# NCERT Class 9 Sound Notes pdf, NCERT Class 9 Science Notes

In the following section, you will get NCERT Class 9 sound notes pdf. These notes are in the form of questions and answers and have been prepared by experts who have a lot of experience in teaching science at various levels. Check these Class 9 science questions and answers below.

## NCERT Class 9 Sound Questions and Answers pdf

These questions and answers for sound chapter of class 9 science will be very helpful to you for your exam preparation. Check these notes out.

### Explain how sound is produced by your school bell?

When a school bell is hit by a hammer, it begins to vibrate and hence sound is produced.

### Why are sound waves called mechanical waves?

Sound waves are called mechanical waves because they need a material medium (like solid, liquid or gas) for their propagation.

### Suppose you & your friend are on the moon. Will you be able to hear any sound produced by your friend?

No, sound can’t be heard directly on the surface of moon because there is no air on the moon to carry the sound wave.

### Which wave property determines 1, Loudness & 2, Pitch.

Loudness is basically determined by Amplitude, whereas pitch is determined by the frequency of wave.

### Guess which sound has higher pitch: Guitar or Car horn?

The sound of guitar is shriller (very high and loud) than that of a car horn. So, the sound of a guitar has a higher pitch.

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

Wave length:- The distance between two consecutive compressions (C)or two consecutive rarefactions (R) in a longitudinal wave (sound wave) is called the wave length. It is usually represented by Greek letter lambda (λ). Its S.I unit is meter (m).

Frequency:- The number of oscillations per unit time is known as frequency of the sound wave. It is denoted by the Greek letter, nu (v) or by f. Its SI unit is hertz (Hz)

Time period:- The time taken to complete one oscillation is known as time period. It is denoted by ‘T’. Its SI unit is second (s).

Amplitude (A):- The maximum displacement of the particles of the medium from their original undisturbed positions, when a wave passes through the medium, is called the amplitude of the wave. Its SI unit is metre (m).

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

Speed = Distance covered / Time taken

Speed = λ / T ——— (1)

Where λ is wave length of sound and It is the distance travelled by the sound wave in one Time Period (T) of the wave

But frequency f= 1/T,

therefore, (1) becomes Speed = λ f,

i.e. speed = wave length x Frequency

### Calculate the wavelength of a sound wave whose frequency is 220Hz and speed is 440m/s in a given medium?

We know that, speed (v) = wave length (λ) x Frequency (f)

or wavelength (λ) = speed (v)/Frequency (f)

Here v = 440m/s & f=220Hz

Therefore wavelength (λ) = 440/220

or wavelength (λ) = 2m.

### A person is listening to a tone of 500 Hz sitting at a distance of 450m from the source of the sound. What is the time interval between successive compressions from the source?

Time interval between two successive compressions is called Time period (T), and

T= 1/f = 1/500 = 0.0025 s.

### Distinguish between loudness and intensity of sound?

The amount of sound energy passing each second through unit area is called intensity of sound while as loudness is the measure of the response of the ear to the sound.

### In which of three media, air, water or iron, does sound travel the fastest at a particular temperature?

Sound travels fastest through iron with a speed of 5950m/s

### An Echo is returned in 3S. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms-1.

Speed of sound is 342 ms-1

Time taken from hearing the echo = 3s

Distance traveled by the sound = velocity (v) x Time (T)= 342 × 3 = 1026 m.

In 3S, sound has to travel twice the distance between the reflecting surface as the source.

Therefore, distance of the reflecting surface from the source = 1026/2 = 513 m

### Why are the ceilings of the concert halls curved?

The ceilings of the concert halls are curved so that sound after reflection reaches all corners of the hall.

### What is the audible ranges of the average human ear?

The audible range of the average human ear is 20 Hz to 20,000 Hz.

### What is the range of frequencies associated with? 1; Infrasound, 2; Ultrasound

1. Range of frequency of Infrasound is less than 20 Hz.
2. Range of frequency of ultrasound is higher than 20Kz.

### A submarine emits a sonar pulse, which returns from an underwater cliff after 1.02s. If the speed of sound in salt water is 1531ms-1 , how far away is the cliff?

Time between transmission and detection of sonar pulse, t = 1.02 S

Speed of sound in salt water, v = 1531ms-1

Distance of the cliff = d (say)

Then, distance traveled by sound = 2d

So, 2d = speed x time = vt

= 1531 × 1.02

or d= 1531 x 1.02/2

or d = 780.81 m

### What is sound and how is it produced?

Sound is a form of energy which gives sensation of hearing and it is produced due to vibrations of different types of object, e.g, a vibrating tuning fork, a bell, wires of a sitar and a violin etc.

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

When a longitudinal wave travels in a medium, then the particles of the medium vibrate back and forth in the same direction in which the wave travels.

When the vibrating particles come closer to one another than they normally are, then there is a momentary reduction in volume, increase in the density & pressure of the medium and a compression is formed.

On the other hand, when the vibrating particles move farther apart form one another than they normally are, then there is a momentary increase in volume, decrease in the density & pressure and a rarefaction is formed.

### Demonstrate an experiment to show that sound needs a material medium for its propagation?

A material medium is necessary for transmitting sound. This can be shown by the following experiment.

An electric bell is suspended by rubber bands RR’ in air-tight glass vessel V called bell jar.

The bell jar is connected to a vacuum pump to remove the air. Initially, the bell jar is full of air. On pressing the switch, the bell starts ringing & its sound can be heard clearly. Thus, when air is present as the medium in the bell jar, sound can travel through it.

We now pump out the air from the bell jar gradually by switching on the vacuum pump attached to it. the sound of the ringing bell becomes fainter and fainter & ultimately diminishes. It is clear from this experiment that sound needs a material medium to travel.

### Why is sound wave called a longitudinal wave?

A sound wave is called longitudinal wave as it travels in a medium in the form of compressions and rarefactions where the particles of the medium vibrate in a direction which is parallel to the direction of propagation of the sound wave.

### Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

The quality (or tone) of sound is that characteristic which enables us to distinguish one sound from the other even when these are of the same pitch and loudness.

Each person has its own quality of sound & it is this characteristic which enables us to identify a person from others even without looking at him.

### Flash and thunder are produced simultaneously, but the thunder is heard a few seconds after the flash is seen, why?

The speed of light (C) is greater than the speed of sound (V) by a factor of the 106. Thus, the flash of light is seen earlier than the thunder of sound even though both are produced simultaneously.

### A person has a hearing range from 20Hz to 20 kHz. What are the typical wavelength of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 34.4ms’?

Here f1 =20Hz and f2 =20KHz = 20000 Hz.

Speed of sound, v = 340ms-1

Therefore λ1 = v/f1 = 340/20 = 17.2m

& λ2= v/f2 = 340/ 20000 = 0.0172m

### Two children are at opposite ends of an aluminum rod. One strikes the end of the rod with a stone. Find the ratio of time taken by the sound wave in air and in aluminum to reach the send child.

Speed of sound in Aluminum = 6420ms-1

Since time taken by sound to travel a given distance in a medium is inversely proportional to its speed in that medium.

Now, Let ‘x’ be the length of the aluminum rod and let ‘t1‘ and ‘t2‘ be the time intervals taken by sound to reach opposite ends via aluminum rod and air respectively.

Then t1 = x/6420

And t2 = x/340

Now Ratio of time intervals = t2 / t1 = (x/340) / (x/6420)

t2 / t1 = 18.55

### The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Frequency f = 100 Hz; T = 1 minute = 60 seconds

Frequency = No. of vibrations in one second = 100

Now number of vibrations in 1 minute (60 s) = Frequency × Time

Now number of vibrations in 1 minute (60 s) = 100 x 60 = 6000

### Does sound follow the same laws of reflection as light does? Explain?

Like light waves, sound waves also get reflected when these fall on the surface of an obstacle. The following simple experiment establishes that reflection of sound follows the same laws as those for reflection of light.

i) Place a large plane board, AB (of a metal, cardboard or wood) in the vertical position.

ii) Take two hollow metallic tubes P & Q (each about 1m long and about 8 – 10 cm in diameter) and place them in the plane of the paper and in positions inclined to the board as shown in fig

iii) Hold a small watch ‘w’ at the free end of the tube P & try to hear the ticking sound of the watch by positioning the ear at E.

iv)Put a card board screen S in between the two tubes so that the sound produced by the watch does not reach the ear directly.

v) Turn the tube Q till the ticking sound of the clock is the loudest. In this position, it is found that the tubes are inclined to S at the same angle, i.e., ∠i = ∠r.

vi) The normal OS at the surface lies in the same plane as that in which the incident & reflected sound wave lie.

From this experiment, we obtain the following two laws for the reflection of sound waves.

These laws are as follows;

1. The angle of reflection (r) is always equal to the angle of incidence (i) i.e., ∠i=∠r.
2. The incident wave, the reflected wave and the normal (at the point of incidence), all lie in the same plane.

### When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remain the same. Do you hear echo on a hotter day?

The minimum distance (d) for the distinct echo to be heard (say at 27°C) is 17.2m (therefore, 2d= vt=344 x 0.1 = 34.4m).

On a hotter day, the temperature increases & the speed of sound in air also increases e.g., at 40°C, speed of sound, i.e., v=356 m/s & as such 2d=356×0.1 =35.6m or d=17.8m.

Thus, if distance of the reflecting surface and the source of sound remains the same (i.e., 17.2m), no echo is heard on the hotter day as the minimum distance now required is 17.8m.

### Give two practical application of reflection of sound waves.

Give two practical application of reflection of sound waves.

1. Megaphone: sometimes we want a given sound to travel a large distance before it becomes inaudible. This can be done if we avoid the wastage of sound energy by its transmission in all directions.

We, therefore, confine the sound waves with the help of a speaking tube or a megaphone so that they travel in a particular direction. Sound waves which are now confined in a particular region by their multiple reflections from the walls of the tube, travel larger distance than without the help of the tube.

2. Ear Trumpet: It is a sort of machine used by persons who are hard of hearing. The sound energy received by the wide end of the trumpet is concentrated into a much smaller area at the narrow end by multiple reflections, which makes the inaudible sound audible to the user.

### A stone is dropped from the top of a tower 500 m high into a pond of water at the base of tower. When is the splash heard at the top.

Here, height through which the stone falls, h =500m

Speed of sound, v=340m/s, and g=10m/s

If t is the time taken by the stone to fall through h, then h=½gt2

t= (2h/g)1/2

therefore t= (2×500/10)1/2 = 10S.

further, if t’ is the time taken by sound to travel to the top of the tower,

t’ = h/v = 500/340 = 1.47s

therefore time after which the splash is heard at the top of the tower = t+ t’

time after which the splash is heard at the top of the tower = 10 s + 1.47s = 11.47s

### A sound wave travels at a speed of 339 ms-1. If its wavelength is 1.5cm, what is the frequency of the wave? Will it be audible?

Here, speed of sound wave, v = 339m/s

Wavelength of sound wave, λ =1.5cm = 1.5×10-2 m

Therefore frequency of the sound wave, f =v/λ = 339 / 1.5×10-2 =22600 Hz.

The sound is not audible as its frequency lies beyond the audible range (20Hz – 20,00Hz)

### What is reverberation? How can it be reduced?

The phenomenon of persistence or prolongation of audible sound after the source has stooped emitting sound is called reverberation.

Since reverberation is due to repeated reflections of sound waves from the ceiling, floor, walls, etc. of a hall or an auditorium; we can reduce reverberation by increasing the absorption of sound energy as:

1. The walls are covered with some sound absorbing material like felt, fibre board, etc. or by heavy curtains with folds.
2. The floor is carpeted.
3. The furniture is upholstered.
4. False ceiling of a suitable sound absorbing material is used.

### What is loudness of sound? What factors does it depend on?

The sensation produced in the ear which enables us to distinguish between a loud and a faint sound is called loudness.

The loudness depends upon:

1. The amplitude of the wave.
2. The surface area of the vibrating body.
3. The density of the medium, etc.

### Explain how bats use ultrasound to catch a prey?

The ultrasonic waves emitted by the bat are reflected from the prey (e.g., an insect) & are detected by its ear.

The nature of reflected waves tells the bat about;

1. the location, and
2. the nature of its prey.

### How is ultrasound used for cleaning?

The object to be cleaned is placed in a cleaning solution. When ultrasonic waves are passed through the solution, due to their high frequency, particles of dirt, dust and grease get detached even from the unreachable portions of the object & drop out in the solution.

### Explain the working and application of a SONAR?

A sonar is a device which measures the distance, direction and speed of objects lying under water using ultrasonic waves.

A sonar which is installed in a ship or a boat, consists of a transmitter & a detector.

The ultrasonic waves produced by the transmitter travel through water. After getting reflected by the object on the seabed, these waves are picked up by the detector.

The detector converts the reflected ultrasonic waves into electrical signals which are properly recorded.

The sonar technique is used to:

1. Determine depth of the sea, called echo depth ranging.
2. Locate under water hills, valleys, icebergs, sunken-ships.
3. locate the position of other ships or submarines, ship-to-ship communication also uses ultrasonic waves.

### A sonar device on a submarine sends out a signal & receives an echo 5s later. Calculate the speed of sound in water if the distance of the object form the submarine is 3625m.

Here, time interval between the transmission of the signal and its reception, t=5s.

Distance of the object from the submarine, d=3625m

If v is the speed of sound in water, then 2d=vt or v=2d/t = 2×3625/5 =1450m/s

### Explain how objects in a metal block can be detected using ultrasound?

Metal blocks are used in the construction of buildings, bridges, machines and scientific equipment. The cracks or holes within the blocks, which are invisible from outside, reduce the strength of a structure.

To detect these flaws, ultrasonic waves are passed through the metal block. Transmitted waves are detected by detectors. Whereas ultrasonic waves pass through the flawless portions of the block, these are reflected back by even a minor defect & do not reach the detector.

### Explain how the human ear works?

The human ear works by the following five step sequential procedure:

1. The outer ear collects sound waves which are conducted through the auditory canal.
2. These waves fall on the ear drum & set it into vibrations.
3. The middle ear consists of three ear ossicles; hammer, anvil & stirrup, amplifies these oscillations about 60 times.
4. The inner ear which contains cochlea & is filled with a fluid converts these pressure variations into electrical signals.
5. These electrical signals are conveyed to the brain via auditory nerve for interpretation.

So, these were NCERT class 9 Sound notes pdf. We are sure that you will find these NCERT Class 9 science solutions very useful.

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