Physics Chapter 4 — Sound | Class 9 CBSE 2026-27

Section 4.1 – 4.2 · The Source

Production of Sound

Every sound begins with a vibration. The sensation our ears feel from these vibrations is what we call sound.

Definition

Sound is a form of energy which produces the sensation of hearing in our ears. It is produced by vibrating objects — a struck bell, a tuning fork, plucked strings, or our vocal cords. The object that produces sound is the source.

🔔 How different sources make sound

A struck school bell keeps vibrating — a pith ball near it gets pushed away, proving the vibration.
A tuning fork's prongs vibrate when struck on a rubber pad.
Humans speak using vibrating vocal cords in the larynx; the tongue, lips and nasal cavity shape the sound.
Crickets and grasshoppers rub their wings or legs; musical instruments vibrate strings, membranes or air columns.

Section 4.3 · Travelling Energy

Propagation of Sound

A sound wave carries energy from the source to your ear — but the particles of the medium do not travel with it. They only vibrate to and fro, passing the disturbance along.

What a wave is

A wave is a disturbance that moves through a medium as the particles set their neighbours into motion. Sound is a mechanical wave: it needs particles of a medium to travel. The direction the wave travels is the direction of propagation.

Two kinds of waves

Longitudinal waves

Particles vibrate parallel to the direction the wave moves. The wave is made of compressions (particles crowded together — high density and pressure) and rarefactions (particles spread apart — low density and pressure). Sound in air is a longitudinal wave. A slinky pushed and pulled along its length shows this beautifully.

Transverse waves

Particles vibrate perpendicular to the wave direction, forming crests (maximum upward displacement) and troughs (maximum downward). There is no pressure variation — only a change of shape. Examples: water-surface waves, a plucked string, and (non-mechanical) light, heat and radio waves, which need no medium.

Longitudinal: compressions (C) & rarefactions (R) along the wave. Transverse: crests & troughs across it. Wavelength λ = distance between two successive C's (or crests).

Section 4.4 · No Vacuum Allowed

Sound Needs a Medium to Travel

Because sound is carried by vibrating particles, it cannot travel through empty space.

🔔 The bell-jar experiment

An electric bell is hung inside a sealed glass jar connected to a vacuum pump. With air in the jar, the ringing is clearly heard. As the air is pumped out, the sound grows fainter and fainter; with a near-perfect vacuum, no sound is heard at all — even though the bell is still ringing. This proves a material medium is essential for sound to propagate.

🚀

There is almost no air in outer space, so astronauts on a spacewalk cannot hear each other directly — they talk through radios, whose signals (electromagnetic waves) need no medium.

Section 4.5 · Describing a Wave

Characteristics of a Sound Wave

A sound wave is fully described by a few quantities. In the graph of a wave, a compression is a crest (high density) and a rarefaction is a trough (low density).

Amplitude (A)

Maximum displacement of particles from the mean position.

Frequency (ν)

Number of complete waves per second. 1 Hz = 1 vibration/s.

Time Period (T)

Time for one complete oscillation.

Wavelength (λ)

Distance between two successive compressions (or rarefactions).

Velocity (v)

m/s

Distance the wave travels per unit time.

Oscillation

↻

One full cycle of density from max → min → max.

Amplitude is the height of a crest; wavelength λ is the distance between two crests (or two troughs).

Frequency & Time Period

ν = 1 / T

Frequency is the reciprocal of the time period (νT = 1). SI unit of frequency: hertz (Hz).

Section 4.5.1 · Telling Sounds Apart

Loudness · Pitch · Quality

In an orchestra, many instruments reach the ear together yet we recognise each one. Three characteristics let us tell sounds apart.

1 · Loudness (Intensity)

The sensation that distinguishes a loud sound from a faint one. Strike a tuning fork harder → bigger amplitude → louder sound. Loudness ∝ (amplitude)², measured in decibels (dB). It also depends on the surface area of the source, the density of the medium, and falls off as the inverse square of the distance. Intensity = power / area = J s⁻¹ m⁻² (W/m²). Loudness is subjective; intensity is an objective, measurable quantity.

2 · Pitch (Frequency)

Pitch tells a shrill sound from a grave one and depends on frequency: higher frequency → higher pitch. A woman's voice is usually shriller (higher pitch) than a man's.

3 · Quality (Timbre)

Quality lets us tell a flute from a violin even at the same pitch and loudness — it depends on the waveform. A tone is a sound of a single frequency; a note is a mixture of several frequencies.

Section 4.6 – 4.7 · The Master Equation

The Wave Equation

In one time period T, a wave travels exactly one wavelength λ. From that single idea comes the equation that governs all waves.

Wave Equation

v = ν λ

velocity = frequency × wavelength. Since v = λ/T and ν = 1/T, we get v = νλ. Knowing any two of v, ν, λ gives the third. This holds for all waves — sound, water and light.

Example A

A source emits sound of wavelength 1.7 × 10⁻² m. If the speed is 343.4 m/s, find the frequency. [2 Marks]

Formula

v = νλ → ν = v/λ

Substitute

ν = 343.4 / (1.7 × 10⁻²)

Answer

ν = 2.02 × 10⁴ Hz

Example B

A bat hears sound of frequency 100 kHz. Find the wavelength in air (speed = 344 m/s). [2 Marks]

Formula

λ = v/ν

Substitute

λ = 344 / (100 × 10³) = 344 / 10⁵

Answer

λ = 3.44 × 10⁻³ m = 3.44 mm

Section 4.8 – 4.9 · How Fast?

Speed of Sound in Different Media

We see lightning before we hear thunder — sound travels much slower than light. Its speed depends on the medium and the conditions.

What the speed depends on

Medium properties: for a gas, speed ∝ 1/√(density).
Nature of medium: solids > liquids > gases (closely packed particles pass on vibrations faster).
Temperature: speed rises with temperature — about 0.6 m/s per 1 °C. In air: 332 m/s at 0 °C, 344 m/s at 20 °C.
Humidity: sound travels faster in humid air than in dry air.
Not on pressure — if the temperature is constant, pressure has no effect.

Medium (at 20 °C)	Speed (m/s)	State
Aluminium	6420	solid
Steel	5960	solid
Iron	5950	solid
Copper	5000	solid
Water (distilled)	1498	liquid
Hydrogen	1284	gas
Air (dry)	344	gas
Oxygen	316	gas

💥 Shock waves & the sonic boom

When a body moves through air faster than the speed of sound (supersonic speed), it leaves behind a cone of highly compressed air called a shock wave. The huge pressure variation produces a loud, explosive noise — the sonic boom — which can even rattle and break window panes.

Section 4.10 · Bouncing Sound

Reflection of Sound

Like light, sound bounces off an obstacle — but unlike light, the surface need not be polished; even a rough wall reflects sound.

Laws of reflection of sound

(1) The angle of reflection equals the angle of incidence (∠r = ∠i). (2) The incident sound, the reflected sound and the normal at the point of incidence all lie in the same plane.

Applications (all use multiple reflections)

Megaphone / horn

📢

A funnel sends sound forward without spreading.

Ear trumpet

📯

Concentrates sound into the ear of a hard-of-hearing person.

Stethoscope

🩺

Heart/lung sounds reach the doctor by multiple reflections.

Sound board

🏛️

Curved ceilings/boards spread sound evenly in halls.

Section 4.11 · The Repeated Sound

Echo

Clap near a distant cliff and you hear the clap again a moment later — that repeated sound is an echo.

Definition

An echo is the repetition of a sound caused by the reflection of the original sound from a large, hard obstacle. To hear it as a separate sound, the reflected sound must arrive at least 0.1 s after the original — the persistence of hearing.

Minimum distance to hear an echo

The sound must travel to the obstacle and back (a distance 2d) in at least 0.1 s. With speed v = 344 m/s in air:

Formula

v = 2d / t → 2d = v × t

Substitute

2d = 344 × (1/10) = 34.4 m

Answer

d = 17.2 m (minimum distance in air at 20 °C)

In water (v ≈ 1500 m/s) the minimum distance is much larger — about 75 m.

Conditions for an echo & echolocation

Time gap between original and reflected sound ≥ 0.1 s.
Distance to the obstacle ≥ about 17 m (in air).
The obstacle must be large and rigid (wall, hill, cliff).

Echolocation: bats and dolphins emit ultrasonic squeaks and judge an object's distance, size and motion from the returning echo. Multiple reflecting surfaces give multiple echoes — as in the Gol Gumbaz whispering gallery, or the rolling of thunder.

Worked Example

A person fires a gun near a cliff and hears the echo after 1.5 s. If the speed of sound is 340 m/s, how far is the cliff? [3 Marks]

Given

t = 1.5 s, v = 340 m/s

Formula

2d = v × t → d = vt/2

Substitute

2d = 340 × 1.5 = 510 m → d = 510/2

Answer

d = 255 m

Section 4.12 · Lingering Sound

Reverberation

In a big hall, sound bounces off walls, ceiling and floor so many times that the echoes overlap and the sound seems to linger.

Definition

Reverberation is the persistence (prolongation) of audible sound after the source has stopped, caused by repeated reflections. The time for which it persists until it becomes inaudible is the reverberation time.

Too much echo? Absorb it

A little reverberation adds "life" to music, but too much makes speech blurred and confusing. Auditoriums reduce it with sound-absorbing materials:

Walls and ceiling covered with compressed fibreboard, rough plaster or acoustic tiles.
Heavy curtains on doors and windows.
Carpets on the floor and padded seats chosen for their absorbing properties.

Section 4.13 – 4.14 · Beyond Hearing

Range of Hearing & Ultrasound

Human ears respond only to a limited band of frequencies. Below and above it lie infrasound and ultrasound.

Infrasound

< 20 Hz

Earthquakes, whales, elephants, rhinos (~5 Hz).

Audible range

20 Hz – 20 kHz

The band humans can hear.

Ultrasound

> 20 kHz

Heard by bats and dolphins; key to many technologies.

🔬 Why ultrasound is so useful

Ultrasound has a very high frequency and great penetrating power, and travels in well-defined narrow beams. Its applications fall into three fields:

Industry: detecting flaws/cracks in metal blocks; cleaning spiral tubes and delicate parts (by "cold boiling"); welding plastics; drilling fragile materials like glass.
Medical science: echocardiography and ultrasonography (imaging organs, gallstones, kidney stones), monitoring a foetus, breaking kidney stones (lithotripsy), and killing bacteria to preserve liquids like milk. Unlike X-rays, it carries no radiation hazard.
Communication (SONAR): measuring the depth of the sea and locating submarines, shoals of fish or shipwrecks.

🚢 SONAR — sound navigation and ranging

A ship sends an ultrasonic pulse downward; it reflects off the seabed (or an object) and returns as an echo. Measuring the time t and knowing the speed v of sound in water, the depth is found from 2d = v × t. SONAR also lets submarines and ships communicate underwater.

Example

A ship's SONAR echo returns from the seabed in 3.42 s; v = 1530 m/s. Find the depth.

Formula

2d = v × t = 1530 × 3.42 = 5232.6 m

Answer

d = 2616.3 m ≈ 2.62 km

Problem-Solving Masterclass

Every Formula, Applied Step by Step

The routine stays the same — Given → Formula → Substitute → Calculate → Answer. Spot whether a problem needs ν = 1/T, v = νλ, or the echo rule 2d = vt.

M1 · ν = 1/T

A sound has frequency 100 Hz. Find the time period of the oscillating particles.

Formula

T = 1/ν

Substitute

T = 1/100

Answer

T = 0.01 s

M2 · v = νλ (find f and T)

Water-surface waves of wavelength 2 cm travel at 16 m/s. Find (a) the frequency and (b) the time period.

Convert

λ = 2 cm = 0.02 m

(a) ν

ν = v/λ = 16/0.02 = 800 Hz

(b) T

T = 1/ν = 1/800 = 0.00125 s

Answer

ν = 800 Hz · T = 0.00125 s

M3 · v = νλ then time

A sound wave has frequency 2 kHz and wavelength 40 cm. How long does it take to travel 1.6 km?

Speed

v = νλ = 2000 × 0.4 = 800 m/s

Time

t = s/v = 1600/800

Answer

t = 2 s

M4 · audible wavelength range

The audible range is 20 Hz to 20 kHz. Find the corresponding wavelength range (speed of sound = 340 m/s).

Longest λ

λ = v/ν = 340/20 = 17 m

Shortest λ

λ = 340/20000 = 0.017 m

Answer

0.017 m to 17 m

M5 · multi-step wave

A source produces 1500 sound waves in 3 s. The combined length of one compression and the adjacent rarefaction is 68 cm. Find the frequency, wavelength and speed.

Frequency

ν = 1500/3 = 500 Hz

Wavelength

one compression + one rarefaction = λ = 68 cm = 0.68 m

Speed

v = νλ = 500 × 0.68 = 340 m/s

Answer

ν = 500 Hz · λ = 0.68 m · v = 340 m/s

M6 · echo (thunder)

A man hears thunder 4 s after seeing the lightning. If the speed of sound is 330 m/s, how far away was the lightning?

Formula

d = v × t (one-way; light arrives instantly)

Substitute

d = 330 × 4

Answer

d = 1320 m

M7 · echolocation (bat)

A bat emits a 30 kHz ultrasonic wave at 350 m/s and hears its echo 0.6 s later. Find (a) the distance to the obstacle and (b) the wavelength.

(a) Distance

2d = v × t = 350 × 0.6 = 210 m → d = 105 m

(b) Wavelength

λ = v/ν = 350/(30 000) = 0.01167 m

Answer

d = 105 m · λ ≈ 0.0117 m

M8 · person between two cliffs

A man midway... actually between two parallel cliffs hears echoes after 1.5 s and 2.5 s (v = 340 m/s). Find the distance between the cliffs and when the third echo is heard.

Cliff A

d₁ = vt₁/2 = 340×1.5/2 = 255 m

Cliff B

d₂ = vt₂/2 = 340×2.5/2 = 425 m

Distance

d = d₁ + d₂ = 255 + 425 = 680 m

Answer

680 m apart · 3rd echo at t₁ + t₂ = 4 s

M9 · stone in a well

A stone is dropped into a 45 m deep well. If the speed of sound is 340 m/s and g = 10 m/s², after how long is the splash heard?

Fall time

h = ½gt₁² → 45 = ½×10×t₁² → t₁ = 3 s

Sound time

t₂ = h/v = 45/340 = 0.13 s

Answer

total ≈ 3 + 0.13 = 3.13 s

Textbook Exercise · Fully Solved

Exercise — Worked with Formula & Alternate Methods

Representative numericals from the chapter, each solved in full with the formula shown and, where useful, a second method to cross-check.

Q1 · 3 Marks

A sound wave has frequency 1000 Hz and wavelength 34 cm. How long will it take to travel 1 km?

Speed

v = νλ = 1000 × 0.34 = 340 m/s

Time

t = s/v = 1000/340

Answer

t ≈ 2.94 s

Q2 · 2 Marks

The frequency of a wave is 40 Hz and its wavelength is 8 m. Find the velocity.

Formula

v = νλ

Substitute

v = 40 × 8

Answer

v = 320 m/s

Q3 · 2 Marks

Calculate the wavelength of a sound wave of frequency 300 Hz travelling at 330 m/s.

Formula

λ = v/ν

Substitute

λ = 330/300

Answer

λ = 1.1 m

Q4 · 3 Marks

A wave of wavelength 0.60 cm is produced in air and travels at 300 m/s. Will it be audible?

Convert

λ = 0.60 cm = 0.006 m

Frequency

ν = v/λ = 300/0.006 = 5 × 10⁴ Hz

Answer

50 000 Hz > 20 000 Hz → NOT audible (it is ultrasound)

Q5 · 5 Marks

A submarine emits a SONAR pulse that returns from an underwater cliff in 1.04 s. If the speed of sound in salt water is 1530 m/s, how far away is the cliff?

Formula

2d = v × t (down and back)

Substitute

2d = 1530 × 1.04 = 1591.2 m

Answer

d = 795.6 m

Q6 · 5 Marks

A person standing between two vertical cliffs, 640 m from the nearer one, hears the first echo after 4 s and the second 3 s later. Find (a) the speed of sound and (b) the distance between the cliffs.

(a) Speed

to the near cliff: 2 × 640 = v × 4 → v = 320 m/s

(b) Far cliff

second echo at 7 s: 2(x − 640) = 320 × 7 → x − 640 = 1120

Answer

v = 320 m/s · distance between cliffs x = 1760 m

Q7 · 3 Marks

In a tank, 10 ripples are produced per second. The distance between a crest and a trough is 10 cm. Find (a) the wavelength, (b) the frequency and (c) the wave speed.

(a) λ

crest-to-trough = ½λ = 10 cm → λ = 20 cm = 0.2 m

(b) ν

ν = 10 Hz (10 ripples per second)

v = νλ = 10 × 0.2 = 2 m/s

Answer

λ = 0.2 m · ν = 10 Hz · v = 2 m/s

Quick Reference

Formula & Fact Sheet

Everything in one place for a 5-minute revision.

Concept	Formula / Fact	Note
Sound	a form of energy	mechanical wave; needs a medium
Sound in air	longitudinal	compressions & rarefactions
Frequency	ν = 1/T	unit hertz (Hz)
Wave equation	v = ν λ	holds for all waves
Loudness	∝ amplitude²	measured in decibel (dB)
Pitch	∝ frequency	high ν → shrill
Quality (timbre)	depends on waveform	tells instruments apart
Intensity	power / area (W/m²)	objective, measurable
Speed order	solids > liquids > gases	↑ with temperature & humidity
Speed in air	344 m/s (20°C), 332 (0°C)	+0.6 m/s per °C
Echo	2d = v t	need t ≥ 0.1 s
Min echo distance	17.2 m (air), ~75 m (water)	persistence of hearing 0.1 s
Reverberation	repeated reflections	reduce with absorbers
Audible range	20 Hz – 20 000 Hz	infra < 20 Hz; ultra > 20 kHz
SONAR	2d = v t	uses ultrasound in water

Smart Study

10 Exam & JEE-Foundation Tips

Where students gain — and lose — marks in this chapter.

Echo uses 2d, not d

The sound goes to the obstacle and back, so 2d = v t. Forgetting the factor of 2 halves your answer.

Convert units first

cm → m, kHz → Hz, km → m. Then apply v = νλ. A unit slip is the most common error.

Loudness vs intensity

Loudness is subjective (depends on the ear); intensity is objective and measurable. Examiners love this difference.

Pitch ≠ loudness

Pitch depends on frequency; loudness on amplitude. A high-pitched sound can be soft, and vice versa.

Sound needs a medium

It cannot travel through vacuum (bell-jar experiment). Light can — that's why we see lightning before thunder.

λ = one C + one R

The combined length of a compression and the next rarefaction equals one wavelength.

Crest-to-trough = ½λ

In a transverse wave, the distance between a crest and the adjacent trough is half a wavelength.

17.2 m is at 20 °C

The minimum echo distance follows from v = 344 m/s and t = 0.1 s; quote the derivation, not just the number.

Audible band: 20–20 000 Hz

Below = infrasound, above = ultrasound. Test whether a calculated frequency is audible.

Well problems have two times

Time to hear the splash = fall time (√(2h/g)) + sound's return time (h/v). Don't forget the second part.

CBSE Pattern Practice

Practice Question Bank

MCQ · Assertion–Reason · VSA (1M) · SA (2M / 3M) · LA (5M) · Application (4M) — covering every question type in the chapter exercise, each with a step-by-step solution.

⚡ Beyond the Textbook

JEE Foundation Challenge Set

Original problems on the same Class 9 toolkit (v = νλ, echo, SONAR, the hearing range) with a little extra reasoning. Try each before opening the solution.

Bonus problems — tap a card's button to reveal the worked solution.