Sound — Concepts, Formulas & Solved Numericals

What Is Sound, Really?

Sound is a form of energy that travels as a mechanical wave — meaning it needs a medium (solid, liquid, or gas) to travel. When a guitar string vibrates, it pushes the air molecules around it back and forth, creating alternating regions of compression and rarefaction that travel outward as a longitudinal wave.

This is the single most important idea in the entire chapter: sound cannot travel through vacuum. Light can — but sound cannot. The famous bell-jar experiment proves this: when air is pumped out of a jar containing a ringing bell, the sound fades even though you can still see the bell vibrating.

At Class 9 level, we go deeper — understanding the wave properties of sound (frequency, amplitude, wavelength, speed) and how they relate to what we actually hear (pitch and loudness). At Class 8, the focus is on production, propagation, and the human ear. Both levels are covered here.

Key Terms & Definitions

Vibration — the back-and-forth motion of an object that produces sound. Every sound source vibrates: vocal cords, tuning forks, speaker cones, tabla membranes.

Longitudinal Wave — a wave where the particles of the medium oscillate parallel to the direction of wave propagation. Sound is always a longitudinal wave (unlike light, which is transverse).

Compression — the region in the medium where particles are pushed close together, creating a zone of high pressure.

Rarefaction — the region where particles are spread apart, creating a zone of low pressure. Compressions and rarefactions together make one full wave cycle.

Frequency (f) — the number of complete oscillations (compressions + rarefactions) per second. Unit: Hertz (Hz).

Amplitude (A) — the maximum displacement of a particle from its mean position. Amplitude determines loudness — higher amplitude = louder sound.

Wavelength (λ) — the distance between two consecutive compressions (or two rarefactions). Unit: metres (m).

Time Period (T) — time taken to complete one full oscillation. T = 1/f.

Pitch — the “highness” or “lowness” of a sound. Determined entirely by frequency — high frequency = high pitch.

Timbre (Quality) — what makes a guitar and a flute sound different even when playing the same note. Determined by the waveform shape (mixture of frequencies).

Core Concepts

How Sound Is Produced and Transmitted

Take a tuning fork. When you strike it, the prongs vibrate. The prong moving forward compresses the air in front of it — that compression travels outward. When the prong moves back, it creates a rarefaction. These compressions and rarefactions travel as pressure variations through the air until they reach your ear.

This is why sound needs a medium: the energy is transferred by collisions between adjacent particles. No particles = no collisions = no sound.

Sound travels through solids > liquids > gases. Particles in solids are closer together, so the energy transfers faster.

The Wave Equation

v = f \lambda

where:

$v$ = speed of sound (m/s)
$f$ = frequency (Hz)
$\lambda$ = wavelength (m)

Also: $T = \dfrac{1}{f}$ and $v = \dfrac{\lambda}{T}$

The speed of sound in air at 0°C is approximately 332 m/s. At room temperature (25°C), it’s about 346 m/s. In water, ~1500 m/s. In steel, ~5100 m/s.

Reflection of Sound — Echo and Reverberation

Sound reflects off hard surfaces just like light reflects off a mirror. The laws of reflection apply: angle of incidence = angle of reflection.

Echo — when you hear the reflected sound distinctly after the original sound. For this, two conditions must be met:

The reflecting surface must be at least 17 metres away (so the echo arrives after 0.1 s, the minimum persistence of hearing)
The original sound must be short and sharp

Reverberation — when multiple reflections cause the sound to persist even after the source has stopped. Concert halls and recording studios are designed to control reverberation.

The 17 m minimum distance for an echo comes directly from: $d = \dfrac{v \times t}{2} = \dfrac{346 \times 0.1}{2} \approx 17.3$ m. The “2” is there because sound travels to the wall AND back. This formula is your friend in numerical problems.

Audible Range, Infrasound, and Ultrasound

Humans can hear frequencies between 20 Hz and 20,000 Hz (20 kHz). This is the audible range.

Infrasound — below 20 Hz. Elephants and whales communicate using infrasound over vast distances. Earthquakes produce infrasound.
Ultrasound — above 20,000 Hz. Bats, dolphins, and dogs can hear ultrasound. Used in SONAR, medical imaging (sonography), and industrial cleaning.

SONAR (Sound Navigation And Ranging) uses ultrasound pulses to detect underwater objects. Ships send a pulse, it reflects off the ocean floor or submarine, and comes back. The time taken tells us the distance.

d = \frac{v \times t}{2}

where $d$ = depth/distance, $v$ = speed of sound in water, $t$ = total time for echo to return

Solved Examples

Example 1 — Easy (CBSE Class 8/9)

A sound wave has a frequency of 440 Hz and a wavelength of 0.75 m. Find the speed of sound.

Using $v = f\lambda$ :

v = 440 \times 0.75 = 330 \text{ m/s}

The speed of sound is 330 m/s.

Example 2 — Easy (CBSE Class 9)

A person claps near a cliff and hears the echo after 3 seconds. Speed of sound in air is 340 m/s. How far is the cliff?

The sound travels to the cliff and back in 3 s. So it travels to the cliff in 1.5 s.

d = v \times t = 340 \times 1.5 = 510 \text{ m}

The cliff is 510 m away.

Most students forget to divide time by 2. The 3 seconds is the total round trip time. The distance is one-way. If you use $d = 340 \times 3$ , you get 1020 m — exactly double the correct answer. This is the most common error in echo problems.

Example 3 — Medium (CBSE Class 9)

A submarine sends an ultrasound pulse. The echo from the ocean floor returns in 0.8 seconds. Speed of sound in seawater is 1500 m/s. Find the depth of the ocean.

d = \frac{v \times t}{2} = \frac{1500 \times 0.8}{2} = \frac{1200}{2} = 600 \text{ m}

The ocean depth is 600 m.

Example 4 — Medium (CBSE Class 9)

The time period of a sound wave is 0.004 seconds. Find its (a) frequency, (b) wavelength if the speed of sound is 330 m/s.

(a) $f = \dfrac{1}{T} = \dfrac{1}{0.004} = 250 \text{ Hz}$

(b) $\lambda = \dfrac{v}{f} = \dfrac{330}{250} = 1.32 \text{ m}$

Example 5 — Hard (CBSE Class 9 / Olympiad)

Two people A and B are 1700 m apart. A fires a gun. B hears the sound twice — once through the air (340 m/s) and once through the ground (1700 m/s). How many seconds apart does B hear the two sounds?

Time for sound through air: $t_1 = \dfrac{1700}{340} = 5 \text{ s}$

Time for sound through ground: $t_2 = \dfrac{1700}{1700} = 1 \text{ s}$

Difference: $5 - 1 = \mathbf{4 \text{ seconds}}$

This problem tests a beautiful concept: since sound travels faster in solids, B hears the ground sound first, then the air sound 4 seconds later.

Exam-Specific Tips

CBSE Class 8: Questions focus on qualitative understanding — how sound is produced, why it can’t travel in vacuum, the bell-jar experiment, and the human ear. Diagrams are high-weightage — draw and label the ear properly.

CBSE Class 9: Numericals using $v = f\lambda$ and the echo formula appear in every board exam. The 3-mark numerical section almost always has one echo problem. Expect one question on SONAR and its applications.

For CBSE boards:

The 17 m minimum distance derivation is a 1-mark frequently asked question. Write it as a formula derivation, not just a statement.
“Give two applications of ultrasound” — always a safe 2-marker. Keep: SONAR, medical sonography, echocardiography, industrial cleaning, detecting flaws in metals.
The human ear diagram is a 5-mark guaranteed question in Class 8. Label: pinna, ear canal, eardrum (tympanic membrane), malleus, incus, stapes, cochlea, auditory nerve.

Common Mistakes to Avoid

Mistake 1: Confusing frequency with amplitude. Frequency → pitch (high/low). Amplitude → loudness (soft/loud). A loud bass note has high amplitude, low frequency. This distinction is tested directly: “What happens to the pitch if frequency is doubled?” — it doubles in pitch, but amplitude is unchanged.

Mistake 2: Echo formula — forgetting to halve the time. Always: $d = \dfrac{v \times t}{2}$ . The $t$ in the formula is total round-trip time. Divide by 2 first, then multiply by $v$ .

Mistake 3: Saying sound is a transverse wave. Sound is always longitudinal in gases and liquids. In some solids, both longitudinal and transverse modes are possible, but for Class 8-9, sound = longitudinal wave. Period.

Mistake 4: Wrong units for frequency. The unit is Hertz (Hz) = cycles per second = s⁻¹. Many students write “m/s” for frequency in a hurry. Hz is never m/s — that’s the unit for speed.

Mistake 5: Assuming speed of sound is always 340 m/s in numericals. Use whatever speed the question provides. If no value is given, use 340 m/s as the standard. The speed of sound depends on the medium and temperature — it’s not a universal constant like the speed of light.

Practice Questions

Q1. The frequency of a tuning fork is 512 Hz and the speed of sound in air is 340 m/s. Find the wavelength.

\lambda = \frac{v}{f} = \frac{340}{512} \approx 0.664 \text{ m}

Q2. A bat emits an ultrasonic pulse and receives the echo after 0.01 s. If the speed of ultrasound is 340 m/s, how far is the obstacle?

d = \frac{340 \times 0.01}{2} = \frac{3.4}{2} = 1.7 \text{ m}

The obstacle is 1.7 m away.

Q3. Why can’t sound travel through outer space but light can?

Sound is a mechanical wave — it requires particles in a medium to transfer energy through collisions. Outer space is a near-perfect vacuum with no particles. Light is an electromagnetic wave and does not need any medium — it can travel through vacuum because it is a disturbance in electric and magnetic fields, not in matter.

Q4. A sound wave has a wavelength of 2 m and a time period of 0.005 s. Find (a) frequency and (b) speed of sound.

(a) $f = \dfrac{1}{T} = \dfrac{1}{0.005} = 200 \text{ Hz}$

(b) $v = f\lambda = 200 \times 2 = 400 \text{ m/s}$

Q5. Two friends stand 850 m apart in an open field. One fires a pistol. After how many seconds does the other hear the sound? (Speed of sound = 340 m/s)

t = \frac{d}{v} = \frac{850}{340} = 2.5 \text{ s}

Note: this is NOT an echo problem — there is no reflection. The sound travels one-way, so no division by 2.

Q6. What is the range of frequencies that (a) humans can hear, (b) a dog can hear, (c) a bat can hear?

(a) Humans: 20 Hz to 20,000 Hz (20 kHz)

(b) Dogs: 40 Hz to 65,000 Hz — they can hear well above the human audible range, which is why a dog whistle (around 23,000 Hz) is inaudible to us but not to them.

(c) Bats: 1 kHz to about 100 kHz — bats are the classic ultrasound users, using echolocation for navigation.

Q7. An echo is heard 2.6 s after a shout near a mountain. Speed of sound = 340 m/s. What is the distance of the mountain from the person?

d = \frac{v \times t}{2} = \frac{340 \times 2.6}{2} = \frac{884}{2} = 442 \text{ m}

Q8. A ship uses SONAR. The ultrasound pulse takes 5 seconds to return from the ocean floor. If speed of sound in seawater is 1450 m/s, find the depth of the ocean.

d = \frac{v \times t}{2} = \frac{1450 \times 5}{2} = \frac{7250}{2} = 3625 \text{ m}

The ocean depth is 3625 m (or 3.625 km).

Frequently Asked Questions

Why does sound travel faster in solids than in gases?

The speed of sound depends on how quickly energy transfers between adjacent particles. In solids, particles are tightly packed and strongly bonded — a disturbance in one particle quickly reaches the next. In gases, particles are far apart and interact less frequently, so energy transfer is much slower.

Can two sounds with different frequencies have the same loudness?

Yes. Loudness depends on amplitude, not frequency. A low-frequency drum beat and a high-frequency flute note can both be equally loud if their amplitudes are equal.

What happens to the wavelength of sound if we double the frequency (keeping speed constant)?

From $v = f\lambda$ , if $v$ is constant and $f$ doubles, then $\lambda$ must halve. Wavelength and frequency are inversely proportional at constant speed.

Why does reverberation make speech hard to understand in large empty halls?

In an empty hall, hard walls reflect sound multiple times. The reflected sounds overlap with the direct sound, blurring words together. Soft materials (curtains, carpets, audience) absorb sound energy and reduce reverberation — that’s why a crowded hall sounds different from an empty one.

What is the difference between echo and reverberation?

An echo is a distinct repetition of the original sound heard after a gap of at least 0.1 seconds (reflecting surface at least 17 m away). Reverberation is the persistence of sound due to multiple rapid reflections — the gap between reflections is less than 0.1 s, so the brain can’t separate them and the sound seems prolonged.

Why are ultrasound waves used in medical imaging instead of audible sound?

Ultrasound has very short wavelengths (since $\lambda = v/f$ and $f$ is very high). Short wavelengths allow detection of smaller structures inside the body. Audible sound waves are too long to resolve fine tissue details. Plus, ultrasound at diagnostic intensities is safe — it doesn’t ionize tissue like X-rays do.

Does the pitch of a train’s horn change as it approaches you?

Yes — this is the Doppler effect (covered in Class 11 Physics). As the train approaches, the sound waves bunch up, increasing the frequency you perceive — so the pitch sounds higher. As it moves away, waves spread out and the pitch drops. You’ve definitely heard this; now you know why.