Class 9 Science Chapter 12 Important Questions
Class 9 Science Chapter 12 Important Questions of Sound. These practice questions include all short and long questions which are important for the school exams. Practice here with class 9 science chapter 12 extra question answers to revise the chapter from NCERT Book.There questions also ensure scoring good marks in exams.
Class 9 Science Chapter 12 Important Questions for 2020-2021
Class 9 Science Chapter 12 Important Extra Questions Set – 1
Why does wave transfer-matter or energy?
What are the mechanical wave?
Waves that are characterised by the motion of particles in a medium are called mechanical waves. Mechanical waves require material medium for their propagation.
Where is the density of air higher-at compressions or at rarefactions?
What is one complete oscillation?
The change in density from one maximum value to the minimum value and again to the maximum value make one complete oscillation.
Ans. Longitudinal waves.
What are transverse waves?
A wave in which the particles of the medium vibrate up and down at right angles to the direction in which the wave is moving.
- The waves produced by moving one end of a long spring (or slinky) up and down rapidly.
- Ripples formed on the surface of water in a pond.
What is a sound board? Explain the working of a soundboard?
The reflection of sound may take place at curved surfaces also. This fact is made use of in the large halls to spread sound evenly throughout the hall. This is done by using sound boards. The speaker is located at the focus of the sound board and the concave reflecting sound boards are placed behind the spreading out of the sound waves in various directions. It sends the sound waves from the speaker at its focus, by reflection towards the audience. This helps in making the speech readily audible even at a distance.
Class 9 Science Chapter 12 Important Extra Questions Set – 2
What are intensity of sound?
The amount of sound energy passing through unit area each second is called the intensity of sound.
What is the relationship between time period and frequency?
Frequency = 1/( Time period )
Name two animals that communicate using infrasound.
Rhinoceroses and whales communicate using infrasound.
Name the sound waves used by bats while flying in the dark.
Bats use ultrasonic waves while flying in the dark.
Earthquake produces which kind of sound before the main shock were begins?
On what factor does the pitch of a sound depend?
The pitch of a sound depends on the frequency of vibrations. Actually the pitch of a sound is directly proportional to its frequency.
Class 9 Science Chapter 12 Important Extra Questions Set – 3
Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the sting suitably. By doing so what is he adjusting?
He is adjusting frequency of the sitar with the frequency of the other musical instruments.
Name the characteristic which helps us to distinguish between a man’s voice and a woman’s voice even without seeing them.
Quality of timbre.
Explain the term crests and troughs of a wave.
The elevation or hump in a transverse wave is called crests. It is that part of the transverse wave which is below the line of zero disturbance.
What is a stethoscope? Name the principle on which a stethoscope works?
Stethoscope is a medical instrument used for listening sounds produced within the body, chiefly in the heart or lungs. Stethoscope works on the principal of multiple reflection of sound.
Why do we hear the sound produces by the humming bees while the sound of vibrations of pendulum is not heard?
Humming bees produce sound by vibrating their wings which is in the audible range. In case of pendulum, the frequency is below 20Hz which does not come in the audible range.
What are longitudinal waves?
The waves in which the particles of medium oscillate to and fro from their mean position in the direction of propagation of waves are called longitudinal waves.
- Sound waves in air.
- The waves which travel along the spring (or slinky) when it is pushed and pulled at one end.
Class 9 Science Chapter 12 Important Extra Questions Set – 4
Write the full form of acronym SONAR. Explain how the method of echo-ranging is used to determine the depth of sea.
Sound Navigation and Ranging: A transmitter producing ultrasonic waves is fitted at the bottom of a ship or a boat. The ultrasound waves emitted by the transmitter go to the bottom of the sea and get reflected from the bottom. These are received back of the ultrasonic waves and the speed of these waves in water, the depth of sea can be calculated.
A girl is sitting in the middle of a park of dimension 12 m × 12 m. On the left side of it there is a building adjoining the park and on right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain your answer.
If the time gap between the original sound and reflected sound received by the listener is around 0.1 s, only then the echo can be heard.
The minimum distance travelled by the reflected sound wave for distinctly listening the echo
= Velocity of sound × Time interval
= 344 × 0.1 = 34.5 m
But in this case the distance travelled by the sound reflected from the building and then recharging to the girl will be (6 + 6) = 12 m which is much smaller than the required distance. Therefore, no echo can be heard.
Give reasons for the followings: (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 like a violin and flute are distinguishable.
(a) If the reverberation time of a hall is long, then the multiple echoes will interfere with original sound. For this reason, nothing will be heard distinctly. So, the reverberation time of the hall should be very short.
(b) A sound is heard only if the body vibrates with a frequency more than 20Hz and less than 20,000Hz. The perpendicular oscillates with a frequency less than 20Hz. Hence, no sound is heard.
(c) This due to the quality or timber of sound waves.
(a) As the sound waves passes a point, it produces regions of higher and lower pressure. State the names of these region. (b) Describe how the movement of the loudspeaker cone produces these regions of different pressure. (c) State the effect on the loudness and pitch of the sound from the loudspeaker when (i) the amplitude increases but the frequency of the sound stays the same, (ii) the amplitude stays the same but the frequency increases.
(a) Region of higher pressure: Compressions
Regions of lower pressure: Rarefactions.
(b) Production of regions of higher pressure: When the loud speaker cone moves forward, i.e. in the direction of propagation of wave, it pushes the layer of air closer. The air layer pushes the next air layer, and process goes on. In this way, the layers of air near the cone are compressed to form a compression, which is a region of higher pressure.
Production of regions of lower pressure: When the cone moves backward, i.e. away from direction of propagation of wave, it leaves a region of low pressure and the air layers move apart to form a rarefaction.
(c) (i) Loudness increases as greater the amplitude of sound waves, louder the sound will be pitch remains same.
(ii) Loudness remains same.
Pitch increases as the pitch of a sound is directly proportional to its frequency.
When we put our ear to a railway track, we can hear the sound of an approaching train even when the train is far off but its sound cannot be heard through air. Why?
Sound travels about 15 times faster in iron than in air. So, sound travels much faster through the railway track made of steel than through air. That is why we can hear the sound of an approaching train even when the train is far off but its sound cannot be heard through air.
State the laws of reflection of sound.
Laws of reflection:
- The angle of incidence is equal to the angle of refection.
- The incident ray, the normal to the reflecting surface at the point of incidence and the reflected ray, all lie in the same plane.
Class 9 Science Chapter 12 Important Extra Questions Set – 5
A radar signal is reflected by an aeroplane and is received 2 × 〖10〗^(-5) s after it was sent. If the speed of these waves is 3 × 10^8 m/s, how far is the aeroplane?
Speed of wave = 3 × 10^8 m/s
Time taken to reach the aeroplane and come back is = 2 × 〖10〗^(-5) s
Therefore, distance of the aeroplane
s = (vt )/2 = (3 × 108× 2 × 10-5 )/2 = 3 × 103 m = 3km
The time taken by the signal to reach the bottom and come back is = 2.4 s
Speed of sound in water = 1,500m/s
Depth of the ocean h = (vt )/2 = (1500 × 2.4 )/2 = 1800m.
A sound wave has a frequency 2kHz and wavelength 40cm. How long will it take to travel 11.6 km?
Given frequency, v = 2kHz = 2 × 〖10〗^3 Hz
Wavelength = 40cm = 0.40m
Speed of sound = Frequency × Wavelength
v = vλ = (2 × 10^3 Hz) × 0.40m
= 0.80 × 103 Hz = 800 m/s
Time, t = (Distance )/(Speed ) = t = (s )/(v )
Given distance s, = 1.6 km = 1.6 × 103 m
Time t = (1.6 × 〖10〗^3)/800 = (1600 )/800 = 2s.
A person standing between two vertical cliffs and 640m away from the nearest cliff should. He heard the first echo after 4 seconds and the second echo 3 seconds later. Calculate. (i) the velocity of sound in air, and (ii) the distance between the cliffs.
(i) Let P be the person standing between the cliffs A and B. Let s1 be the distance of nearest cliff A form P and s2 the distance of second cliff B from P. The first echo is heard when sound reaches the person after being reflected from cliff A.
Given s1 = AP = 640m
Time interval of echo, t1 = 4 seconds
From relation, 2s1 = vt1, we have
Speed of sound, v = 2s1/(2t1 ) = (2 ×640 )/(4 ) = 320 m/s
Speed of sound in air, v = 320 m/s
(ii) The second echo is heard when sound reaches the person after being reflected from the cliff B.
Time interval of second echo t2 = 4 + 3 = 7 seconds
2s2 = vt2
We have s2= vt2/2 = (320×7 )/(2 ) = 1120m
Distance between cliffs A and B,
s = s1 + s2 = 640 + 1120 = 1760m
A boy hears the echo of his own voice from a distance hills after 0.8 seconds. If the speed of sound in air is 340m/s, calculate the distance of hill from the boy.
Let s be the distance of the hill from the boy and t the time of to and fro journey of sound waves, then from relation
Distance = Velocity × Time
We have 2s = vt
s = vt/2
Here v = 340 m/s
t = 0.8 s
s = (340 ×0.8 )/2 = 340 × 0.4m = 136 m
Describe an activity to study the reflection of sound.
To observe the reflection of sound, take a drawing board and fix it on the floor. Put two metallic or cardboard tubes. These tubes are making some angle with each other. Put a clock near the end of one tube and a screen between the two tubes so that sound of clock may not be heard directly. The sound (like tick-tick) waves pass through the tubes are reflected by the drawing board. The reflected sound waves enter the second tube and are heard by the placed in front of the sound tube.
Class 9 Science Chapter 12 Important Extra Questions Set – 6
A sound wave of wavelength 0.332 m has a time period of 〖10〗^(-3) s. If the time period is decreased to 10-4s, calculate the wavelength and frequency of new wave.
Give wavelength, = 0.332m
Time period of wave, T = 〖10〗^(-3)
Frequency of wave v= 1/T = 1/〖10〗^(-3) = 1000 Hz
Velocity of wave, u = v = 1000 × 0.332 = 332 m/s
In a given medium the velocity of sound wave remains same.
New time period of new wave T’ = 〖10〗^(-4)
Frequency of the new wave v’ = 1/T’ = 1/〖10〗^(-4) = 10,000Hz
Wavelength of new wave, = u/v’ = (332 )/10000 = 0.0332m
The wavelength of waves produced on the surface of water is 20cm. If the wave velocity is 24 m/s calculate (a) the number of waves produced in one second (b) the time required to produce one wave.
Given wavelength = 20cm = 0.20m, wave velocity, v = 24 m/s
(a) From the relation, v = vλ
v = v/λ
v = v/λ = 24/0.20 = 120 waves per second
(b) Time period T = 1/v = 1/120 second = 8.33 × 10^-3 seconds.
If the velocity of sound in air is 340m/s calculate (i) wavelength when frequency is 256 Hz (ii) frequency when wavelength is 0.85 m.
Using the formula v = vλ
Given velocity v, = 340m/s
(i) 340 = 256 λ
λ = 340/256 = 1.33m.
(ii) 340 = v(0.85)
v = 340/0.85 = 400Hz.
A person is listening to a tone of 500Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compression from the source?
Given frequency v = 500 Hz,
Time interval between successive compression is called time period.
Time period, T = 1/v = 1/(500s^(-1) ) = 0.002s.
Difference between Transverse Waves and Longitudinal Waves
|Transverse Waves||Longitudinal Waves|
|1. The particles of the medium oscillates up and down about their mean position.||1. The particles of the medium move in the parallel to the direction of propagation of the disturbance.|
|2. They propagate as crests and troughs.||2. They propagate as compression and rarefaction.|
|3. The propagation of waves is possible in solid or surface of liquid but not in gases. Examples: Light wave.||3. The propagation of these waves is possible in solids, liquids and gases. Examples: Sound wave|
What is meant by reflection of sound?
Reflection is a phenomenon of reversion of a wave going from one medium to the same medium after striking the second medium. Sound wave also experience reflection like other waves.