SK TUITIONS • CLASS 9 CBSE SCIENCE

Sound Waves: Characteristics and Applications – 100 Question Bank

NCERT • Exemplar Level • Numericals • Competency Based • HOTS • Olympiad

Complete chapter practice with detailed answers and progressively increasing difficulty.

30MCQs
152 Markers
203 Markers
154 Markers
5Case Studies
15Exemplar/HOTS

Complete Chapter Coverage

Production of Sound
Vibration & Oscillation
Sources of Sound
Vocal Cords
Tuning Fork
Propagation of Sound
Sound in Solids
Sound in Liquids
Sound in Gases
Medium & Vacuum
Vacuum Bell Jar
Slinky Analogy
Compression
Rarefaction
Longitudinal Waves
Mechanical Waves
Energy of Sound Waves
Microphone & Speaker
Density–Distance Graph
Crest & Trough
Wavelength
Frequency
Time Period
ν = 1/T
Amplitude
Intensity
Energy & Amplitude
Speed of Sound
v = λν
Sound in Different Media
Temperature & Humidity
Pitch
Loudness
Audible Range
Infrasound
Ultrasound
Tone & Musical Note
Fundamental & Overtones
Timbre
Octave
Reflection of Sound
Echo
Reverberation
Echolocation
SONAR
Ultrasonography
Noise & Hearing
Human Ear

Section A – 30 Multiple Choice Questions

Fundamental concepts gradually progress into graph interpretation and numerical reasoning.

Q1
Sound is generally produced when an object:
  • (A) becomes hot.
  • (B) vibrates.
  • (C) changes colour.
  • (D) remains completely stationary.
View Detailed Answer

Correct option: (B).

Sound is produced by vibrating objects. Vibration is a periodic to-and-fro motion about a mean position.

Q2
Human speech is produced mainly by vibration of:
  • (A) the tongue only.
  • (B) vocal cords.
  • (C) teeth.
  • (D) lungs.
View Detailed Answer

Correct option: (B).

The vocal cords are stretched muscular flaps inside the larynx or voice box.

Q3
The two sides of the U-shaped part of a tuning fork are called:
  • (A) coils
  • (B) diaphragms
  • (C) prongs or tines
  • (D) membranes
View Detailed Answer

Correct option: (C).

Q4
The material through which sound travels is called:
  • (A) source
  • (B) medium
  • (C) vacuum
  • (D) echo
View Detailed Answer

Correct option: (B).

Q5
Sound cannot normally propagate through:
  • (A) steel
  • (B) water
  • (C) air
  • (D) vacuum
View Detailed Answer

Correct option: (D).

Sound is a mechanical wave and needs particles of a material medium to transfer the disturbance.

Q6
A sound wave in air is primarily:
  • (A) longitudinal and mechanical.
  • (B) transverse and electromagnetic.
  • (C) longitudinal and electromagnetic.
  • (D) transverse and non-mechanical.
View Detailed Answer

Correct option: (A).

Air particles vibrate parallel to the direction in which the sound wave propagates.

Q7
A region of a sound wave where air density is greater than average is called:
  • (A) rarefaction
  • (B) compression
  • (C) trough only
  • (D) vacuum
View Detailed Answer

Correct option: (B).

Q8
A rarefaction corresponds to a region where particles are:
  • (A) more closely packed than average.
  • (B) more spread out than average.
  • (C) completely absent.
  • (D) permanently at rest.
View Detailed Answer

Correct option: (B).

Q9
When a sound wave travels through air, the air particles:
  • (A) travel continuously from source to listener.
  • (B) vibrate about their mean positions.
  • (C) disappear after transferring energy.
  • (D) move only upward.
View Detailed Answer

Correct option: (B).

It is the disturbance and energy that propagate; the particles themselves oscillate locally.

Q10
Which property most directly shows that sound is a mechanical wave?
  • (A) Sound has frequency.
  • (B) Sound needs a material medium.
  • (C) Sound can be loud.
  • (D) Sound can be pleasant.
View Detailed Answer

Correct option: (B).

Q11
What is mainly transported by a sound wave from one region to another?
  • (A) Air particles
  • (B) Energy
  • (C) Matter from the source
  • (D) Tuning-fork metal
View Detailed Answer

Correct option: (B).

Q12
A microphone primarily converts:
  • (A) electrical energy to sound energy.
  • (B) sound energy to electrical signals.
  • (C) light energy to sound energy.
  • (D) heat energy to light energy.
View Detailed Answer

Correct option: (B).

Q13
A loudspeaker does approximately the reverse of a:
  • (A) microphone
  • (B) thermometer
  • (C) prism
  • (D) spring balance
View Detailed Answer

Correct option: (A).

A loudspeaker uses electrical signals to make its cone or diaphragm vibrate and produce sound.

Q14
The distance between two consecutive crests in the density graph of a sound wave is its:
  • (A) frequency
  • (B) amplitude
  • (C) wavelength
  • (D) time period
View Detailed Answer

Correct option: (C).

Q15
A point in a medium undergoes 12 complete density oscillations in 3 s. The frequency is:
Numerical
  • (A) 3 Hz
  • (B) 4 Hz
  • (C) 12 Hz
  • (D) 36 Hz
View Detailed Answer
ν = number of oscillations / time = 12/3 = 4 Hz

Correct option: (B).

Q16
The time period of a 50 Hz sound wave is:
Numerical
  • (A) 50 s
  • (B) 5 s
  • (C) 0.02 s
  • (D) 0.2 s
View Detailed Answer
T = 1/ν = 1/50 = 0.02 s

Correct option: (C).

Q17
On a density–distance graph, a crest corresponds most closely to:
  • (A) maximum-density compression.
  • (B) minimum-density rarefaction.
  • (C) zero frequency.
  • (D) absence of particles.
View Detailed Answer

Correct option: (A).

Q18
A sound wave of larger amplitude generally carries:
  • (A) less energy.
  • (B) more energy.
  • (C) no energy.
  • (D) exactly the same energy in every case.
View Detailed Answer

Correct option: (B).

Q19
Sound intensity is best described as:
  • (A) sound energy crossing unit area per unit time.
  • (B) frequency divided by wavelength.
  • (C) number of particles in the source.
  • (D) subjective perception of pitch.
View Detailed Answer

Correct option: (A).

Q20
A sound wave has wavelength 2 m and frequency 170 Hz. Its speed is:
Numerical
  • (A) 85 m/s
  • (B) 170 m/s
  • (C) 340 m/s
  • (D) 680 m/s
View Detailed Answer
v = λν = 2×170 = 340 m/s

Correct option: (C).

Q21
In general, sound travels fastest through:
  • (A) gases
  • (B) liquids
  • (C) solids
  • (D) vacuum
View Detailed Answer

Correct option: (C).

The chapter gives the general order: solids > liquids > gases.

Q22
The speed of sound in air generally increases when:
  • (A) temperature increases.
  • (B) temperature falls to absolute zero only.
  • (C) all air is removed.
  • (D) the source becomes softer.
View Detailed Answer

Correct option: (A).

The chapter also notes that increasing humidity increases the speed of sound in air.

Q23
Pitch is most directly associated with:
  • (A) frequency
  • (B) speed alone
  • (C) wavelength alone
  • (D) distance from source only
View Detailed Answer

Correct option: (A).

Higher frequency is generally perceived as higher pitch.

Q24
The approximate human audible range is:
  • (A) 0–20 Hz
  • (B) 20 Hz–20 kHz
  • (C) 20 kHz–200 kHz
  • (D) 200 Hz–2 kHz only
View Detailed Answer

Correct option: (B).

Q25
A sound of frequency 35 kHz is:
  • (A) infrasonic.
  • (B) audible to all humans.
  • (C) ultrasonic.
  • (D) necessarily silent because it has no energy.
View Detailed Answer

Correct option: (C).

35 kHz is above 20 kHz.

Q26
A sound of frequency 12 Hz is classified as:
  • (A) ultrasonic
  • (B) infrasonic
  • (C) audible sound
  • (D) electromagnetic radiation
View Detailed Answer

Correct option: (B).

Q27
A nearly single-frequency sound is called a:
  • (A) tone
  • (B) echo
  • (C) reverberation
  • (D) sonic boom
View Detailed Answer

Correct option: (A).

Q28
If one musical note has frequency 200 Hz, the note one octave higher has frequency:
Numerical
  • (A) 100 Hz
  • (B) 200 Hz
  • (C) 300 Hz
  • (D) 400 Hz
View Detailed Answer

Correct option: (D).

An octave corresponds to doubling the fundamental frequency.

Q29
Taking the speed of sound as 340 m/s, the minimum distance of a large reflecting surface for a clearly separated echo is approximately:
Echo Numerical
  • (A) 8.5 m
  • (B) 17 m
  • (C) 34 m
  • (D) 68 m
View Detailed Answer

The sound must make a round trip in at least 0.1 s.

2d = vt = 340×0.1 = 34 m
d = 17 m

Correct option: (B).

Q30
A SONAR pulse travels in seawater at 1500 m/s and returns after 2 s. The object is:
SONAR Numerical
  • (A) 750 m away.
  • (B) 1500 m away.
  • (C) 3000 m away.
  • (D) 375 m away.
View Detailed Answer
Distance = vt/2 = 1500×2/2 = 1500 m

Correct option: (B).

Section B – 15 Two-Mark Questions

Short explanations, observations and numerical applications.

Q31
What conclusion is drawn from plucking a stretched rubber band and observing it before and after the vibrations stop?
View Detailed Answer

Sound is heard while the stretched rubber band is vibrating. When the vibration stops, the sound also stops.

Hence, the activity supports the conclusion that sound is produced by vibrations.

Q32
How does the tuning-fork-and-water experiment demonstrate that a tuning fork is vibrating?
View Detailed Answer

When a struck tuning fork is touched gently to the water surface, the vibrating prong produces visible disturbances or waves in the water.

This provides visible evidence that the prongs are vibrating while producing sound.

Q33
Give one experimental observation each to show that sound travels through solids and liquids.
View Detailed Answer

Solid: Knocking a desk can be heard clearly when the ear is placed against the desk.

Liquid: Metal spoons struck while submerged in water can still be heard.

Q34
Explain the main observation and conclusion of the vacuum bell-jar experiment.
View Detailed Answer

As air is removed from the bell jar, the sound of the ringing bell becomes progressively fainter even though the bell can still be seen vibrating.

This shows that sound requires a material medium and cannot propagate through vacuum.

Q35
Differentiate between a compression and a rarefaction.
View Detailed Answer

Compression: region of higher-than-average density where particles are relatively closer together.

Rarefaction: region of lower-than-average density where particles are relatively farther apart.

Q36
Do the particles of air travel from a loudspeaker all the way to your ear? Explain.
View Detailed Answer

No.

The particles oscillate about their own mean positions. The sequence of compressions and rarefactions, and therefore the energy, travels through the medium.

Q37
A source completes 1200 oscillations in one minute. Calculate its frequency and time period.
Numerical
View Detailed Answer

1 minute = 60 s.

ν = 1200/60 = 20 Hz
T = 1/20 = 0.05 s
Q38
Two consecutive compressions on a density graph are located at 1 m and 4 m. Find the wavelength.
Graph Numerical
View Detailed Answer

Distance between successive compressions equals one wavelength.

λ = 4−1 = 3 m
Q39
A sound wave has frequency 200 Hz and wavelength 1.7 m. Calculate its speed.
Numerical
View Detailed Answer
v = λν = 1.7×200 = 340 m/s
Q40
Using 340 m/s in air, 1500 m/s in water and 5000 m/s in steel, arrange them in increasing order of speed and calculate water-to-air speed ratio.
View Detailed Answer

Increasing order:

Air < Water < Steel
Water/Air = 1500/340 ≈ 4.41

Sound travels about 4.4 times faster in water than in air using these values.

Q41
Differentiate between pitch and loudness.
View Detailed Answer

Pitch: human perception associated mainly with frequency. Higher frequency generally gives higher pitch.

Loudness: subjective perception associated mainly with amplitude. Larger amplitude is generally heard as a louder sound.

Q42
Why should loudness and intensity not be treated as exactly the same quantity?
View Detailed Answer

Intensity is a physically measurable rate of sound-energy flow per unit area.

Loudness is a subjective perception and depends partly on the hearing ability of the listener.

Q43
Define audible, infrasonic and ultrasonic sound ranges for humans.
View Detailed Answer
  • Infrasonic: below 20 Hz.
  • Audible: approximately 20 Hz to 20,000 Hz.
  • Ultrasonic: above 20,000 Hz or 20 kHz.
Q44
Differentiate between echo and reverberation.
View Detailed Answer

Echo: a reflected sound heard separately from the original, typically when the time gap is at least about 0.1 s.

Reverberation: persistence of sound caused by multiple reflections arriving so quickly that they overlap with one another.

Q45
State the energy conversion occurring in a microphone and in a speaker.
View Detailed Answer

Microphone: sound energy → electrical signal.

Speaker: electrical signal → vibration of diaphragm/cone → sound energy.

Section C – 20 Three-Mark Questions

Numerical-intensive practice on frequency, wavelength, sound speed, echoes and applications.

Q46
A vibrating source completes 150 oscillations in 5 s. Calculate its frequency and time period.
Numerical
View Detailed Answer
ν = 150/5 = 30 Hz
T = 1/30 ≈ 0.0333 s
Q47
An oscillating piston produces a 20 Hz sound wave. How many complete oscillations does it make in one minute?
Textbook-Style Numerical
View Detailed Answer
Number = νt = 20×60 = 1200

Answer: 1200 oscillations.

Q48
A sound wave in air has wavelength 0.86 m and travels at 344 m/s. Calculate its frequency and time period.
Numerical
View Detailed Answer
ν = v/λ = 344/0.86 = 400 Hz
T = 1/400 = 0.0025 s
Q49
A 680 Hz sound travels in air at 340 m/s. Find its wavelength.
Numerical
View Detailed Answer
λ = v/ν = 340/680 = 0.50 m
Q50
Using 340 m/s as the speed of sound, estimate wavelengths corresponding to 20 Hz and 20 kHz.
Human Hearing Numerical
View Detailed Answer

20 Hz:

λ = 340/20 = 17 m

20 kHz = 20,000 Hz:

λ = 340/20000 = 0.017 m = 1.7 cm
Q51
Lightning is seen and thunder is heard 6 s later. Estimate the distance of the lightning strike if sound travels at 340 m/s.
Thunder Numerical
View Detailed Answer

The travel time of light is negligible for this estimate.

d = vt = 340×6 = 2040 m
d = 2.04 km
Q52
Two friends are 340 m apart along a steel fence. Sound speed is 5000 m/s in steel and 340 m/s in air. Calculate the arrival times and time difference.
Textbook-Type Numerical
View Detailed Answer

Through air:

tair = 340/340 = 1.00 s

Through steel:

tsteel = 340/5000 = 0.068 s
Difference = 1.000−0.068 = 0.932 s

Since the time gap is far greater than 0.1 s, the arrivals can be distinguished separately.

Q53
A sound wave propagating through steel has wavelength 50 m. If the speed of sound in steel is 5000 m/s, find its frequency and time period.
Graph-Style Numerical
View Detailed Answer
ν = 5000/50 = 100 Hz
T = 1/100 = 0.01 s
Q54
Sound travels at 331 m/s at 0°C and 344 m/s at 22°C. For a distance of 1720 m, approximately how much extra time is required at 0°C?
NCERT Exercise-Type
View Detailed Answer

At 0°C:

t₀ = 1720/331 ≈ 5.196 s

At 22°C:

t₂₂ = 1720/344 = 5.000 s
Extra time ≈ 5.196−5.000 = 0.196 s

Approximately 0.20 s extra.

Q55
A student claps and hears an echo 0.60 s later. If sound speed is 340 m/s, find the distance of the reflecting wall.
Echo Numerical
View Detailed Answer
2d = vt
d = 340×0.60/2 = 102 m
Q56
What minimum distance from a wall is needed for a distinct echo when sound speed is 343 m/s and the minimum separable time is 0.10 s?
Echo Numerical
View Detailed Answer
d = vt/2 = 343×0.10/2 = 17.15 m
Q57
An experiment requires the reflected sound to arrive at least 0.20 s later. Find the minimum wall distance if v = 343 m/s.
Textbook Numerical
View Detailed Answer
d = 343×0.20/2 = 34.3 m
Q58
A SONAR signal in seawater returns after 0.90 s. If sound speed is 1530 m/s, find the object’s distance.
SONAR Numerical
View Detailed Answer
d = 1530×0.90/2
d = 688.5 m
Q59
A SONAR pulse used to measure ocean depth returns after 4 s. Find the depth if sound travels at 1500 m/s in seawater.
Textbook Numerical
View Detailed Answer
Depth = 1500×4/2 = 3000 m

Depth = 3.0 km.

Q60
A parking sensor detects an obstacle 1.2 m away. Find the round-trip time of its ultrasonic pulse if sound speed is 345 m/s.
Application Numerical
View Detailed Answer

Total distance:

2×1.2 = 2.4 m
t = 2.4/345 ≈ 0.00696 s
t ≈ 6.96 ms
Q61
A source in air changes its frequency from 250 Hz to 500 Hz. If sound speed remains constant, what happens to its wavelength?
View Detailed Answer
λ = v/ν

When frequency doubles while speed remains constant, wavelength becomes half.

Q62
Why does sound intensity normally decrease as we move farther from a small source?
View Detailed Answer

Sound spreads outward over an increasingly larger area.

The available wave energy is therefore distributed over a larger area, so the sound energy crossing each unit area per unit time decreases.

Q63
Differentiate among a tone, a musical note and timbre.
View Detailed Answer

Tone: nearly single-frequency sound.

Musical note: combination of a fundamental frequency and higher overtones.

Timbre: characteristic quality that allows different instruments to sound different even when playing the same note at similar loudness.

Q64
Briefly explain how the human ear converts incoming sound into a sensation perceived by the brain.
View Detailed Answer
  1. Sound entering the ear causes the eardrum to vibrate.
  2. Tiny bones amplify and transmit the vibrations.
  3. The cochlea converts them into electrical signals.
  4. These signals travel to the brain, which interprets them as sound.
Q65
Give three applications each of ultrasonic or infrasonic waves discussed in the chapter.
View Detailed Answer

Ultrasonic applications include:

  • ultrasonography,
  • breaking kidney stones,
  • detecting defects inside metal blocks,
  • ultrasonic cleaning or welding,
  • SONAR and echolocation.

Infrasonic applications include:

  • detecting earthquakes,
  • monitoring volcanic activity,
  • detecting severe storms over long distances.

Section D – 15 Four-Mark Questions

Detailed explanation, derivation, graph analysis and advanced application.

Q66
Describe the vacuum bell-jar experiment and explain why it proves that sound requires a material medium.
View Detailed Answer
  1. An electric bell is placed inside a bell jar and switched on.
  2. With air present, the bell is clearly heard.
  3. A vacuum pump gradually removes air from the jar.
  4. The sound becomes progressively fainter even though the bell continues vibrating visibly.
  5. When a near vacuum is produced, almost no sound is heard.
  6. When air is allowed back into the jar, the sound becomes louder again.

The vibrating source remains active, but sound transmission decreases as the number of particles in the medium decreases. Hence a material medium is necessary for sound propagation.

Q67
Use the slinky analogy to explain how a longitudinal sound wave propagates without transferring matter from source to listener.
View Detailed Answer

When one end of a stretched slinky is repeatedly pushed and pulled, alternating regions of closely spaced and widely spaced turns appear.

These are analogous to compressions and rarefactions in air.

A marked turn of the slinky does not travel to the opposite end. Instead, it moves back and forth about its original position while the disturbance travels along the slinky.

Similarly, air particles oscillate parallel to the direction of propagation, while the sound disturbance and energy move forward.

Q68
Explain how a sound wave can be represented using a density–distance graph. Identify crest, trough, wavelength and amplitude.
View Detailed Answer
  • The horizontal axis represents distance from the source.
  • The vertical axis represents density of the medium.
  • The dashed central level represents average density.
  • A crest represents maximum density and corresponds to a compression.
  • A trough represents minimum density and corresponds to a rarefaction.
  • Wavelength is the distance between two consecutive crests or two consecutive troughs.
  • Amplitude is the maximum density change above or below the average value.
Q69
Differentiate between longitudinal and transverse waves and explain why sound in air is longitudinal while light can travel through vacuum.
View Detailed Answer
Longitudinal Wave Transverse Wave
Particles vibrate parallel to wave propagation. Particles vibrate perpendicular to wave propagation.
Sound in air is an example. A transverse representation has displacement perpendicular to propagation.

Sound is a mechanical wave and requires particles to transfer compressions and rarefactions. Therefore it cannot travel through vacuum.

Light is not a mechanical wave and can propagate through vacuum.

Q70
Derive the relation v = λν for a sound wave.
View Detailed Answer

In one complete time period T, a particular phase of the wave such as a crest travels one wavelength λ.

Speed = distance/time
v = λ/T

Since:

ν = 1/T

therefore:

v = λν

Thus speed equals wavelength multiplied by frequency.

Q71
Two waves A and B travel through the same medium at 300 m/s. A has wavelength 1.5 m and B has wavelength 0.75 m. Find their frequencies and compare their pitches.
Graph/Concept Numerical
View Detailed Answer

Wave A:

νA = 300/1.5 = 200 Hz

Wave B:

νB = 300/0.75 = 400 Hz

Wave B has twice the frequency and therefore would generally be perceived as having the higher pitch.

Q72
A sound wave has wavelength 3.44 m and propagates at 344 m/s. Find its frequency, time period and number of oscillations in 30 s.
Multi-Step Numerical
View Detailed Answer
ν = 344/3.44 = 100 Hz
T = 1/100 = 0.01 s
N = νt = 100×30 = 3000

Answers: 100 Hz, 0.01 s, 3000 oscillations.

Q73
Using speeds 340 m/s in air, 1500 m/s in water and 5000 m/s in steel, calculate the ratios water:air, steel:water and steel:air.
Data Numerical
View Detailed Answer
Water/Air = 1500/340 ≈ 4.41
Steel/Water = 5000/1500 ≈ 3.33
Steel/Air = 5000/340 ≈ 14.71

These ratios illustrate the chapter’s general statement that sound travels fastest in solids, slower in liquids and slowest in gases.

Q74
Two sound sources are at equal distances from the same cliff. One is in air and the other in water. The return time in air is 4.5 times the return time in water. Find the ratio of their speeds.
NCERT Exercise-Level
View Detailed Answer

For equal round-trip distance:

v ∝ 1/t

If:

tair = 4.5 twater

then:

vair / vwater = twater / tair = 1/4.5
vair : vwater = 1 : 4.5
Q75
Explain echo and reverberation, and describe how an auditorium can be designed to control unwanted reverberation.
View Detailed Answer

An echo is a reflected sound heard distinctly after the original sound, usually when the time separation is at least about 0.1 s.

Reverberation is the persistence of sound due to repeated reflections arriving too quickly to be heard separately.

Excess reverberation makes speech and music unclear. It can be controlled by using sound-absorbing materials such as:

  • curtains,
  • upholstered chairs,
  • soft porous wall or ceiling panels.

Good auditorium design preserves useful reflection while reducing unwanted multiple reflections.

Q76
Explain the meanings of pitch, loudness and timbre and identify the main physical factor associated with each.
View Detailed Answer
Perception Main Association Description
Pitch Frequency Higher frequency generally sounds shriller or higher.
Loudness Amplitude Larger amplitude is generally heard as louder.
Timbre Pattern and intensity of overtones Gives different voices or instruments their distinctive sound quality.
Q77
A parking sensor emits ultrasound of frequency 40 kHz. Find its wavelength in air if v = 345 m/s. If an obstacle is 2.0 m away, find the round-trip echo time.
Ultrasound Numerical
View Detailed Answer

Frequency:

40 kHz = 40000 Hz

Wavelength:

λ = 345/40000 = 0.008625 m
λ = 8.625 mm

Round-trip distance:

2d = 4 m
t = 4/345 ≈ 0.0116 s
Q78
A bat emits ultrasound of frequency 50 kHz. Its echo returns after 0.040 s. Taking sound speed as 340 m/s, find the wavelength and distance of the obstacle.
Echolocation Numerical
View Detailed Answer

Wavelength:

λ = 340/50000 = 0.0068 m = 6.8 mm

Distance:

d = vt/2 = 340×0.040/2 = 6.8 m
Q79
Explain the principle of SONAR and why the measured travel time must be divided by two when calculating distance.
View Detailed Answer

SONAR sends ultrasonic sound waves through water. These waves reflect from an underwater object and return to a detector.

By analysing the reflected sound, the system can determine properties such as the object’s distance and direction.

The recorded time represents the journey:

Ship → Object → Ship

Therefore the one-way distance is:

d = vt/2
Q80
Explain how excessive noise may affect hearing and describe the basic working of a hearing aid.
View Detailed Answer

Prolonged exposure to high sound levels may affect health, sleep and hearing, and can lead to hearing loss.

A hearing aid generally includes:

  • a microphone to capture sound,
  • an amplifier to strengthen the electrical signal,
  • a speaker to reproduce the amplified sound for the user.

Reducing unnecessary exposure to loud sound is therefore important for protecting hearing.

Section E – 5 Competency-Based Case Studies

Integrated CBSE-style applications involving experiments, graphs, echoes and technology.

Q81 – Musical Notes and Frequency

Frequency • Pitch • Octave • Time Period
A music student uses an audio-spectrum app. One note has a fundamental frequency of 220 Hz, while a second note has a fundamental frequency of 440 Hz. Both are played through the same speaker in the same room.

(a) Which note has higher pitch?

(b) What is the relation between these two notes?

(c) Find the time period of the 220 Hz note.

(d) Find the time period of the 440 Hz note.

(e) If both travel at the same speed, which has the shorter wavelength?

View Detailed Case Study Solution

(a) The 440 Hz note has higher pitch.

(b) 440 Hz is double 220 Hz, so the notes are separated by an octave.

(c)

T = 1/220 ≈ 0.00455 s

(d)

T = 1/440 ≈ 0.00227 s

(e) The 440 Hz wave has the shorter wavelength because λ = v/ν.

Q82 – Astronauts During a Spacewalk

Vacuum • Mechanical Waves • Communication
Two astronauts are outside a space station. One astronaut strikes a metal tool against the station structure. Both astronauts are wearing sealed spacesuits equipped with communication systems.

(a) Can sound from the strike travel normally through the vacuum between them?

(b) Why not?

(c) What type of wave is sound?

(d) How can they communicate despite the vacuum?

View Detailed Case Study Solution

(a) No, not directly through the near-vacuum.

(b) Sound requires a material medium containing particles.

(c) Sound is a longitudinal mechanical wave.

(d) Their communication devices convert speech to electrical/electromagnetic signals and reproduce it as sound inside the other spacesuit.

Q83 – Auditorium Acoustics

Reflection • Echo • Reverberation
A stage is 8 m from one reflecting wall and 25 m from another. Take the speed of sound as 340 m/s.

(a) Find the round-trip reflection time from the 8 m wall.

(b) Would it normally be heard as a distinct echo?

(c) Find the round-trip time from the 25 m wall.

(d) Would that reflection be more likely to be heard separately?

(e) Name two materials used to reduce unwanted reverberation.

View Detailed Case Study Solution

(a)

t = 2×8/340 ≈ 0.047 s

(b) No. It arrives too quickly to be separated clearly from the original sound.

(c)

t = 2×25/340 ≈ 0.147 s

(d) Yes. The delay exceeds about 0.1 s.

(e) Curtains, upholstered chairs, porous acoustic panels or similar soft materials.

Q84 – Searching for a Shipwreck

SONAR • Ultrasound • Reflection
A ship sends an ultrasonic SONAR pulse downward. It receives the reflected pulse after 5.0 s. The speed of sound in seawater is 1525 m/s.

(a) Why is ultrasound suitable for SONAR?

(b) Calculate the total distance travelled by the pulse.

(c) Find the depth of the wreck.

(d) Why is the result divided by two?

View Detailed Case Study Solution

(a) Ultrasonic waves can propagate through water and can be reflected from underwater objects.

(b)

Total distance = vt = 1525×5 = 7625 m

(c)

Depth = 7625/2 = 3812.5 m

(d) The measured time includes both the downward and upward journeys.

Q85 – Thunder on a Cold and Warm Day

Speed • Temperature • Time Delay
Lightning occurs 1720 m from an observer. The speed of sound is 331 m/s at 0°C and 344 m/s at 22°C.

(a) Find the sound travel time at 0°C.

(b) Find the sound travel time at 22°C.

(c) Calculate the difference.

(d) At which temperature does thunder arrive sooner?

View Detailed Case Study Solution

(a)

t₀ = 1720/331 ≈ 5.196 s

(b)

t₂₂ = 1720/344 = 5.000 s

(c)

Difference ≈ 0.196 s

(d) Thunder reaches the observer sooner at 22°C because sound travels faster in the warmer air.

Section F – 15 NCERT Exemplar-Level + Olympiad/HOTS Challenges

Advanced numerical and reasoning questions combining multiple sound-wave concepts.

Q86
Exemplar-Level: Twenty-five compressions pass a fixed point in 0.10 s. If the speed of sound is 340 m/s, calculate frequency, time period and wavelength.
Exemplar Level Numerical
View Challenge Solution
ν = 25/0.10 = 250 Hz
T = 1/250 = 0.004 s
λ = 340/250 = 1.36 m
Q87
Two sound waves travel at the same speed. Their wavelengths are in the ratio 3:2. Find the ratio of their frequencies.
Exemplar Level Ratio Reasoning
View Challenge Solution
v = λν

For constant v:

ν ∝ 1/λ

Therefore:

ν₁ : ν₂ = 2 : 3
Q88
A sound wave has a time period of 2.5 × 10−3 s and wavelength 0.85 m. Find its frequency and speed.
Exemplar Numerical
View Challenge Solution
ν = 1/T = 1/(2.5×10−3) = 400 Hz
v = λν = 0.85×400 = 340 m/s
Q89
A source in air changes frequency from 500 Hz to 1000 Hz. The speed of sound remains 344 m/s. Calculate both wavelengths and explain what changes and what remains constant.
Exemplar Level Numerical
View Challenge Solution

At 500 Hz:

λ₁ = 344/500 = 0.688 m

At 1000 Hz:

λ₂ = 344/1000 = 0.344 m

Frequency doubles and wavelength halves. The speed remains essentially fixed because the medium and conditions are unchanged.

Q90
Why would music become severely distorted if different frequencies travelled through ordinary air at very different speeds?
HOTS
View Challenge Solution

Music contains many frequencies produced simultaneously.

If those frequencies travelled at substantially different speeds, they would arrive at a distant listener at different times. The timing and relation between the components of the musical sound would be altered.

Ordinary air does not show this large frequency-dependent speed variation under the conditions discussed in the chapter, so musical sounds preserve their temporal relationship much better.

Q91
A density–distance graph has consecutive crests at x = 2 cm and x = 6 cm. Its density amplitude is 3 arbitrary units. Find wavelength and describe where the next crest and intervening trough would occur.
Graph Challenge
View Challenge Solution
λ = 6−2 = 4 cm

Next crest:

6+4 = 10 cm

A trough occurs halfway between consecutive crests:

x = 4 cm

The amplitude remains 3 density units above or below the average-density line.

Q92
Two listeners stand at the same location. One reports a sound as very loud while the other reports it as less loud. Can the physical intensity at their location still be the same? Explain.
Exemplar Reasoning
View Challenge Solution

Yes.

Intensity is a measurable physical property of the sound wave at that location. Loudness is a subjective perception that also depends on the listener’s hearing sensitivity.

Thus the same physical intensity may not necessarily produce exactly the same perceived loudness for every person.

Q93
At 340 m/s the minimum echo distance is about 17 m. What does this distance become if the speed rises to 344 m/s while the minimum distinguishable time remains 0.1 s?
Echo Numerical
View Challenge Solution
d = vt/2
d = 344×0.1/2 = 17.2 m

The minimum echo distance rises slightly because sound travels farther during the same 0.1 s interval.

Q94
For a thunder source 1720 m away, calculate precisely the approximate extra travel time at 0°C compared with 22°C, using 331 m/s and 344 m/s respectively.
Exemplar Numerical
View Challenge Solution
t₀ = 1720/331 ≈ 5.1964 s
t₂₂ = 1720/344 = 5.0000 s
Δt ≈ 0.1964 s

Approximately 0.20 s extra.

Q95
A sound travels 1500 m through water and the same distance through steel. Using speeds 1500 m/s and 5000 m/s respectively, find the travel times and difference.
Olympiad Numerical
View Challenge Solution

Water:

t = 1500/1500 = 1.0 s

Steel:

t = 1500/5000 = 0.30 s
Difference = 1.0−0.30 = 0.70 s
Q96
Starting with a 200 Hz musical note, write the frequencies of notes one, two and three octaves higher. Also find their time periods.
Olympiad Numerical
View Challenge Solution

One octave:

400 Hz, T = 1/400 = 0.0025 s

Two octaves:

800 Hz, T = 1/800 = 0.00125 s

Three octaves:

1600 Hz, T = 1/1600 = 0.000625 s
Q97
A bat emits a 50 kHz pulse in air. Calculate its wavelength at 340 m/s and the obstacle distance if the echo returns after 0.040 s. State why a human would not hear the emitted pulse.
Olympiad Integrated Numerical
View Challenge Solution
λ = 340/50000 = 0.0068 m = 6.8 mm
d = 340×0.040/2 = 6.8 m

50 kHz is above the upper human hearing limit of approximately 20 kHz, so the sound is ultrasonic for humans.

Q98
A ship searching for a wreck receives its SONAR echo 5 s after transmission. The speed of sound in seawater is 1525 m/s. Calculate the depth and express it in kilometres.
Olympiad SONAR Numerical
View Challenge Solution
d = vt/2 = 1525×5/2 = 3812.5 m
d = 3.8125 km
Q99
An outdoor sound-speed experiment places two students 171 m apart. The measured average delay between seeing a balloon burst and hearing it is 0.50 s. Estimate the speed of sound and explain why the visual signal is used as the starting reference.
Experimental HOTS Numerical
View Challenge Solution
v = d/t = 171/0.50 = 342 m/s

The burst is seen almost immediately because light travels enormously faster than sound over this distance.

Therefore the interval between seeing the flash/burst and hearing the sound is a good approximation to the sound travel time.

Q100
Olympiad Final Challenge: A 680 Hz source produces sound in air where the speed is 340 m/s. A wall produces an echo after 0.40 s. The source is then submerged in water where sound speed is 1500 m/s, but its frequency remains 680 Hz. Find: (a) wavelength in air, (b) time period, (c) distance of the wall in air, (d) wavelength in water, (e) explain why frequency remains unchanged while wavelength changes.
Olympiad Final Challenge Integrated Numerical
View Challenge Solution

(a) Wavelength in air:

λair = 340/680 = 0.50 m

(b) Time period:

T = 1/680 ≈ 0.00147 s

(c) Distance of wall:

d = 340×0.40/2 = 68 m

(d) Wavelength in water:

λwater = 1500/680 ≈ 2.21 m

(e) Explanation:

The frequency is fixed by the vibrating source. When the sound enters a different medium, its speed changes. Since:

v = λν

the wavelength adjusts to the new speed while the source frequency remains unchanged.

Final answers:
λ in air = 0.50 m
T ≈ 1.47 ms
Wall distance = 68 m
λ in water ≈ 2.21 m

Complete Formula & Concept Revision

Concept Formula / Key Idea
Frequency ν = Number of oscillations / Time
Time period T = Time / Number of oscillations
Frequency–period relation ν = 1/T and T = 1/ν
Wavelength Distance between two consecutive crests or troughs
Wave speed v = λν
Wavelength λ = v/ν
Frequency ν = v/λ
Echo distance d = vt/2
Minimum clear-echo delay Approximately 0.1 s
Minimum echo distance at 340 m/s Approximately 17 m
Compression Region of greater-than-average density
Rarefaction Region of lower-than-average density
Sound wave Longitudinal mechanical wave
Particle motion Parallel to propagation direction for sound
Amplitude Maximum density change relative to average density
Intensity Sound energy crossing unit area per unit time
Pitch Mainly associated with frequency
Loudness Mainly associated with amplitude and human perception
Audible range 20 Hz to 20 kHz approximately
Infrasonic Below 20 Hz
Ultrasonic Above 20 kHz
Octave One note has twice the fundamental frequency of another
Speed order Solids > liquids > gases, generally
Typical speed in air at 15°C ≈ 340 m/s
Typical speed in water ≈ 1500 m/s
Typical speed in steel ≈ 5000 m/s

High-Yield Exam Rules & Common Traps

  • Sound is produced by vibration. A vibrating object acts as the source of sound.
  • Do not say that air travels from the source to the listener. Air particles only oscillate about their mean positions.
  • It is the disturbance and energy that propagate through the medium.
  • Sound cannot travel through vacuum because it is a mechanical wave.
  • Sound can travel through solids, liquids and gases.
  • In a longitudinal sound wave, particle vibration is parallel to wave propagation.
  • Compression means higher density; rarefaction means lower density.
  • On a density graph, a crest corresponds to maximum density and a trough to minimum density.
  • Do not confuse wavelength with amplitude. Wavelength is measured horizontally between successive identical points; amplitude describes the maximum density variation.
  • Frequency tells how many complete oscillations occur each second.
  • Always convert minutes into seconds before calculating frequency in hertz.
  • Frequency and time period are reciprocals.
  • For a fixed medium, if frequency increases, wavelength decreases because v = λν.
  • Changing the source frequency does not normally change the speed of sound in the same ordinary medium under unchanged conditions.
  • Sound usually travels fastest in solids, then liquids, then gases.
  • The speed of sound in air increases with increasing temperature and humidity.
  • Pitch is not the same as frequency, although frequency is the main physical factor associated with pitch.
  • Loudness is subjective; intensity is physically measurable.
  • Larger amplitude generally means greater wave energy and greater perceived loudness.
  • The human audible range is approximately 20 Hz to 20 kHz, but it varies among people and generally decreases with age.
  • Ultrasound means frequency above 20 kHz, not merely a very loud sound.
  • Infrasound means frequency below 20 Hz.
  • For echo and SONAR problems, the measured time usually represents a round trip.
  • Therefore use d = vt/2 for the source-to-obstacle distance.
  • A separate echo generally requires a delay of about 0.1 s or more.
  • Multiple reflections that arrive too quickly to separate may produce reverberation.
  • Soft, porous materials help reduce unwanted reverberation by absorbing sound.
  • A microphone and a speaker perform opposite energy-conversion roles.
  • A tone is nearly a single-frequency sound, whereas a musical note contains a fundamental and overtones.
  • Timbre allows a flute, tabla, sitar or another source to sound different even when playing the same note.
  • An octave corresponds to a doubling of fundamental frequency.

Recommended Numerical Strategy

Step 1: Write every given quantity with its SI unit.

Step 2: Convert kHz to Hz, milliseconds to seconds, centimetres to metres and minutes to seconds where required.

Step 3: For oscillation questions, start with ν = N/t.

Step 4: For period questions, use T = 1/ν.

Step 5: For wave-speed questions, choose the correct form of v = λν.

Step 6: In a density–distance graph, measure wavelength between successive crests or successive troughs.

Step 7: For echo, echolocation and SONAR, check whether the given time is a round-trip time.

Step 8: If it is a reflected signal returning to the source, normally use d = vt/2.

Step 9: For lightning/thunder problems, the light travel time over ordinary terrestrial distances can usually be neglected in the chapter’s approximation.

Step 10: Check whether your answer has a physically sensible unit: Hz, s, m, m/s or km.

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