1
Sound is produced —
💡 Hint:
A tuning fork.
Explanation:
A vibrating object produces sound.
Class 9 · Science · Chapter 12 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Sound
Sound
How to use this page:
1. Read — 12.1 tuning fork · 12.2 water · 12.3 instruments · 12.4 slinky · 12.5 reflection, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The picture to remember is one wave, with the wavelength λ marked between two neighbouring crests.
In the Bihar book this is chapter 12, while in the Exploration book sound is chapter 10. Progress stays in this browser.
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- 1 How sound is produced — Activities 12.1, 12.2, 12.3
- 2 Sound needs a medium
- 3 Reading a wave — Activity 12.4
- 4 Reflection, echo, reverberation — Activity 12.5
- 5 The audible range and ultrasound
- 6 The human ear
- Chapter winner — every lesson at mastery ★
🎬 Shorts
This chapter has no Short on the channel yet, so ten empty cards are not shown. The notes, diagrams and checks are complete.
📄 PDF download — notes + answer key
How sound is produced — Activities 12.1, 12.2, 12.3
ध्वनि का उत्पन्न होना — क्रियाकलाप 12.1, 12.2, 12.3 · Section 12.1 · Activities 12.1 to 12.3
Vibration is the sourceNotes
Sound is produced by a vibrating object. Strike a tuning fork on a rubber pad and bring it near the ear. Touch a prong with a finger and both the vibration and the sound stop.
A stretched rubber band also twangs. In a musical instrument the drum skin, the string of a sitar or the air column of a flute moves. A part that does not move is not the source.
Activities 12.1, 12.2 and 12.3Activity
In 12.1 touch a table-tennis ball hung on a thread with a vibrating prong. The ball jumps. The jump shows that the prong is really moving, not only that a sound is imagined.
In 12.2 touch the water in a full glass with a prong. The surface shivers. Dip the prong in the water and drops splash up. In 12.3 list instruments and write the part that moves in each one.
Worked exampleExample
Question: A vibrating tuning fork became still after it touched the ball. Why did the sound stop?
Answer: The ball and the finger took up the vibrations. The source stopped moving, so the sound stopped too. The cause of sound is vibration, not merely the name of the metal.
10-second revision
- Sound is produced by vibration
- The jump of the ball is evidence of vibration
- The moving part of an instrument is the source
Board tip · BSEBBoard tip
In the answer write both the name of the object and the part that moves.
Board tip · CBSEBoard tip
The splash of water also shows vibration. Only the ear is an incomplete answer.
Check your understandingall correct = mastery ★
2
Touching a prong of a tuning fork can stop the vibrations.
Explanation:
True — when the vibrations stop, the sound stops too.
3
A hanging ball ______ when a vibrating prong touches it.
💡 Hint:
Activity 12.1.
Explanation:
The ball jumps.
4
In a flute the source of sound is —
💡 Hint:
Activity 12.3.
Explanation:
In a flute the column of air vibrates.
5
Why does the water splash in Activity 12.2?
Model answer (board format)
The vibrating prong shakes the water. The splash is evidence that the prong is vibrating.
Check your answer:
Sound needs a medium
ध्वनि को माध्यम चाहिए · Section 12.2 · bell jar
The disturbance travels, not the particlesNotes
The substance through which sound passes is called a medium. It can be a solid, a liquid or a gas. A particle near the source moves about its place and pushes the next particle. The particles do not travel the whole way to the ear.
A source moving forward squeezes the nearby air into a compression, a region of high pressure. Moving back, it makes a rarefaction, low pressure. The chain of these is the sound wave.
The bell jar — the experiment in the bookActivity
Hang an electric bell in a closed glass jar and join the jar to a vacuum pump. Press the switch and the bell is heard. As the pump runs, the sound grows faint while the air decreases, even though the current stays the same.
This means sound does not travel in a vacuum. Two friends on the Moon cannot hear each other, because there is no air as a medium. This experiment does not have a separate activity number. Number 12.4 belongs to the slinky.
Worked exampleExample
Question: Why does the sound of the bell fall as the air in the jar decreases?
Answer: The particles of the medium became fewer. Fewer particles carry the disturbance forward, so the sound grows faint. The bell does not stop vibrating, because the current is the same.
10-second revision
- The medium can be a solid, a liquid or a gas
- Sound does not travel in a vacuum
- A compression is high pressure, a rarefaction is low pressure
Board tip · BSEBBoard tip
In the Moon answer write the missing medium, not the distance.
Board tip · CBSEBoard tip
The particles do not travel across. That sentence is needed before the longitudinal wave.
Check your understandingall correct = mastery ★
1
Sound cannot travel —
💡 Hint:
The bell jar.
Explanation:
A vacuum has no medium.
2
In a sound wave the particles of the medium travel the whole distance from the source to the ear.
Explanation:
False — the particles move near their own place. The disturbance moves on.
3
A region of high pressure is called a ______.
💡 Hint:
A rarefaction is the opposite.
Explanation:
A compression.
4
Why can two friends standing on the Moon not hear each other?
Model answer (board format)
There is no air on the Moon. Sound needs a medium, so the voice does not arrive.
Check your answer:
Reading a wave — Activity 12.4
तरंग की पहचान — क्रियाकलाप 12.4 · Sections 12.2.2–12.2.4 · Activity 12.4
Activity 12.4 — a slinkyActivity
A friend holds one end of a slinky. Stretch the other end and give it a sharp push toward the friend. The coils bunch into a compression and then spread into a rarefaction. A dot marked on it moves back and forth, parallel to the disturbance.
That is a longitudinal wave. Sound too is a longitudinal wave of compressions and rarefactions in the medium.
Frequency, amplitude and speedNotes
The number of vibrations passing a point each second is the frequency. The unit is the hertz. The brain reads frequency as pitch. A higher frequency is a higher pitch.
Amplitude decides loudness. A light hit makes a soft sound and a hard hit a loud one. Quality separates two sounds of the same pitch and loudness. A single frequency is a tone. A mix of frequencies is a note.
The distance between two neighbouring compressions is the wavelength λ. The time of one vibration is the time period T. The speed is v = fλ. The book also writes ν in place of f.
Worked exampleExample
Question: In a medium the frequency is 220 Hz and the speed is 440 m/s. Find the wavelength.
Formula: v = fλ, so λ = v / f.
Substitution: λ = 440 m/s ÷ 220 Hz = 2 m.
In air the speed is about 331 m/s at 0°C and 344 m/s at 22°C. Every speed in the table does not have to be memorised.
10-second revision
- Sound is a longitudinal wave
- v = fλ and λ = v / f
- Pitch from frequency, loudness from amplitude
Board tip · BSEBBoard tip
At 220 Hz and 440 m/s the answer is 2 m. Do not square anything.
Board tip · CBSEBoard tip
The hertz is the unit of frequency. Do not mix it with the joule.
Check your understandingall correct = mastery ★
1
If the frequency is 220 Hz and the speed is 440 m/s, the wavelength is —
💡 Hint:
λ = v / f.
Explanation:
440 / 220 = 2 m.
2
A sound wave is longitudinal.
Explanation:
True — the particles move parallel to the disturbance. The slinky shows this.
3
Loudness depends mainly on the ______.
💡 Hint:
Pitch is the frequency.
Explanation:
Amplitude.
4
A higher pitch means —
💡 Hint:
The brain reads the frequency.
Explanation:
A higher frequency means a higher pitch.
5
The speed in air at 22°C is 344 m/s. If the frequency is 172 Hz, find the wavelength.
Model answer (board format)
λ = v / f = 344 / 172 = 2 m.
Check your answer:
6
Quality distinguishes sounds of the same pitch and loudness.
Explanation:
True — this is how two instruments are recognised as different.
Reflection, echo, reverberation — Activity 12.5
परावर्तन, प्रतिध्वनि, अनुरणन — क्रियाकलाप 12.5 · Section 12.3 · Activity 12.5
Activity 12.5 — two pipes and a wallActivity
Place two long identical pipes of chart paper on a table near a wall. Keep a clock at the open end of one and listen for the tick through the other. Set the pipes so the sound is clearest.
Measure the angle of incidence and the angle of reflection. See the relation. Lift the right-hand pipe a little and the sound worsens, because the reflected ray no longer reaches the ear.
0.1 second and 17.2 metresNotes
The same sound returned by a wall or a mountain is an echo. The sensation of hearing lasts about 0.1 s. A distinct echo needs at least that gap.
If the speed is 344 m/s, the whole path = 344 × 0.1 = 34.4 m. The distance from the source to the wall is half of that, 17.2 m. If the temperature changes, the speed and this distance change. Echoes one after another from clouds and the ground make the roll of thunder.
If repeated reflection keeps a sound alive in a large hall, that is reverberation. It is reduced by putting absorbent material on the ceiling and the walls. A megaphone, a stethoscope, a curved ceiling and a soundboard are uses of multiple reflection.
Worked exampleExample
Question: The speed in air is 344 m/s. Find the least distance of a wall for a distinct echo.
Formula: d = v t / 2.
Substitution: d = 344 m/s × 0.1 s / 2 = 34.4 / 2 = 17.2 m.
10-second revision
- An echo is a reflected sound
- The least gap is 0.1 s and the distance is 17.2 m
- Reverberation is repeated reflection
Board tip · BSEBBoard tip
17.2 m is the one-way distance. 34.4 m is the whole path. Do not write them as one number.
Board tip · CBSEBoard tip
Reverberation and an echo are different. The hang of a hall is reverberation.
Check your understandingall correct = mastery ★
1
At a speed of 344 m/s the least distance for a distinct echo is —
💡 Hint:
Half of the whole path.
Explanation:
d = 344 × 0.1 / 2 = 17.2 m.
2
Reverberation is another name for a distinct echo made by one reflection.
Explanation:
False — reverberation is the persistence of sound by repeated reflection.
3
The sensation of hearing lasts about ______ s.
💡 Hint:
The time used for an echo.
Explanation:
0.1 s.
4
A stethoscope works by —
💡 Hint:
Repeated bouncing in the tube.
Explanation:
The sound of the heart reaches the ear by multiple reflection.
5
How is reverberation reduced in a hall?
Model answer (board format)
Sound-absorbing materials are put on the ceiling and the walls so that repeated reflection decreases.
Check your answer:
The audible range and ultrasound
श्रव्य सीमा और पराश्रव्य ध्वनि · Sections 12.4–12.5 · SONAR
Three bandsNotes
Humans hear from about 20 Hz to 20,000 Hz (20 kHz). A lower frequency is infrasound. A rhinoceros communicates with it. A higher frequency is ultrasound. Bats, dolphins and porpoises produce it.
Ultrasound still travels on a clear path among obstacles. In a cleaning bath it shakes dirt loose. It returns from a crack in a metal block. An image of the heart is echocardiography. Stones in the liver region, the gall bladder and the kidney are also seen with it.
SONAR means Sound Navigation And Ranging. It sends an ultrasonic wave and measures distance, direction and speed under water.
Worked exampleExample
Question: Place 25 kHz and 15 Hz as audible, infrasound or ultrasound.
Answer: 25 kHz = 25000 Hz, which is above 20 kHz, so it is ultrasound. 15 Hz is below 20 Hz, so it is infrasound. 500 Hz is in between, so it is audible.
The echo a bat uses
A bat sends ultrasound and reads the distance of prey or an obstacle from the wave that returns. The same idea is in SONAR.
10-second revision
- Audible range 20 Hz to 20 kHz
- Infrasound below, ultrasound above
- SONAR measures distance with ultrasound
Board tip · BSEBBoard tip
Do not write 20 kHz as 20 Hz. Kilo means a thousand.
Board tip · CBSEBoard tip
Antibiotics are not in this chapter. Cleaning and finding cracks are uses of ultrasound.
Check your understandingall correct = mastery ★
1
The audible range of a human is —
💡 Hint:
Infrasound below, ultrasound above.
Explanation:
About 20 Hz to 20 kHz.
2
A sound of 15 Hz is infrasound for a human.
Explanation:
True — below 20 Hz is infrasound.
3
SONAR in full is Sound Navigation And ______.
💡 Hint:
Measuring range.
Explanation:
Ranging.
4
A suitable wave for finding an inner crack in metal is —
💡 Hint:
The wave returns from the crack.
Explanation:
Ultrasound is reflected from a crack.
5
Write two uses of ultrasound.
Model answer (board format)
Cleaning a hard-to-reach place, and finding a crack in metal. An image of the heart or SONAR are also correct uses.
Check your answer:
6
A bat measures the distance of prey by the speed of light.
Explanation:
False — it reads the distance from the echo of ultrasound.
The human ear
मानव कान · Section 12.6 · eardrum to the brain
From the outside to the insideNotes
The outer ear, the pinna, gathers sound from around. It passes along the auditory canal and falls on the eardrum. A compression pushes the drum inward and a rarefaction moves it outward. In this way the drum vibrates.
Three bones of the middle ear — the hammer, the anvil and the stirrup — multiply these vibrations and carry them to the inner ear. The cochlea turns the pressure changes into electrical signals. The auditory nerve carries them to the brain, and the brain takes them as sound.
Worked exampleExample
Question: Why does the eardrum move, and how does the signal reach the brain?
Answer: Compressions and rarefactions change the pressure on the drum, so it moves. The three bones amplify the vibration. The cochlea makes an electrical signal and the auditory nerve gives it to the brain.
Do not reverse the order
Pinna, canal, eardrum, three bones, cochlea, nerve, brain. The cochlea does not come before the bones.
10-second revision
- The pinna collects sound
- Three bones amplify the vibration
- The cochlea makes an electrical signal
Board tip · BSEBBoard tip
Write the names hammer, anvil and stirrup. Only the words middle ear stay incomplete.
Board tip · CBSEBoard tip
The brain interprets. The eardrum itself does not find the meaning.
Check your understandingall correct = mastery ★
1
Sound is first gathered from outside by —
💡 Hint:
The outer ear.
Explanation:
The pinna gathers the sound from around.
2
The cochlea turns pressure changes into electrical signals.
Explanation:
True — then the auditory nerve carries the signals to the brain.
3
The three bones of the middle ear are the hammer, the anvil and the ______.
💡 Hint:
The third bone.
Explanation:
The stirrup.
4
Write the path of sound in the ear from the pinna to the brain, in order.
Model answer (board format)
Pinna, auditory canal, eardrum, hammer-anvil-stirrup, cochlea, auditory nerve, brain.
Check your answer:
❓ Full question bank — with answers and explanations — 66 questions
Pick a type. The 30 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ130 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA120 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/13Pick one option.
1
The source of sound is —
💡 Hint:
A tuning fork.
Explanation:
A vibrating object.
2
When air leaves the bell jar —
💡 Hint:
The medium decreases.
Explanation:
The sound grows faint.
3
A compression is —
💡 Hint:
A rarefaction is the opposite.
Explanation:
A region of high pressure.
4
In v = fλ, λ is —
💡 Hint:
Between two compressions.
Explanation:
The wavelength.
5
At 440 m/s and 220 Hz, λ is —
💡 Hint:
Division.
Explanation:
2 m.
6
Pitch depends on —
💡 Hint:
Faster vibration.
Explanation:
On frequency.
7
The least distance for a distinct echo, at 344 m/s, is —
💡 Hint:
vt/2.
Explanation:
17.2 m.
8
Reverberation is reduced by —
💡 Hint:
Repeated reflection should fall.
Explanation:
With absorbent material.
9
A frequency above 20 kHz is —
💡 Hint:
The upper band.
Explanation:
Ultrasound.
10
SONAR measures —
💡 Hint:
Under water.
Explanation:
Distance, direction and speed.
11
The job of the cochlea is —
💡 Hint:
The inner ear.
Explanation:
To turn pressure into an electrical signal.
12
The bones of the middle ear are —
💡 Hint:
Three.
Explanation:
Hammer, anvil and stirrup.
13
The speed of sound in air at 22°C is about —
💡 Hint:
The echo figure uses this speed.
Explanation:
344 m/s.
True or false
0/81
Sound travels faster in a vacuum.
Explanation:
False — a vacuum has no medium.
2
A dot on the slinky moves parallel to the disturbance.
Explanation:
True — the wave is longitudinal.
3
Loudness is decided mainly by amplitude.
Explanation:
True.
4
A gap of less than 0.1 s is enough for a distinct echo.
Explanation:
False — at least about 0.1 s is needed.
5
A megaphone is made to send sound in one direction.
Explanation:
True — multiple reflection toward the front.
6
15 Hz is inside the human audible range.
Explanation:
False — it is infrasound.
7
The pinna is a bone of the inner ear.
Explanation:
False — the pinna is the outer ear.
8
Quality separates sounds of the same pitch and loudness.
Explanation:
True.
Fill in the blanks
0/81
The type of a sound wave is called ______.
💡 Hint:
Longitudinal.
Explanation:
The slinky.
2
The unit of frequency is the ______.
💡 Hint:
The hertz.
Explanation:
Vibrations per second.
3
v = f × ______.
💡 Hint:
λ.
Explanation:
Wavelength.
4
In air at 22°C the speed is about ______ m/s.
💡 Hint:
344.
Explanation:
The echo figure.
5
The least echo distance is ______ m if the speed is 344 m/s.
💡 Hint:
17.2 m.
Explanation:
Half the path.
6
The upper audible limit is about 20 ______.
💡 Hint:
kHz.
Explanation:
A thousand hertz.
7
The outer ear is called the ______.
💡 Hint:
The pinna.
Explanation:
Gathering.
8
The ______ nerve carries the signal to the brain.
💡 Hint:
The auditory nerve.
Explanation:
After the ear.
Match
0/21
Match the word with its meaning.
Column B: A. Amplitude · B. Repeated reflection · C. Frequency · D. Distance of two compressions
1. Pitch
2. Loudness
3. Wavelength
4. Reverberation
Explanation:
Pitch is frequency, loudness is amplitude, wavelength is a distance, reverberation is repeated reflection.
2
Match the frequency with its band.
Column B: A. Infrasound · B. Audible · C. Ultrasound · D. Lower edge of audible
1. 15 Hz
2. 500 Hz
3. Above 20 kHz
4. 20 Hz
Explanation:
15 is infrasound, 500 is audible, above 20 kHz is ultrasound, and 20 Hz is the lower edge.
Assertion–reason
0/51
Assertion (A): Sound is not heard on the Moon.
Reason (R): Sound needs a material medium.
Explanation:
Both are true and R is the reason.
2
Assertion (A): A distinct echo needs a gap of about 0.1 s.
Reason (R): Sound is a longitudinal wave.
Explanation:
Both are true, but R is not the reason for this gap. The reason is the persistence of the sensation.
3
Assertion (A): A loud sound has a larger amplitude.
Reason (R): Loudness is another name for frequency.
Explanation:
A is true. R is false — frequency is pitch.
4
Assertion (A): Ultrasound is below 20 Hz.
Reason (R): A frequency above 20 kHz is called ultrasound.
Explanation:
A is false. R is true.
5
Assertion (A): The eardrum vibrates.
Reason (R): Compressions and rarefactions change the pressure on it.
Explanation:
Both are true and R explains A.
Coefficient practice
0/4These equations are not results of this chapter. They are only coefficient practice on H2 + O2 → H2O.
1
This is coefficient practice. This equation is not a result of this chapter. Balance it by filling coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
This is coefficient practice. This equation is not a result of this chapter. Balance it by filling coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
3
This is coefficient practice. This equation is not a result of this chapter. Balance it by filling coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
4
This is coefficient practice. This equation is not a result of this chapter. Balance it by filling coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
Classify
0/21
Place each frequency in the right band.
15 Hz
500 Hz
20 kHz
40 kHz
Explanation:
15 is infrasound, 500 and 20 kHz are in the audible range, and 40 kHz is ultrasound.
2
Place each part in its region of the ear.
Pinna
Stirrup
Eardrum
Cochlea
Explanation:
The pinna and the eardrum are on the outer path, the stirrup is in the middle, and the cochlea is inner.
Very short answer
0/51
How is sound produced?
Model answer (board format)
By a vibrating object.
Check your answer:
2
Write one difference between a compression and a rarefaction.
Model answer (board format)
A compression is high pressure. A rarefaction is low pressure.
Check your answer:
3
Write the meaning of each quantity in v = fλ.
Model answer (board format)
v is speed, f is frequency and λ is wavelength.
Check your answer:
4
Write the full name of SONAR.
Model answer (board format)
Sound Navigation And Ranging.
Check your answer:
5
Write the audible range.
Model answer (board format)
About 20 Hz to 20 kHz.
Check your answer:
Short answer
0/41
The frequency is 172 Hz and the speed is 344 m/s. Find the wavelength.
Model answer (board format)
λ = v / f = 344 / 172 = 2 m.
Check your answer:
2
Find the least distance for a distinct echo at 344 m/s.
Model answer (board format)
d = vt / 2 = 344 × 0.1 / 2 = 17.2 m.
Check your answer:
3
What is reverberation, and write one remedy.
Model answer (board format)
Reverberation is the persistence of sound by repeated reflection. Put absorbers on the walls.
Check your answer:
4
Write the job of the bones of the middle ear.
Model answer (board format)
The hammer, anvil and stirrup multiply the vibrations of the eardrum and carry them to the inner ear.
Check your answer:
Long answer
0/31
Write how sound is produced, why a medium is needed, and why the wave is longitudinal. Add the conclusion of the bell jar.
Model answer (board format)
Sound is produced by vibration. Particles of the medium pass the disturbance on and do not themselves travel the whole way. A dot on the slinky moves parallel, so the wave is longitudinal. When air leaves the jar the sound falls, so sound does not travel in a vacuum.
Check your answer:
2
Write the difference between an echo and reverberation. Show the calculation of 17.2 m. Write one use of multiple reflection.
Model answer (board format)
An echo is one clear returned sound. Reverberation is sound that persists by repeated reflection. d = 344 × 0.1 / 2 = 17.2 m. A stethoscope is a use of multiple reflection.
Check your answer:
3
Distinguish audible sound, infrasound and ultrasound. Write two uses of ultrasound and the path of the signal in the ear.
Model answer (board format)
20 Hz to 20 kHz is audible. Below that is infrasound and above that is ultrasound. Uses: a crack in metal and SONAR. Path: pinna, canal, eardrum, three bones, cochlea, nerve, brain.
Check your answer:
BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination.
1
A table-tennis ball jumps when a vibrating fork touches it. This shows —
💡 Hint:
Activity 12.1.
Explanation:
Evidence of vibration.
2
The speed of sound in air at 0°C is about —
💡 Hint:
344 at 22°C.
Explanation:
331 m/s.
3
The lower frequency of the audible range is about ______ Hz.
💡 Hint:
20 Hz.
Explanation:
The upper limit is 20 kHz.
4
In Activity 12.5, why does the sound fall when a pipe is lifted?
Model answer (board format)
The reflected sound no longer enters the second pipe, so the tick of the clock is not heard clearly.
Check your answer:
5
This is coefficient practice. This equation is not a result of this chapter. Balance H2 + O2 with H2O. A blank means 1.
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
6
Assuming the wave figure, write the link of λ, frequency and speed. Give the numerical example of 220 Hz and 440 m/s.
Model answer (board format)
In the figure λ is between two crests. v = fλ. λ = 440 / 220 = 2 m.
Check your answer:
CBSE-style questions
0/6These are competency-based practice questions. They are not copies of a CBSE paper.
1
A student expects a loud sound even after all the air is removed from the bell jar. The best reason against that hope is —
💡 Hint:
The conclusion of the experiment.
Explanation:
Without a medium the disturbance does not travel on.
2
A cliff is 34.4 m away and the speed is 344 m/s. The time for the echo is —
💡 Hint:
Going and returning.
Explanation:
The path is 68.8 m. Time = 68.8 / 344 = 0.2 s.
3
Assertion (A): The cochlea is the outer ear.
Reason (R): The pinna gathers sound from the surroundings.
Explanation:
A is false. The cochlea is in the inner ear. R is true.
4
Why do two instruments still sound different at the same frequency and the same amplitude? Write the meaning of quality.
Model answer (board format)
Quality is the feature that gives a different identity to sounds of the same pitch and loudness. A note of mixed frequencies is formed differently.
Check your answer:
5
A hall has too much ringing and there is no board behind the stage. Give two suggestions.
Model answer (board format)
Put absorbers on the walls and the ceiling so reverberation falls. Place a soundboard behind the stage so the sound spreads evenly in the hall.
Check your answer:
6
Assertion (A): A bat can read the distance of an obstacle in the dark.
Reason (R): It sends a ray of light from the ear.
Explanation:
A is true. R is false — it sends ultrasound.
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🧠 What you learned + equation sheet
What you learned
- Sound is produced by vibration and it needs a medium in order to travel.
- A compression is high pressure and a rarefaction is low pressure. The wave is longitudinal.
- Speed is v = fλ. Pitch is tied to frequency and loudness to amplitude.
- For a distinct echo the reflector is about 17.2 m away if the speed is 344 m/s.
- Repeated reflection causes reverberation. Absorbers are put on the ceiling and the walls.
- The audible range is 20 Hz to 20 kHz. The ear has an eardrum, three bones and a cochlea.
| What | Keep this |
|---|---|
| Production | vibration |
| Speed | v = fλ |
| Air, 22°C | 344 m/s |
| Echo | at least 17.2 m |
| Audible | 20 Hz to 20 kHz |
| SONAR | Sound Navigation And Ranging |
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