NCERT Solutions for Class 9 Science Chapter 11 - Sound

Page No. 129

Q1. How does the sound produced by a vibrating object in a medium reach your ear?
Ans: When an object vibrates, it pushes and pulls the neighbouring particles of the medium (for example, air) about their mean positions. These particles repeatedly compress and rarefy, passing the disturbance from one particle to the next as a series of compressions and rarefactions. The particles themselves do not travel from the source to the ear; only the disturbance (energy) moves through the medium. When this disturbance reaches the ear, it causes the eardrum to vibrate and the brain interprets these vibrations as sound.

Q2. Explain how sound is produced by your school bell.
Ans: When the hammer strikes the gong of the bell, the gong begins to vibrate. These vibrations cause alternating compressions and rarefactions in the air around the bell. This disturbance travels through the air as longitudinal sound waves in all directions. When these waves reach our ears, they make our eardrums vibrate and we hear the ringing of the bell.
Q3. Why are sound waves called mechanical waves?
Ans: Sound waves are called mechanical waves because they are produced by the oscillations of particles of a material medium and therefore require a medium (such as air, water or a solid) for their propagation. They cannot travel through a vacuum.
Q4. Suppose you and your friends are on the moon. Will you be able to hear any sound produced by your friend?
Ans: No. There is practically no atmosphere on the Moon, so there is no material medium to carry the vibrations. Since sound is a mechanical wave that needs a medium, it cannot travel through the vacuum-like conditions on the Moon and you would not be able to hear your friend.

Page No. 132

Q1. Which wave property determines
(a) loudness, (b) pitch?
Ans: (a) Loudness is determined mainly by the amplitude (or intensity) of the sound wave. A larger amplitude means greater energy delivered to the ear and thus a louder sound.
(b) Pitch is determined by the frequency of the sound wave. A higher frequency is heard as a higher pitch, while a lower frequency is heard as a lower pitch.
Q2. Guess which sound has a higher pitch: guitar or car horn?
Ans: Guitar, because the strings of a guitar produce vibrations of higher frequency than the typical frequency of a car horn. Hence, a guitar note generally has a higher pitch.
Q3. What are the wavelength, frequency, time period and amplitude of a sound wave?
Ans: Wavelength: The distance between two consecutive compressions or two consecutive rarefactions in a sound wave is called the wavelength (denoted by λ). It is the distance over which the wave pattern repeats itself.
Frequency: The number of complete waves (or vibrations) produced per second is called the frequency (denoted by f). Its SI unit is hertz (Hz).
Time Period: The time taken to complete one full vibration is called the time period (T). It is the reciprocal of frequency: T = 1/f.
Amplitude: The maximum displacement of particles of the medium from their mean positions during the propagation of the wave is called the amplitude. For sound, larger amplitude corresponds to greater loudness.
Q4. How are the wavelength and frequency of a sound wave related to its speed?
Ans: The speed (v) of a sound wave in a given medium is related to its frequency (f) and wavelength (λ) by the relation: v = f × λ.
Q5. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Ans: Given that,
Frequency of sound wave = 220 Hz
Speed of sound wave = 440 m/s
We know that
Speed = Wavelength × Frequency
v = λ × f
440 = λ × 220
λ = 440 / 220
λ = 2
Therefore, the wavelength of the sound wave = 2 metres

Q6. A person is listening to a tone of 500 Hz, sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Ans: The time interval between successive compressions is the time period of the wave. Time period T = 1 / frequency = 1 / 500 = 0.002 s. This time interval is a property of the source and does not depend on the distance of the listener.

Page No. 133

Q1. Distinguish between loudness and intensity of sound.
Ans: Loudness and intensity are related but different concepts:
- Intensity is a physical quantity defined as the sound energy passing per unit time through unit area perpendicular to the direction of propagation. Its SI unit is W m^-2. Intensity depends on the power of the source and the distance from it.
- Loudness is a perceptual (subjective) measure of how strong a sound appears to the human ear. Loudness depends on intensity and on the sensitivity of the ear to different frequencies. Loudness is commonly measured in decibels (dB).
In short, intensity is an objective measure of sound energy, while loudness is how loud that sound seems to a listener.

Q2. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Ans: Sound travels fastest in solids. Among the three media given - air, water and iron - sound travels fastest in iron. This is because particles in solids are more tightly bound and the medium transmits the disturbance more quickly, giving a higher speed of sound in solids than in liquids and gases.

Page No. 134

Q1. 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?
Ans: If d is the distance of the reflecting surface from the source, the sound travels a total distance 2d (to the reflector and back). Using 2d = v t we get d = v t / 2.
Here v = 342 m s^-1 and t = 3 s.
d = (342 × 3) / 2 = 1026 / 2 = 513 m.
Therefore, the reflecting surface is 513 metres away from the source.

Page No. 135

Q1. Why are the ceilings of concert halls curved?
Ans: The ceilings of concert halls are curved so that sound reflected from the stage is distributed evenly throughout the hall. Curved surfaces help to scatter and direct sound waves to reach all parts of the audience, improving clarity and loudness uniformly and avoiding dead spots or strong echoes.

Page No. 136

Q1. What is the audible range of the average human ear?
Ans: Audible range for the average human ear is approximately 20 Hz to 20 kHz (20,000 Hz).
Q2. What is the range of frequencies associated with: 
(a) Infrasound
(b) Ultrasound?
Ans: (a) Infrasound: Frequencies below about 20 Hz (sounds lower than the audible range).
(b) Ultrasound: Frequencies above about 20 kHz (20,000 Hz), above the audible range for humans.

Page No. 138

Q1. What is sound and how is it produced?
Ans: Sound is a form of energy that gives the sensation of hearing. It is produced when an object vibrates; these vibrations cause neighbouring particles of the medium to oscillate, creating compressions and rarefactions that travel as longitudinal waves through the medium until they reach a receiver such as the ear.
Q2. Describe with the help of a diagram, how compressions and rarefactions are produced in the air near a source of the sound. 
Ans: 

The forward motion of a vibrating object pushes air molecules together to form regions of higher pressure called compressions. When the object moves back, it creates regions where air molecules are spread apart called rarefactions. As the object continues to vibrate, a sequence of compressions and rarefactions travels away from the source as a longitudinal sound wave.

Q3. Why is sound wave called a longitudinal wave?

Ans: In a longitudinal wave the particles of the medium vibrate parallel to the direction of wave propagation. In sound waves in air, the particles oscillate back and forth along the direction the wave moves, hence sound waves are longitudinal.

Q4. Which characteristic of the sound helps you to identify your friend by his voice while sitting with friends in a dark room?

Ans: It is the quality (also called timbre) of the sound that helps us to identify a friend's voice. Quality depends on the waveform and the combination of frequencies present in the voice; even if pitch and loudness are similar, the timbre makes each voice distinctive.

Q5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Ans: The speed of light in air is about 3 × 10⁸ m s^-1, while the speed of sound in air is roughly 330-344 m s^-1. Light reaches the observer almost instantly, whereas sound takes much longer to travel the same distance. Because sound travels far more slowly than light, thunder is heard some seconds after the flash is seen. The actual delay depends on the distance of the lightning from the observer.

Q6. 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 ms¹.
Ans: For sound waves, v = λ × f where v = 344 m s^-1.
(a) For f = 20 Hz: λ₁ = v / f = 344 / 20 = 17.2 m.
(b) For f = 20,000 Hz: λ₂ = v / f = 344 / 20,000 = 0.0172 m.
Therefore, the wavelengths corresponding to human hearing range are approximately 17.2 m to 0.0172 m.

Q7. 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.
Ans: Let the length of the aluminium rod be d.
Speed of sound in aluminium (at 25 °C), v_Al = 6420 m s^-1.
Time for sound to travel along the rod: T_Al = d / 6420.
Speed of sound in air, v_air = 346 m s^-1.
Time for sound to travel the same distance in air: T_air = d / 346.
Ratio T_air : T_Al = (d/346) : (d/6420) = 6420 : 346 ≈ 18.55 : 1.
Thus, sound takes about 18.55 times longer to travel the same distance in air than in aluminium.

Q8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute
Ans: Frequency = Number of oscillations per second. For 100 Hz, there are 100 vibrations each second.
Number of vibrations in 1 minute (60 s) = 100 × 60 = 6,000.
The source vibrates 6,000 times in one minute.
Q9. Does sound follow the same laws of reflection as light does? Explain.
Ans: Yes. Sound follows the same two laws of reflection as light:

• The angle of incidence is equal to the angle of reflection (both measured from the normal to the reflecting surface).
• The incident wave, the normal at the point of incidence and the reflected wave all lie in the same plane.

These laws explain how sound is reflected from surfaces to produce echoes or to direct sound in rooms and halls.

Q10. 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?

Ans: An echo is heard when the time interval between the original sound and the reflected sound is at least about 0.1 s. On a hotter day the speed of sound in air increases (because speed of sound rises with temperature). As a result, the reflected sound returns sooner and the time gap between original and reflected sound decreases. If this reduced time interval becomes smaller than about 0.1 s, the echo will not be perceived distinctly. Thus, on a hotter day an echo is less likely to be heard for the same fixed distance to the reflector.

Q11. Give two practical applications of reflection of sound waves.
Ans: Applications of reflection of sound waves include:

• Use of SONAR (Sound Navigation and Ranging) to locate underwater objects such as submarines, rocks and icebergs by sending sound pulses and detecting their echoes.
• Ultrasonic testing to detect defects and cracks in metal parts and structures; sound pulses are reflected from flaws and the echoes are analysed to find and locate the defect.

Q12. 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 ms^-2 and speed of sound = 340 ms^-1.
Ans: Height of tower s = 500 m.
Acceleration due to gravity g = 10 m s^-2, initial velocity u = 0.
Time t₁ for the stone to fall:
s = ut₁ + (1/2) g t₁²
500 = 0 + (1/2) × 10 × t₁²
t₁² = 100 ⇒ t₁ = 10 s.
Time t₂ for sound to travel up from the pond: t₂ = 500 / 340 ≈ 1.47 s.
Total time t = t₁ + t₂ = 10 + 1.47 = 11.47 s.
The splash is heard approximately 11.47 seconds after the stone is dropped.

Page No. 139

Q13. A sound wave travels at a speed of 339 ms^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Ans: Given: v = 339 m s^-1, λ = 1.5 cm = 0.015 m.
Frequency f = v / λ = 339 / 0.015 = 22,600 Hz ≈ 22.6 kHz.
This frequency is higher than 20 kHz, so it is not audible to the average human ear (it is in the ultrasonic range).

Q14. What is reverberation? How can it be reduced?

Ans: Reverberation is the persistence of sound in an enclosed space after the sound source has stopped, due to multiple reflections from walls, ceiling and floor. The reverberation time is the period until the sound loudness falls to a low level. Reverberation can be reduced by using sound-absorbent materials (such as curtains, carpets, acoustic panels and cushioned seats), by designing ceilings and walls to scatter rather than strongly reflect sound, and by adding baffles or other absorbers to break up reflected waves.

Q15. What is loudness of sound? What factors does it depend on?

Ans: Loudness is the subjective measure of how strong or intense a sound appears to a listener. It depends on:
(i) The amplitude (or intensity) of the sound wave - larger amplitude delivers more energy to the ear and sounds louder.
(ii) The sensitivity of the human ear to different frequencies - the ear responds more to some frequencies than others.
Loudness is measured in decibels (dB). Because intensity depends on the square of amplitude, changes in amplitude strongly affect perceived loudness.

Q16. How is ultrasound used for cleaning?

Ans: Ultrasound is used to clean objects that have hard-to-reach places (for example, coils, small machine parts and electronic components). The object is placed in a liquid cleaning solution and high-frequency ultrasonic waves are passed through the liquid. The rapid oscillations loosen and dislodge dust, grease and dirt from the surface and from crevices, giving a thorough cleaning without mechanical scrubbing.

Q17. Explain how defects in a metal block can be detected using ultrasound. 
Ans: An ultrasonic pulse is sent into the metal block. If the block is sound (without defects), the pulse travels through and only small predictable reflections appear. If there is a crack or void, a strong reflection (echo) will be produced at the defect. By measuring the time taken for the reflected pulse to return and knowing the speed of ultrasound in the metal, the position and size of the defect can be estimated. This method is widely used for non-destructive testing of metals and welded joints.