By the end of this chapter you'll be able to…

  • 1Define wave; distinguish transverse and longitudinal waves with examples
  • 2Identify compressions and rarefactions in a sound wave; state which property of the wave each represents
  • 3Define and compute wavelength, frequency, time period, amplitude, and wave speed for a given sound
  • 4Apply v = fλ to numerical problems
  • 5List speed of sound in air, water, iron — compare across media
  • 6Define echo; state the minimum distance for a clear echo and derive it from the 0.1-s rule
  • 7Describe SONAR operation and solve range-finding problems with the d = vt/2 formula
  • 8Categorise sounds by audible/infrasonic/ultrasonic range; list applications of ultrasound
  • 9Describe the structure of the human ear and trace a sound's path from external air to the brain
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Why this chapter matters
Sound is the most accessible wave phenomenon — you experience it constantly. Master wave properties here and the rest of wave physics (light, EM, optics) builds on the same vocabulary: frequency, wavelength, amplitude, speed = fλ. The applications (ultrasound, SONAR, echolocation) span medicine, defence and biology.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Sound — Class 9 (CBSE)

Sound is everywhere — but you've never seen it. This chapter is about how something invisible can travel from a singer's vocal cord to your eardrum, how a bat hunts in pitch dark, and how ultrasound machines build a 3D image of an unborn baby. It's all wave physics.


1. The story — sound as wave

Sound was, for centuries, a puzzle. People knew it could be loud or soft, high or low, close or distant. But what IS sound? In 1660, Robert Boyle did the decisive experiment: he placed a ringing bell inside a chamber and pumped the air out. As the air thinned, the bell's sound got fainter and fainter, until in a near-vacuum it became inaudible. Conclusion: sound needs a medium to travel.

Today we understand sound as a mechanical wave — a disturbance that propagates through a medium (gas, liquid, or solid) by causing particles to vibrate. Sound CANNOT travel through vacuum (which is why space movies showing explosions with sound are physically wrong — explosions in space would be silent).

This chapter is the wave properties of sound, the math of its speed, and the technology built on it (ultrasonic imaging, SONAR, hearing aids).


2. Sound is a longitudinal wave

A wave is a disturbance that travels through a medium without the medium moving as a whole.

Two types:

  • Transverse wave: particles vibrate PERPENDICULAR to the wave's direction of motion. Example: a wave on a stretched string, light, water surface waves.
  • Longitudinal wave: particles vibrate PARALLEL to the wave's direction of motion. Example: SOUND, compression of a slinky spring.

Compressions and rarefactions

A sound wave is alternating regions of:

  • Compression — particles pushed close together; high pressure, high density.
  • Rarefaction — particles spread out; low pressure, low density.

Both move in the same direction as the wave. The particles themselves oscillate forward and backward but don't move with the wave.


3. Properties of a sound wave

Wavelength ()

The distance between two consecutive compressions (or rarefactions). SI unit: metre (m).

Frequency ( or )

The number of waves passing a point per unit time. SI unit: hertz (Hz) = 1/s = waves per second.

Time period ()

The time taken for one complete wave to pass a point.

If , then .

Amplitude ()

The maximum displacement of a particle from its mean position. For sound, also: the maximum compression/rarefaction. Determines loudness.

Wave speed ()

How fast the disturbance travels through the medium. Connects and :

For air at 20°C: ≈ 1235 km/h.

Pitch and timbre

  • Pitch = perceived "highness" or "lowness" of a sound. Determined by FREQUENCY. Higher frequency → higher pitch. A whistle has higher pitch than a tuba.
  • Loudness = how loud the sound is. Determined by AMPLITUDE. More amplitude → louder.
  • Timbre (also called quality) = what makes a violin and a piano playing the same note sound different. Determined by the WAVE SHAPE (harmonic content).

4. Speed of sound in different media

Sound travels faster in DENSER media (mostly because the particles transmit the disturbance more efficiently).

MediumSpeed (m/s)
Vacuum0 (sound can't travel)
Air (0°C)331
Air (20°C)343
Hydrogen (0°C)1284 (much faster because particles are lighter)
Water (25°C)1493
Sea water1531
Iron5950
Steel5960
Granite6000
Diamond12000

Key observations:

  • Solids > Liquids > Gases (in general).
  • Speed depends on the elasticity and density of the medium.
  • In air, speed INCREASES with temperature (~ 0.6 m/s per °C of warming). Hot summer day → sound travels slightly faster.

5. Reflection of sound — echoes

When sound hits a hard surface, it bounces back. This is reflection of sound, just like light.

Laws of reflection (apply to sound too)

  1. The angle of incidence equals the angle of reflection.
  2. The incident wave, reflected wave, and normal at the point of reflection all lie in the same plane.

Echo

An echo is a reflected sound heard distinctly (i.e., separable from the original sound). For a clear echo:

  • The reflected wave must reach your ear at least 0.1 s after the original. (Below this, our brain merges them into a single sound — called reverberation.)
  • Sound travels at ~340 m/s in air. In 0.1 s, sound covers 34 m. Round trip means the reflecting surface must be at least 17 m away.

So if you stand within 17 m of a wall, you don't hear an echo (you hear reverberation). Beyond 17 m, you hear a distinct echo.

Why concert halls have rounded interiors

To reduce echo (which would muddle the music) but increase reverberation (which "fills out" the sound). Walls and ceilings absorb high-frequency reflections; the audience absorbs additional sound. Master designers balance the two for clarity + warmth.


6. SONAR — Sound Navigation and Ranging

SONAR uses ultrasonic waves to measure distances under water (where light doesn't penetrate well but sound does).

How it works:

  1. A SONAR device on a ship sends out an ultrasonic pulse.
  2. The pulse hits an object (sea bed, fish, submarine) and reflects back.
  3. The device detects the returning pulse and measures the round-trip time .
  4. If speed of sound in water = , the distance to the object is:

The factor of 2 is because the pulse goes out AND comes back — the total distance traveled is .

Applications

  • Mapping the sea floor (geology, oceanography).
  • Finding shoals of fish (commercial fishing).
  • Detecting submarines (naval defence).
  • Finding sunken ships and aircraft.
  • Navigation in foggy or cloudy conditions.

7. Range of hearing

Human ears can perceive sounds with frequencies in a specific range:

  • Infrasonic (below 20 Hz) — not audible. Whales, elephants, earthquakes produce infrasonic waves.
  • Audible range (20 Hz to 20,000 Hz = 20 kHz) — what humans can hear.
  • Ultrasonic (above 20 kHz) — not audible to humans. Used in industry, medicine.

Animals' ranges

  • Dogs: up to ~40 kHz (can hear dog whistles).
  • Bats: up to ~120 kHz (used for echolocation).
  • Dolphins: up to ~150 kHz.

Why old people lose high-frequency hearing first

Inner-ear hair cells degenerate with age. The high-frequency cells (closer to the entrance of the cochlea) are exposed to more noise over a lifetime and degenerate first. This is presbycusis — age-related hearing loss.


8. Ultrasound — applications

Medical imaging

Send ultrasonic pulses into the body. They reflect at boundaries between different tissues (skin/fat, fat/muscle, muscle/bone). The pattern of reflections is converted into an image.

  • Pregnancy scans — see the fetus without harmful radiation.
  • Cardiac ultrasound (echocardiogram) — see the heart in motion.
  • Tumour detection — abnormal tissue reflects differently.

Why ultrasound vs X-rays for pregnancies? Ultrasound is non-ionising — no radiation damage to the fetus.

Industrial cleaning

Ultrasonic baths clean jewellery, surgical instruments, electronic components by creating tiny bubbles (cavitation) that scrub away dirt.

Echolocation (in nature)

Bats and dolphins emit ultrasonic clicks and listen for returning echoes. This gives them a 3D "sound map" of their environment — used for navigation and hunting in darkness or murky water.

Range-finding

Ultrasonic sensors in cars (parking sensors), security systems, and robotic devices use the principle: time × speed = distance.


9. Structure of the human ear

The ear has three parts:

1. Outer ear

  • Pinna (visible ear) — funnels sound into the ear canal.
  • Auditory canal — channels sound to the eardrum.

2. Middle ear

  • Eardrum (tympanic membrane) — vibrates when sound reaches it.
  • Three small bones (hammer/malleus, anvil/incus, stirrup/stapes) — amplify the vibrations ~ 20×.
  • Eustachian tube — equalises pressure with the atmosphere.

3. Inner ear

  • Cochlea (spiral, fluid-filled) — converts vibrations to electrical signals.
  • Hair cells lining the cochlea — different cells respond to different frequencies.
  • Auditory nerve — sends the electrical signals to the brain.

Why we have two ears

To localise sound. Differences in:

  • Arrival time at each ear (sound from the left arrives at the left ear slightly earlier).
  • Intensity at each ear (the head shadows one ear).

… allow the brain to compute the sound's direction. Lose one ear and you lose most direction sense.


10. Worked example — SONAR ranging

A SONAR device on a ship emits an ultrasonic pulse. The reflected pulse from a submarine is detected 1.5 s later. If the speed of sound in sea water is 1500 m/s, find the distance to the submarine.

Step 1 — Total distance traveled = speed × time = 1500 × 1.5 = 2250 m.

Step 2 — Pulse went OUT and came BACK, so it traveled twice the distance to the submarine.

Step 3 — Distance to submarine = 2250 / 2 = 1125 m.


11. Closing thought

Sound is a workhorse of nature and technology:

  • Music = harmonics of sound waves arranged in pleasing ratios.
  • Speech = your vocal cords vibrating at 100–300 Hz, producing the longitudinal pulses we call language.
  • Echolocation = millions of years of evolution before humans invented SONAR.
  • Earthquakes = enormous infrasonic waves that travel through the Earth, used by seismologists to map the planet's interior.

You can't see a sound wave but you can hear it, feel it (a bass beat in your chest), measure it (oscilloscope), produce it (string, drum, voice), reflect it (echo), focus it (parabolic dish), absorb it (carpet, ceiling tiles), and use it (ultrasound). All from the simple idea that a disturbance can propagate through a medium without the medium moving as a whole. That's wave physics — and after this chapter, you'll see waves everywhere.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Wave equation
v = f × λ
v in m/s, f in Hz, λ in m. Most-used formula.
Time period
T = 1 / f
Inverse relationship.
Frequency from period
f = 1 / T
Echo (round trip)
Distance = v × t / 2
t is round-trip time. Used for SONAR and echo measurements.
Minimum distance for echo
d_min = (v × 0.1) / 2
For air at 340 m/s: d_min ≈ 17 m.
Speed of sound (air, 20 °C)
v_air ≈ 343 m/s
Increases with temperature by ~0.6 m/s per °C.
Range of hearing
20 Hz ≤ f_audible ≤ 20,000 Hz
Below: infrasonic; above: ultrasonic.
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Saying sound is a transverse wave
Sound in fluids (air, water) is LONGITUDINAL — particles vibrate parallel to wave propagation. Only in solids can sound also be transverse (shear waves). For Class 9, always say 'sound is longitudinal'.
WATCH OUT
Saying compressions and rarefactions move with the particles
Particles oscillate in place. The PATTERN of compressions and rarefactions moves through the medium — that's the wave. Particles don't travel; the disturbance does.
WATCH OUT
Forgetting the factor of 2 in SONAR distance calculation
Pulse goes OUT and comes BACK. So distance to target = (v × t)/2, not v × t. Half the round-trip distance is the one-way distance.
WATCH OUT
Saying pitch = loudness
Pitch = frequency (high vs low). Loudness = amplitude (soft vs loud). Two SEPARATE sound properties controlled by different wave properties.
WATCH OUT
Saying sound can travel in vacuum
Sound REQUIRES a medium (gas, liquid, or solid). No medium = no sound. (Light, by contrast, can travel through vacuum.)
WATCH OUT
Saying ultrasonic sounds are loud high-pitch sounds
Ultrasonic = above 20 kHz frequency. They are INAUDIBLE to humans, regardless of intensity. A bat's ultrasonic clicks are loud, but you can't hear them.
WATCH OUT
Using cm or km in v = fλ without converting
v = fλ requires SI units: v in m/s, f in Hz, λ in m. Convert other units first.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Definition
What kind of wave is sound — transverse or longitudinal? Justify your answer.
Show solution
Step 1 — Recall the distinction. Transverse: particles vibrate PERPENDICULAR to wave propagation. Longitudinal: particles vibrate PARALLEL to wave propagation. Step 2 — Sound mechanism. Sound in a gas: air particles vibrate back-and-forth in the same direction as the wave moves (creating compressions and rarefactions). ✦ Answer: Sound is a LONGITUDINAL wave. Particles in the medium oscillate parallel to the direction of wave propagation, alternately compressing and rarefying the medium.
Q2EASY· v = fλ
A wave has frequency 500 Hz and wavelength 0.5 m. Find its speed.
Show solution
Step 1 — Use v = fλ. Step 2 — v = 500 × 0.5 = 250 m/s. ✦ Answer: 250 m/s.
Q3EASY· Concepts
Differentiate pitch and loudness of sound. On which wave property does each depend?
Show solution
Step 1 — Pitch. • Perceived 'highness' or 'lowness' of sound. • Depends on FREQUENCY. • Higher frequency → higher pitch. A whistle (high f) is shrill; a tuba (low f) is bassy. Step 2 — Loudness. • Perceived intensity / volume. • Depends on AMPLITUDE. • Higher amplitude → louder. A drum can be played softly or loudly at the SAME pitch. ✦ Answer: Pitch is determined by FREQUENCY (Hz). Loudness is determined by AMPLITUDE (the maximum displacement). They are independent — same loudness can have different pitches, and vice versa.
Q4EASY· Range
Classify each as audible, infrasonic, or ultrasonic for humans: (a) 10 Hz, (b) 500 Hz, (c) 25,000 Hz, (d) 18,000 Hz.
Show solution
Step 1 — Recall the human range. Audible: 20 Hz to 20,000 Hz (20 kHz). Below 20 Hz: INFRASONIC. Above 20 kHz: ULTRASONIC. Step 2 — Classify each. (a) 10 Hz: below 20 Hz → INFRASONIC. (b) 500 Hz: in 20 Hz to 20 kHz → AUDIBLE. (c) 25,000 Hz: above 20 kHz → ULTRASONIC. (d) 18,000 Hz: below 20 kHz → AUDIBLE. ✦ Answer: (a) Infrasonic, (b) Audible, (c) Ultrasonic, (d) Audible (high-pitched but still audible).
Q5EASY· Echo
Why do we hear an echo only when we are at least 17 m away from a wall?
Show solution
Step 1 — Recall: for a distinct echo, reflected sound must reach our ears at least 0.1 s after the original sound (otherwise our brain merges them). Step 2 — In 0.1 s, sound travels: 340 × 0.1 = 34 m. Step 3 — This is the ROUND TRIP distance (to the wall and back). The wall is at half this: 34/2 = 17 m. ✦ Answer: At 17 m, the round trip is 34 m, taking 0.1 s — the minimum gap for our brain to distinguish original from reflected. Closer than 17 m, the two merge as reverberation; farther than 17 m, distinct echo.
Q6MEDIUM· v = fλ inverse
A radio station broadcasts at frequency 100 MHz. Find the wavelength. (Speed of EM waves in air = 3 × 10⁸ m/s)
Show solution
Step 1 — Convert MHz to Hz. 100 MHz = 100 × 10⁶ Hz = 10⁸ Hz. Step 2 — Use v = fλ ⇒ λ = v/f. λ = (3 × 10⁸) / 10⁸ = 3 m. ✦ Answer: Wavelength = 3 m. (Note: this example uses light/radio speed, not sound speed. Same v = fλ formula applies to all waves.)
Q7MEDIUM· SONAR
A SONAR device sends an ultrasonic pulse into the sea. The reflected pulse from the ocean floor is received 4 s later. If sound speed in sea water is 1500 m/s, find the depth.
Show solution
Step 1 — Apply SONAR formula. d = (v × t) / 2. Step 2 — Substitute. d = (1500 × 4) / 2 = 6000 / 2 = 3000 m. ✦ Answer: Depth = 3000 m = 3 km. Reminder: divide by 2 because the pulse travels DOWN and BACK — the time t covers the round trip.
Q8MEDIUM· Wave numerical
A tuning fork produces 256 vibrations per second. If sound travels at 340 m/s in air, find the wavelength of the sound.
Show solution
Step 1 — Identify givens. Frequency f = 256 Hz. Speed v = 340 m/s. Step 2 — Use v = fλ ⇒ λ = v/f. λ = 340 / 256 ≈ 1.33 m. ✦ Answer: Wavelength ≈ 1.33 m.
Q9MEDIUM· Echo distance
A man fires a gun and hears the echo from a cliff 2 s later. If the speed of sound is 340 m/s, find the distance of the cliff from the man.
Show solution
Step 1 — Use d = v × t / 2. Step 2 — Substitute. d = (340 × 2) / 2 = 340 m. ✦ Answer: Distance to the cliff = 340 m. Why divide by 2: the sound travels TO the cliff and BACK. 2 s is the round-trip time; the one-way distance is half.
Q10MEDIUM· Speed
Why does sound travel faster in iron than in air?
Show solution
Step 1 — Recall: speed of sound depends on the elasticity and density of the medium. v = √(B / ρ) for fluids, where B is the bulk modulus (a measure of stiffness) and ρ is density. For solids: v = √(E / ρ), where E is Young's modulus (rigidity). Step 2 — Compare iron and air. • Iron is FAR stiffer (Young's modulus ~10¹¹ Pa) than air (bulk modulus ~10⁵ Pa). • Iron is also denser, but the elasticity dominates. • The ratio Y/ρ for iron >> B/ρ for air. Step 3 — Result. Sound in iron ≈ 5960 m/s. In air ≈ 343 m/s. About 17× faster. ✦ Answer: Iron's particles are tightly bound by very strong inter-atomic forces — they transmit vibrations far more efficiently than the loosely-connected air molecules. The stiffness (elasticity) of iron vastly exceeds air, so vibrations propagate much faster.
Q11HARD· Echo location
A boy stands between two parallel cliffs and claps his hands. He hears two distinct echoes after 2 s and 4 s respectively. Find the distance between the two cliffs. (Speed of sound = 340 m/s)
Show solution
Step 1 — Distance to closer cliff. Round-trip 2 s at 340 m/s = 680 m. One-way distance = 680/2 = 340 m. Step 2 — Distance to farther cliff. Round-trip 4 s at 340 m/s = 1360 m. One-way distance = 1360/2 = 680 m. Step 3 — Total distance between cliffs. 340 + 680 = 1020 m. ✦ Answer: Distance between the two cliffs = 1020 m.
Q12HARD· Time period
A wave has time period 0.005 s and travels at 300 m/s in a medium. Find its frequency and wavelength.
Show solution
Step 1 — Frequency from time period. f = 1/T = 1/0.005 = 200 Hz. Step 2 — Wavelength from v = fλ. λ = v/f = 300/200 = 1.5 m. ✦ Answer: Frequency = 200 Hz; Wavelength = 1.5 m.
Q13HARD· Compare media
A sound wave has frequency 500 Hz. Find its wavelength in (a) air, (b) water, (c) steel. (Speeds: air 340 m/s, water 1500 m/s, steel 5960 m/s).
Show solution
Step 1 — Frequency is the SAME in any medium. Wavelength changes because v changes. Use λ = v/f. Step 2 — Compute each. (a) Air: λ = 340 / 500 = 0.68 m = 68 cm. (b) Water: λ = 1500 / 500 = 3.0 m. (c) Steel: λ = 5960 / 500 = 11.92 m. Step 3 — Interpret. Same 'note' (same frequency, same pitch) has very different wavelengths in different media. But the EAR detects frequency (which stays constant) — so the note sounds the same to you whether the sound travels through air or water. ✦ Answer: (a) 0.68 m in air, (b) 3.0 m in water, (c) 11.92 m in steel. Frequency conserved; wavelength scales with the medium's sound speed.
Q14HARD· HOTS
Bats use ultrasonic waves for navigation. Explain how a bat can find its prey in the dark.
Show solution
Step 1 — Bat emits ultrasonic clicks. Frequency 20 kHz to 120 kHz (mostly ultrasonic; some clicks audible to humans as the bat exhales). Short pulses, several per second. Step 2 — Pulses reflect off objects. When a click hits a flying insect or a tree branch, it reflects back. The bat's specialized ears detect the echo. Step 3 — Bat measures the time delay. Distance = (v × t)/2 (same SONAR formula). With ultrasound at 340 m/s in air, and time resolution of microseconds, the bat can locate a mosquito-sized insect within centimetres. Step 4 — Bat measures the Doppler shift. If the insect is moving toward the bat, the echo's frequency is slightly higher than the emitted click (Doppler effect). The bat reads this to estimate the prey's velocity. (Beyond Class 9, but a useful concept.) Step 5 — Bat builds an internal 3D 'map' from clicks and echoes. Many pulses per second + slight angular separation between the two ears = a continuous 3D sound image. The bat can fly through dense forest in pitch dark. ✦ Answer: Bats emit ultrasonic clicks, listen for echoes, and use time-of-flight + intensity + frequency-shift to triangulate prey position and velocity in 3D — a biological version of SONAR (called ECHOLOCATION). Insects (especially moths) have evolved counter-measures (ultrasonic listening, evasive turns) — a continuous evolutionary arms race.
Q15HARD· HOTS
If sound takes 6 s to travel through 2 km of water and 1.0 s in air for 340 m, calculate the ratio of speeds of sound in water and air.
Show solution
Step 1 — Speed in water. v_water = distance / time = 2000 m / 6 s ≈ 333 m/s. Wait, this seems too slow. Let me reread... Reinterpret: 'takes 6 s' might be a typo for the wrong number. Let me use what's given. Step 1 — Per the given numbers. v_water = 2000/6 ≈ 333 m/s. v_air = 340/1.0 = 340 m/s. Step 2 — Ratio. v_water / v_air = 333 / 340 ≈ 0.98. Step 3 — Note. In reality, v_water ≈ 1500 m/s and v_air ≈ 340 m/s, so v_water / v_air ≈ 4.4. The numbers in this problem appear to use different data — possibly distance in water was meant to be 8 km. Let me solve with both interpretations. Likely intended: 8 km in 6 s → v_water = 8000/6 ≈ 1333 m/s. Ratio = 1333/340 ≈ 3.9. Closer to the expected ratio of ~4. ✦ Answer (using given numbers): v_water = 333 m/s, v_air = 340 m/s. Ratio ≈ 0.98 (close to 1). Likely intended answer (8 km, 6 s): v_water = 1333 m/s, v_air = 340 m/s. Ratio ≈ 3.9. Note: Always check whether the given numbers are physically plausible.
Q16HARD· Ultrasonic
Give two medical and two industrial applications of ultrasound.
Show solution
Step 1 — Medical applications. (i) Diagnostic imaging — Pregnancy scans, echocardiograms, organ examinations. Ultrasound penetrates soft tissue, reflects at boundaries (e.g., between fluid and solid), and is non-ionising (unlike X-rays, so it's safe for pregnant women). (ii) Lithotripsy — Treating kidney stones by focusing high-intensity ultrasound to break stones into small pieces that can pass naturally. Step 2 — Industrial applications. (iii) Ultrasonic cleaning — Jewellery, surgical instruments, electronics components are cleaned in ultrasonic baths. Tiny bubbles (cavitation) created by the ultrasound scrub away dirt at microscopic level. (iv) Non-destructive testing — Detecting cracks in metal parts (aircraft, pipelines, bridges) using ultrasonic pulses. Faults show up as unexpected reflections. (Bonus: Echolocation in bats and dolphins; ultrasonic welding of plastics; SONAR for fisheries and sea floor mapping.) ✦ Answer: Medical — diagnostic imaging (pregnancy/cardiac scans), lithotripsy for kidney stones. Industrial — ultrasonic cleaning of jewellery/instruments, non-destructive testing for cracks in metal.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Sound is a LONGITUDINAL mechanical wave: particles oscillate parallel to wave propagation.
  • Sound needs a medium (gas, liquid, solid). Vacuum: zero sound.
  • Compressions = high pressure / density; Rarefactions = low pressure / density.
  • Wave properties: wavelength (λ), frequency (f), time period (T = 1/f), amplitude (A), speed (v = fλ).
  • Pitch ↔ frequency. Loudness ↔ amplitude. Timbre ↔ wave shape.
  • Speed of sound: solids > liquids > gases. Air ≈ 343 m/s at 20°C. Iron ≈ 5960 m/s.
  • Echo: reflected sound heard distinctly. Min distance ≈ 17 m from reflector (so round-trip ≥ 0.1 s).
  • SONAR: distance = v × t / 2. Used for sea-floor mapping, fish-finding, submarine detection.
  • Human hearing: 20 Hz to 20 kHz. Below: infrasonic. Above: ultrasonic.
  • Ultrasound applications: medical imaging (non-ionising), cleaning, NDT, echolocation.
  • Ear structure: outer (pinna, canal) → middle (eardrum, 3 bones) → inner (cochlea, hair cells, auditory nerve).

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Sound is a MECHANICAL wave — a vibration propagated by particles. No particles, no sound. Light is an ELECTROMAGNETIC wave — oscillating electric and magnetic fields can propagate through empty space. Different physical natures, different requirements.

Speed of sound in air increases with temperature (~ 0.6 m/s per °C). Hot air molecules move faster on average → they transmit vibrations more quickly to their neighbours. On a 40 °C summer day, sound is about 12 m/s faster than at 20 °C.

Yes — that's TIMBRE. A violin and a piano playing the same note (e.g., A above middle C, 440 Hz) sound different because of their harmonic content (how strong each multiple-of-440 Hz overtone is). Pitch is the same; timbre differs.

Air pressure decreases as the plane climbs. The Eustachian tube (connecting middle ear to throat) lets air escape to equalise pressure across the eardrum. The 'pop' is the eardrum suddenly adjusting position when the pressure equalises. Yawning or swallowing helps the Eustachian tube open.

Shorter string → higher frequency of vibration → higher pitch. The wavelength of the string's fundamental vibration depends on its length. v = fλ; with v (controlled by tension and density of string) roughly fixed, shorter λ means higher f.

No — space is mostly vacuum (very few particles per cubic metre). Sound NEEDS a medium of particles to propagate. Movie depictions of explosions with sound in space are physically impossible (though they sound dramatic). Light from explosions DOES reach you, because light needs no medium.
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Last reviewed on 18 May 2026. Written and reviewed by subject-matter experts — read about our process.
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