Compression and rarefaction medium

In a sound wave travelling through air, the region where particles are crowded together and pressure/density is high is called:

Q4 Characteristics of sound wave easy

The SI unit of wavelength of a sound wave is:

Q5 Time period medium

If the frequency of a sound wave is 500 Hz, its time period is:

Q6 Speed of sound (v = fλ) medium

A wave has a wavelength of 2 m and a frequency of 150 Hz. Its speed is:

Q7 Loudness and amplitude easy

The loudness of a sound mainly depends on its:

Q8 Audible range easy

The audible range of frequency for a normal human ear is about:

Q9 Ultrasonic sound easy

Bats mainly use which type of sound waves to locate obstacles and prey in the dark?

Q10 Echo hard

Taking the speed of sound in air as 344 m/s and the minimum time gap needed for a distinct echo as 0.1 s, the reflecting surface must be at least this far from the source:

Q11 Human ear medium

The three small bones found in the middle ear, joined end to end, are commonly called:

Q12 Human ear medium

The part of the human ear that converts sound vibrations into electrical signals for the brain is the:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 A pendulum oscillates 10 times in 4 seconds. Will you be able to hear the sound produced by this pendulum? Give a reason.Audible range / infrasonic sound

Given: Number of oscillations = 10, time taken = 4 s.

Formula: Frequency, f = number of oscillations ÷ time taken.

Substitution: f = 10 ÷ 4 s = 2.5 Hz.

Result: The frequency of oscillation of the pendulum is 2.5 Hz, which is far below the human audible range of 20 Hz to 20,000 Hz. So this is an infrasonic vibration, and we will not be able to hear the sound produced by this pendulum.

2 Sound travels at a speed of 339 m/s in a certain medium. If the wavelength of the sound is 1.5 cm, what is its frequency? Will this sound be audible to a human ear?Speed, frequency and wavelength

Given: Speed of sound, v = 339 m/s; wavelength, λ = 1.5 cm = 0.015 m.

Formula: v = f × λ, so f = v ÷ λ.

Substitution: f = 339 m/s ÷ 0.015 m = 22,600 Hz.

Result: The frequency of the wave is 22,600 Hz (22.6 kHz). Since this is greater than 20,000 Hz, it lies in the ultrasonic range and is not audible to a normal human ear.

3 An echo returned in 3 s. What is the distance of the reflecting surface from the source, if the speed of sound is 342 m/s?Echo

Given: Speed of sound, v = 342 m/s; time taken for the echo to return, t = 3 s.

Formula: Total distance travelled by sound = v × t; this equals twice the distance to the reflecting surface, so distance, d = (v × t) ÷ 2.

Substitution: d = (342 m/s × 3 s) ÷ 2 = 1026 m ÷ 2.

Result: d = 513 m. The reflecting surface is 513 m away from the source.

4 Why are the ceilings of concert halls and conference halls generally curved?Reflection of sound

The ceilings of large concert halls, cinema halls and conference halls are generally curved so that sound, after reflecting off the ceiling, gets evenly distributed to every corner of the hall, including seats far from the stage.

A curved (usually concave) ceiling reflects sound waves in a controlled way, directing them down towards the audience instead of allowing sound to reflect unevenly and create loud spots, echoes, or weak-sound regions. This design is based on the laws of reflection of sound (angle of incidence = angle of reflection) and ensures that everyone in the hall hears the sound clearly.

5 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 after which he will hear the second successive compression?Time period

Given: Frequency of the tone, f = 500 Hz; distance from the source = 450 m (this value is not needed for this part of the question).

Formula: Successive compressions of a sound wave are separated in time by exactly one time period, T = 1/f.

Substitution: T = 1 ÷ 500 Hz = 0.002 s.

Result: The listener will hear the second successive compression 0.002 s (2 milliseconds) after the first. This time gap does not depend on how far the listener is from the source.

6 How do bats use ultrasonic sounds to catch their prey in the dark?Ultrasound / echolocation

Bats emit very high-frequency ultrasonic squeaks (above 20,000 Hz) as they fly. These sound waves travel outward, strike nearby objects such as flying insects, walls or trees, and bounce back as an echo.

The bat's ears are extremely sensitive to these reflected ultrasonic waves. By sensing the direction the echo comes from and the time delay between sending the squeak and receiving its echo, the bat can work out the direction, distance and even size of the prey or obstacle. This technique of using reflected ultrasound to sense the surroundings is called echolocation, and it lets bats hunt and navigate accurately even in complete darkness.

Previous-year board questions 4

Q1 Define the wavelength of a sound wave and state its SI unit. CBSE 2020 1 mark

The wavelength of a sound wave is the distance between the centres of two consecutive compressions or two consecutive rarefactions. Its SI unit is the metre (m).

Q2 Differentiate between infrasonic sound and ultrasonic sound, giving one example of each. CBSE 2019 2 marks

Infrasonic sound has a frequency below 20 Hz, which is lower than the human audible range, so it cannot be heard by the human ear. Example: the very low-frequency sound produced by elephants and whales to communicate over long distances, and low-frequency vibrations associated with earthquakes.

Ultrasonic sound has a frequency above 20,000 Hz, higher than the human audible range, so it also cannot be heard by the human ear. Example: the high-frequency squeaks emitted by bats for echolocation, and the ultrasound waves used in SONAR and medical ultrasonography.

Q3 A person clapped his hands near a hillside and heard the echo after 4 s. If the speed of sound in air is 340 m/s, calculate the distance of the hillside from the person. CBSE 2022 3 marks

Given: Speed of sound, v = 340 m/s; time taken for the echo to be heard, t = 4 s.

Formula: Distance of reflecting surface, d = (v × t) ÷ 2, since the sound travels to the hillside and back.

Substitution: d = (340 m/s × 4 s) ÷ 2 = 1360 m ÷ 2.

Result: d = 680 m. The hillside is 680 m away from the person.

Q4 What is SONAR? Explain briefly how it works and state two of its applications. Also: a SONAR device on a submarine sends an ultrasonic signal that returns as an echo after reflecting off an underwater object after 2.4 s. If the speed of sound in water is 1500 m/s, calculate the distance of the object from the submarine. CBSE 2023 5 marks

SONAR stands for SOund Navigation And Ranging. It is a device that uses ultrasonic waves to determine the distance, direction and speed of underwater objects.

Working: A transmitter in the SONAR device sends out ultrasonic waves into the water. These waves travel through the water, strike an underwater object (such as the sea bed, a submarine, or a shoal of fish), and get reflected back as an echo. A receiver in the device detects this reflected wave and records the total time taken for the round trip. Since the speed of sound in water is known, the distance of the object can be calculated from this time.

Applications: (i) Measuring the depth of the sea and mapping the sea bed. (ii) Detecting the position of underwater objects such as submarines, shipwrecks, or shoals of fish.

Numerical part — Given: Speed of sound in water, v = 1500 m/s; time for echo to return, t = 2.4 s.

Formula: Distance of the object, d = (v × t) ÷ 2.

Substitution: d = (1500 m/s × 2.4 s) ÷ 2 = 3600 m ÷ 2.

Result: d = 1800 m. The underwater object is 1800 m from the submarine.

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