Sound - Class 9 Physics - Extra Questions

Find the distance between a surface and the source of sound, if speed of sound is $$334\ m/s$$ and echo returns from the surface in $$1.5\ s$$.

Given
Speed of sound $$(u) = 334\ m/s$$
Time taken for hearing the echo $$(t) = 1.5\ s$$
Now
$$speed = \dfrac {distance}{time}$$

$$\Rightarrow 334 = \dfrac {distance}{1.5}$$

$$\Rightarrow$$ Distance travelled by the sound $$= 334\times 1.5 = 501\ m$$.

In $$1.5\ s$$, sound has to travel twice the distance between the source and surface. Therefore, the distance between the source and surface is $$= \dfrac {501}{2} = 250.5\ m$$
Distance of source from surface $$= 250.5\ m$$.


An echo is returned in $$6$$ seconds. What is the distance of reflecting surface from source? [given that speed of sound is $$342\ m/s$$.]


In $$6s$$, sound has to travel twice the distance between the source and reflecting surface. 
Distance travelled in $$6 \ s$$= $$speed \times time=342 \times 6=2052 \ m$$
Therefore, the distance of the reflecting surface from the source $$= \dfrac {2052}{2}m = 1026\ m$$
Answer: Distance of reflecting surface $$= 1026\ m$$.


Name two animals which produce ultrasound.

Ultrasonic waves are the sound waves with frequency more than $$20\ kHz$$.
Animals like dolphins, bats produce ultrasound.


Can SONAR technique be used to determine the depth of the sea ? 

Yes, SONAR technique can be used to determine the depth of the sea. SONAR is a technique that uses ultrasonic waves to measure the distance, direction and speed of under water objects. Sonar can also be used for determining the depth of the sea.
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Give three medical uses of ultrasound.

  1. Use to break the stone in the kidney into fine grains.
  2. Use to investigate the internal organ of the human body.
  3. Used to monitor the development of an unborn baby inside the mother's uterus. 


What is sound ? 

Sound is a form of energy, which is produced due to vibration of an object and it is transferred by medium particles.


A girls hears an echo of her own voice from a distant tall building after $$2\,s$$ . What is the distance of the girl from the building ? (Speed of sound in air=$$344m/s$$)

Let distance of girl from building be $$d$$
Speed of sound = $$v=344m/s$$
Distance travelled by sound from girl to building is $$d$$ and distance travelled by echo from building to girl is also $$d$$.
So total distance travelled=$$2d$$
Time taken  $$t=2s$$
$$Speed=\dfrac{Distance}{Time}$$
$$v=\dfrac{2d}{t}$$
$$2d=vt$$
$$2d=344\times2$$
$$2d=688m$$
$$d=344m$$


A person clapped his hands near a cliff and heard the echo after $$5$$ seconds. What is the distance of the cliff from the person if the speed of sound V is taken as $$ 346\,m/sec $$.

Given $$ V = 346 \, m/s , t = 5 $$ second 
Distance travelled $$ = V \times t $$ 
$$ S = 346 \times 5 = 1730\,m $$ i.e., in $$5$$ second , the sound has to travel twice the distance between the cliff and the person. Thus , the distance between the cliff and the person
$$ = \dfrac{1730}{2} = 865 $$ metres


Megaphone works on the prinicple of____________of sound.

Megaphone works on the principle of $$multiple$$ $$reflection$$ of sound. Megaphones are designed to send sound in a particular direction without spreading it in all directions. In these instruments, a tube followed by a conical opening reflects sound successively to guide most of the sound waves from the source in the forward direction towards the audience.
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Pitch of sound depends upon__________of vibrating body.

Pitch of sound depends upon frequency of vibrating body. A high frequency produce a high pitched sound and a low frequency produces a low pitched sound.


The sound heard after reflection from a rigid surface is called_____________

The reflection of a sound is known as echo and it is a reflection of sound that arrives at the listener with a delay after the direct sound. 
So, the sound heard after reflection from a rigid surface is called echo.


Sound travels ________times faster in water than in air.

In water sounds travels faster than the air. The reason is that the particles are much closer in water and so they can quickly transmit vibration energy from one particle to next particle. Generally, Sound travels four times faster in water than in air.


Sound travels fast in _______than liquids.

The speed of sound increases by increase in the density of the medium. As a solid has more density than that of a liquid, so sound travels faster in solids than liquids.


Sound can be visualised as a wave. Explain.

A wave is a disturbance that moves through a medium when the particles of the medium set neighbouring particles into motion. They in turn produce similar motion in others. The particles of the medium do not move forward themselves, but the disturbance is carried forward. This is what happens during propagation of sound in a medium, hence sound can be visualised as a wave. Sound waves are characterised by the motion of particles in the medium and are called mechanical waves.
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The separation between two consecutive crests in a transverse wave is $$100\ m$$. If wave velocity is $$20\ ms^{-1}$$, find the frequency of wave.

Separation between two consecutive crest = Wavelength
Wavelength = 100 m
Velocity = 20 m/s

Frequency = Velocity / wavelength
= 20 m/s / 100 m
= 0.2 Hz

Frequency is 0.2 Hz


What change, if any, would you expect in the characteristic of a musical sound when we increase its amplitude?

The magnitude of the maximum disturbance in the medium on either side of the mean value is called the amplitude of the wave. The loudness or softness of a sound is determined by its amplitude.
If we increase the amplitude of a musical sound then the loudness will also increase.


What change, if any, would you expect in the characteristic of a musical sound when we increase its frequency?

Pitch

A musical sound has the characteristics of loudness, pitch, and quality. Out of them, pitch depends upon the frequency of sound, therefore if the frequency is increased, its pitch will also increase.


State two properties of ultrasound that make it useful to us.

Listed below are the two properties of ultrasound:
(a) The energy carried by ultrasound is high.
(b) The ultrasound can travel along a well defined straight path.it does not bend appreciably at the edges of an obstacle because of its small wavelength
The above two properties of ultrasound makes it very useful to us for many purposes.


State two applications of ultrasound.

The two applications of ultrasound are as follows:
(a)It is used to drill holes or make cuts of desired shape in materials such as glass
(b)It is used to detect defects in metals. Ultrasound passes through the object if there is no defect in the object. But if there is some defect, a part of ultrasound gets reflected back


Explain the terms crest and trough in relation to a transverse wave.

A transverse is composed of a crest and trough. Crest is the position of maximum upward displacement while trough is the position of maximum downward displacement. 
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Complete the following sentence:
Wave velocity = ______ x wavelength.

Wave velocity = Frequency x wavelength

$$v = \nu \times \lambda$$


What is being transferred when a wave is traveling?

A wave is produced by some kind of disturbance that is associated with energy. When the wave travels, it carries that energy with it. Therefore, energy transportation is realized in a wave motion.


Complete the following sentences:
A longitudinal wave is composed of compression and ______.

A longitudinal wave is composed of successive compression and $$rarefaction$$.


Fill in the blanks :
The region in a sound wave, with higher pressure and density, is called ________ and that with low pressure and density, is called ________.

The region in a sound wave,  with higher pressure and density, is called compression and that with low pressure and density is called rarefaction.


Answer the following question in your words.
What should be the dimensions and the shape of classrooms so that no echo can be produced there?

For no echo to be produced in a classroom, the classroom should be square shaped with distance between the walls to be less than $$17.2\,m$$. Also, the ceiling should be curved shaped so that the reflection of sound wave is uniform throughout the room.


Justify the statement:
Sound is carried by longitudinal waves.

A longitudinal wave is that in which the vibrations are parallel to the direction of propagation of the wave. When a disturbance is produced in a medium, medium particles are expanded and compressed alternatively, resulting in the formation of a sound wave, these expansions and compressions happen in the direction of wave propagation i.e. vibrations are in wave direction, therefore sound is carried by longitudinal waves. 


Have you ever sung in the shower ? Why does your voice sound so much better there.

Because showers aren't usually symmetrical cubes, some of those waves travel further than others. Also the ceramic doesn't absorb sound well, gives your singing the effect of being stretched out, or reverberated. Reverb also helps to even out your pitch. Your voice tends to gets blurry as it reverberates off many surfaces. So if you quite hit the exact note, it sounds closer than it would outside the shower.


Echoes are the reflected sound waves that we hear. Explain.

When we produce a sound wave and it strikes an obstacle, it comes back to us after reflection, this reflected sound wave is called echo. For example, when we shout or clap near a suitable reflecting objects such as a tall building or a mountain, we will hear the same sound again a little later. This sound which we hear is called an echo.


Quality of two different string instruments cannot be the same. Explain.

    Quality of sound is that characteristic of sound which enables to distinguish between two sounds from two sources even of the same pitch and same intensity . Therefore quality of the two sources (string instruments) cannot be same .


Explain why a person at a rock concert will not feel gusts of wind coming out of the speakers.

Waves transfer energy, not matter, thus air particles will not travel with the sound wave. Instead, they will merely vibrate in the direction of propagation to create sound. Hence a person at a rock concert will not feel gusts of wind coming out of the speakers


Justify the statement :
A non-vibrating body cannot produce sound.

Sound is produced by a disturbance in the medium by a vibrating body. A non-vibrating body cannot produce disturbance in the medium and thus cannot produce sound.



Define the wavelength of a sound wave. How is it related to the frequency and the wave speed?

Wavelength is the total length of one complete wave oscillation.
The wave speed, wavelength and frequency can be related by the equation
$$v=\nu .\lambda$$
Where $$v$$ is the wave speed.
$$\nu$$ is the frequency.
$$\lambda$$ is the wavelength.



Describe an experiment to show that each source of sound is a vibrating body.

Stretch a string by holding one end in mouth between the teeth and the other end in one hand. Pluck it by the other hand near the middle.

It is noticed that the string starts vibrating and simultaneously a sound is heard. After some time when the string stops vibrating, no sound is heard.



If velocity of sound waves is $$330$$ $${m/s}$$ and wave length of wave is $$0.5$$ $$m$$, frequency of the wave is                  .

Speed of sound is $$V=330\ m/s$$
Wavelength is $$\lambda=0.5\ m$$
Frequency $$f=?$$
We know that, $$V=f\lambda$$
$$\therefore f=\cfrac{V}{\lambda}=\cfrac{330}{0.5}=660$$ $$Hz$$


What is reverberation ?

The persistence of sound in a big hall due to repeated reflection from the walls, ceiling and floor of the hall is called reverberation.


A dolphin in $$25^{\circ}$$C sea water emits a sound directed toward the bottom of the ocean 150 m below. How much time passes before it hears an echo?
(Take speed of sound in sea water at $$25^{\circ}$$ C = 1530 m/s)

To hear an echo , sound will come back to dolphin after reflection from the bottom of sea ,
 total distance travelled by sound , $$d=150+150=300m$$ ,
given speed of sound in sea water , $$v=1530m/s$$ ,
therefore , time after which dolphin hears an echo ,
                  $$t=d/v=300/1530=0.196=0.20s$$


What are mechanical waves?

The waves which require a material medium to travel are called as mechanical waves.
Eg. Sound waves.


Loudness of a note increases with the increase in _________.

$$Amplitude$$

Sound consists of longitudinal waves of the medium of propagation. These waves  are produced by a vibrating object present in or near the medium.
In any wave, the particles are vibrating about each one's mean position.
The maximum displacement of such a particle from its mean position is known as amplitude.
A note is a sound wave with definite pitch.
The greater the amplitude of the particles, louder is the note and vice versa.


(i) Name the waves used for echo depth sounding.
(ii) Give one reason for their use for the above purpose.
(iii) Why are the waves mentioned by you not audible to us?

(i) The waves used for echo depth sounding are ultrasonic waves. 
(ii) Ultrasonic waves are used for echo depth sounding because they can travel undeviated through long distances and can be confined to a narrow beam. 
(iii) Ultrasonic waves are not audible to us because those waves have frequency more than 20,000 Hz and our audible range is between $$20 Hz$$ to $$20,000 Hz$$.


Define wavelength of a wave.


The distance between two consecutive compressions (C) or two consecutive rarefactions (R) is called the wavelength. The wavelength is usually represented by λ (Greek letter lambda). Its SI unit is metre (m).
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Technique of echocardiography is based on reflection property of .......... (ultrasonic or radio) waves.

Technique of echocardiography is based on reflection property of ultrasonic waves.
An echocardiogram (also called an echo) is a type of ultrasound test that uses high-pitched sound waves that are sent through a device called a transducer. The device picks up echoes of the sound waves as they bounce off the different parts of your heart. These echoes are turned into moving pictures of your heart that can be seen on a video screen.


The distance between any two successive crests or compressions in a longitudinal wave is called .......... .

The distance between any two successive crests or compressions in a longitudinal wave is called $$wavelength $$.


Name the waves which are used in 'SONAR' to find the depth of a sea. State one property of waves for their use.

SONAR uses sonic waves (sound). Typically, sonar use sounds with frequencies inaudible to human ear. 
Actually, sonar uses a wide range of frequencies. Very low frequencies (100 Hz to 1 kHz) are used for long range detection and deep sea floor sounding. Mid-range frequencies (1 kHz to 10 kHz) are used for target identification. Ultrasonic frequencies (50 kHz and up) are used in navigation and commercial sonar devices like fish finders.
Properties of Ultrasonic waves:
  • They produce heating effect when passes through the medium.
  • They get reflected, refracted and absorbed by the medium similar to the ordinary sound waves.
  • They act as catalytic agents to accelerate chemical reactions.
  • They produce stationary wave pattern in the liquid while passing through it.


What is 'SONAR' ? State the principle on which it is based.

Sonar (SOund Navigation And Ranging) is a technique that uses sound propagation (usually underwater, as in submarine navigation) to navigate, communicate with or detect objects on or under the surface of the water, such as other vessels.
The echolocation principle is used in SONAR.
Echolocation is the ability to locate objects by bouncing sound waves off of them, and then measuring the time taken for an echo to return, and calculating the direction the echo came from. These sound waves are very high-pitched, and most humans are unable to hear them. 


What in meant by an echo? What is the condition necessary for an echo to be heard distinctly ?

Like all waves, sound waves can be reflected. Sound waves suffer reflection from the large obstacles. As a result of reflection of sound wave from a large obstacle, the sound is heard which is named as an echo. 
The sensation of any sound persists in our ear for about 0.1 seconds. This is known as the persistence of hearing. If the echo is heard within this time interval, the original sound and its echo cannot be distinguished. So the most important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 second after the original sound dies off. As the speed of sound is 340 m/s, the distance travelled by sound in 0.1 second is 34 m. This is twice the minimum distance between a source of sound and the reflector. So, if the obstacle is at a distance of 17 m at least, the reflected sound or the echo is heard after 0.1 second, distinctly.


State the use of echo by a bat, dolphin and fisherman.

Bats send out sound waves using their mouth or nose.  When the sound hits an object an echo comes back. The bat can identify an object by the sound of the echo. They can even tell the size, shape and texture of a tiny insect from its echo. Most bats use echolocation to navigate in the dark and find food.
Dolphin's echolocation was the result of evolution over time. This process allows them to send out sound waves that are like a click. When those sounds hit an object it bounces back vibrations to the dolphins. This allows them to identify where objects are located. It also gives them information about the location of the object and some indication of the shape and size of it. The amount of time it takes for the sound waves to come back help them to identify the distance. The longer it takes for the sound waves to return, the more distance between them and that given object. 
Fishermen's machine sends a loud beep. The machine starts listening again and picks up the echo. This is repeated many times as the fisher moves the boat across the sea. When the echo changes, that means the beep is bouncing off a shoal of fish. So that's the place to drop the nets in.


State three factors on which loudness of sound heard by a listener depends.

Loudness of sound is a measure of response of sound to our ear. Loudness of sound is not simply the energy reaching the human ear, but it also tells about the sensitivity of human ear detecting this energy. Loudness of sound is measured in decibel (dB). As energy reaching the ear depends on square of amplitude, loudness of sound depends on various factors namely,
(i) Amplitude of sound waves
(ii) Sensitivity of ear
(iii) Distance from the source of the sound and the listener.


With the help of the given clues, unscramble the letters to make a meaningful word.
HCEO (A reflected sound wave)

ECHO
The repetition of sound caused by the reflection of sound waves. For example, if we shout or clap near a suitable reflecting object such as a tall building or a mountain, we will hear the same sound again a little later. This sound which we hear is called an echo.


What devices are used to locate underwater objects and to determine depths of water beneath the ship?

SONAR (sound navigation and ranging) is used to determine the depth of water and to locate the underwater objects. By SONAR ultrasonic waves are transmitted towards the bottom of water, these waves are reflected by any object or bottom, total time is recorded and multiplied with the speed of ultrasonic waves, to find the depth.
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Sound is produced by ___________.

Ans) $$Vibrations$$
When we give energy to the particles of a medium they vibrate, and due to the vibrations of medium particles sound waves are produced .


With the help of the given clues, unscramble the letters to make a meaningful word.
NASRO (A device used in the ship to locate unseen underwater objects)

NASRO , can be arranged as SONAR. SONAR stands for sound navigation and ranging , used to find the position of underwater objects .
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Why are sound waves in air characterized as longitudinal?

Sound travels as a longitudinal wave through a material medium. In sound waves the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles do not move from one place to another but they simply oscillate back and forth about their position of rest. Hence sound waves are longitudinal waves.


The .......... of a sound depends on its amplitude.

The loudness of a sound depends upon the intensity of sound and we know that intensity of sound depends upon amplitude as,
     $$I\propto A^{2}$$,
So we can say that loudness depends upon its amplitude.


Reflected sound waves are called ......... .

Sounds heard by a listener after the reflections of sound waves  from any surface is called echoes . 


Explain why (or how):
(a) in a sound wave, a displacement node is a pressure antinode and vice versa, 
(b) bats can ascertain distances directions, nature and sizes of the obstacles without any "eyes"
(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and 
(e) the shape of a pulse gets distorted during propagation in a dispersive medium. 

(a) A node (N) is a point where the amplitude of vibration is the minimum and pressure is the maximum.
An antinode (A) is a point where the amplitude of vibration is the maximum and pressure is the minimum.
Therefore, a displacement node is nothing but a pressure antinode, and vice versa.

(b) Bats emit ultrasonic waves of large frequencies. When these waves are reflected from the obstacles in their path, they give them the idea about the distance, direction, size and nature of the obstacle.

(c) The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

(d) This is because solids have both, the elasticity of volume and elasticity of shape, whereas gases have only the volume elasticity.

(e) A sound pulse is a combination of waves of different wavelength. As waves of different λ travel in a dispersive medium with different velocities, therefore, the shape of the pulse gets distorted.


What is a tuning fork?

Tuning Fork - Tuning fork is a U-Shaped fork of steel provided with a steel handle The two arms of U are called prongs while the handle is called the stem. We can produce sound of definite frequency by striking the tuning fork with a rubber pad.
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What do you understand by the term Echolocation?

Echolocation is the process where sound waves and echoes are used to determine objects in space and a sensory perception by which certain animals orient themselves to their surroundings detect obstacles, communicate with others, find food and escape from their enemies


What do you understand by the term Echo?

Echo is a reflection of sound that arrives at the listener with a delay after the direct sound. The delay is proportional to the distance of the reflecting surface from the source and the listener. Typical examples are the echo produced by the bottom of a well, by a building, or by the walls of an enclosed room and an empty room.


What do you understand by timber of a sound?

The property due to which two notes of the same pitch and loudness produced by two different vibrating bodies can be distinguished is called timbre or quality of sound.


What do you understand by the term Sonar?

SONAR - Sonar is the short form of Sound Navigation and Ranging.
It is a device fitted in a ship to find the depth of the sea which is based on the principle of reflection of sound (echo). It is also used for locating underwater objects like reefs submarines shoal of fish etc 
A sonar unit consists of an ultrasonic transmitter and receiver mounted on the bottom of a ship.


Why is sound wave called a longitudinal wave?

The vibration of medium that travel along or parallel to the direction of wave is called a longitudinal wave. In a sound wave, the particles of the medium vibrate in the direction parallel to the direction of propagation of disturbances.
Hence, a sound wave is called a longitudinal wave.


Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound?

$$\textbf{Part1: Diagram}$$

$$\textbf{Part2: Explanation}$$
$$\bullet$$Air is the most common medium through which sound travels. 
$$\bullet$$When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called a compression (c) as shown in Fig. This compression starts to move away from the vibrating object, 
$$\bullet$$When the vibrating object moves backward, it creates a region of low pressure called rarefaction $$(R)$$ as shown in Fig. As the object moves back and forth rapidly, a series of compressions and rarefactions is created in the air.


Explain how defects in a metal block can be detected using ultrasound.

Ultrasound is passed through one end of the metal and a detector is installed at the other end. When the metal is perfect, the ultrasonic waves travel as expected and hence are detected on the other end. For a metal with a defect, the waves are reflected back and hence don't reach the other end. When no detection is received, it can be concluded that the metal has a defect.


Explain the working and application of a sonar.

Working: Sonar consists of transmitter and a detector and is installed in a boat or a ship as shown in Fig. The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector.
The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the soundwave can be calculate by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound. Let the time interval between transmission and reception of ultrasound signal be tt and the speed of sound through sea water by vv. The total distance, 2d2d travelled by the ultrasound is then, 2d=vxt2d=vxt.
Applications : The sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunkenship etc.

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What is reverberation? How can it be reduced?

Persistence of sound (after the source stops producing sound) due to the repeated reflection is called reverberation. Reverberation can be reduced by absorbing the sound using some materials as it reaches the wall and ceiling of the room and thus prevent the sound from getting reflected. Some materials which are used to reduce reverberation are fibre board, heavy curtains, plastics etc.


State True or False.
Two sounds can differ only by the difference in their loudness.

False. The difference between two sounds can depend on their frequencies (shrillness) and amplitude (loudness) too. There can also be a difference in two sounds on the basis of their quality of timbre.


A sound wave traveling in the air is made to propagate through a liquid such that the velocity of sound is quadrupled. If the frequency of the sound wave is constant, find the change in the wavelength of the sound wave.

Let the speed of sound in air be $$v$$ and frequency of sound be $$f$$.
Wavelength of sound in air  
$$\lambda_a = \dfrac{v}{f}$$
Speed of sound in liquid  
$$v_l = 4v$$
Thus wavelength of sound in liquid  
$$\lambda_l = \dfrac{v_l}{f}$$
$$\Rightarrow \lambda_l = \dfrac{4v}{f}$$
Change in wavelength of sound wave 
$$\Delta \lambda  = \lambda_l - \lambda_a$$
$$\therefore$$   $$\Delta \lambda  = \dfrac{4v}{\nu} - \dfrac{v}{\nu}$$

$$\Rightarrow \Delta \lambda = \dfrac{3v}{\nu}$$


How are multiple reflections of sound helpful to doctors and engineers?

A stethoscope is a medical instrument used for listening to sounds produced within the body chiefly in heart or lungs. In stethoscope, the sound of the patient's heart beat reaches the doctors ears by multiple reflections and by amplifying sound.
Generally, the ceilings of concert halls, conference hall,s and cinema halls are designed such that sound after reflection reaches all corners of the hall. in some halls, a curved ceiling is arranged in such a way that the sound after reflecting from ceiling spreads evenly across the hall. 


Give the differences between a transverse wave and a longitudinal wave.

Longitudinal waves - In these waves, the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles do not move from one place to another but they simply oscillate back and forth about their position of rest. 
Sound waves are longitudinal waves. 

Transverse waves - In a transverse wave particles do not oscillate along the line of wave propagation but oscillate up and down about their mean position as the wave travels. Thus a transverse wave is the one in which the individual particles of the medium move about their mean positions in a direction perpendicular to the direction of wave propagation. 
Light is a transverse wave.


The quality of music from a group of instruments is independent of the listener's distance from the instruments. Why?

There are 3 properties of sound, based on which we can distinguish one sound note from the other.
They are Loudness, Pitch(or shrillness), and Quality.
Quality. 
The loudness of a sound depends on the amplitude of the sound wave. The greater the amplitude, the louder is the sound.
Pitch or shrillness depends on the frequency. Greater the frequency, shriller is the sound.
Quality or Timbre depends on the overtones present in the sound, along with the fundamental note and their amplitudes. Quality is the property by which two sounds may appear different even if they are equally loud and shrill.

When a sound wave travels, the waves split and spread around the direction of propagation. The further the sound travels, the more is the splitting. The amplitude of the original wave gets divided equally among the split waves.
Hence the loudness of the sound heard becomes less and less with an increase in distance.
The split waves however have their fundamental frequencies and the overtone frequencies same as those of the original wave.
Hence the pitch and quality of sound don't change with distance.


What is an echo? Mention the condition for the echo to be heard.

Echo is a reflection of sound that arrives at the listener with a delay after the direct sound. The delay is proportional to the distance of the reflecting surface from the source and the listener.

The conditions necessary for hearing the echo:

The distance between the sound source and the reflecting surface must not be less than 17 metres where the time period between hearing the original sound and its echo should not be less than 0.1 of a second .

The human ear can not distinguish between two successive sounds if the period between them is less than 0.1 second , Wide and big reflecting surface must be presented to hear the echo such as the walls , the mountains or the water bodies .

The velocity of sound through the air is 340 m / sec , so the distance travelled by the sound and its echo in 0.1 sec = 34 metres , The distance travelled by the sound is 17 metres from the sound source to the reflecting surface and 17 metres from the reflecting surface to the ear ( echo ) .

The echo can not be heared if the distance between the sound source and the reflecting surface is less than 17 metres because the time between hearing the main sound and its echo will be less than 0.1 of a second , and the human ear can not distinguish between the two successive sounds .

When the distance between hearing the main sound and the reflecting surface is multiplies of 17 metres ( twice or three times ) , the echo is heard in the form of the last two or three phrases of the whole produced sound .



Explain how echoes are used by bats to judge the distance of an obstacle in front of them.

Bats search out prey and fly in dark night by emitting and detecting reflections of ultrasonic waves. The high pitched ultrasonic squeaks of the bat are reflected from the obstacles ( prey ) and returned to bat's ear. The nature of reflections tells the bat where the obstacle (prey) is and what it is like. Porpoises also use ultrasound for navigation and location of food in the dark.



How does an ultrasound scanner work? Explain. Write one use of it.

An ultrasound scanner works by sending high frequency sound pulses into a patient’s body. The sound waves travel through the patient’s body, passing through different types of tissue.
The parts of an ultrasound scanner are given below:-
1. The transducer probe:- Is the part of the machine that produces the sound waves and receives the echoes. It consists of one or more crystals in a plastic housing.
2. The transducer pulse controls:-It allow the ultrasonographer operating the machine to alter the frequency and length of the ultrasound pulses to suit any given conditions.
3. The Central processing unit:- Is a computer. It sends the electrical signals to the transducer probe and receives signals back from it.
4. The display:- Is usually a high definition computer monitor. Ultrasound images are displayed using a grey scale.
5. The keyboard and cursor:- allow the ultrasonographer to add notes to the images as they are displayed.
6. Disc storage:- allows scans to be stored digitally and kept with a patient’s medical records. 
7. A printer:- is usually attached to the machine so that hard copies of images can be printed off as required.


Explain the working and applications of SONAR.

The acronym SONAR stands for Sound Navigation And Ranging.
Sonar is a device that uses ultrasonic waves to measure the distance, direction, and speed of underwater objects.
PRINCIPLE:
It uses the phenomenon of echos in determining the sea depth and locating the pressure of water-water objects.
WORKING:
 (i) Sonar consists of a transmitter and a detector and is installed in a boat or a ship.
(ii) The transmitter produces and transmitted strong ultrasonic waves.
(iii) These waves travel through the water and after striking the objects the beam is reflected from the sea beds and is received by an underwater detector which is also mounted on the ship.
(iv) The detector converts the ultrasonic waves into electrical signals which are approximately interpreted.
(v) The time interval between transmission and receiving of the ultrasonic signal is noted.
The sonar method is also called echo ranging. This technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarines, icebergs, etc.

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A man is lying on the floor of a large, empty hemispherical hall, in such a way that his head is at the centre of the hall. He shouts "Hello!" and hears the echo of his voice after $$0.2 s$$. What is the radius of the hall?

Given,
Time period $$T=0.2 Sec$$
Velocity of sound $$V=340 m/s$$
Radius of the hall $$d=?$$
Distance travel by the sound $$2d$$
since, $$2d=Vt$$

$$d=\cfrac{340\times 0.2}{2}=34 m$$

Hence, the Radius of hemispherical hall is $$349m$$


The device used to measure the distance and direction of underwater substances using ultrasonic waves is _____________.

The device is sonar. It stands for Sound Navigation and Ranging. It is made up of 2 parts transmitter and detector. Transmitter produces ultrasound waves and detector reflects it and converts it to the electrical signal. It is used to measure the depth of the sea and to locate underground objects.  


With the help of a diagram describe how compression and rarefaction pulses are produced in the air near a source of the sound.


When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called compression. This compression starts to move away from the vibrating object. When the vibrating object moves backward, it creates a region of low pressure called rarefaction. As the object moves back and forth rapidly, a series of compressions and rarefactions are created in the air.

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In certain auditoriums, due to multiple reflections, the sound persists for some time even after the source has actually stopped producing the sound.
(a) What is this phenomenon known as?
(b) Suggest two ways to overcome this disturbance.
(c) Cite two devices that utilize multiple reflections.

(a) Reverberation:  It is the persistence of sound after a sound is produced.

(b) To reduce reverberation, a large number of windows and ventilators should be provided or the walls of the auditoriums should be made rough.

(c) Megaphone and Stethoscope


The human ear can detect continuous sounds in the frequency range from $$20$$Hz to $$20,000$$Hz. Assuming that the speed of sound in air is $$330ms^{-1}$$ for all frequencies, calculate the wavelengths corresponding to the given extreme frequencies of the audible range.

Speed of sound in air   $$v = 330 \ m/s$$
Wavelength of sound wave   $$\lambda = \dfrac{v}{\nu}$$
where  $$\nu$$ is the frequency of sound wave.
Minimum frequency heard by human ear  $$\nu_{min} = 20 \ Hz$$
So, maximum wavelength of sound wave  $$\lambda_{max} = \dfrac{v}{\nu_{min}}$$
$$\therefore$$  $$\lambda_{max} = \dfrac{330}{20} = 16.5 \ m$$
Maximum frequency heard by human ear  $$\nu_{max} = 20 \ kHz =20000 \ Hz$$
So, minimum wavelength of sound wave  $$\lambda_{min} = \dfrac{v}{\nu_{max}}$$
$$\therefore$$  $$\lambda_{min} = \dfrac{330}{20000} = 16.5  \times 10^{-3} \ m = 16.5 \ mm$$


In an auditorium or a cinema hall, the roof and walls are covered with draperies or compressed fibreboard. Why ?

The roof and walls of the auditorium or cinema hall are generally covered with sound absorbent materials like draperies or compressed fibreboard to reduce reverberation. These materials reduce the formation of echoes by absorbing sound waves.


A man is standing between hill A and hill B, claps louder. He hears an echo after $$4$$ seconds from hill A and after $$6$$ seconds from hill B. The speed of sound in air is $$340ms^{-1}$$. Calculate the distance between the two hills.

Given,
Speed of sound $$v=340\ m/s$$

Time to hrar echo      $$t_1=4\ s\\t_2=6\ s$$

Total distance travel by sound to hear echo from hill A $$d_a=2x$$

Total distance travel by sound to hear echo from hill B $$d_b=2y$$

Distance of Hill A from man is, $$x=\dfrac{v\times t}{2}=\dfrac{4\times340}{2}=680m$$

Distance of Hill B from man is, $$y=\dfrac{v\times t}{2}=\dfrac{6\times340}{2}=1020m$$

So distance between Hill A and B is, $$d=x+y=680+1020=1700\ m$$

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How are loudness and amplitude related?

In acoustics, loudness is the subjective perception of sound pressure. More formally, it is defined as, "That attribute of auditory sensation in terms of which sounds can be ordered on a scale extending from quiet to loud."

Amplitude  the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path



How does the amplifier  effect the loudness of vibration?

An explanation of sound waves. The pitch of a sound is dictated by the frequency of the sound wave, while the loudness is dictated by the amplitude. When a drum is beaten, the air particles around the drum skin vibrate in the form of a compression wave



For hearing the loudest ticking sound heard by the ear, find the angle x in the figure.
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Incident line is making an angle of $$50^o$$ with reflecting surface.
The angle of incidence is measured from the normal
So angle of incidence $$\angle i=90^o - 50^o =40^o$$
The directions in which the sound is incident and is reflected make equal angles with the normal to the reflecting surface at the point of incidence, and the three are in the same plane.
$$\angle i = \angle r$$
So Angle of reflection $$\angle r =$$ angle of incidence $$\angle i= 40^o$$
Hence, $$\angle x=90^o-50^o=40^o$$


How will you reduce reverberation in public halls or buildings ?

To reduce reverberation, the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are also selected on the basis of their sound absorbing properties.


Name the characteristic which differentiates two sounds of the same pitch and same loudness.

$$Quality$$ of a sound is the characteristic property which differentiates two sounds of having the same pitch and same loudness.Quality of sound is also known as timbre of sound.


How does a man's voice differ from a woman's voice ?

Adult men and women typically have different sizes of vocal fold; reflecting the male-female differences in larynx size. Adult male voices are usually lower-pitched and have larger folds. 
The female vocal folds are between 12.5 mm and 17.5 mm in length. The folds are within the larynx.
Hence female voice is more shrill than male voice, i.e the female voice is high pitched and the male voice is low pitched.



How are the wavelength and frequency of a sound wave related to its speed?

The relation between wavelength $$(\lambda)$$, frequency $$(\nu)$$ and speed of wave $$(v)$$ is 
$$v = \lambda \nu$$.


Calculate the wavelength of a sound wave whose frequency is $$220\ Hz$$ and speed is $$440\ ms^{-1}$$ in a given medium.

Given
Frequency $$f = 220\ Hz$$,
Speed $$u = 440\ ms^{-1}$$
Wavelength $$\lambda = ?$$

From the relation $$v= f\lambda$$
$$440 = 220\lambda $$.

$$\therefore \lambda = \dfrac {440}{220} = 2.0\ m$$.


What is sonar ? Explain the working of sonar.


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Calculate the wavelength of a sound wave whose frequency is $$22 0

\ \mathrm { Hz }$$ and speed is $$440 \ \mathrm { m/s } $$ .

The wavelength $$\lambda,$$ speed $$v$$ and frequency $$f$$ are related as:
$$ \lambda \ (in \ m) = \dfrac{v \ (in \ m/s)}{f \ (in \ Hz)} $$
Given:
$$v=440 \ m/s$$
$$f=220 \ Hz$$
$$ \lambda = \dfrac{v}{f} = \dfrac{440}{220}= 2\ m $$


State two differences between sound waves and light waves. 

(i) The light waves can travel in a vacuum while sound waves need a material medium for their propagation. 
(ii) The light waves are electromagnetic waves while sound waves are mechanical waves.
(iii) Light waves are transverse in nature while sound waves are mostly longitudinal.


An echo returned in $$3\ S$$. What is the distance of the reflecting surface from the source, given that the speed of sound is $$342\ ms$$?

Echo returns in $$3S$$. Therefore time taken by sound to travel from the source to the reflecting surface
$$t = \dfrac {3}{2} = 1.5\ S$$
Speed $$v = 342\ ms^{-1}$$
$$\therefore$$ Distance of reflecting surface from the source $$= Speed \times Time$$
$$= 342\times 1.5 = 513\ m$$.


A musician recognizes the musical instrument by hearing the sound produced by it, even without seeing the instrument. Which characteristic of sound makes this possible?

A musician can recognize the musical instrument by hearing sound but not seeing it. this is possible because of the characteristic of sound called quality. The quality off sound depends on the frequencies present in it and their relative loudness.


A student standing in front of a huge building claps his hands. He Heard its echo after $$2$$ seconds.[Speed of sound in air is $$340$$m/sec]
Calculate the distance between the student and building.

Let’s assume that the student is standing at a distance x from the building.

The sound of his clap travels x metres to reach the building, and x metres again to reach the ears of the man, which leads to him hearing the echo.

So, total distance travelled by the waves (sound) is 2x. Time taken is 2s. Speed of sound is 340 m/s.

The data we have is:

Distance : 2x meters 

Speed:340 m/s.

Time: 2s.

Distance=Speed $$ \times $$ time

$$2 x =340 \times 2$$

$$x = 340 m$$

Therefore, the distance is 340m.



A stone is dropped on the surface of water in a pond. Name the type of waves produced.

Transverse waves are produced when stone is dropped on the surface of water in a pond.


Name the two types of waves which can be generated in a long flexible spring.

The two types of waves are longitudinal waves and transverse waves.


What will be the change in the wavelength of a sound wave in air if its frequency is doubled?

The velocity of sound is the product of frequency and wavelength. 
The speed of sound in the air is constant. Therefore, when the frequency is doubled, the wavelength reduces to half.
$$\lambda = \dfrac{v}{f}$$
$$\lambda$$; wavelength of the wave
$$v$$; velocity of the wave which remains constant in air
$$f$$; frequency of the wave.
Now frequency is doubled,
$$f' = 2f$$
So wavelength,
$$\Rightarrow \lambda ' = \dfrac{v}{2f}$$

$$\Rightarrow \lambda ' = \dfrac{\lambda}{2}$$


Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger then the diameter of the loudspeaker.
Sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one tenth of the diameter of the speaker described above. Take the speed of sound to be $$340\ m\ s^{-1}$$, Diameter of loudspeaker $$20$$cm.   

According to the question,
Wavelength $$\lambda=\dfrac d {10}=\dfrac{20}{10}cm=2\times 10^{-2}\ m$$
Frequency $$f=\dfrac{speed}{wavelength}=\dfrac{v}{\lambda} \Rightarrow \dfrac{340}{2\times 10^{-2}}$$

$$f=1.7\ Hz$$ ( because $$\lambda=2\ cm=2\times 10^{-2}\ m)$$ 


State the general name of the waves in which the particles of the medium vibrate:

a) in the same direction as the wave

b) at right angles to the direction of wave

a) Longitudinal waves; Longitudinal waves are waves in which the vibration of the medium is parallel to the direction the wave travels and displacement of the medium is in the same direction of the wave propagation.
b) Transverse waves: a transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance. This is in contrast to a longitudinal wave that travels in the direction of its oscillations.


What type of waves are illustrated by the movement of a rope whose one end is fixed to a pole and the other end is moved up and down?

Transverse waves
A transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance. This is in contrast to a longitudinal wave which travels in the direction of its oscillations.


A wave is moving in air with a velocity of $$340 \,m/s.$$ Calculate the wavelength if its frequency is $$512 \,vibrations/sec.$$

Velocity , $$v = 340\,m/s$$
Frequency , $$f = 512\,Hz$$

Using the relation, $$v = f \lambda$$
Wavelength , $$\lambda =\dfrac vf=\dfrac {340}{512}$$
$$\lambda= 0.66\,m$$


The frequency of the sound emitted by the loudspeaker is $$1020\,Hz.$$ Calculate the wavelength of the sound wave in air in cm where its velocity is $$340\, m/s.$$

Frequency , $$f = 1020\,Hz$$
Velocity , $$v = 340\,m/s$$
Wavelength , $$\lambda = \dfrac vf$$
$$ \lambda=\dfrac{340}{1020}= 33.3 \,cm$$


a) Define the terms frequency, wavelength, and velocity of a sound wave. What is the relation between them?

b) A body vibrating with a time period of $$1/256\, s$$ produces a sound wave which travels in air with a velocity of $$350 \,m/s.$$ Calculate the wavelength.

a) Frequency is defined as the no.of vibrations per second.

Wavelength is the minimum distance in which the wave repeats.

Velocity is the speed with which one wave can travel in one second.

Following is the relation between frequency, wavelength, and velocity:

$$v = f\,\lambda$$

b) Time period $$= 1/256\,s$$

Velocity of wave $$= 350\,m/s$$

$$Frequency = 1 / time \,period = 256\,Hz$$

We know that,

$$v = f\lambda$$

therefore , $$wavelength = 1.36\,m$$



 If a ringing bicycle bell is held tightly by hand, it stops producing sound. Why?

The ringing bicycle bell does not produce sound when it is held tightly with hand because the vibrations are stopped and so the sound production stops.


Define the frequency and time period of a wave. What is the relation between the two?

Frequency is defined as the no.of complete waves that are produced in one second.

Time period is defined as the time taken by the wave to complete one cycle.

The relation between frequency and time period is:

$$Frequency = 1/ time\, period$$



Give reason for the following :
Why sound wave is called a longitudinal wave ?

In a longitudinal wave, the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles do not move from one place to another but they simply oscillate back and forth about their position of rest. This is how a sound wave propagates, hence sound waves are longitudinal waves.

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Give reason for the following : 
Why sound waves are called mechanical waves ? 

The waves that require a medium to propagate are called mechanical waves. They require a medium to transport energy from one point to another. Sound waves are characterized by their motion in a medium. They cannot travel in vacuum (without any medium). So, they are called mechanical waves.


Define - Compression and rarefaction. 

Compressions: The particles are closest to each other in this. The density of the medium is maximum at compression. 

Rarefaction: The particles are farthest from each other. The density of the medium is minimum for rarefaction.


Give reason for the following :
Why echoes in the long gallery and big halls can be heard ? 

A sound persists in our ear for $$ 0.1\,s $$. At temperature $$ 22^{\circ} \,C $$ in air and speed of sound $$ 344\,ms^{-1} $$ , the total distance covered by sound between the point of generation and reflecting surface should be at least $$ (344\, m/s) \times 0.1 \,s = 34.4 \,m $$ . Thus, the minimum distance between the point of generation and reflecting surface should be half of it i.e. $$ 17.2\,m $$. So, in a long gallery and big halls, we can hear the echo due to the distance between the point of generation and reflecting surface more than the minimum distance required in given conditions.


When we hear a sound, does any part of our body vibrate? Name the part. 

Our eardrum vibrates whenever we hear a sound.


The sound produced by a thunderstorm is heard $$10\ s$$ after the lightning is seen. The approximate distance of the thunder cloud:
(Given speed of sound $$=340\ ms^{-1}$$)

Given:-
speed of sound $$v=340\ m/s$$ and time $$t=10\ s$$

$$Distance = v \times t =340 \times 10=3400\ m$$


Aditi clapped her hands near a cliff and heard the echo after $$4$$ seconds. What is the distance of the cliff from her if the speed of sound is taken as $$346\ ms^{-1}$$.

Given
Speed of sound $$(u) = 346\ m/s$$
Time taken for hearing the echo $$(t) = 4s$$
Now
$$speed = \dfrac {distance}{time}$$

$$\Rightarrow 346 = \dfrac {distance}{4}$$

$$\Rightarrow distance$$ $$= (346 \times 4) = 1384$$.

Total distance covered by sound is $$1384\ m$$
In $$4\ s$$, sound has to travel twice the distance between the cliff and aditi.
Therefore, the distance between the cliff and aditi is $$= \dfrac {1384}{2}\ m = 692\ m$$
Distance of cliff from aditi $$= 692\ m$$.


Describe two uses of multiple reflections of sound.

Multiple reflection of sound is used in various appliances and devices. The examples are as follows:
1. $$Stethoscope$$ $$\Rightarrow$$ A medical instrument used for listening the sound produced by heart or lungs. The sounds of heartbeat reach the doctor ear by multiple reflections and amplify the sound. The figure below describes multiple reflections in stethoscope.

2. $$Designing\ of\ Concert\ Halls$$ $$\Rightarrow$$ The ceilings of concert halls, conference hall and cinema halls are designed such that sound undergoes multiple reflections and reaches all corners of the hall. Some halls have a curved ceiling for the same. The figure below depicts the multiple reflections in concert hall.


Velocity of sound in air is $$340 m/s^{-1}$$. Calculate wavelength when frequency is $$256\ Hz$$.

Given, velocity $$v=340 \ m/s$$
frequency $$\nu = 256 \ Hz$$

We know that, $$\upsilon =\nu \lambda $$
$$340=256 \lambda$$

Hence, wavelength 
$$\lambda =\dfrac{340}{256}=1.33\ m$$


State the factors on which the pitch of a sound depends .

The pitch of a sound depends on its frequency
 (i.e., on the frequency of the vibrating body). 
It is also depend upon the wavelength and time period .


Why is the loudness of sound heard by a plucked wire increased when mounted on a sound board ?

  • Loudness of sound depends on the surface area of the vibrating body

  • When a wire mounted on a sound board is plucked, the surface area of vibrating air increases and it sends forth a greater amount of energy so the sound appears more loud.




List the uses of ultrasonic waves.

Ultrasound is sound waves with frequencies higher than the upper audible limit of human hearing. Ultrasound is not different from "normal" sound in its physical properties, except that humans cannot hear it. This limit varies from person to person and is approximately 20 kilohertz in healthy young adults.

Ultrasounds are used extensively in industries and for medical purposes. Ultrasounds are high frequency waves (higher than $$20\ kHz$$).

The ultrasonic waves are used in:
  • Ultrasonic waves are used in echocardiography It is used to construct the image of the internal body parts.
  • They are also used in the detection of cracks in metal blocks.
  • It is also used in Ultrasonic soldering and welding.


If the amplitude of a wave is doubled, what will be the effect on its loudness ?

  • The loudness of a sound depends on the amplitude of vibration of the vibrating body producing sound. Greater the amplitude of vibrations, louder is the sound produced.

  • The loudness of sound is directly proportional to the square of amplitude of wave.

Loudness $(Amplitude)^2$


  • Hence,  on doubling the amplitude of the wave, the loudness becomes 

$(2)^2$ = 4 times.




State the factors on which loudness of sound heard by a listener depends. 

Three factors on which loudness of sound depends

  • Surface area of sounding body. i.e. is directly proportional to the surface area of the vibrating body.

  • On the distance of the source of sound: Loudness decreases with increase in distance.

  • On amplitude of wave: Increases with amplitude.


 




Name and define the characteristic which enables one to distinguish two sounds of same loudness, but of different frequencies, given by the same instrument.

Two sounds of same loudness (amplitude) of different frequencies given by the same instrument can be distinguished by the characteristics called Pitch.


Additional Information:

Pitch is the sensation (brain interpretation) of the frequency of an emitted sound.

  • Faster the vibration of the source, higher is the frequency and higher is the pitch. Similarly, low pitch sound corresponds to low frequency.



You recognize your friend by hearing his voice on a telephone . Explain .

We can recognize our friend by hearing his voice on a telephone.
 
This is due to quality of sound and pitch of sound.


Define the following terms :
(a) Amplitude 
(b) Frequency
(c) Time period.

(a) Amplitude (A) : The maximum displacement of a wave on either side of its mean position is called Amplitude A = XY = is amplitude.

(b) Frequency (f) or n :Number of oscillations made by a wave in one second is known as its frequency .

(c) Time Period (T) : Time taken to complete one vibration is called Time period , i.e from A to B.


How is loudness of sound related to the intensity of wave producing it?

The loudness of sound is decided by the amplitude ($$A$$) of the sound wave. On the other hand, the intensity ($$I$$) of the sound wave is proportional to the square of the amplitude $$(I\propto A^2)$$


Give five uses of ultrasonic waves ?

Ultrasonic wave is sound waves with frequencies higher than the upper audible limit of human hearing. Ultrasound is not different from "normal" sound in its physical properties, except that humans cannot hear it. This limit varies from person to person and is approximately 20 kilohertz in healthy young adult.

  1. Ultrasonic waves are used in echocardiography It is used to construct the image of the heart .
  2. Ultrasonic waves are used in the detection of cracks in metal blocks.
  3. Ultrasonic waves are used to kill bacteria in liquid.
  4. Ultrasonic waves are used for determining the depth of the sea.
  5. Ultrasonic waves are also used to see what is happening to the components inside a nuclear reactor.


Write a full form of SONAR. How do determine the depth of the sea using echo ranging ?

SONAR: Full form of SONAR is 'Sound Navigation and Ranging'. This technique is used by ships to detect the presence of submarines, icebergs, sunken ships etc. in the sea. SONAR is a device that uses ultrasonic waves to measure the distance, direction and speed of underwater objects. Sonar can also be used for determining the depth of the sea.

Determination of the depth of a sea using SONAR Let the time interval between transmission and reception of ultrasound signal be $$t$$ and the speed of sound through sea-water be $$v$$. The total distance, $$2d$$ travelled by the ultrasound is $$ 2d = v \times t $$  

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Name the medium in which longitudinal waves can be produced.

Longitudinal waves can be produces in all the three mediums i.e., (solid , liquid and gases).


Write the S.I unit of wave length ?

The S.I unit of wavelength is Metre(m)


Calculate the frequency of sound wave whose wavelength is $$2\,m$$ and the speed is $$ 440\,m/sec $$ in a given medium ?


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What do you mean by reflection of sound waves ? State the laws of reflection.

When a sound wave strikes a hard surface, it changes its direction and goes back to the same medium. This phenomenon is called a reflection of sound.
Laws of reflection:
  1. The angle of incidence is equal to the angle of reflection of sound. 
  2. The incident sound wave, the normal and the reflected sound wave, all lie in the same plane.


Write the relation between the frequency and wavelength of the sound wave ?

The relationship between frequency and wavelength is given by 
$$ v = \dfrac{\lambda}{T} $$ 
where , $$v$$ = frequency , $$\lambda$$ = wavelength , $$T$$ = time period.


Name the waves used to break small stones formed in the kidneys into fine grains.

Ultrasound was used to break small stones formed in the kidneys into fine grains. These grains later get flushed out with urine.


Why do auditoriums have curved roofs, curtains , carpets and false ceiling ?

The roofs of auditoriums are curved, so that sound after reflection reaches all corners of the auditorium. In auditorium excessive reverberation is highly undesirable. So, sound absorbent material like curtains, carpets and false ceiling are used to reduce reverberation.


Differentiate between echo and reverberation. 

Echo: 
  1. Echo is a reflection of sound which arrives at the listener sometime after the sound is produced.
  2. Echo is a single reflection of sound. 
Reverberation: 
  1. Reverberation is defined as persistence of sound in an enclosure after the sound producing source is removed. 
  2. Reverberation is multiple reflections of sound.


What produces sound?

Sound is produced when an object vibrates, creating a pressure wave. This wave causes particles in the surrounding medium to vibrate. The particles transmit this energy to nearby particles , which is perceived by human ears as sound if the frequency of vibration lies in the audible range.
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Explain uses of ultrasonic sound.

The uses of ultrasonic sounds are:
1. They are used in medical field for diagnosis and treatment of internal illnesses of the human body.
2. Using ultrasound, kidney stones can be broken into small grains which later on get flushed out through the urine.
3. Ultrasonography is used to examine the foetus during pregnancy.
4. Ultrasound is also used in heart checks (ECG).
5. Industrially, ultrasonic sounds are useful in examining metals for defects like cracks etc. and for cleaning machine parts.


The frequency of a sound wave with velocity $$340\ m/s$$ is $$5\ kHz$$. Calculate its wavelength. 

Given,
Frequency, $$f = 5\ kHz = 5000\ Hz$$
Velocity of sound, $$v = 340\ m/s$$
Wavelength, $$\lambda$$
We know,
Wavelength, $$\lambda = \dfrac{v} {f}$$

$$\therefore \lambda = \dfrac{340} {5000}$$

$$\lambda = 0.68 \ m$$


When do materials produce sound?

Materials generally produce sound when they move back and forth rapidly. We call such motion as 'Vibration'.


A jerks is given to a slinky to produce a longitudinal wave. The wave travels at a speed of $$30 \,cm/s$$ and it has a frequency of $$20\,Hz $$. Find the minimum separation distance between consecutive rarefactions of the slinky.


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Find the minimum and maximum wavelengths of sound in water that is in the audible range $$(20 - 20000\ Hz)$$ for an average human ear. Speed of sound in water $$= 1450\ ms^{-1}$$.

a) for minimum wavelength $$n$$ should be maximum $$n=20\ KHz$$
$$\Rightarrow v=n\lambda \Rightarrow \lambda =\left(\dfrac{1450}{20\times 10^3}\right)=7.25\ cm$$.
b) for maximum wavelength $$n$$ should be minimum 
$$\Rightarrow v=n\lambda \Rightarrow \lambda =v/n\Rightarrow 1450/20=72.5\ m$$.


If we felt a boom of sound in empty rooms, then:
(a) Which are the regions where sound waves in a room get reflected?
(b) Do these repeatedly reflected sound waves reach the ear of a listener simultaneously?
(c) Will you be able to hear all these sounds clearly due to the persistence of audibility? Won’t you be hearing only a boom of all the sound?

a) The sound waves get reflected form the walls, ceilings, floor etc.
b) No, the repeatedly reflected sound waves reach the ear of a listener at different instances of time.
c) We will not be able to hear all these sounds clearly and distinctly, due to the persistence of audibility. We will be able to hear only a boom of all the sounds. This is a result of reverberation.


The distance traveled by a wave can be calculated by knowing the speed of ultrasonic sound in seawater and the time taken for the wave to return. Explain.

For calculating the distance travelled by sound wave,
$$Distance = speed \times time $$
Hence, if we know the speed of ultrasonic sound and time taken by wave to return we can calculate distance. And the object (or obstacle) is one-half of this calculated distance travelled by the sound wave.


Which of the following media can pass longitudinal waves only air, water, or Iron?

Air media can pass longitudinal waves.


Wavelength of a sound wave having frequency $$2 kHz$$ is $$35 cm$$. How much time will it take to travel a distance $$1500 m$$ ?


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Bats make use of ultrasonic sounds for catching prey. How do bats catch prey? Observe above figure and write down in your science diary.

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Bats produce ultrasonic sounds. These waves reflect after striking the prey and return back to bat's ears. This gives bats an idea about the location of prey.
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For a person with normal hearing, the limit of audibility is $$20 Hz$$ to $$20000 Hz$$. If so, what will be the limit of wavelength of sound waves that are audible to human beings? Assume that the speed of sound is $$340 m/s$$.

For a person with normal hearing, the limit of audibility is $$20\ Hz$$ to $$20000\ Hz$$.
Let the frequencies $$f_1=20\ Hz$$ and $$f_2=20000\ Hz$$
Velocity of sound, $$v= 340\ m/s$$
Wavelength, $$\lambda$$

We know,
$$\lambda=\dfrac vf$$

For $$f_1=20\ Hz$$,
$$\lambda_1=\dfrac {v}{f_1}$$

$$\lambda_1=\dfrac {340}{20}$$

$$\lambda_1 = 17\ m$$

For $$f_2=20000\ Hz$$,
$$\lambda_2=\dfrac {v}{f_2}$$

$$\lambda_2=\dfrac {340}{20000}$$

$$\lambda_2 = 0.017\ m$$

The limit of the wavelength of sound waves that are audible to human beings is from $$0.017\ m$$ to $$17\ m$$.


What should be the minimum distance to the reflecting surface if the velocity of sound in air is $$340 m/s$$?

Due to the persistence of audibility, we can hear the first sound and reflected sound separately, only if there is a time gap second between them.
$$Speed =\dfrac{ Distance\ travelled }{ time}$$

Speed of sound in air = $$340\ m/s$$

time $$=0.1\ s$$

$$Distance = speed \times  time$$
$$=340 \times 0.1$$
$$=34\ m$$
So the minimum distance to the reflecting surface is half of $$34\ m$$.
i.e. = $$17\ m$$.


Explain why (or how):
bats can ascertain distances, directions, nature, and sizes of the obstacles without any "eyes",

Bats emit ultrasonic waves of large frequencies. When these waves are reflected from the obstacles in their path, they give them the idea about the distance, direction, size and nature of the obstacle.


A sound wave enters water from air. What happens to its wavelength? Why?

Wavelength increases. 

Speed of sound in water is $$1482 \ m/s$$ and speed in air is $$343 m/s$$. We know that the frequency of the sound wave doesn't change as the medium differs. 

Using, $$v = f\lambda$$, where $$v$$ is speed, $$f$$ is frequency and $$\lambda$$ is the wavelength
So, as the speed increases when wave enters water from air, its wavelength should increase.


The wavelength of sound traveling in air with the velocity $$330 \ m/s$$ was found to be $$66 \ m$$. If so,
a) Find the frequency of sound.
b) By what name are sounds of such frequency known as?


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Velocity of sound in air is $$340 m/s$$. Sound waves of wavelength $$0.01 m$$ from a vibrating body reach your ear through air. Will you be able to hear the sound? Justify your answer.

$$v = 340 m/s$$,
$$\lambda = 0.01 m$$
$$v = f \lambda$$
$$f = v/\lambda$$
$$f = 340/0.01$$
$$= 34000 Hz > 20,000 Hz$$
Human beings cannot hear these sounds. Audible frequency range is $$20 Hz$$ to $$20000 Hz$$.


If a person claps his hands stands at a distance of $$99\ m$$ from a wall hears the echo after $$0.6\ s$$, what is the velocity of sound?

Total distance traveled by sound wil be double of wall distance because first sound start from clap and reflect from wall and reaches to the person.
$$d = 2 \times 99\ m$$
$$\Rightarrow d = 198\ m$$
$$t = 0.6\ s$$; time for echo
So velocity of the sound-
$$V = \dfrac{\text{distance}}{\text{time}}$$

$$\Rightarrow V = \dfrac{ 198 }{ 0.6 }$$
$$\Rightarrow V = 330\ ms^{-1}$$


Why is a pulse on a string considered to be transverse?

As the pulse moves down the string, the elements of the string itself move up and down. Since the medium - here, the string - moves perpendicular to the direction of wave propagation, the wave is transverse by definition.
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A person standing at a distance x in front of a cliff fires a gun. Another person B standing behind the person A at distance y from the cliff hears two sounds of the fired shot after 2s and 3s respectively. Calculate x and y (take speed of sound $$320 mx^{-1}$$)

Hint: The person B hears two of the fired shots, the first one is direct from the gun while other sound comes after reflection from the cliff.

Step 1: Write the equation using formula of speed when shot sound goes directly from A to B 

$$Speed = \dfrac{distance}{time}$$

$$320 = \dfrac{y – x}{2}$$

$${y – x} = 640$$ ……… $$(1)$$

Step 2: Write the equation using formula of speed when shot sound goes from A to B after reflection 

$$Speed = \dfrac{distance}{time}$$

$$320 = \dfrac{y + x}{3}$$

$${y + x} = 960$$ ……… $$(2)$$

Step 3: Solving equations 

Adding equation $$(1)$$ and $$(2)$$

$$2y = 1600$$

$$y = 800 m$$

From equation $$(1)$$,

$${800 – x} = 640$$ 

$$x = 160 m$$

Hence, the value of $$x$$ is $$160m$$ and $$y$$ is $$800m$$.




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How do we hear : Match the parts of a human ear with their functions :

Match the columns :

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Following are the functions of human ear :
A. Pinna               -      collects sound from the air and directs them down the ear canal to the ear drum.
B. Ear drum          -      picks up vibrations of air in the canal .  
C. Ear bone          -      magnify the vibrations before passing them on to the cochlea. 
D. Auditory nerve -       sends electrical signals to the brain .
E. Cochlea             -       inside it vibrating fluid moves then produce electrical signal .



............. is produced by animals such as bats, dogs, insects etc. and cannot be heard by humans.

Ultrasonic sound is produced by animals such as bats, dogs, insects etc. and cannot be heard by humans.Because audible frequency range of human ear is 20Hz to 20,000Hz.And ultrasonic sound has frequency more than 20,000Hz.



Explain how the 'Echo' is produced. What should be the minimum distance between the listener and the reflector to hear an echo of sound propagating with a speed $$ v \, ms^{-1} $$?

The repetition of sound caused by the reflection of sound waves. For example, if we shout or clap near a suitable reflecting object such as a tall building or a mountain, we will hear the same sound again a little later. This sound which we hear is called an echo.
The sensation of sound persists in our brain for about $$0.1\ s$$. To hear a distinct echo the time interval between the original sound and the reflected one must be at least $$0.1\ s$$. 
Given,
Sound is propagating with speed, $$v\ m/s$$
Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least,
$$D=speed\times time$$
$$D= v\times 0.1$$
$$D=\dfrac{v}{10}$$
Thus, for hearing distinct echoes, the minimum distance of the obstacle from the source of sound must be half of this distance, that is, 
$$\dfrac D2= \dfrac{v}{20}$$


What happens to a particle velocity, When a sound wave is reflected from rarer medium and denser medium?


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Propagation of sound can be visualised as propagation of density variations or pressure variations in the medium. Comment.

Air is the most common medium through which sound travels. A vibrating object moves back and forth rapidly, a series of compression and rarefaction is created in the air. These make the sound wave that propagates through the medium. Compression is the region of high pressure and rarefaction is the region of low pressure. Pressure is related to the number of particles of a medium in a given volume. More density of the particles in the medium gives more pressure and vice versa. Thus, propagation of sound can be visualised as propagation of density variations or pressure variations in the medium

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What causes reverberation of thunder sound?

It is due to the successive and multiple reflections of the sound from a number of reflecting surfaces, such as the clouds and the land.


A man standing  $$25 { m }$$  away from a wall produces a sound and receives the reflected sound.
(a) Calculate the time after which he receives the reflected sound if the speed of sound in air is  $$350 { m }  { s } ^ { - 1 }.$$ 
(b) Will the man be able to hear a distinct echo ? Explain .

For echo, total distance travel by sound = $$2D$$
(a) Velocity $$=\dfrac{2D}{Time}$$
      
     Time $$= \dfrac{2 \times 25}{350} = 0.143 \ seconds$$

(b) Yes, because the reflected sound reaches the man $$0.1$$ second after the original sound is heard and the original sound persists only for $$0.1$$ second.


What are wavelength, frequency, time period and amplitude of a sound wave?

1. Wavelength : The distance travelled by the wave during one complete oscillation is called wavelength of the wave.
2. Frequency : The number of oscillations made by the wave in one second is called the frequency of the wave.
3. Time period : The time taken by the wave to complete one oscillation is called the amplitude of the wave.
4. Amplitude : The maximum displacement of the wave from its mean or equilibrium position is called the amplitude of the wave.


A person can hear sound waves in the frequency range $$20\ Hz$$ to $$20\ kHz$$. Find the minimum and the maximum wavelengths of sound that is audible to the person. The speed of sound is $$360\ m/s$$.

The relation between wavelength $$\lambda$$ frequency $$\nu$$ and speed $$v$$ is given by:
$$v=\nu\times \lambda$$
$$\nu=\frac{v}{\lambda}$$
$$\lambda=\frac{v}{\nu}$$

For maximum wavelength $$\nu=20\ Hz$$.
as $$\left(\nu \propto \dfrac{1}{\lambda}\right)$$

$$\therefore \lambda_{max} =$$ $$\frac{v}{\nu}$$ = $$\frac{360}{20}$$ =$$18m$$

For minimum wavelength, $$\nu=20\ kHz$$

$$\Rightarrow \lambda_{mini}=$$ $$\frac{v}{\nu}$$ = $$\frac{360}{20 \times 10^3}$$ = $$18 \times 10^{-3}m$$


Describe a simple experiment to show that the prongs of a sound-producing tuning fork are vibrating.

Fill the beaker with water till its brim. Now touch the surface of the water with the prongs of the vibrating tuning forks. It can be seen that water starts to splash. This shows that the tuning fork is vibrating.
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(a) What is an echo?
(b) How is an echo different from a reverberation?
(c) Describe an activity to demonstrate that sound follows the same laws of reflection as light.

$$(a)$$ The repetition of the sound, which is reflected from a high building or any such surface, is called an echo. An echo can be heard only when the distance between the source of sound and the reflecting body is at least $$17\ m$$.
Figure below illustrates it more clearly.
For Example when a person shouts in a big empty hall listens to his sound repeatedly.
$$(b)$$ Following are the difference between echo and reverberation.

                              $$Echo$$                 $$Reverberation$$
1. Echo is a single reflection of a sound wave off a surface
1. Reverberation is the sound or the pattern created by the superposition of such echoes.

2. An echo can be heard only when the distance between the source of sound and the reflecting body is at least $$17\ m$$
2. Reverberation can occur when sound wave is reflected by a nearby wall also.
3. An echo is usually clear and can be clearly distinguished
3.Reverberation is not a clear replica of the original sound sample.

4. Echo can be used to determined the distance of a reflecting object such as a large building or a mountain, if the ambient temperature is known
4. Reverberation cannot be utilized for distance measurement applications
5. An echo can be heard both in open and closed spaces
5.Reverberation is usually experienced in closed spaces with multiple reflecting objects.
$$(c)$$ The laws of reflection are the following:
1. The angle of incidence is equal to the angle of reflection.
2. The incident ray, the normal at the point of incidence and the reflected ray all lie in the same plane.

Activity:
Material required : a drawing board, a white sheet of paper, a few common pins, a protractor, pencil, eraser and a plane mirror.

Method:
1. Fix the white sheet of paper firmly on the drawing board. Place the plane mirror on it and trace its outline on the paper.
2. Remove the mirror and draw the normal.
3. Now place the mirror again on the outline. The normal will be reflected clearly on the mirror.
4. Next place two pins in a straight line on one side of the normal on the white sheet of paper.
5. Next place two pins on the other side of the normal in such a way that these two pins is in a straight line with the reflection of the two pins on the other side of the normal.
6. Now remove the mirror and the pins and join the pin marks to the normal.
7. Measure the angle of incidence and the angle of reflection. Both will be equal, proving the first law of reflection.

Result : since the lines representing the normal, the incident ray and the reflected ray are all represented on the same sheet of paper, the second law is also verified.

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 a) What are longitudinal waves and transverse waves? Explain with the help of labelled diagrams.

b) Give two examples each of longitudinal waves and transverse waves.

a) In longitudinal waves, the particles vibrate in the back and forth in the same direction. These waves can be produced in solids, liquids, and gases.

Transverse waves have particles vibrating perpendicular to the direction of wave propagation. These waves can be produced in solids and liquids. 


b) Examples of transverse wave:

  1. Waves moving in the spring when pushed and pulled at one end. 
  2. Striking the strings of guitar have transverse waves.
  3. Light is a transverse wave.

Examples of longitudinal wave:

  1. Production of ripples on the surface of the water.
  2. Waves production when the spring is move up and down have transverse waves. 
  3. Sound is a longitudinal wave.


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If a wavelength is much larger than the diameter of a loudspeaker, the speaker radiates in all directions, much like a point source. On the other hand, if the wavelength is much smaller than the diameter, the sound travels in an approximately straight line in front of the speaker. Find the frequency of a sound wave that has a wavelength ten times the diameter of a $$30\ cm$$ speaker. Take the speed of sound to be $$340\ m\ s^{-1}$$.

The wavelength of sound is ten times the diameter of a $$30 \ cm$$ speaker.
$$λ = 10 \times d_{speaker}  = 10 \times 30 =300 \ cm= 3 \ m$$

The speed $$v,$$ frequency $$f$$ and wavelength $$\lambda$$ are related as:
$$v = f  \times λ$$ 
$$\Rightarrow f=\dfrac{v}{\lambda}$$
Given:
Speed $$v=340 \ m/s$$
So
Frequency $$ f = \dfrac{340 \ m/s}{3 \ m} = 113.33$$ sec$$^{-1}$$ $$=113.33 \ Hz$$ 


a) How is sound produced? Explain with the help of an example.

b) How does sound from a sound producing body travel through air to reach our ears? Illustrate your answer with the help of a labelled diagram.

a) Sound is produced when the object vibrates. For example, the production of sound from the strings of the guitar is possible because when the strings of the guitar are struck, there is vibrations generation which travels through air molecules and reaches our ear as sound.

b) Sound from a sound producing body travel through air to reach our ears is explained with the help of a tuning fork. When a tuning fork is struck against a hard surface, there is production of vibrations that travel through air. These waves are longitudinal and move in the same direction as the movement of the sound. This is how the sound reaches to our ears.


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What is reverberation ? What will happen if the reverberation time in a big hall is too long ? How can we reduce it ?

Reverberation : The repeated reflection of sound waves that results in the persistence of sound for some time is called reverberation. 
  • In a big hall or an auditorium, along reverberation is highly undesirable. It makes the sound distorted, blur red and confusing.
Methods for reducing reverberation:
  1. To reduce the undesirable effects due to reverberation, roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies. 
  2. The seat materials are also selected, having sound absorbing properties, Carpets are used to absorb the sound. Curtains are used on doors and windows to absorb the sound.


Establish the relationship between speed of sound, its wavelength and frequency. Velocity of sound in air is $$340 m/s^{-1}$$. Calculate frequency when wavelength is $$0.85\ m$$.

The speed of sound is defined as the distance which a point on a wave, such as a compression or a rarefaction, travels per unit time.
We know,
speed, $$v = distance / time$$
$$v=\dfrac \lambda T$$
Here $$\lambda$$ is the wavelength of the sound wave. It is the distance traveled by the sound wave in one time period $$(T)$$ of the wave. Thus,
$$v = \lambda f$$
where $$f=$$ frequency
$$f=\dfrac 1T$$

Given,
Speed of sound in air, $$v=340\ m/s$$
Wavelength, $$\lambda=0.85\ m$$
$$v = \lambda f$$
$$\implies f=\dfrac{v}{\lambda}$$

$$f=\dfrac{340}{0.85}$$

$$f=400\ Hz$$


Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger then the diameter of the loudspeaker.
Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is $$20\ cm$$. Take the speed of sound to be $$340\ m\ s^{-1}$$

Given,
Diameter of the speaker, $$D= 20\ cm=0.2\ m$$
The wavelength of sound in air is ten times the diameter of the speaker
$$\therefore \  \lambda =10\times D$$
$$\  \lambda =10\times 0.2$$
$$ \lambda =2\ m$$
Velocity of the sound, $$v=340\ m/s$$
frequency, $$f$$

We know,
$$v=\lambda f$$

$$f=\dfrac{v}{\lambda}$$

$$f=\dfrac{340}{2}$$

$$f= 170\ Hz$$


What is a longitudinal wave?

Longitudinal waves - In these waves the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles do not move from one place to another but they simply oscillate back and forth about their position of rest. 
Sound waves are longitudinal waves.

Longitudinal waves can be produced in solids, liquids as well as gases. 


What is a transverse wave? In which medium: solid, liquid or gas, can it be produced?

In a transverse wave particles do not oscillate along the line of wave propagation but oscillate up and down about their mean position as the wave travels. Thus a transverse wave is the one in which the individual particles of the medium move about their mean positions in a direction perpendicular to the direction of wave propagation.
Light waves are transverse waves.

Transverse waves can only be produced in solids and on the surface of liquids. They cannot be produced inside liquids and in gases.


What is longitudinal wave?

  • when particle of a wave in the medium start oscillating about their mean position (back and forth) in the direction in which sound travels is called longitudinal wave.

  •  solids, liquids and gases can produce such type of waves.

 




State two applications of echo.

The applications of echo: 
(i) Dolphins detect their enemy and obstacles by emitting the ultrasonic waves and hearing their echo.             
(ii) In medical science, the echo method of ultrasonic waves is used for imaging the human organs such as the liver, gall bladder, uterus, womb etc. This is called ultra sonography .


State the factors that determine:
(a) the pitch of a note.
(b) the loudness of the sound heard.
(c) the quality of the note.

Part (a): Frequency determines the Pitch of a note.

Pitch is the sensation (brain interpretation) of the frequency of an emitted sound. 

  • Faster the vibration of the source, higher is the frequency and higher is the pitch. Similarly, low pitch sound corresponds to low frequency.



Part (b): Amplitude of waveform determines the loudness of the sound heard.

The loudness of a sound depends on the amplitude of vibration of the vibrating body producing sound. Greater the amplitude of vibrations, louder is the sound produced.

  • Soft sound notes (faint) have smaller Amplitude.

  • Loud sound has larger Amplitude (noise amplitude) 



Part (c): Waveform determines the quality of the note.

  • Quality or timbre is a characteristic of a sound which depends upon the waveform of sound waves produced by instruments.




Explain the meaning of terms compression and rarefaction in relation to a longitudinal wave.

Compression - When a vibrating object advances, it pushes and compresses the air that is in front of it, thus creating a region of high pressure, the region is called compression C as observed in the figure. The compression begins to drift away from the object that vibrates. When the vibrating object moves backwards, it creates an area of low pressure known as rarefaction as observed in the figure.
Rarefaction is the region of low density, wherein the particles of the medium move away from each other, compression on the other hand, are the regions of high density where the particles of the medium are very close to each other.
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Which property of sound leads to the formation of echo ? Briefly explain.

Reflection of sound leads to the formation of echo. When a sound is repeatedly reflected from a number of obstacles, such as two parallel distant buildings or hills, more than one echo can be heard.


By what name is the outer ear called ? What is its function ? Where are the vibrations in the eardrum amplified the ear ?

The outer ear is called pinna. It collects the sound from the surroundings and passes on to the eardrum through the auditory canal. The vibrations in the eardrum are amplified by Three bones, (the hammer, anvil and stirrup) in the middle ear.


With a diagram explaining the use of curved sound board behind the speaker in conference halls ?

The arched ceiling and walls of large halls often reflect the sound waves. These reflected sound waves interfere with the words of the speaker. This problem is solved by hanging curtains, putting up screens or by using soundboards. A soundboard is often a concave rigid surface. The speaker is located at the focus of the sound board, placed behind the speaker. The soundboard renders the reflected sound waves parallel. This enables the sound to reach large distance. The sound board prevents the spreading out of the sound waves in a different direction.
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A stone dropped from the top of a tower of height $$ 300 \ {m} $$ high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is $$ 340 \ {m} \ {s}^{-1} ?\left(\ {g}=9.8 \ {ms}^{-2}\right) $$

Time taken to hear the splash at the top = Time taken by the stone to reach the pond + Time taken by the sound to travel to the top from the base of tower, 
$$ \mathrm{ie}_{.,} \mathrm{T}=\mathrm{T}_{1}+\mathrm{T}_{2} \ldots \ldots . .(1) $$
$$ \mathrm{T}_{1}: $$ 
We know that $$ \mathrm{s}=\mathrm{ut}+1 / 2 \mathrm{a} \mathrm{t}^{2} $$
$$ \mathrm{u}=0, \mathrm{s}=300 \mathrm{m}, \mathrm{a}=\mathrm{g}=9.8 \mathrm{ms}^{-2} $$
$$ \Rightarrow \mathrm{t}=\sqrt{\dfrac{2 \mathrm{s}}{\mathrm{a}}}=\sqrt{\dfrac{2 \times 300}{9.8}}=7.82 \mathrm{s} $$
$$ \therefore T_{1}=7.82 \mathrm{s} $$
$$ T_{2}: \mathrm{s}=v \mathrm{t} \Rightarrow \mathrm{t}=\dfrac{\mathrm{s}}{\mathrm{v}}=\dfrac{300}{340}=0.88 \mathrm{s} $$
$$ \therefore \mathrm{T}_{2}=0.88 \mathrm{s} $$
$$ \therefore \mathrm{T}_{1}+\mathrm{T}_{2}=7.82+0.88=8.70 \mathrm{s} $$


A person who bursts a cracker hears its echo after $$1 s$$. What is the distance to the reflecting surface if the speed of sound in air is $$340 m/s$$ ?


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How is sound produced in human body?

In humans, sound is produced by the the vocal cords which are attached within the larynx, present in the upper part of human body. They produce sound when they get stretched and the opening becomes narrower. They then vibrate as air passes through them during exhalation of air from lungs.


Is a sound wave with wavelength $$1.32\ cm$$ and wave velocity $$330\ ms^{-1}$$ audible to a human ear? Why?

Given,
Wavelength = 1.3 cm 
$$\Rightarrow \lambda = 1.3 \times 10^{-2} \space m$$
Velocity of sound = 330 m/s

$$\therefore$$ Frequency = $$\dfrac {Velocity \space of \space sound} {Wavelength}$$

$$\therefore$$ Frequency = $$\dfrac {340} {1.3 \times 10^{-2}}$$ Hz

$$\therefore$$ Frequency = $$26153.8 \space Hz > 20000 \space Hz$$, the maximum range of audible frequency.
Sound of such high frequency cannot be heard by the human ear. 


The product of frequency and wavelength of a sound wave is called ______.

The relation between velocity$$(v)$$, wavelength$$(\lambda)$$ and frequency$$(f)$$ of a wave-
$$v = f \lambda$$
So, the product of frequency and wavelength of a sound wave is called velocity.


With respect to the reflection of sound, what are the problems if the distance between the walls in a room is more than $$17.2\ m$$?

The sensation of sound persists in our brain for about $$0.1\ s$$. To hear a distinct echo the time interval between the original sound and the reflected one must be at least $$0.1\ s$$. 
Speed of sound in air, $$v=340\ m/s$$
Hence, the total distance covered by the sound from the point of generation to the reflecting surface and back should be at least, 
$$d=v\times t$$
$$d=340\times 0.1$$
$$d=34\ m$$
Thus, for hearing distinct echoes, the minimum distance of the obstacle from the source of sound must be half of this distance, 
$$D=\dfrac d2$$

$$D=\dfrac{34}{2}$$

$$D=17\ m$$
If the distance between the room is more than $$17\ m$$, the sound created in the room will persist by repeated reflection from the walls until it is reduced to a value where it is no longer audible. The repeated reflection that results in this persistence of sound is called reverberation. Excessive reverberation is highly undesirable.


Explain an experiment showing sound reflection.

There are two identical card board tube $$P$$ and $$Q$$ with metal surface $$AB$$ and screen $$S$$ between the two tube as shown in figure. Keep a small clock at the other end of the tube $$P$$ and ear at the end of the tube $$Q$$. The tick-tack of the clock can be heard if the angle are same.Which denotes that sound of clock is reflected at an angle from surface$$AB$$ and reach the ears.
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A hospital uses an ultrasonic scanner to locate tumors in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is $$ 1.7 \ km/s$$? The operating frequency of the scanner is $$4.2 \ MHz.$$

Given: 
Frequency $$ f=4.2 \ MHz= 4.2 \times 10^{6} \ Hz$$
Speed $$ v=1.7 \ km/s=1.7 \times 10^{3} m/s$$

We know that, wavelength $$\lambda,$$ speed $$v$$ and frequency $$f$$ are related as:
$$ \lambda=\dfrac{\mathrm{v}}{\mathrm{f}} $$

$$ =\dfrac{1.7 \times 10^{3}}{4.2 \times 10^{6}}=4.05 \times 10^{-4} \mathrm{m} $$

$$ \therefore $$ wavelength of sound in tissue $$ =4.05 \times 10^{-4} \mathrm{m} $$


A sound signal of $$50 kHz$$ is sent to the bottom of a sea. It returned back after $$4s$$. The velocity of sound in seawater is $$1500 m/s$$. Then,
(a) Calculate the depth of the sea.
(b) What is the wavelength of the wave?

(a) Given:
Velocity of sound in seawater $$v=1500 \ m/s$$
Time taken by sound to return $$t= 4 \ s$$
Suppose the depth of the sea $$=d$$
Then, distance travelled by the wave $$= 2d$$

Distance  $$(2d)=velocity \ (v) \times time \ (t)$$
                         $$= 1500 \times 4 = 6000m$$
$$\therefore$$ Depth $$d= \dfrac{6000}{2}$$
                   $$= 3000 m$$.

b) Relation between velocity $$v$$, frequency $$f$$ and wavelength $$\lambda$$ is:
$$v = f \lambda$$ 
$$\Rightarrow \lambda \ (in \ m)=\dfrac{v \ (in \ m/s)}{f \ (in \ Hz)}$$
Given:
$$v = 1500 \ m/s$$,
$$f=50 \ kHz=50000 \ Hz$$

$$\lambda =\dfrac{1500}{50000}=0.03m$$.


A pendulum has a frequency of $$4$$ vibrations per second. An observer starts the pendulum and fires a gun simultaneously. He hears the echo from the cliff after $$6$$ vibrations of the pendulum. If the velocity of sound in air is $$340 m s^{-1}$$, find the distance between the cliff and the observer.

Hint: 
speed is $$\frac{Distance}{Time}$$ formula is used.
$$\textbf{Step1:Time taken to complete 6 vibrations}$$
$$4$$ vibrations completed by pendulum in $$1 sec$$
So $$6$$ vibrations completed in $$6/4$$ seconds $$= 1.5 sec $$
$$\textbf{Step2:Distance berween cliff and observer}$$
Total distance is $$2D$$ to hear an echo.
$$\Rightarrow$$Speed  $$= 2 × D/ time$$ 
   $$340 = 2 × D/ 1.5 $$
   $$D = 340 × 1.5 / 2 = 255 m$$
The distance between the cliff and the observer is $$255 m$$


What happens to the ultrasonic waves after striking the object on the seabed?

Ultrasonic waves that are reflected back after striking the object reach the detector. The detector converts the ultrasonic waves into electrical signals.


A room contains air in which the speed of sound is $$ 343 \mathrm{m} / \mathrm{s} . $$ The walls of the room are made of concrete in which the speed of sound is $$ 1850 \mathrm{m} / \mathrm{s} $$.
"A bare concrete wall is a highly efficient mirror for sound." Give evidence for or against this statement.

Most of the energy gets reflected. The concrete wall reflects almost all the sound waves. Suppose a person is standing in front of a wall and starts clapping. Sound waves produced have energy with them; all the energy will not get transmitted through the wall. Sound waves will reflect from the wall. An echoing sound may be heard. So, the concrete wall is a highly efficient mirror for sound


A sound was produced. Three seconds later, its echo was heard.
a) What is the distance between the reflecting surface and source? (Speed of sound in air is $$340\ ms^{-1}$$)
b) Write two methods, by which you can reduce the harmful effects of reflection of sound in big halls.

a) Distance of the reflecting surface
$$\text{distance}= Velocity \times time$$
$$\Rightarrow d= 340 \times 3 $$
$$\Rightarrow d = 1020\ m$$
So, the distance between the source and reflecting surface
$$D = \dfrac d2$$
$$\Rightarrow D= \dfrac{1020}{2}$$
$$\Rightarrow D = 510\ m$$.

b) Make the walls rough, provide a large number of ventilators.
Make ceiling curved.


A sound signal from a ship floating on water hits a rock at the bottom of the sea and comes back to the ship after $$4s$$. Calculate the distance of the rock from the surface of water. The speed of sound in water is estimated to be $$1500 m/s$$.

$$V = s/t$$
Velocity of sound $$= v = 1500 m / s$$.
Time, $$t = 4s$$.
Distance travelled, $$S = v * t =1500 * 4 = 6000 m$$
$$= 6000 m$$.
Distance of the rock from the surface of water $$= 6000/2 m$$
$$= 3000 m$$.


Class 9 Physics Extra Questions