Selina Concise Class 9 Physics Solution Chapter No. 8- ‘Propagation of Sound Waves’ For ICSE Board Students.
Exercise-A Solution:
1.) Solution:
Ans: The vibrations produced in a body leads to production in sound.
2.) Solution:
Ans:
- Sound is nothing but the longitudinal mechanical waves which requires medium for their propagation and they carry mechanical energy i.e. vibrations which can be propagated through particles of the medium and we get sensation of hearing to our ear.
- Sound is mainly produced due to the vibrations of particles of the medium in a direction parallel to the direction of propagation.
For example:
- When we stretched a rubber string with one end in our mouth tightly closed by teeth and other is in our hand. Now we plucked the rubber string in the middle and released then vibrations are created in the rubber string and at the same time we hear some sound.
- So from this activity we can say that sound waves are produced when there are vibrations in particles of medium which transfers mechanical energy from particle to particle without changing their position.
3.) Solution:
Sound is produced by a vibrating body.
4.) Solution:
The following experiment demonstrate the sound produced by tuning fork is due to the vibrations in its arms.
- We have taken a tuning fork which is a rectangular rod made from steel or aluminium and it is bent in U shape with the metallic stem at the bent.
- Now we strikes the one end of tuning fork with the rubber pad and we brought that vibrating arm of tuning fork near the tennis ball which is suspended by thread as shown in figure.
- We observed that, when the vibrating tuning fork is kept near the tennis ball then that ball also vibrates or jumps in to and fro motion as shown in figure. At the same time we heard the sound produced by the vibrating tuning fork. The ball jumps till the vibrations are there in the arm of the tuning fork and sound is also there.
- When the vibrations in the arm of tuning fork are stopped then the ball comes in a state of stationary and the sound also stopped.
- From this experiment we demonstrate that, the sound produced in the tuning fork is due to the vibrations produced in its one arms.
5.) Solution:
Ans:
- Sound is nothing but the longitudinal mechanical waves which requires material medium for their propagation. Sound waves are produced due to the vibrations produced in a body and which are traveling from one part to another part of body through vibrating particles in the form of mechanical waves.
- Thus we can say that sound waves requires material medium. But this can be experimentally demonstrated by performing following bell jar experiment.
- We have taken an electric bell jar and another air tight glass bell jar. We have suspended the electric bell jar in side the glass jar which is connected to vacuum pump as shown in figure.
- As we completed the circuit when we press the key the hammer of the bell starts striking with the gong continuously and repeatedly and we hear a sound.
- Now key is pressed and the air from the glass jar is removed by using vacuum pump. Now we observe that the loudness of sound is decreasing with the air is removed from the jar and the we cant hear any sound when the entire air from the bell jar is removed. But, still the hammer of electric bell jar is repeatedly striking the gong that means the vibrations are still generated but we can’t hear them.
Explanation:
- When we press the key, the hammer of jar strikes with the find repeatedly and gong produces vibrations in it. These vibrations produced are travelled through air near the walls of jar and hence walls of jar are also in vibrations due to which we hear the sound outside the jar also.
- But, when we remove the entire air from the jar, although the gong is set in vibrations but due the absence of air the vibrations of gong are not travelled to the walls of jar and hence we can’t hear any sound outside the jar.
- Thus, from this experiment we demonstrate that for the propagation of sound waves material medium is necessary and sound waves cannot pass through the vacuum.
6.) Solution:
Ans:
- As we know that, the sound waves requires material medium for their propagation and they cannot pass through the vacuum.
- Because of this, as there is no atmosphere on moons surface hence we cannot hear other people sound on moon’s surface.
7.) Solution:
Ans:
The characteristics of the medium required for the propagation of sound are as follows:
1) The medium should be elastic due to which particles of the medium will come back to their initial position after vibrations. In short, particles of the medium should perform vibrations about their mean position.
2) The medium should possess inertia because it has capacity to store the mechanical energy.
3) The medium should be frictionless so that the energy will not be lost during the propagation of sound waves through that medium.
8.) Solution:
Ans:
- When the particles of the medium vibrates then disturbances are produced near to it. These disturbances are travelled or transferred through the medium in the form of waves.
- To know propagation of sound waves in a medium in detail see the following example.
- We have taken a thin metal strip which is kept vertical with lower end is fixed. Now we pushed its upper end to another side and released it. Due to this the metal strip starts vibrating and we heard a sound.
- In following figure a) we can see that when the strip is in normal position then the air layers near to it are in undisturbed position.
- But, as we moved the strip from a to b as shown in figure b) then the air layers near to it get pushed in front of it. Due to which air particles in that air comes closer to each other that means we can say that the air layers get compressed here and forms the compression C.
- While moving towards right, the particles of the air layer push and compress the layer next to them which again compress the layer next to them and the process is continuously going.
- Thus, we can say that the disturbance produced is moving in forward direction through the compression. In this way particles of the air layer get displaced but they don’t get displaced along with the compression.
- When the metal strip starts returning from b to a as shown in figure c) due to pushing particles in front of it the particles of the air layer comes to their initial position or mean position due to the elastic nature of the medium.
- But, when the strip moves from a to c as shown in figure d) it pushes the air layers near to it in backward direction in left due to which a space of very low pressure is created in right side of strip. Due to which the air layers in the right side expands and forms rarefied air layers. This region of low pressure is called as rarefaction R.
9.) Solution:
Ans:
When sound travels in medium, the disturbance travels in the form of waves.
10.) Solution:
Ans:
Following are the two kinds of wave in the form of which sound travels in a medium.
- Longitudinal waves
- Transverse waves
11.) Solution:
Ans:
- Longitudinal waves are the mechanical waves in which particles of the medium are vibrating in the direction parallel to the direction of propagation of wave.
- In longitudinal waves, particles does not shifted from one position to other but they only oscillate about their mean position in order to transfer mechanical energy in the form of vibrations.
- The sound waves are the longitudinal waves. And longitudinal waves can be produced in solids, liquids and gases also.
12.) Solution:
Ans:
- Transverse waves are the mechanical waves in which particles of the medium are vibrating in a direction perpendicular to the direction of propagation.
- In transverse waves particles are not shifted from one position to other but particles only vibrate about their mean position so that transverse mechanical energy will be transferred through them.
- Transverse waves does not require material medium for their propagation.
- Light waves are the transverse waves.
- Transverse waves are produced in solids and only on the surface of liquid.
- They can not be produced inside the liquid and gases.
13.) Solution:
Ans:
- Sound waves are the longitudinal waves.
- Longitudinal waves are traveling through the medium on the form of compression and rarefaction.
- Longitudinal waves requires material medium for their propagation.
- When a vibrating object is moving in forward direction then it pushes and also compress the air near to it and creates a region of high pressure near to it as shown in figure. And this region of high pressure is called as the compression. And in this way compression is moved in forward direction.
- When the vibrating object moves in backward direction it creates a region of low pressure which we called as rarefaction as shown in figure.
- Compression is the region where particles of the medium comes closer and this region is having high density.
- While rarefaction is the region where particles of the medium are moving away from each other. And this region is having low density.
14.) Solution:.
Ans:
- The transverse waves are the waves which propagates in the form of crest and troughs.
- Crest is the point of the transverse wave where the displacement of the wave is maximum.
- And trough is that point of the medium where there is minimum displacement of the wave occurs.
- The following figure shows transverse waves are traveling or propagated in the form of alternate crests and trough as shown in figure.
15.) Solution:
Ans:
- To show in wave motion only energy is transferred but particles of the medium are not leaving their mean position we described the following experiment.
- If we have dropped the stone in the still water of pond then suddenly we hear a sound of striking the stone with the water surface. In actual, there is a disturbance will be produced at the point where stone is striking the water surface. And this disturbance is spreading over all the water surface in the form of circular waves as shown in figure.
- Now, we have placed the piece of cork at a some distance from the point where stone strikes the water surface.
- We here observed that, the cork is not moving ahead but still it is moving up and down only although the wave get moved ahead. This is because the cork along with the particles of the water medium vibrates up and down only.
- And we see that, the vibrating particles gives their energy to neighbouring particles and they comes to rest at their mean position. In this way, the process is continuing and the only disturbance produced is moving ahead on the surface of water in the form of waves.
- This shows us that, the in wave motion only energy is transferred but particles of the medium are not leaving their position in the propagation of wave.
16.) Solution:
Ans:
- Amplitude of the wave is the maximum displacement of the particle of the medium on either side of its mean position. It is represented by letter a.
- The SI unit of amplitude is meter.
17.) Solution:
Ans:
- Frequency of the wave is nothing but the number of vibrations performed by the particle of the medium in one second.
- It is represented by letter f.
- The SI unit of frequency is second-1 or hertz.
18.) Solution:
Ans:
- The time period of the wave is the reciprocal of frequency of the wave.
- If T is the time period of wave and f is the frequency of the wave then relation between them is given by,
- Frequency= 1/Time period
- Hence, f= 1/T
19.) Solution:
Ans:
- Wave velocity or wave speed is the distance travelled by the wave in one second.
- It is represented by letter V.
- The SI unit of wave velocity is m/s.
20) Solution:
21.) Solution:
Ans:
22.) Solution:
Ans:
Derivation for relation between wavelength, velocity and frequency of the wave:
Let V be the velocity of the wave, f be the frequency of the wave and lambda be the wavelength of the wave.
Then according to the definition of wavelength we can write,
Wavelength= distance travelled by the wave in one time period
= Wave velocity*time period
= V*T
Thus, wavelength= V*T
But, f=1/T
Hence, wavelength= V*(1/f)
Or V = f*wavelength
Thus,
Wave velocity= wavelength*frequency
This is the relation between wave velocity, wavelength an frequency of the wave.
23.) Solution:
Ans:
The speed of sound Inna medium depends on the elasticity and density of the medium.
24.) Solution:
Ans:
- The velocity of sound in gas, liquid and solid is in the following ascending order.
Vg< Vl < Vs
- That means, velocity of it more in solids than liquids.
- And velocity of sound is more in liquids than gas.
25.) Solution:
Ans:
- The speed of light in air is 3*108m/s
- The speed of sound in air is 330m/s.
26.) Solution:
Ans:
The ratio of speed of sound in air, water and steel will be found as 1:4:15.
27.) Solution:
Ans:
a)
- The sound waves are the longitudinal waves which requires material medium for their propagation.
- Hence, sound waves cannot pass through the vacuum.
b)
The speed of sound is maximum in solids, less in liquids and least in the gases.
28.) Solution:
Ans:
- We know that, velocity of sound in air is 330m/s and velocity of light in air is 3*108m/s.
- That means velocity of light in air is greater than the velocity of sound in air.
- Hence, when thunderstorms happens the light initially reaches to us than the sound of thunder only because of the less velocity of sound than light in air.
29.) Solution:
Ans:
- When we placed our ear near to an iron railing which struck some distance away , then we heard the sound twice.
- Because, the velocity of sound is more in solid than gases. The sound we heard first is coming through the iron rail and the sound second time we heard is the sound coming through air.
30.) Solution:
Ans:
a)
The diver which is 100m below the point of explosion will heard the sound first.
b)
The diver is in water and we know that, the velocity of sound in water is greater than the velocity of sound in air.
c)
If t is the time required by the sound to reach the boat man then the time required by the sound to reach the diver must be less than t.
Hence, approximately the sound will reach the diver in time 0.25*t.
31.) Solution:
Ans:
a)
There is no effect of frequency of sound in air medium.
b)
The speed of sound is increasing with increase in temperature.
c)
There is no effect of pressure in air on the speed of sound in air.
d)
The speed of sound increases as the moisture in air increases.
32.) Solution:
Ans:
The speed of sound doesn’t depends on the amplitude and wavelength of the wave.
Hence, although there is change in amplitude and wavelength of the wave there will be no change in the speed of sound in air.
33.) Solution:
Ans:
- The speed of sound is more in humid air than in dry air.
- Because, as the moisture in the air increases the density of the air decreases.
- And we know that, the speed of sound is inversely proportional to the square root of density of the medium.
- Thus, V a 1/√q
- Hence, humid air is having less density and hence speed of sound is maximum in humid air than dry air.
34.) Solution:
Ans:
- The speed of sound increases with increase in temperature.
- Per 1°C rise in temperature increases the speed of sound by 0.61m/s.
35.) Solution:
Ans:
The following experiment is demonstrated to find the speed of sound in air.
- We have chosen two places at higher altitudes as A and B which are facing to each other and distance d apart from each other.
- At each place A and B there is a gun and stopwatch.
- Now, initially the observer at place A fires the gun and observer at B starts it’s stopwatch as he see the firing and stop the stopwatch when he heard the sound of firing.
- The time t1 be required to reach the sound from A to B noted by the observer at place B.
- Now, the observer B fires the gun and when firing is seen by observer at place A it starts it’s stopwatch and stop the stopwatch when he heard the sound.
- The observer at A thus noted time t2 which is required by the sound to reach from place B to A.
- Now, the average of the two times is given by,
t= (t1 + t2)/2
- Thus, t is the time taken by sound to travel the distance d between the places A and B. Hence, we can find the speed of sound as follows.
- Speed of sound = distance/time = d/t
- In this way, we can determine the speed of sound in air.
- The approximation which is taken here is the speed of sound remains same when the experiment is performing.
36.) Solution:
Ans:
a)Sound cannot travel through vacuum, it requires a medium.
b) When sound travels in a medium, the particles of the medium do not moves but the disturbance
c) A longitudinal wave is composed of compression and rarefaction.
d) A transverse wave is composed of crest and trough.
e) Wave velocity= frequency*wavelength
Multiple choice type:
1.) The correct statement is
Ans: c) sound needs medium but light does not need medium for its propagation.
2.) The speed of sound in air at 0°C is nearly
Ans: d) 330m/s
3.) Sound in air propagates in the form of
Ans: a) longitudinal waves
4.) The speed of light in air is
Ans: a) 3*108m/s
Numerical:
1.) Solution:
Ans:
Given that,
Heart beats 75 times in a minute.
We know that, frequency is the number of times heart beats in one second
Hence, f= 75/60= 1.25s-1
The time period is the reciprocal of frequency.
T= 1/f = 1/1.25= 0.8s
2.) Solution:
Ans:
Given that, time period of simple pendulum is 2s.
Hence, frequency is given by
Frequency=1/T = 1/2= 0.5Hz
3.) Solution:
Ans:
Given that,
Wavelength = 100m
Velocity = 20m/s
We know that the relation between wavelength, velocity and frequency of the wave is
V= wavelength*frequency
Thus, frequency= velocity/wavelength
= 20/100= 0.2 Hz
4.) Solution:
Ans:
Given that,
Velocity of wave = 0.3m/s
Frequency= 20Hz
Separation between two consecutive compressions is nothing but the wavelength of the wave.
Thus, we know that
Wave velocity= frequency*wavelength
Wavelength=wave velocity/ frequency
= 0.3/20= 1.5*10-2m
5.) Solution:
Ans:
Given that,
40 waves are produced in 0.4 seconds
Thus, the frequency of the wave is given by
Frequency= number of waves/ time required
= 40/0.4= 100Hz
6.) Solution:
Ans:
Given that,
The distance between the observers A and B = 1650m
Speed of sound in air= 330m/s
The time required to heard the sound by B = distance/speed = 1650/330= 5s
Thus, we can say that the observer B will hear the sound 5s after the gun is fired.
7.) Solution:
Ans:
Given that,
Speed of sound in air= 330m/s
Time interval between a lightning a flash and the first sound of thunder = 5s
Hence, the distance of the flash from the observer= v*t= 330*5= 1650m
8.) Solution:
Ans:
Given that,
Speed of sound in air = 340m/s
Time in which the sound of firing is heard after flash = 2.5 seconds
Hence, we can find the distance between flash and observer that means the distance between the two boys= V*t= 340*2.5= 850m
Thus, the distance between two boys is 850m.
9.) Solution:
Ans:
Given that,
Distance between two tanks = 510m
Time difference between the two shots heard after flashing= 3.5 – 2 = 1.5 seconds
Thus the speed of sound is given by,
Speed = distance/time = 510/1.5= 340m/s
Thus, the speed of sound will be 340m/s.
10.) Solution:
Ans:
Given that,
Length of iron rail = 3.3 km= 3300m
Speed of sound in iron = 5280m/s
Hence, the time required to travel the iron rod length by sound is given by,
Time= distance /velocity
Time t= 3300/5280= 0.625s
Thus, in 0.625s the sound travels through an iron rod.
Speed of sound in air=330m/s
Hence, the time required for the sound to travel in air is given by,
Time= distance/velocity= 3300/330= 10s
Thus, in 10s the sound will travel in the air.
11.) Solution:
Ans:
Given that,
Speed of sound in air = 340m/s
Speed of sound in water= 1360m/s
Distance required to travel= 1700m
Time taken in air= 1700/340= 5s
Time taken in water= 1700/1360= 1.25s
Thus, time taken to travel distance 1700m by sound in air is 5s and in water it is 1.25 s.
Exercise-B Solution:
1.) Solution:
Ans:
The range of frequencies of the sound within which human ear can hear the sound is called as the audible range of frequency.
2.) Solution:
Ans:
The audible frequency range for humans is 20Hz to 20KHz.
3.) Solution:
Ans:
For the range of frequencies 2000Hz to 3000Hz, the human ears are most sensitive.
4.) Solution:
Ans:
Ultrasonic sound has higher frequency than the infrasonic sound.
5.) Solution:
Ans:
a) An average person can hear sound of frequencies in the range 20Hz to 20KHz.
b) Ultrasound is of frequency above 20KHz.
c) Infrasonic sound is of frequency below 20Hz.
d) Bats can produce and hear ultrasonic sound.
e) Elephant produce infrasonic sound.
6.) Solution:
Ans:
a) 10Hz- Infrasonic sound
b) 100Hz – Audible sound
c) 1000Hz – Audible sound
d) 40KHz – Ultrasonic sound
7.) Solution:
Ans:
We can’t hear the sound produced by the seconds pendulum.
Because the frequency of sound produced by the seconds pendulum is 0.5Hz which is the infrasonic frequency of sound and that cannot be audible to human ears.
8.) Solution:
Ans:
The sound of frequencies greater than 20KHz is called as Ultrasound.
Ultrasound cannot be heard by human ears.
9.) Solution:
Ans:
The approximate speed of ultrasound in air is 330m/s.
10.) Solution:
Ans:
The following two properties of the ultrasound that make it useful to us.
- High energy content
- High directivity
11.) Solution:
Ans:
Bats locate obstacles and prey in their way by producing and hearing the ultrasonic sound.
When they produces ultrasonic sound it get incident on the obstacle. After reflection of ultrasound from the obstacle it further reaches to bats. By hearing this reflected ultrasound and from the time interval of reflection of ultrasound they may judge or decide the direction and location of the obstacle in their way.
12.) Solution:
Ans:
Following are the applications of ultrasound:
- Ultrasounds are used for drilling holes in glass and also used for making desired shaped glass materials.
- To detect the defects in metals ultrasounds are used. When ultrasonic sound waves are passed through metal, if all the waves are passed through the metal then there will be no defect in the metal.
- But, if some ultrasonic sound waves are reflected back then there will be defect in the metal.
Multiple choice type:
1.) A man can hear the sound of frequency
Ans: b) 1000Hz
2.) The properties of ultrasound that make it useful are
Ans: b) high power and good directivity
3.) Sonar makes use of
Ans: b) ultrasound