NCERT Solutions for Class 9 Science Chapter 12 Sound

1. What is sound and how is it produced?

Answer: Sound is produced due to the vibration. When a body vibrates, it allows the the surrounding particles of medium to vibrate. This causes disturbance in the medium and travels to the ear in the form of waves. This disturbance produces sound.

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

Answer:

A vibrating body creates high pressure region in its surrounding, when it moves forward. This region of high pressure is known as compression. When vibrating body moves backward, it creates low pressure region in its surrounding. This region is known as rarefaction. During vibration body continuously moves forward and backward, thereby creating simultaneous compressions and rarefactions. This can be well understood from figure given below.

3. Cite an experiment to show that sound needs a material medium for its propagation.

Answer: To demonstrate that sound need material medium for its propagation, we will use bell jar method.Arrange the setup as shown in the figure.

Now allow the bell to ring. One can easily hear the sound of the bell. Now attach the vacuum pump to the neck of jar such that air in the jar is removed. While doing this keep the bell ringing. One can observe that as the air inside the jar is removed out, sound of the bell decreases and at a point sound of bell ultimately stops. It is observed that bell keep on ringing, but sound of bell cannot be heard. This happens because air inside the jar is removed. This suggest that sound requires medium to travel.

4. Why is sound wave called a longitudinal wave?

Answer: Sound waves show same mechanism as that of longitudinal waves as particles of medium travels parallel to direction of propagation of sound waves.

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

Answer: Characteristics that helps us to identify person by the voice while sitting in dark room is quality or timbre. Quality of sound produced by two persons is always different.

6. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Answer: During thunderstorm, flash and thunder are produced simultaneously. Light travels faster than sound. Thus, flash is always seen few seconds before the sound is heard.

7. A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m/s.

Answer: Given : ?₁ = 20 Hz

?₂ = 20 kHz = 20,000 Hz

v = 344 m/s

Speed of sound wave is given by the expression,

v = λ x ?

For ?₁ = 20 Hz,

λ₁ = v / ?₁= 344 / 20 = 17.2 m

For ?₂ = 20,000 Hz

λ₂ = v / ?₂ = 344 / 20,000 = 0.0172 m

Wavelengths of sound waves in air corresponding to two frequencies are 17.2 m and 0.0172 m.

8. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

Answer: Let l be the length of the rod.

Speed of sound in aluminium rod at 25 °C, v_al = 6420 m/s

Let t_al be the time taken to reach other end.

t_al = l / v_al = l / 6420

Speed of sound in air at 25 °C, v_air = 346 m/s

Therefore, time taken by soundwave to reach other end,

t_air = l / v_air = l / 346

∴ The ratio of sound wave in air and aluminium

t_air / t_al = l / 346 / l / 6420 = 6420 / 346 = 18.55

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

Answer: Frequency is expressed as-

Frequency = number of oscillations / time

∴ Number of oscillations = frequency x time

Vibrations = number of oscillations = frequency x time

= 100 x 1 min = 100 x 60 s = 6000 vibrations

Thus, the source vibrates 6000 time in one minute.

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

Answer: Sound follows same laws of reflections as that of light. Angle of incidence of sound wave is same as angle of reflection of sound wave. Incident wave, reflected wave and normal to the point of incident all lie in same plane.

11. 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 remains the same. Do you hear echo sound on a hotter day?

Answer: When time interval between original sound and echo is 0.1 s or more, then one can hear echo of the sound. With increase in temperature, speed of sound also increases for a given medium. Thus, time interval between original sound and reflected sound decreases during hot days. Therefore, to hear an echo, it is necessary that time interval between original sound and echo should be greater than 0.1 s.

12. Give two practical applications of reflection of sound waves.

Answer: Following are the applications of reflection of sound waves.

• Reflection of sound is used in SONAR, which finds out distance and speed of objects underwater.
• Stethoscope used by doctors, receives the sound from body organs, by multiple reflections, and finally reaches to doctors ear.

13. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m/s2 and speed of sound = 340 m/s.

Answer: Given – Height of the tower s = 500 m

Velocity of sound v = 340 m/s

Acceleration due to gravity, g = 10 m/s2

Initial velocity u of the stone = 0

Let t1 be the time taken for the stone to fall to the base of the tower.

Using second equation of motion,

s = ut₁² + ½ gt₁²

500 = 0 + ( ½ ) 10 x t₁²

1000 = 10t₁²

t₁² = 100

t₁ = 10 s

Let t₂ be the time taken by the sound to reach the top of the tower from its base.

t₂ = 500 / 340 = 1.47 s

Thus, splash is heard after time t

Where, t = t₁ + t₂ = 10 + 1.47 = 11.47 s

14. A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Answer: Given – Speed of sound waves = v = 339 m/s

Wavelength of sound waves λ = 1.5 cm = 0.015 m

Frequency of sound waves = ? = ?

As we know,

Speed of sound waves = Wavelength x Frequency

∴ Frequency( ? ) = Speed of sound waves ( v ) / wavelength( λ )

= 339 / 0.015

= 22600 Hz

We know that audible range for humans is 20 Hz to 20000 Hz. Obtained frequency is above the audible range, therefore, sound waves will not be audible.

15. What is reverberation? How can it be reduced?

Answer: Reverberation is continuation of sound due to multiple reflection. Consider a source producing sound in a room. Sound waves travels in all direction and reaches the wall, where it gets reflected back. These reflected waves again strikes to other walls and are reflected back again. This causes hearing of sound even when the source has stopped producing the sound. Reverberation is generally found in big halls. To reduce reverberation, sound absorbent like curtains, fibre board, etc are used.

16. What is loudness of sound? What factors does it depend on?

Answer: High energy sound is called loudness of sound. It depends on amplitude of vibrations. Loudness of sound is directly proportional to square of amplitude of vibration.

17. Explain how bats use ultrasound to catch a prey.

Answer: Bats emit ultrasound to detect and catch prey. This sound gets reflected from prey present in surrounding and reaches to bat’s ear. Intensity of reflected sound helps the bat to judge distance between bat and prey.

18. How is ultrasound used for cleaning?

Answer: Ultrasound is passed through parts of object which are out of human reach. Waves removes the dirt and cleans the object.

19. Explain the working and application of a sonar.

Answer:

Working: SONAR technique is also known as echo ranging. SONAR consist of transmitter and receiver. Transmitter transmits ultrasound waves. Waves strikes with objects under water or seabed and reflects back. Reflected waves are received by receiver, and gets converted to electrical signals. Later signal are inspected to know underwater condition.

Total distance (D) travelled by ultrasound is = 2 × depth of sea (d) = 2 d

We know that, speed = distance (D) / time (t)

∴ v = 2d /t

∴ Depth of sea is d = ( v × t ) / 2

Applications: SONAR is used to detect underwater objects, to measure the depth of sea, locating submarines underwater, sunken ships, etc.

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

Answer: Given – Time t = 5 s

Distance d of object from the submarine = 3625 m

Also, distance travelled by sonar waves = 2d

∴ Velocity of sound in water = v = 2d/t = ( 2 x 3625 ) / 5

v = 1450 m/s

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

Answer: Ultrasound is passed through the metal block, if there is presence of defect then the sound waves are reflected back. From One end of the metal block ultrasound is passed and at other end detectors are placed. Defective area of the block doesn’t allow the passage of ultrasound. Therefore, detector cannot detect the waves which helps to easily locate the defect in the block.

22. Explain how the human ear works.

Answer:

Human ear is divided into 3 parts, outer ear, middle ear and inner ear.

• Outer Ear: Pinna collects sound from surrounding and passes through auditory canal which further joins the eardrum.
• Middle Ear: When sound waves reaches eardrum, it pushes eardrum inward and forms compression, when eardrum moves outward it forms rarefactions. This creates vibration of ear drum. Vibrations are amplified by hammer, anvil and stirrup.
• Inner Ear: Amplified signals are passed to cochlea, which converts it into electrical signals. These signals are passed to brain through auditory nerve.

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