Important Questions Sound Class 9 Science

Important Questions Sound Class 9 Science

Important Questions Class 9

Please refer to Important Questions Sound Class 9 Science for Chapter 12 below. These Questions and Answers have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students will be able to learn the important problems and solutions which will help them to get more marks in their class tests and examinations. We have provided Class 9 Science Question Answers for all chapters in your NCERT Book for Class 9 Science

Sound Class 9 Science Chapter 12 Question Answers

Sound Class 9 Science Important Questions and solutions for upcoming examination will help you to get better marks.

1 Mark Questions

Question. Name two animals which can produce infrasonic waves.
Answer. Hippopotamus and whale

Question. What is persistence of hearing?
Answer. The time period till which the sensation of sound persists in our brain i.e. 0.1 second is known as persistence of hearing.

Question. Define pulse.
Answer. Pulse is a short duration wave generated by a single disturbance in a medium.

Question. What is wave number?
Answer. The no. of waves contained in unit length of the medium is known as wave number. It is the reciprocal of wavelength.

Question. What is an echo?
Answer. Echo is the recurring sound produced when the sound waves are reflected by an obstacle.

Question. Name one natural phenomenon caused by reflection of sound.
Answer. Thunder produced during lightning is a natural phenomenon caused by reflection of sound.

Question. Which type of surfaces reflects the sound waves better?
Answer. The substances with hard surfaces reflect the sound waves better.

Question. What is sound?
Answer. Sound is a form of energy emitted by a vibrating object producing sensation of hearing.

Question. Which device is used to measure intensity of earthquake?
Answer. Seismometers

Question. What is decibel?
Answer. The intensity of a sound is measured in a unit known as decibel which is denoted by dB.

Question. Sound waves cannot travel through vacuum. Explain.
Answer. Sound waves cannot travel through vacuum as these are mechanical waves and needs a material medium to travel.

Question. Name the characteristic of sound involved when a baby distinguishes her mother’s voice with others.
Answer. Quality or timber is the characteristic of sound involved here.

Question. What do you mean by oscillation?
Answer. Oscillation is the repetitive variation of the density about a central value or between two or more different values

Question. What is wavelength?
Answer. The distance between two consecutive compressions and two consecutive rarefaction is known as wavelength. It is denoted by λ.

Question. What is reverberation?
Answer. Reverberation is defined as persistence of sound after the source has stopped emitting sound.
This is due to multiple reflections of sound waves.

Question. What is the frequency of wave with time period 0.025 s?
Answer. Frequency (f) = 1/ time period (T)

Therefore, frequency of the wave = 40 Hz.

Question. A baby recognizes her mother by her voice. Name the characteristic of sound involved.
Answer. The characteristic of sound involved in uniqueness of the sound is quality of sound or timber.

Question. What are infrasonic and ultrasonic sounds ?
Answer. Infrasonic sound has frequency less than 20 Hz and ultrasonic sound has frequency higher than 20 kHz.

Question. What is the audible range of human ear?
Answer.
 The audible range of human ear is 20 Hz to 20 kHz.

Question. Why do we hear sound of an approaching car before the car reaches us?
Answer. 
This is because velocity of sound is much greater than the velocity of car.

Question. Why are sound waves called mechanical waves? Explain.
Answer. Sound waves are called mechanical waves because they travel by passing energy through particles in the form of compression and rarefaction.

Question. Describe two uses of multiple reflections of sound.
Answer.
 The phenomenon of multiple reflections of sound is used in Stethoscope and Megaphones.

2 Marks Questions

Question. Ocean waves have speed 15 m/s and time period 0.01. Find out the wavelength and the distance between the adjacent crest and the trough.
Answer. Wavelength (λ) = Time period (T) x speed of sound (v) = 15 x 0.01 = 0.15m
Distance between crest and trough is half the Wavelength = λ / 2 = 0.075m

Question. In the figure given here, a displacement-distance graph for a wave is shown. The wave velocity is 320 m/s. Find
(a) Wavelength
(b) Frequency
(c) Amplitude

Answer. (a) Wavelength = 0.4 m
(b) Frequency v = v / λ = 320 / 0.4 = 800 Hz
(c) Amplitude = 2 m

Question. If a source of sound produces 500 compressions and 500 rarefactions in air in 25 seconds, find the frequency of sound produced.
Answer. 
500 compressions are produced in 25 s.
Therefore in 1 sec, no. of compression = 500 / 25 = 20
Frequency = 20 Hz

Question. What do you mean by seismic waves?
Answer. 
Seismic waves are produced on the earth’s layer due to volcanic activity or earthquake. They are low frequency waves which travel through the ground.

Question. How do bats locate their prey?
Answer. 
Bats emit ultrasound. These are reflected by various obstacles and return to the bat’s ear. The nature of reflection tells the bat where the obstacles or prey is and accordingly the bat is able to catch its prey.

Question. How is ultrasound used in cleaning?
Answer. 
Ultrasound are high frequency waves. Objects to be cleaned are placed in a cleansing solution and ultrasonic waves are passed. The continuous high frequency vibration cause the dust, grime etc. to detach from the object and can then be easily washed away.

Question. How is noise different from music?
Answer. 
Sound which is pleasant to hear is called music. Noise is unpleasant to hear. A sound which is produced due to a mixture of frequencies is pleasant whereas noise has no periodicity and creates no recognized pitch or tone quality.

Question. What is SONAR? For what it is used?
Answer. SONAR is Sound Navigation And Ranging. It is a technique used to measure the depth of the sea, locate the sunken ships or icebergs and submarines.

Question. What is a supersonic speed?
Answer. When the speed of an object exceeds the speed of the sound, then the speed with which the body travels is supersonic speed. For example, aircrafts, bullet, etc. which travels at a higher speed than the speed of sound.

Question. When does a body produce sound?
Answer. Sound is produced through the to and fro motion of the particles of the medium in which sound travels.
These particles vibrate about their mean position forming disturbances in the form of waves.

Question. (a) How does a sound wave propagate?
(b) What is the audible range of sound for human beings?
(c) It is observed that some animals get disturbed before earthquakes. How?
Answer. (a) Sound propagates by the vibrations of the particles in a medium in the form of longitudinal waves.
(b) 20 Hz – 20,000 Hz
(c) Some animals can hear low frequency sound produced by earthquakes and get disturbed.

Question. Give significance of ultrasound in metal industry.
Answer. 
Ultrasound is used in metal industry to detect flaws in metals . The cracks and holes inside metal blocks reduce its strength. Ultrasound is passed through the metal and if there is a flaw , it is detected by the reflection of these sound waves.
Hence, ultrasound has a significant role to play in the metal industry as it ensures the strength of the material by detection of flaws.

Question. Give the function of (a) ear drum (b) cochlear fluid in human ear.
Answer. 
(a)The eardrum is a membrane which vibrates when sound reaches it . It helps in amplifying sound by means of 3 small bones attached to it.
(b) The cochlear fluid is set into vibrations. There are tiny hair like structures in it which convert these
vibrations into electrical impulses that are carried to the brain by the auditory nerve.

Question. Why do we see streaks of lightning much before we hear the sound of thunder?
Answer. 
Thunder is heard after the flash of light is seen because sound and light travel at different speed.
Light travels at the speed of about
3 x 108 m / s whereas sound travels at about 300 m/s.

Question. What is reverberation? Suggest two methods to reduce it in big halls.
Answer. The persistence of sound due to its repeated reflection until it is not audible is called reverberation. It can be reduced by covering walls and roof of halls by sound absorbent material like compressed fiberboard, rough plaster or draperies.

Question. Write two points of difference between sound wave and light wave.
Answer. Differences between sound and light waves:

3 Marks Questions

Question. Draw a graph showing a person with soft and loud voice.
Answer. 

Question. Explain the working of SONAR ?
Answer. SONAR stand for Sound, Navigation And Ranging. It is a device used to measure distance, direction and speed of underwater objects. It has a transmitter and detector near its base. The transmitter transmits ultrasonic signals which get reflected by various underwater objects. These are received by the detector which can convert these waves into appropriate electrical signals and give us the required information.

Question. The successive crest and trough of a wave 30 cm apart. Find the wavelength. Also, find the frequency of wave if 10 crests and 10 troughs are produced in 2 s.
Answer. Distance between crest and trough = 30 cm.
Distance between crest and crest = 60 cm
Hence wavelength λ = 60 cm = 0.60 m
In 2 sec, 10 crest or 10 waves are formed
Therefore in 1 sec. 5 crest or waves are formed
Hence frequency = 5 Hz

Question. A Ship emits ultrasound waves from SONAR fitted down. This wave returns from an underwater iceberg in 1.02 seconds. Calculate the distance of iceberg from the ship when the speed of sound in sea water is 1531 m/s.
Answer. Given,
Time taken by the ultrasonic wave to return back to ship after reflection from the iceberg, t = 1.02 seconds
Speed of sound in sea water = 1531 m/s
Total distance travelled by the ultrasonic waves = 2 × Distance between the ship and the iceberg, d Distance = Speed × Time
2 d = 1531 × 1.02
∴ d = 1531 x 1.02 / 2 = 780.81m
Therefore, the distance between the ship and iceberg is 780.81 m.

Question. In the old age people lost their hearing ability. Which device can be useful for them to support them hearing.
Answer. People which have problem in hearing, or which hear hard are given hearing aid for solving the problem. This is an electronic, battery operated device. It consists of a microphone and a speaker. The microphone converts the sound into electric signals which are then amplified by the amplifier. Now, these amplified signals are directed to the speaker where they are again converted to sound. This sound can now be heard by the person distinctly and clearly.

Question. Rahul went to Russia for holidays there he watched an opera performance in Novosibirsk Opera and Ballet Theatre, in Novosibirsk. He admired the architecture and furnishing of the auditorium. There were curved ceiling, swags, curtains and cushions positioned all around. The floor was also carpeted. After the performance was over all were taken to watch the auditorium. During that he noticed a sound board behind the stage. After watching all this he doubted whether all these things where just for increasing the beauty of the hall or also had a scientific reason. 
On the basis of above passage answer the following questions:
(a) What could be the reason for placing of curtains, cushions and carpets in the opera house?
(b) Discuss the use of curved ceiling and sound board?
(c) Which values are possessed by Rahul?
Answer. (a) To avoid reverberation (multiple reflection of sound) in the auditoriums, materials which can absorb sound energy is used.
(b) Sound boards and curved ceilings are present in the auditorium to generate maximum reflections to facilitate the sound in every corner of the auditorium.
(c) Rahul has a keen interest in art but also has scientific approach towards the happenings around.

Question. Shikha went to a Himachal Pradesh to meet her friend Rita. She was astonished to watch high mountain ranges, waterfalls, and the terrific greenery all around during her journey. After reaching their Rita took her to a peak and asked her to call upon her name at the top of her voice. By doing this Shikha could hear the echoes of her own voice which filled her with joy. After returning to Rita’s home, she tried to hear her echo in the same way in the room but failed. She searched about this on internet and got answers to her questions and went to bed.
(a) What is Echo?
(b) Why Shikha could not heard the echo at Rita’s room?
(c) Which qualities do Shikha have?
Answer. (a) Echo is a phenomenon which produces recurrences of sound from a source by reflections through the obstacle.
(b) For hearing an echo, there should be some requirements to be fulfilled like the distance between the source of sound and obstacle should be minimum 17.2m. Also most of the sound is being absorbed by the furnishing in the room.
(c) She is curious to learn about the natural phenomenon for which she makes use of latest technology.

Question. Vartika was conceiving for the first time. Her mother -in -law did not wanted the first child of her to be a girl. Vartika’s mother -in -law took her to a gynecologist for ultrasonography to determine whether the child is a boy or a girl. Doctor denied telling the gender of the child.
(i) Define ultrasonography? Why is ultrasonography used in pregnancy?
(ii) Which principle is involved behind its working?
(iii) Why did Doctor denied the in telling the gender of the child?
Answer. (i) Ultrasonography is a technique in which ultra sound waves are used for imaging of internal organs of body, like heart, uterus, etc. to detect any dis functioning in the organ. It is used in the diagnosis of congenital defects and abnormalities which could be present in the child.
(ii) Reflection of sound.
(iii) To reveal the gender of a child before birth is against the law. Under this law the parent of the child and the doctor are charged with a fine and also could be put in jail.

Question. In a cancer treatment hospital an ultrasound scanner is used to locate tumors in a tissue. It operates at frequency of 4.2 MHz. Calculate the wavelength of sound in a tissue? (Speed of sound in the tissue is 1.7 km/s)
Answer. Speed of sound, v = 1.7 km/s,
Frequency of wave, v = 4.2 MHZ = 4.2 x 106 Hz

Question. The following figures shows the wave shapes of two sounds of same frequency. Which of these is likely to represent the sound produced by a car-horn?
Answer. 

Question. Draw a diagram of transverse wave and label it.
Answer. 

Question. Draw a diagram of longitudinal wave and label it.
Answer. 

Question. Graphically distinguish between sounds of low pitch and high pitch.
Answer. 

Question. An echo is returned in 6 seconds. What is the distance of reflecting surface from source? [Given that speed of sound is 342 m/s.]
Answer. Given,
Time in which echo returned, t = 6 s,
Speed of sound, v = 342 m/s
Distance = Speed × Time = 342 × 6 = 2052m
As this distance is twice the distance of reflecting surface from source.
So,
The distance of reflecting surface from source = 2052 / 2 = 1026 m.

Question. What will be the wavelength and velocity of the wave produced when a stone is thrown in a pond. After the fall, stone produces 12 full ripples in 1 second in water. Also, the distance between a crest and a trough is 10 cm.
Answer. Given,
Frequency = No of waves / Time = 12 / 1 = 12
Distance between the crest and next trough, λ /2 = 10 cm
Therefore, wavelength, λ = 20cm = 0.20 m
Since, velocity = frequency × wavelength
Therefore, velocity = 12 × 0.20 = 2.40 m/s

Question. How does the ear drum vibrates?
Answer. Sound wave is a longitudinal wave i.e. the particles of the medium vibrate to and fro parallel to the direction of wave. Through this there generates a low pressure and high pressure areas alternately known as rarefaction and compression respectively. When this sound wave of different pressure reaches the ear drum, the low pressure region or the rarefaction pulls the ear drum outward whereas the high pressure region or compression pushes the ear drum inwards. This impinges by alternate regions on the drum sets the ear drum in vibratory motion.

Question. What is echo- ranging or sound ranging? State any one application of this technique.
Answer. The technique of finding the distance of an object with the concept behind as Echois called echoranging or sound ranging. Under this technique, the time taken by the echo to return back is used in finding the positions of substances. The depth of the sea bed can be determined by this technique.

Question. Why are ultrasonic waves used in SONAR?
Answer. • These waves are high frequency and very short wavelength so can easily penetrate in sea water to locate the objects beneath the water while audible range of sound cannot be too penetrative.
 These waves cannot be misinterpreted by the sounds produced by ships or engine as this range of sound could not be heard by the humans.

Question. Why do auditoriums have:
(a) Curved roofs?
(b) Curtains, carpets and false ceilings?
Answer. (a) The ceilings of an auditorium are curved so that the sound produced reaches all corners of the hall by reflection from the curved surface. This enables the audience sitting in the hall at any place clear voice of the speaker. Principle behind this is ‘reflection of sound waves’.
(b) In big halls, sound is sustained in the hall due to repeated reflections from the walls, ceilings and floor of the hall. This is known as reverberation. A small reverberation is preferred but if it is increased the sound becomes blurred, and could not be easily audible. For removing this unnecessary reverberation there are certain substances like curtains, carpets and false ceilings used to absorb sound as a result reverberation is decreased. This is because the soft and porous materials are good absorbers and bad reflectors of sound.

Question. State reason:
(a) Bats cannot see, then how do they fly in dark and even catch their Prey.
(b) There are Moths of certain families which are able to escape capture from the bat.
Answer. (a) Bats produce high frequency squeaks in the direction they fly. The reflected ultrasonic waves are then heard back by them after striking their prey. By this process the position of an obstacle or a prey is judged. So the bat prevents collapsing with the obstacle, on the other hand eats up the prey. 
(b) There are certain families of moth which can hear the squeaks produced by the nearby flying bat and could easily escape from being captured.

Question. When there are some natural calamities to be occurred then some animals escape from the site, before the calamities had happened.
Answer. There are some animals like dog, elephant, etc. which could hear the low frequency sound waves known as infrasound. These sounds cannot be heard by humans as it low frequency sound less than 20 Hz. As earthquakes produce low- frequency infrasound before the main shock waves begin. Now these sounds can be heard by these animals and they get alert and escape from the site much before.

Question. (i) Define the time period of a wave.
(ii) Give the relation among speed of sound v, wavelength λ and its frequency ν.
(iii) A sound wave travels at a speed of 339 ms . If its wavelength is 1.5 cm, what is the frequency of the wave?
Answer. (i) Time period (T) – It is defined as the time required to complete one wave.
(ii) Speed of sound (v) = Wavelength (λ) × Frequency (ν)
(iii) Given,
Speed of sound, v = 339 ms
Wavelength, λ = 1.5 cm= 0.015m
Since, Velocity = Wavelength × Frequency
Therefore, frequency = velocity / wavelemgth
= 339m/s x 0.015m
= 22600 Hz
The Frequency of wave = 22600 Hz

Question. (a) What is one vibration in a second called as?
(b) A tuning fork produces 256 waves in four seconds. Calculate the frequency of the tuning fork.
Answer.
 (a) One Hertz (Hz) is defined as one vibration in one second.

Therefore, the frequency of the tuning fork is 64 Hz.

Question. (a) Suppose a person whistles standing on the moon. Will the person standing nearby hear the sound? Explain giving reasons.
(b) What kind of wave needs a material medium to propagate?
Answer. (a) No, the person standing nearby would not be able to hear the whistle as there is no atmosphere on the moon. Sound waves needs medium for travelling, but the moon has no medium for the propagation of sound. Therefore, no sound could be heard.
(b) Mechanical waves need material medium to propagate. For example: sound waves.

Question. A nail was gently touched by the hammer and then was hit harder.
(a) When will be the sound created louder?
(b) Which characteristic of sound here is responsible for change in sound?
(c) Give the SI unit of loudness.
Answer.
 (a) Sound is produced only when the nail is hit harder not when it is gently touched.
(b) Amplitude of the pulsating body.
(c) The SI unit of loudness is decibel (dB)

Question. Define the term ‘tone’. A person is listening to a sound of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions reaching his ears from the source?
Answer. Tone is the sound of single frequency.
Given,
Frequency, ν = 500 Hz;
Distance, s = 450 m
Time period, T=?
T = 1 / v = 1 / 500 = 0.002s
Therefore, the time interval will be 0.002 s

Question. (a) State a condition for an echo to be heard.
(b) Bats cannot see, then how do they catch their Prey.
Answer. (a) For echo to be heard the minimum distance between the source and the obstacle should be 17.2m.
(b) Bats produce ultrasonic waves in the direction of their way. They hear back the echoes produced by the reflection of waves while striking their prey. By this the position of prey is judged and is being eaten by the bat.

Question. In the figure given here, a displacement-distance graph for a wave is shown. The wave velocity is 320 m/s. Find
(a) Wavelength
(b) Frequency
(c) Amplitude

Answer. (a) Wavelength, λ = Distance between two consecutive crests = 0.4 m
(b) Frequency, v = v / λ = 320 / 0.4 = 800 Hz
(c) Amplitude = During wave propagation, the maximum displacement of particles in a medium from mean position = 2 cm

Question. Explain a technique to find fault in a block of metal using reflection of sound.
Answer. Ultrasound waves are used to detect a fault in a block of metal using reflection of sound. These waves are made to pass through one side of the metal block. The detectors placed on the other side detect the transmitted waves. If there is any flaw or defect like crack or hole, etc. then ultrasound gets reflected back and does not reach the detector.

Question. A construction worker’s helmet slips and falls when he is 78.4 m above the ground. He hears the sound of the helmet hitting the ground 4.23 seconds after it slipped. Find the speed of sound on air.
Answer. h = 78. 4 m (height from which helmet fell)
t1 + t2 = 4.23s (total time)
u = 0
h = ut + 1/2 gt12 (if t1 is the time for the helmet to reach the ground)
78.4 = 0 + 12 x 9.8 x t12
t12 = 16
Therefore t1 = 4s
Therefore t2 = 4.23 – 4 = 0.23s
Speeds of sound h/t2 = 78.4 / 0.23 = 340.9 m/s

Question. Define the term ‘tone’. A person is listening to a sound of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions reaching his ears from the source.
Answer. Sound of single frequency is called tone
frequency v = 500 Hz
Time interval between successive compression = T = 1 / v
1 / v = 1 / 500 = 0.002s

Question. What does SONAR stand for? Name its two main parts. List two uses of SONAR technique.
Answer. SONAR stands for Sound, Navigation, and Ranging. Its two main parts are transmitter and receiver.
SONAR technique is used for –
(i) Detection of underwater objects such as ice bergs , rocks etc
(ii) Detection of depth of sea.

Question. (a) State a condition for an echo to be heard.
(b) Bats cannot see, then how do they catch their Prey.
Answer. (a) The echo should be heard after 0.1 s after the sound is produced.
(b) Bats emit ultrasonic waves, which are reflected from obstacles and prey and reach the bat’s ear. These reflected waves make the bat know where the prey is located.

Question. A sound wave travels at a speed of 339 m/s if the wavelength is 1.2 cm, what is the frequency of the wave.
Answer. Speed V = 339 m/s
Wavelength λ = 1.2cm = 1.2 x 10-2 m
frequency v = v/λ = 339 / 1.2 x 10-2 = 28250 Hz = 2.8250 x 104 Hz

Question. Write three medicinal applications of ultrasound.
Answer. Three medicinal applications of ultrasound are :
(1) Used in echocardiography to get image of heart
(2) Used in ultrasonography to get images of parts of body
(3) Used to break small ‘stones’ formed in the kidney.

Question. (a) The sound of an explosion on the surface of lake is heard by a boatman 100 m away and a driver 100 m below the point of explosion. Of the two persons mentioned (boatman or driver) who would hear the sound first? And why?
(b) Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer. (a) The driver will hear the sound of explosion first because sound will travel faster in water than in air.
(b) ν = 220 Hz
V = 440 m/s
Therefore λ v / v = 440 / 220 = 2m

Question. (a) What is audible range of the average human ear?
(b) Explain how ultrasound is used to clean spiral tubes and electronic components?
Answer. (a) Audible range 20 Hz−20,000 Hz
(b) Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out. The objects thus get thoroughly cleaned.

Question. (a) Why the stage of an auditorium has curved background curtains, carpets and false ceiling?
(b) The sound of a ringing bell inside a vacuum chamber cannot be heard. Why?
Answer. (a) The stage of an auditorium has curved background, so that sound after reflection reach all corners of the auditorium curtains, carpets and false ceiling are used to reduce reverberation.
(b) The sound of a ringing bell inside a vacuum chamber cannot be heard because there is no medium through which sound could travel as sound is a mechanical wave.

Question. Ocean waves of time period 0.01s have a speed of 15 m/s. Calculate the wavelength of these waves. Find the distance between the adjacent crest and the trough.
Answer. T = 0.001 s
V = 15 m/s
Wavelength λ = V × T = 0.01 × 15 = 0.15 m
the distance between adjacent crest and trough = λ / 2 = 0.075m

Question. State the relationship between frequency and time period of a wave. The wavelength of vibrations produced on the surface of water is 2 cm. If the wave velocity is 16 m/s, find its frequency and Time period.
Answer. Frequency v = 1 / T where T the time period
Wavelength λ = 2cm = 2 x 10-2m
Velocity v = 16 m/s

Question. A sound wave has a frequency of 2 kHz and a wavelength of 45 cm. It takes 4 s to travel. Calculate the distance it travels.
Answer. Frequency ν = 2 Khz =2000 Hz
Wavelength λ = 45 cm = 0 .45 m
Therefore Velocity V = 2000 × 0.45 = 900 m/s
Distance travelled in 4 s = V × t = 900 × 4 = 3600 m

Question. Define (a) pitch (b) quality (c) loudness of sound.
Answer. (a) Pitch is that characteristic of sound which gives sensation of shrillness. High pitch is more shrill. Higher the frequency, higher is the pitch.
(b) Quality of sound is that characteristic which enables us to distinguish one sound from another. It depend on the mixture of frequency and their relative amplitude
(c) Loudness is a measure of response of the ear to the sound .It depends on amplitude.

Question. Give reason for the following:
Answer. (a) We hear the sound of a horn of approaching car before the car reaches us.
(b) We hear the sound produced by the humming bees while the sound of vibration of pendulums is not heard.
(c) This is so as the speed of car is much less than that of the sound. So, sound of horn travels faster than the car itself and we can hear the sound earlier than actually see the car.
(d) This is so as the sound produced by the humming of a bee is in the audible range of our hearing unlike the pendulum, which is so low to be heard by humans.

Question. (a) Draw diagrams showing soft and louder sound.
(b) An orchestra, different musical instruments produce their own sounds. Do these sounds reach us with the same speed or different speed. Give reasons.
Answer.

(b) All sound reach at the same time as all frequency have same speed.

Question. Waves of frequency 100 Hz are produced in a string as shown in the figure. Give its :
(A) amplitude
(B) wavelength
(C) velocity
(D) nature

Answer. Given Frequency =100 Hz
(a) Amplitude = 5 cm
(b) Wavelength = 20 cm
(c) Velocity = Frequency × Wavelength = 100 x 20 / 100 = 20 m/s
(d) Transverse waves

Question. Give reason:
(a) Astronauts cannot talk to each other in space.
(b) A bomb explosion on moon cannot be heard.
Answer. (a) Since sound waves being mechanical in nature, needs a material medium to travel and in space there is no atmosphere present. So, sound could not be heard in space. To communicate in this environment they use radio waves, a type of electromagnetic wave which do not use medium for its propagation.
(b) Since moon does not have atmosphere or a medium in which sound can travel so, if there occurs a bomb explosion on moon, it could not be heard by the persons even present on the moon.

Question. (a) What is reverberation? How is it reduced?
(b) If the velocity of sound in air is 340 m/s, calculate the frequency of a wave whose wavelength is 1 m. Will it be audible to us?
Answer. (a) The persistence of sound due to multiple reflection within a room or hall is called reverberation. It
is reduced by putting sound absorbing material in the room like curtains, cushion, false ceiling etc.
(b) V = 340 m/s
λ = 1 m
v = v / λ = 340 / 1 = 340 Hz
It will be audible

Question. (a) Moths of certain families are able to escape capture when a bat is flying nearby. Explain how?
(b) What should be the minimum distance of the obstacle from the source of sound for hearing distinct echoes?
Answer. (a) Some moths can hear the high frequency squeaks of the bats. So when they hear the bats, they escape by flying away.
(b) Human beings can hear the echo only if sound returns after 0.1 s, if v is the velocity of sound, then minimum distance,
d = n x t / 2 = n x 0.1 / 2
if we take V = 340 m/s
d = 340 x 0.1 / 2 = 17m

Question. (a) Why is the ceiling and wall behind the stage of good conference halls or concert halls made curved?
(b) Which property of sound leads to the formation of echoes? Briefly explain.
(c) What is reverberation? What will happen if the reverberation time in a big hall is too long? How can we reduce it?
Answer. (a) The walls and ceilings are made curved so that sound reaches to all parts of the hall after reflection.
(b) The property which leads to the reflection of echoes is reflection. When sound is produced the wave can be produced by any large obstacle like wall or cliff. If the reflected sound reaches the observer after 0.1 s, a clear repetition of sound is heard which is called echo.
(c) The persistence of sound due to repeated reflection is called reverberation, echo may also be heard. To reduce reverberation we can cover the wall and roof of hall by sound absorbing material.

Question. Give two practical applications of reflection of sound waves.
Answer. 
Two practical applications of reflection of sound waves are –
(a) Megaphones or loud hailers are designed to send sound in a particular direction.
(b) Stethoscope are based on the principal of multiple reflection of sound within the stethoscope tube enabling the doctor to hear a patient’s heartbeat.

5 Mark Questions

Question. State reason for the following statements:
(a) Ceiling of good conference halls and concert halls are curved.
(b) We hear sound produced by the humming bees while that of moving pendulum is not heard.
(c) Sound waves are mechanical waves.
(d) Sometimes we hear echo of sound.
(e) People in their old age suffering from hearing loss, wear hearing aids.
Answer. (a) Ceiling of good conference halls and concert halls are curved so that the sound waves after reflecting from these walls reaches every part of the hall and can be easily heard by the listeners.
(b) We hear sound produced by the humming bees while that of moving pendulum is not heard. This is because the frequency of vibration of wings of bees is in the audible range of Humans (20Hz to 20 kHz) whereas frequency of vibrations of pendulum is less than 20 Hz so cannot be heard.
(c) Sound waves are mechanical waves as they need material medium for propagation which is the characteristic of the mechanical waves.
(d) We sometimes hear the echo of a sound produced because the distance between the source of the sound and the obstacle is at least 17.2.
(e) People in their old age suffering from hearing loss, wear hearing aids because hearing aid amplifies the electrical signal and converts it into sound which can then be heard much clearly by the old people.
(f) People in their old age suffering from hearing loss, wear hearing aids. This is because hearing aid amplifies the electrical signal and converts it into sound which can then be heard much clearly by the old people.

Question. What are ultrasonic waves? Name one animal which emits ultrasonic waves and explain how it uses the waves? What is the role of ultrasonic waves in medical science?
Ultrasonic waves are those waves which have frequency more than 20 kHz.
Answer. Bats use ultrasonic waves to fly and search for their prey at night without colliding.
Bat produces ultrasound waves in the direction of its way and hear back the echoes produced by reflection of the waves while striking the object in the path. Through this process they judge the position of the object and change their path.
Role of ultrasound waves in Medical science:
 Imaging of internal body structure and functioning of heart in humans.
 Breaking of stones produced in kidney into smaller ones to flush out through urination.
 Observing the foetus development in the mother’s womb for any abnormality or defect.

Question. Give reasons for the following:
(a) The reverberation time of a hall used for speeches should be very short.
(b) A vibrating body produces sound. However, no sound is heard when a simple pendulum oscillates in air.
(c) Sounds of same loudness and pitch, but produced by different musical instruments such as violin and flute are distinguishable.
Answer. (a) The time for which the sound persists in the atmosphere after the sound has stopped producing from the source is known as Reverberation time. If reverberation time of a hall is long, then the multiple echoes produced would interfere with the original sound, and the speeches would not be heard clearly and distinctly.
(b) Sound is produced only when the frequency of the wave is greater than 20 Hz. As a simple pendulum produces waves less than 20Hz they cannot be heard.
(c) The quality or timbre is the characteristic which helps in distinguishing two sounds of same pitch and loudness. So, the musical instruments like violin and flute produces distinguishable sound.

Question. (a) The stone is dropped from a tower of 500 m height into a pond of water at the base of the tower. When is the splash heard at the top? (Given g = 10 ms-2 and speed of sound = 340 ms-1 ).
(b) How do the sound waves cause vibrations in the eardrum of human ear?
Answer. (a) Given,
Height of the tower, h = 500m
Time taken by stone to reach water surface from the top of the tower:

(b) Sound waves are longitudinal waves which vibrate back and forth in the direction of its movement. It consists of compressions and rarefactions which results in pressure variation in the medium. This causes vibrations in the eardrum of human ear with required frequency and hence the sound becomes audible.

Question. (a) The stone is dropped from a tower of 500 m height into a pond of water at the base of the tower. When is the splash heard at the top?
(given g = 10 m/s-2 and speed of sound = 340 m/s-1
(b) How do the sound waves cause vibrations in the eardrum of human ear?
Answer. (a) Height of tower = 500 m
Acceleration due to gravity = 10m / s2
Speed of sound = 340 m/s
Time for the stone to reach the water surface.

Question. Write a short note on propagation of sound.
                                                        Or
Write a short note on how does the sound produced by a vibrating object in a medium reaches our ear.
Answer. – When sound is produced by the source, for listener to listen the sound passes through certain medium.
– The vibrating object sets the particles of the medium around it into vibration, to and fro motion.
– A particle in contact with the vibrating object moves towards a particular direction. This particle exerts a force on the nearby particles which displaces from its equilibrium position and starts moving towards the same direction in which the original molecule was travelling. After displacing the second particle, the
first particle returns to its original position. Similarly, particle second exerts a force on third particle and so on.
– Like this all the particle in between the source and listener vibrates and the sound is conducted to the listener.
Note – the particles of medium do not actually move from the vibrating body, they just vibrates the particles.

Question. Describe one application of ultrasound or ultrasonic waves in industry.
Answer. In industries, the large metal blocks used for the construction of bridges, buildings are checked before use. Ultrasound waves are used to detect any fault present in a block of metal by using the principle, ‘reflection of sound’. This is a technique in which flaws in a block can be detected without being damaging.
Working – The sound waves are made to pass through one side of the metal block. The detectors are placed on the other side which detects the transmitted waves. If there is any flaw or defect like crack or hole, etc. inside the block then ultrasound gets reflected back and does not reach the detector.

Question. (a) What is amplitude?
(b) Name the characteristics of sound.
Answer. (a) Amplitude of a wave is the maximum displacement of the particles of the medium from their mean position when a wave passes through the medium. It is denoted by A. the SI unit is metre (m).

(b) The characteristics of sound are:
(i) Loudness
(ii) Pitch or Shrillness
(iii) Quality or timbre

Question. Sound waves follow the same laws of reflection as light waves. State the laws of reflection of sound.
Answer. The sound wave and light waves follow the similar laws of reflection.

The Two laws of reflection of sound are:
1. First law: The angle of incidence (i) is always equal to the angle of reflection(r). i = r
2. Second law: The incident sound wave, the reflected sound wave and the normal at the point of incidence, all lie in the same plane.

Question. In a concert a pianist played some music on piano. Compare the Two Note A and B played when the speed of sound is 340 m/s and their wavelengths are 1.5 m and 1.33 m respectively.
Answer. Given,
Speed of sound, v = 340 m/s
Wavelength of note A, λA = 1.5m
Wavelength of note B, λA = 1.33m
Frequency of Note A, vA = ?
Frequency of Note B, vA = ?
Frequency of the sound wave, v = v/λ

The Frequency of Note B is more than Frequency of Note A

6 Mark Questions

Question. (a) Draw a diagram depicting soft sound and a loud sound. What is the main difference between the two?
(b) Why are ceiling of concert halls and conference halls made curved? Explain by giving a diagram.
(c) Can two astronauts talk on the surface of the moon as they do on the surface of the earth? Why?
Answer. (a)

The main difference is that soft sound has small amplitude and loud sound has large amplitude.
(b)The walls and ceilings are made curved so that sound reaches all parts of the hall after reflection.

(c) No, because there is no atmosphere on the moon and sound cannot travel in vacuum.

Question. Graphically represent:
Answer. (i) Two sound waves with the same amplitude but different frequencies.
(ii) Two sound waves with the same frequency but different amplitudes.
(iii) Two sound waves with different amplitudes and also different wavelengths.

Question. Define the term:
(a) Loudness
(b) Intensity
(c) Pitch
(d) Quality

Answer. (a) Loudness: it is the feeling produced in the ear which helps us to differentiate between a loud and a dim sound. The loudness or softness of sound depends upon the amplitude of the wave. The soft sound has small amplitude and louder sound has large amplitude.

(b) Intensity: it is the amount of energy passing at every second through a unit area.
(c) Pitch: it is the characteristic of sound which helps in differentiating between a harsh sound and a dull sound. It depends upon the frequency of vibration. Low pitch sound have low frequency and high pitch sound have high frequency.
(d) Quality: it is the characteristic of sound which differentiate between the two waves of having same pitch and loudness.

Question. Write a short note on Human ear.
Answer. Definition: Ear in humans is the sense organs which helps them to hear sound. It converts pressure variations in air of audible frequencies into electric signals which travel to brain through the auditory nerve.
Structure of ear:
(a) Outer part –
Pinna collects sound waves from surroundings.
Ear canal acts as long passage.
Ear drum (tympanum) is present at the end of ear canal which is a circular and elastic membrane
(b) Middle part –It consist of three bones, hammer, anvil and stirrup,
(c) Inner Part –It has a coiled tube, a cochlea which is filled with liquid containing nerve cells that are sensitive to sound.
Working:
 Pinna collects sound waves from the surroundings and made to fall on the ear drum, after which the eardrum starts vibrating back and forth quickly due to compression and rarefaction.
 These vibrations are then amplified many times by three bones in the middle ear.
 These vibrations are then passed through the liquid in cochlea which begins to vibrate and the pressure variations are turned into electric signals.
 These electric signals are carried by auditory nerve to brain. The brain interprets them as sound and humans get the sensation of hearing.

Important Questions Sound Class 9 Science