Understanding sound waves, their properties, and applications
Sound is a form of energy that produces a sensation of hearing in our ears. It is produced by vibrating objects and travels in the form of waves through a medium.
Pluck a stretched rubber band - it moves back and forth rapidly.
While it's vibrating, you hear sound. When it stops vibrating, the
sound stops too!
Same with:
• Guitar strings vibrate → produce music
• Drum skin vibrates → produces beats
• Bell vibrates → produces ringing sound
• Vocal cords in throat vibrate → produce voice
No vibration = No sound! It's that simple.
Sound CANNOT travel through vacuum (empty space with no matter).
It needs a material medium like air, water, or solid to travel.
This is why:
• There is no sound in space (no air)
• Astronauts use radio waves to communicate (not sound)
• You can't hear someone shouting in space even if they're next to
you
• Sound travels through walls, water, metals, etc.
Experiment: An electric bell is placed inside a
glass bell jar connected to a vacuum pump.
Observation:
• Initially (with air): Bell rings, sound is heard
• As air is pumped out: Sound becomes fainter
• When complete vacuum: Bell is still ringing (can see it), but NO
sound heard!
Conclusion: Sound needs a medium to travel.
Without air (in vacuum), sound cannot reach our ears even though
bell is vibrating.
Sound travels in the form of longitudinal waves. When an object vibrates, it creates regions of compression (high pressure) and rarefaction (low pressure) in the medium, which travel as waves.
Imagine a long line of dominoes. When you push the first one, it
doesn't directly touch the last one, but the push travels through
all the dominoes in between!
Similarly, when a sound source vibrates:
• It pushes nearby air particles
• These particles push the next particles
• The push (compression) travels through air
• But the air particles themselves don't travel far - they just
vibrate in place!
Sound is energy traveling through particles, not the particles
themselves traveling!
Compression is the region of high pressure where particles are crowded together or compressed. This is formed when the vibrating object moves forward.
Rarefaction is the region of low pressure where particles are spread apart or rarefied. This is formed when the vibrating object moves backward.
| Medium | Speed of Sound (approx.) | Examples |
|---|---|---|
| Solids | Fastest (~ 5000 m/s in steel) | Wood, metal, glass, concrete |
| Liquids | Moderate (~ 1500 m/s in water) | Water, oil, alcohol |
| Gases | Slowest (~ 340 m/s in air at 20°C) | Air, oxygen, nitrogen, helium |
Particles in solids are very close together, so they can pass
vibrations to each other very quickly - like tightly packed
dominoes fall faster!
In gases, particles are far apart, so it takes longer for one
particle to hit the next - like widely spaced dominoes take longer
to fall.
That's why you can hear a train coming by putting your ear to the
railway track long before you hear it through air!
Wavelength is the distance between two consecutive compressions or two consecutive rarefactions. It is the length of one complete wave. Symbol: λ (lambda).
SI Unit: meter (m)
Wavelength = Distance between two consecutive compressions
OR
Wavelength = Distance between two consecutive rarefactions
Frequency is the number of complete waves (or oscillations) produced per second. It tells us how fast the object is vibrating. Symbol: ν (nu) or f.
Frequency = Number of oscillations / Time taken
ν = n / t
SI Unit: Hertz (Hz)
1 Hertz = 1 oscillation per second
1 Hz = 1/s = 1 s⁻¹
Larger units:
1 kilohertz (kHz) = 1000 Hz
1 megahertz (MHz) = 1,000,000 Hz
Time Period is the time taken to produce one complete wave (one compression + one rarefaction). It is the time for one complete oscillation.
Time Period = Time taken / Number of oscillations
T = t / n
SI Unit: second (s)
Relationship with frequency:
T = 1 / ν
ν = 1 / T
Frequency and time period are reciprocals!
A sound source produces 500 oscillations in 2 seconds.
Calculate:
(a) Frequency
(b) Time period
Solution:
Number of oscillations (n) = 500
Time (t) = 2 seconds
(a) Frequency:
ν = n / t
ν = 500 / 2
ν = 250 Hz
(b) Time period:
T = 1 / ν
T = 1 / 250
T = 0.004 s
T = 4 milliseconds
Answer: (a) 250 Hz, (b) 0.004 s or 4 ms
Amplitude is the maximum displacement of particles from their mean position during vibration. It represents the "strength" or intensity of the vibration.
When you shout, your vocal cords vibrate with LARGE amplitude →
loud sound
When you whisper, your vocal cords vibrate with SMALL amplitude →
soft sound
Same frequency (same pitch), but different amplitude (different
loudness)!
Think of a guitar string:
• Pluck it gently (small amplitude) → soft sound
• Pluck it hard (large amplitude) → loud sound
• But the note (frequency) remains the same!
Speed of sound is the distance traveled by sound per unit time. It depends on the properties of the medium through which sound travels.
Speed = Frequency × Wavelength
v = ν × λ
This is the wave equation - one of the most important formulas!
Where:
v = Speed of sound (m/s)
ν = Frequency (Hz)
λ = Wavelength (m)
At 20°C (room temperature): v ≈ 340 m/s
At 0°C: v ≈ 331 m/s
At 25°C: v ≈ 346 m/s
For every 1°C increase in temperature, speed increases by about
0.6 m/s.
Why? Higher temperature → particles move faster → sound travels
faster!
A sound wave has frequency 500 Hz and wavelength 0.68 m. Calculate
the speed of sound.
Solution:
Frequency (ν) = 500 Hz
Wavelength (λ) = 0.68 m
Speed = Frequency × Wavelength
v = ν × λ
v = 500 × 0.68
v = 340 m/s
Answer: 340 m/s (this is the speed of sound in
air at 20°C!)
A sound wave of frequency 1000 Hz travels in air at speed 340 m/s.
Find its wavelength.
Solution:
Frequency (ν) = 1000 Hz
Speed (v) = 340 m/s
Using v = ν × λ
λ = v / ν
λ = 340 / 1000
λ = 0.34 m
λ = 34 cm
Answer: 0.34 m or 34 cm
When sound waves strike a surface, they bounce back. This phenomenon is called reflection of sound. The sound waves follow the laws of reflection, just like light.
An echo is the repetition of sound caused by reflection from a distant surface. We hear the original sound, then after a short delay, we hear the reflected sound (echo).
To hear an echo distinctly, two conditions must be met:
1. Minimum Distance: The reflecting surface
should be at least 17 m away (in air at 20°C)
2. Time Gap: There should be at least 0.1 second
gap between original sound and reflected sound
Why 0.1 second?
Human ear can distinguish two sounds only if they reach at least
0.1 second apart. This is called
persistence of hearing.
Calculation: If sound travels at 340 m/s, in 0.1 s it travels:
Distance = speed × time = 340 × 0.1 = 34 m
But sound goes to wall and comes back (travels twice the
distance)
So minimum distance to wall = 34 / 2 = 17 m
A person claps and hears an echo after 1.5 seconds. If speed of
sound is 340 m/s, how far is the reflecting surface?
Solution:
Time for echo (t) = 1.5 s
Speed of sound (v) = 340 m/s
Total distance traveled by sound = v × t
Total distance = 340 × 1.5 = 510 m
This is the distance to the wall and back (sound travels twice)
Distance to reflecting surface = 510 / 2 = 255 m
Answer: 255 m
1. Mountains and Valleys: Shout in a valley, hear
echo from mountains
2. Empty Halls: Clap in empty auditorium, sound
echoes from walls
3. Tunnels: Sound in tunnel echoes multiple
times
4. Bats: Use echolocation - produce ultrasonic
sounds and listen to echoes to navigate and hunt in darkness
5. Dolphins: Use echo to locate fish
underwater
6. Sonar: Ships use sound echoes to measure depth
of ocean and detect submarines
Reverberation is the persistence of sound after the source stops producing sound, caused by multiple reflections from walls, ceiling, and floor of an enclosed space.
In auditoriums, cinema halls, and conference rooms, reverberation
is reduced by:
• Carpets on floor: Absorb sound, reduce
reflection
• Curtains on walls: Soft fabric absorbs sound
• Acoustic ceiling tiles: Specially designed to
absorb sound
• Upholstered seats: Cushions absorb sound
• Perforated boards: Trap sound in holes
These materials absorb sound instead of reflecting it, making
speech clearer!
The audible range for humans is 20 Hz to 20,000 Hz (20 kHz). Sounds with frequencies in this range can be heard by the human ear.
Sound with frequency less than 20 Hz is called infrasonic sound or infrasound. Humans cannot hear it, but some animals can.
Sound with frequency between 20 Hz and 20,000 Hz is called audible sound. This is the range humans can hear.
Sound with frequency greater than 20,000 Hz (20 kHz) is called ultrasonic sound or ultrasound. Humans cannot hear it, but some animals can.
| Type of Sound | Frequency Range | Can Humans Hear? | Examples |
|---|---|---|---|
| Infrasonic | Less than 20 Hz | No | Earthquakes, elephants |
| Audible | 20 Hz to 20,000 Hz | Yes | Human speech, music |
| Ultrasonic | Above 20,000 Hz | No | Bats, medical ultrasound |
Bats are mostly blind but can fly perfectly in complete darkness
using echolocation:
1. Bat produces ultrasonic sound (high frequency, can't be heard
by humans)
2. Sound travels and hits obstacles (trees, insects, buildings)
3. Sound reflects back as echo
4. Bat hears the echo with its sensitive ears
5. Brain calculates distance and direction from time taken for
echo
6. Bat navigates, catches prey, and avoids obstacles!
This natural "sonar" is so precise that bats can catch tiny flying
insects in complete darkness!
SONAR is a device that uses ultrasonic sound waves to detect and locate objects underwater. It works on the principle of reflection of ultrasound.
2d = v × t
Distance to object: d = (v × t) / 2
Where:
d = Distance to object (m)
v = Speed of sound in water (≈ 1500 m/s)
t = Time between transmission and reception of echo (s)
We divide by 2 because sound travels to the object and back (twice
the distance)
A SONAR device sends ultrasound and receives echo after 4 seconds.
If speed of sound in water is 1500 m/s, find the depth of the
ocean.
Solution:
Time for echo (t) = 4 s
Speed in water (v) = 1500 m/s
Total distance = v × t = 1500 × 4 = 6000 m
This is distance to seabed and back (twice the depth)
Depth of ocean = 6000 / 2 = 3000 m = 3 km
Answer: 3000 m or 3 km
The human ear is a complex organ that converts sound waves into electrical signals that the brain can understand. It has three main parts: Outer Ear, Middle Ear, and Inner Ear.
Step 1: Sound waves enter through pinna (outer
ear) - like a satellite dish collecting signals
Step 2: Waves travel through ear canal to eardrum
- like signal traveling through cable
Step 3: Eardrum vibrates like a drum skin -
converts sound waves to mechanical vibrations
Step 4: Three tiny bones (hammer, anvil, stirrup)
amplify vibrations - like a mechanical amplifier
Step 5: Vibrations reach cochlea (fluid-filled
spiral) - vibrations create waves in fluid
Step 6: Tiny hair cells in cochlea move with
fluid - like seaweed moving in ocean waves
Step 7: Hair cells generate electrical signals -
converting mechanical to electrical
Step 8: Auditory nerve carries signals to brain -
like electric wire
Step 9: Brain interprets signals - "Ah! That's
music!" or "That's a car horn!"
Avoid:
• Very loud sounds (can damage eardrum and hair cells)
• Listening to music at high volume for long time
• Inserting sharp objects in ear
• Exposure to sudden very loud noise (crackers, gunshots)
Remember:
• Hair cells in cochlea, once damaged, CANNOT regenerate
• Permanent hearing loss can occur from prolonged loud noise
• Use ear protection in noisy environments
• Keep volume moderate when using headphones
A sound wave has frequency 256 Hz and speed 340 m/s. Calculate its
wavelength.
Solution:
Frequency (ν) = 256 Hz
Speed (v) = 340 m/s
Using v = ν × λ
λ = v / ν
λ = 340 / 256
λ = 1.33 m
Answer: 1.33 m
A source produces 1500 waves in 5 seconds. Calculate:
(a) Frequency
(b) Time period
Solution:
Number of waves = 1500
Time = 5 s
(a) Frequency:
ν = Number of waves / Time
ν = 1500 / 5
ν = 300 Hz
(b) Time period:
T = 1 / ν
T = 1 / 300
T = 0.00333 s
T = 3.33 milliseconds
Answer: (a) 300 Hz, (b) 0.00333 s or 3.33 ms
A person standing between two cliffs fires a gun. He hears first
echo after 3 seconds and second echo after 5 seconds. If speed of
sound is 340 m/s, find distance between the two cliffs.
Solution:
Speed (v) = 340 m/s
Time for first echo (t₁) = 3 s
Time for second echo (t₂) = 5 s
Distance to first cliff = (v × t₁) / 2 = (340 × 3) / 2 = 510 m
Distance to second cliff = (v × t₂) / 2 = (340 × 5) / 2 = 850 m
Distance between cliffs = 510 + 850 = 1360 m = 1.36 km
Answer: 1360 m or 1.36 km
A submarine uses SONAR to detect a sunken ship. The ultrasound
signal returns after 6 seconds. If speed of sound in seawater is
1500 m/s, find the distance of the ship from submarine.
Solution:
Time (t) = 6 s
Speed in water (v) = 1500 m/s
Distance = (v × t) / 2
d = (1500 × 6) / 2
d = 9000 / 2
d = 4500 m
d = 4.5 km
Answer: 4500 m or 4.5 km
Calculate wavelength of sound waves in air of:
(a) Lowest audible frequency (20 Hz)
(b) Highest audible frequency (20,000 Hz)
(Speed of sound in air = 340 m/s)
Solution:
Speed (v) = 340 m/s
(a) At 20 Hz:
λ = v / ν = 340 / 20 = 17 m
(b) At 20,000 Hz:
λ = v / ν = 340 / 20000 = 0.017 m = 1.7 cm
Answer: (a) 17 m, (b) 0.017 m or 1.7 cm
Note: Low frequency = large wavelength; High
frequency = small wavelength
At 20°C, sound travels at 340 m/s in air, 1500 m/s in water, and
5000 m/s in steel. A sound wave has frequency 500 Hz. Calculate
its wavelength in:
(a) Air
(b) Water
(c) Steel
Solution:
Frequency (ν) = 500 Hz (same in all media)
(a) In air:
λ = v / ν = 340 / 500 = 0.68 m = 68 cm
(b) In water:
λ = v / ν = 1500 / 500 = 3 m
(c) In steel:
λ = v / ν = 5000 / 500 = 10 m
Answer: (a) 0.68 m, (b) 3 m, (c) 10 m
Note: Same frequency, but wavelength increases
with speed!
For Wave Equation (v = ν × λ):
• If two values are given, find the third
• Remember: speed (m/s), frequency (Hz), wavelength (m)
• Check units before calculation
For Frequency and Time Period:
• They are reciprocals: T = 1/ν and ν = 1/T
• If given oscillations and time, find frequency first
• Time period is always in seconds
For Echo Problems:
• Sound travels to surface and back (twice the distance)
• Distance = (speed × time) / 2
• Minimum distance for echo = 17 m (in air)
• Always divide by 2 for one-way distance!
For SONAR Problems:
• Use d = (v × t) / 2
• Speed in water ≈ 1500 m/s
• Time is for echo (round trip)
• Divide by 2 to get one-way distance
For Range of Hearing:
• Infrasound: < 20 Hz
• Audible: 20 Hz to 20 kHz
• Ultrasound: > 20 kHz
Common Mistakes to Avoid:
• Forgetting to divide by 2 in echo/SONAR problems
• Mixing up frequency (Hz) with wavelength (m)
• Using wrong speed value (air vs water vs solid)
• Not converting units (kHz to Hz, km to m)
• Confusing time period with time taken
• Not including units in final answer