CBSE - Sound

When an object vibrates, it sets the particles of the medium around it vibrating. The particles in the medium in contact with the vibrating object displace from its equilibrium position. It then exerts force on the adjacent particles. After displacing the adjacent particle the first particle of medium comes back in its original position. This process continues in the medium till the sound reaches your ear.

When the bell continues to move forward and backward, it creates a series of compressions and rarefactions making production of sound.

Sound waves needs material medium to propagate therefore, they are called mechnical waves. Sound waves propagate through a medium because of theinteraction of the particles present in that medium.

No, because sound waves needs a medium through which they can propagate. Since there is no material medium on the moon due to absence of atmosphere, you cannot hear any sound on the moon.

(a) Amplitude

(b) Frequency

Guitar has a higher pitch than car horn, because sound produced by the strings of guitar has high frequency than that of car horn. High the frequency higher is the pitch.

→ Wavelength: The distance between two consecutive compressions or two consecutive rarefactions is known as the wavelength. Its SI unit is metre (m).

→ Frequency: The number of complete oscillations per second is known as the frequency of a sound wave. It is measured in hertz (Hz).

→ Amplitude: The maximum height reached by the crest or trough of a sound wave is called its amplitude.

Speed, wavelength, and frequency of a sound wave are related by the following equation:
Speed (v) = Wavelength (λ) x Frequency (ν)
v = λ x ν

Frequency of the sound wave, ν= 220 Hz
Speed of the sound wave, v = 440 m s^-1
For a sound wave,
Speed = Wavelength x Frequencyv = λ x ν
∴ λ= v / ν = 440 / 220 = 2m
Hence, the wavelength of the sound wave is 2 m.

The time interval between two successive compressions is equal to the time period of the wave. This time period is reciprocal of the frequency of the wave and is given by the relation:T= 1 / Frequency = 1/ 500 = 0.002 s

Intensity of a sound wave is defined as the amount of sound energy passing through a unit area per second. Loudness is a measure of the response of the ear to the sound. The loudness of a sound is defined by its amplitude. The amplitude of a sound decides its intensity, which in turn is perceived by the ear as loudness.

The speed of sound depends on the nature of the medium. Sound travels the fastest in solids. Its speed decreases in liquids and it is the slowest in gases. Therefore, for a given temperature, sound travels fastest in iron

Speed of sound, v = 342 m s⁻¹
Echo returns in time, t = 3 s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
In the given time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source.
Hence, the distance of the reflecting surface from the source= 1026 / 2 m = 513 m.

Ceilings of concert halls are curved so that sound after reflection (from the walls) spreads uniformly in all directions.

The audible range of an average human ear lies between 20 Hz to 20,000 Hz.

(a) Infrasound has frequencies less than 20 Hz.
(b) Ultrasound has frequencies more than 20,000 Hz.

Time taken by the sonar pulse to return, t = 1.02 s
Speed of sound in salt water, v = 1531 m s - 1
Distance of the cliff from the submarine = Speed of sound x Time taken
Distance of the cliff from the submarine = 1.02 x 1531 = 1561.62 m

Distance travelled by the sonar pulse during its transmission and reception in water = 2 x Actual distance = 2d

Actual Distance, d= Distance of the cliff from the submarine/2
= 1561/2
= 780.31 m

Sound is a form of energy which gives the sensation of hearing. It is produced by the vibrations caused in air by vibrating objects.

When a vibrating body moves forward, it createsa region of high pressure in its vicinity. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure in its vicinity. This region is known as a rarefaction. As the body continues to move forward and backwards, it produces a series of compressions and rarefactions. This is shown in below figure. 

Take an electric bell and an air tight glass bell jar connected to a vacuum pump. Suspend the bell inside the jar, and press the switch of the bell. You will be able to hear the bell ring. Now pump out the air from the glass jar. The sound of the bell will become fainter and after some time, the sound will not be heard. This is so because almost all air has been pumped out.This shows that sound needs a material medium to travel.

Sound wave is called longitudinal wave because it is produced by compressions and rarefactions in the air. The air particles vibrates parallel to the direction of propagation.

The quality or timber of sound enables us to identify our friend by his voice.

The speed of sound (344 m/s) is less than the speed of light(3 x 10⁸ m/s). Sound of thunder takes more time to reach the Earth as compared to light. Hence, a flash is seen before we hear a thunder.

For a sound wave,
Speed = Wavelength x Frequencyv = λ x ν
Speed of sound in air = 344 m/s (Given)
(i) For, ν= 20 Hz
λ₁= v/ν = 344/20 = 17.2 m

(ii) For, ν= 20000 Hz
λ₂= v/ν = 344/20000 = 0.172 m

Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

Velocity of sound in air= 346 m/s
Velocity of sound wwave in aluminium= 6420 m/s
Let length of rode be 1

Time taken for sound wave in air, t₁= 1 / Velocity in air
Time taken for sound wave in Aluminium, t2= 1 / Velocity in aluminium

Therefore, t1 / t2 = Velocity in aluminium / Velocity in air = 6420 / 346 = 18.55 : 1

Frequency = 100 Hz (given)
This means the source of sound vibrates 100 times in one second.
Therefore, number of vibrations in 1 minute, i.e. in 60 seconds = 100 x 60 = 6000 times.

Sound follows the same laws of reflection as light does. The incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Also, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane.

An echo is heard when the time for the reflected sound is heard after 0.1 s
Time Taken= Total Distance / Velocity
On a hotter day, the velocity of sound is more. If the time taken by echo is less than 0.1 sec it will not be heard.

Two practical applications of reflection of sound waves are:
→ Reflection of sound is used to measure the distance and speed of underwater objects. This method is known as SONAR.
→ Working of a stethoscope is also based on reflection of sound. In a stethoscope, the sound of the patient's heartbeat reaches the doctor's ear by multiple reflection of sound.

Height of the tower, s = 500 m
Velocity of sound, v = 340 m s⁻¹
Acceleration due to gravity, g = 10 m s⁻²
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower, t₁
According to the second equation of motion:

Now, time taken by the sound to reach the top from the base of the tower, t2= 500 / 340 = 1.47 s
Therefore, the splash is heard at the top after time, t
Where, t= t1 + t2 = 10 + 1.47 = 11.47 s

Speed of sound, v= 339 m s - 1
Wavelength of sound, λ= 1.5 cm = 0.015 m
Speed of sound = Wavelength x Frequencyv= λ x v
∴ v= v / λ = 339 / 0.015 = 22600 Hz
The frequency range of audible sound for humans lies between 20 Hz to 20,000 Hz. Since the frequency of the given sound is more than 20,000 Hz, it is not audible.

The repeated multiple reflections of sound in any big enclosed space is known as reverberation.
The reverberation can be reduced by covering the ceiling and walls of the enclosed space with sound absorbing materials, such as fibre board, loose woollens, etc.

The effect produced in the brain by the sound of different frequencies is called loudness of sound.
Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Bats produce high-pitched ultrasonic squeaks. These high-pitched squeaks are reflected by objects such as preys and returned to the bat's ear. This allows a bat to know the distance of his prey.

Objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

SONAR is an acronym for Sound Navigation And Ranging. It is an acoustic device used to measure the depth, direction, and speed of under-water objects such as submarines and ship wrecks with the help of ultrasounds. It is also used to measure the depth of seas and oceans.

Sonar Working

A beam of ultrasonic sound is produced and transmitted by the transducer (it is a device that produces ultrasonic sound) of the SONAR, which travels through sea water. The echo produced by the reflection of this ultrasonic sound is detected and recorded by the detector, which is converted into electrical signals. The distance (d) of the under-water object is calculated from the time (t) taken by the echo to return with speed (v) is given by 2d = v � t. This method of measuring distance is also known as �echo-ranging�.

Time taken to hear the echo, t= 5 s
Distance of the object from the submarine, d= 3625 m
Total distance travelled by the sonar waves during the transmission and reception in water= 2d

Velocity of sound in water, v= 2d / t = 2 x 3625 / 5 = 1450 ms^-1

Defects in metal blocks do not allow ultrasound to pass through them and they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. The defective part of the metal block does not allow ultrasound to pass through it. As a result, it will not be detected by the detector. Hence, defects in metal blocks can be detected using ultrasound.

The human ear consists of three parts � the outer ear, middle ear and inner ear.
→ Outer ear: This is also called �pinna�. It collects the sound from the surrounding and directs it towards auditory canal.
→ Middle ear: The sound reaches the end of the auditory canal where there is a thin membrane called eardrum or tympanic membrane. The sound waves set this membrane to vibrate. These vibrations are amplified by three small bones- hammer, anvil and stirrup.
→ Inner ear: These vibration reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound.

1. Note is a sound
(a) of mixture of several frequencies
(b) of mixture of two frequencies only
(c) of a single frequency
(d) always unpleasant to listen
Ans. (a) of mixture of several frequencies
Explanation: Note is a sound of mixture of several frequencies and is pleasant of hear.

2. A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case
(a) sound will be louder but pitch will not be different
(b) sound will be louder and pitch will also be higher
(c) sound will be louder but pitch will be lower
(d) both loudness and pitch will remain unaffected
Ans. (a) sound will be louder but pitch will not be different
Explanation: The pitch depends on frequency of the particular key which is being hit and hence there would be no change in pitch of sound. Loudness depends on amplitude which will be more if key is struck harder.

3. In SONAR, we use
(a) ultrasonic waves
(b) infrasonic waves
(c) radio waves
(d) audible sound waves
Ans. (a) ultrasonic waves

4. Sound travels in air if
(a) particles of medium travel from one place to another
(b) there is no moisture in the atmosphere
(c) disturbance moves
(d) both particles as well as disturbance travel from one place to another.
Ans. (c) disturbance moves
Explanation: Propagation of sound does not require movement of particles. It only requires movement of disturbance from one particles to the next.

5. When we change feeble sound to loud sound we increase its
(a) frequency
(b) amplitude
(c) velocity
(d) wavelength
Ans. (b) amplitude
Explanation: Loudness of sound depends on amplitude and it increases when amplitude is increased.

6. In the curve (Fig.12.1) half the wavelength is

(a) AB
(b) BD
(c) DE
(d) AE
Ans. (b) BD
Explanation: The distance between two consecutive peaks or two consecutive troughs is called wavelength. The figure shows one peak and one trough. Sum of widths of a peak and trough is also same as wavelength; which is AE in this case. Hence, BD is half the wavelength.

7. Earthquake produces which kind of sound before the main shock wave begins
(a) ultrasound
(b) infrasound
(c) audible sound
(d) none of the above
Ans. (b) infrasound
Explanation: It is due to infrasound that some animals get advanced warning of earthquake and show strange changes in their behaviour.

8. Infrasound can be heard by
(a) dog
(b) bat
(c) rhinoceros
(d) human beings
Ans. (c) rhinoceros
Explanation: Rhinoceros communicate using infrasound of as low frequency as 5 Hz.

9. Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting
(a) intensity of sound only
(b) amplitude of sound only
(c) frequency of the sitar string with the frequency of other musical instruments
(d) loudness of sound
Ans. (c) frequency of the sitar string with the frequency of other musical instruments Explanation: Frequency of a particular musical instrumental should be in tune with that of other musical instruments. This helps in production of pleasant music. That is why artists usually adjust the frequencies at the start of a musical concert.

Ans. From the graph

Time period, T = 2 � 10^-6 s.

Frequency, v = 1/T = 5 � 10⁵ Hz.

Wavelength, λ υ / v =    5 � 10⁵ m .

Which of the two graphs (a) and (b) (Fig.12.3) representing the human voice is likely to be the male voice? Give reason for your answer.

Ans. Graph (a) represents the male voice. Usually the male voice has less pitch (or frequency) as compared to female.

If the time gap between the original sound and reflected sound received by the listener is around 0.1 s, only then the echo can be heard.

The minimum distance travelled by the reflected sound wave for the distinctly listening the echo. = velocity of sound × time interval

= 344 × 0.1

 = 34.4m

But in this case the distance travelled by the sound reflected from the building and then reaching to the girl will be (6 + 6) = 12 m, which is much smaller than the required distance. Therefore, no echo can be heard.

Humming bees produce sound by vibrating their wings which is in the audible range. In case of pendulum the frequency is below 20 Hz which does not come in the audible range.

 Longitudinal waves; because sound waves are longitudinal waves.

 Speed of sound = 340 m/s and time = 10 s
Distance = speed × time = 340 × 10 = 3400 m

For hearing the loudest ticking sound heard by the ear, find the angle x in the Fig.12.4

Ans. Incident line is making an angle of 50^o with reflecting surface.

So, angle of incidence = 90^o � 50^o = 40^o

Angle of reflection = angle of incidence Hence,

<x= 40^o

Ceiling and walls are made curved so that sound after reflection reaches the target audience.

Represent graphically by two separate diagrams in each case

(i) Two sound waves having the same amplitude but different frequencies?

(ii) Two sound waves having the same frequency but different amplitudes.

(iii) Two sound waves having different amplitudes and also different wavelengths.

ANS

Establish the relationship between speed of sound, its wavelength and frequency. If velocity of sound in air is 340 ms^-1, calculate 

(i) wavelength when frequency is 256 Hz.

(ii) frequency when wavelength is 0.85 m.

Ans. Derivation of formula υ = vλ 

(i) 340 = 256  λ

λ = 1.33 m

(ii) 340 = v (0.85)

v = 400 Hz

Wavelength is the distance between two consecutive compressions or two consecutive rarefactions. Time period is the time taken to travel the distance between any two consecutive compressions or rarefactions from a fixed point.