NCERT Solutions Class 9 Science Chapter 12

Q.1 What is sound and how is it produced?

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Vibration in the particles causes sound. When a body vibrates, it forces the nearby particles of the medium to vibrate. Hence, a disturbance is created in the medium, which travels in the form of waves. This disturbance is called sound.

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

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When a body vibrates, it creates a region of high pressure and low pressure in its surrounding. These regions of high pressure and low pressure are called compressions and rarefactions. As the body continues to move forward and backward, it produces a series of compressions and rarefactions.

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

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For this, take an electric bell and hang it inside an empty bell-jar fitted with a vacuum pump. Now, start removing the air from the bell-jar with the help of a vacuum pump. It can be observed that the sound of the ringing bell decreases. On further pumping off air, vacuum will be created inside the jar. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating. This experiment shows that sound cannot travel through vacuum. Hence, sound requires a material medium for its propagation.

Q.4 Why is sound wave called a longitudinal wave?

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The vibration of the medium that moves along the direction of the wave is known as longitudinal wave. In the case of sound wave, the particles of the medium vibrate in the direction parallel to the direction of the propagation of disturbance. Therefore, a sound wave is called a longitudinal wave.

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

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Quality is the characteristic of the sound which helps us to recognise a particular person. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different.

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

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The speed of sound is very less as compared to the speed of light. Hence, sound of thunder takes more time to reach the Earth as compared to light. Therefore, a flash is seen before than the thunder is heard.

Q.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^–1.

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\begin{array}{l}\text{Given,}{\text{v}}_{\text{1}}\text{=}\text{20}\text{Hz,}{\text{v}}_{\text{2}}\text{=}\text{20}\text{kHz}\text{=}\text{20000}\text{Hz}\\ {\text{Speed of sound in air, v = 344 ms}}^{\text{-1}}\\ \text{On using the relation,}\\ \text{v = v×}\text{λ}\\ \text{or,}\text{λ}\text{=}\frac{\text{v}}{\text{v}}\\ \text{Let,}\text{the corresponding wavelengths for frequencies}\\ \text{20}{\text{Hz and 20 kHz are λ}}_{\text{1}}\text{and}{\text{λ}}_{\text{2}}\text{.}\\ \therefore {\text{λ}}_{\text{1}}\text{=}\frac{\text{v}}{{\text{v}}_{\text{1}}}\text{=}\frac{\text{344}}{\text{20}}\text{=}\text{17}\text{.2}\text{m}\\ {\text{λ}}_{\text{2}}\text{=}\frac{\text{v}}{{\text{v}}_{\text{2}}}\text{=}\frac{\text{344}}{\text{20000}}\text{=}\text{0}\text{.0172}\text{m}\end{array}

Q.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.

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\begin{array}{l}\text{Consider the length of the aluminium rod = l}\text{.}\\ \text{Speed of sound wave in aluminium at 25 °C,}\\ v_1=6420{\text{ms}}^{\text{-1}}\\ Hence,\text{time taken by the sound wave to reach}\\ {\text{the other end, t}}_1=\frac{d}{v_1}=\frac{d}{6420}\mathrm{\dots }\text{(i)}\\ \text{Speed of sound in air at 25 °C,}\\ v_2={\text{346 ms}}^{\text{-1}}\\ Hence,\text{the time taken by sound wave to reach the}\\ {\text{other end, t}}_2=\frac{d}{v_2}=\frac{d}{\text{346}}\mathrm{\dots }\text{(ii)}\\ \text{From (i) and (ii),}\\ \frac{{\text{t}}_{\text{2}}}{{\text{t}}_{\text{1}}}=\frac{\frac{d}{346}}{\frac{d}{6420}}=\frac{6420}{346}=18.55\text{.}\end{array}

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

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\begin{array}{l}\text{Given,}\text{Frequency}\text{of}\text{sound}\text{=}\text{100}\text{Hz}\\ \text{Total}\text{time}\text{=}\text{1}\text{min}\text{=}\text{60}\text{s}\\ \text{On using the relation,}\\ \text{Frequency =}\frac{\text{Number}\text{of}\text{oscillations}}{\text{Total}\text{time}}\\ \text{Here,}\text{Number}\text{of}\text{oscillations}\text{=}\text{Frequency}\text{×}\text{Total}\text{time}\\ \text{=}\text{100 Hz}\text{×}\text{60}\text{s\hspace{0.17em}}\text{=}\text{6000}\\ \text{Thus,}\text{the}\text{source}\text{vibrate}\text{6000}\text{times}\text{in}\text{a}\text{minute}\text{.}\end{array}

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

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Yes, the incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Moreover, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane.

Q.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?

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An echo can be heard when the time interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Therefore, echo cannot be heard on a summer day as the time interval between the original sound and the reflected sound decreases.

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

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(i) Sound board: It is used to send the sound towards audience in a big hall or auditorium. This works on the basis of laws of reflection of sound waves.

(ii) Stethoscope: It is also based on reflection of sound. In a stethoscope, the sound of a patient’s heartbeat reaches to the doctor’s ear by multiple reflection of sound.

Q.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 s^–2 and speed of sound = 340 m s^–1.

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\begin{array}{l}{\text{Given, h = 500 m, g = 10 ms}}^{\text{-2}}\text{,}\\ {\text{Speed of sound, v = 340 ms}}^{\text{-1}}\\ \text{The intial velocity (u) of the stone when it is dropped}\\ \text{is zero, i}\text{.e}\text{., u = 0}\\ \text{Let, it takes (t) time to reach on the surface of water}\\ \text{On using the relation, s = ut +}\frac{\text{1}}{\text{2}}{\text{gt}}^{\text{2}}\\ \therefore \text{500 = 0+ 0}{\text{.5 ×10×t}}^{\text{2}}\\ {\text{or, 100 = t}}^{\text{2}}\\ \text{or, t = 10 s}\\ \text{Now, let time taken by the sound to reach at the top the}\\ \text{tower be t’}\\ \text{then, h =v×t’}\\ \text{or, 500 = 340×t’}\\ \text{or, t’ =1}\text{.47 s}\\ \therefore \text{Total time = t+t’ = 10+1}\text{.47 = 11}\text{.47 s}\text{.}\\ \text{Hence, after 11}\text{.47 s later, the splash will be heard at}\\ \text{the top of the tower}\text{.}\end{array}

Q.14 A sound wave travels at a speed of 339 m s–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

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\begin{array}{l}{\text{Given, Speed of sound, v = 339 ms}}^{\text{-1}}\\ \text{Wave length, λ = 1}\text{.5 cm = 0}\text{.015 m}\\ \text{On using the relation,}\\ \text{v = λ × v}\\ \therefore \text{v =}\frac{\text{v}}{\text{λ}}\text{=}\frac{\text{339}}{\text{0}\text{.015}}\text{= 22600 Hz}\\ \text{As the frequency range of audible sound for humans lies between 20 Hz to 20000 Hz}\text{.}\\ \text{Since, the frequency of the given sound is more than 20000 Hz, it is not audible}\text{.}\end{array}

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

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The reverberation is the persistence of sound due to repeated reflection. When the source produces sound, it starts moving in all the directions. It is partly reflected back from the wall. This reflected sound reaches the other wall and again gets reflected partly. Thus, sound can be heard even after the source has stopped producing the sound.

To minimise reverberation, sound should be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fibre board, heavy curtains, and cushioned seats can be used to reduce reverberation.

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

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Loudness depends on the amplitude of the sound. The louder a sound, the more energy it has. Loudness is proportional to the square of the amplitude of vibrations.

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

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Bats generate high-pitched ultrasonic shrill. These high-pitched shrill are reflected by objects such as preys and comes back to the bat’s ear. Thus, a bat can easily determine the distance of a prey to catch it.

Q.18 How is ultrasound used for cleaning?

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To clean an object when it is put in a cleaning solution, the ultrasonic sound waves are passed through the solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

Q.19 Explain the working and application of a sonar.

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It is a device that uses ultrasound propagation to navigate, communicate or detect underwater objects such as submarine, a sunken ship, an iceberg, etc. A SONAR apparatus consists of two parts:

1. A transmitter (for emitting ultrasonic waves).
2. A receiver (for detecting the reflected ultrasonic waves).

Both these parts are installed in a ship or a boat. The transmitter sends ultrasonic waves towards the ocean floor.

These waves when reflected back by an object or the ocean floor in form of ‘echo’ are detected by a detector.

Depth of an object can be calculated using time interval between generation of wave and reception of its echo and the speed of sound in water. This method is known as sonar.

Q.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.

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\begin{array}{l}\text{Given, Time taken to hear to echo, t = 5 s}\\ \text{Distance of the object from the submarine, s = 3625 m}\\ \therefore \text{Total distance travelled = 2 s}\\ \therefore \text{Velocity, v =}\frac{2\text{s}}{t}\text{=}\frac{2\times \text{3625 m}}{5}\\ {\text{= 1450 ms}}^{-1}\end{array}

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

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The ultrasounds cannot pass through a defective metal block but 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. Thus, defects in metal blocks can be detected by using ultrasound.

Q.22 Explain how the human ear works.

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Pinna collects different sounds produced in our surroundings and sends these sounds to the ear drum through ear canal. At the end of ear canal, there is a thin membrane called the eardrum or tympanic membrane. When the compression of sound waves strikes the eardrum, it is pushed inward. When the rarefaction of sound waves strikes the ear drum, it moves outward. Thus, eardrum vibrates. Middle ear consists of three bones – hammer, anvil and stirrup. These bones amplify the vibrations from the eardrum and transmit these vibrations to the inner ear. Inner ear consists of cochlea that converts vibrations or pressure variations into electrical signals. These electrical signals sent to brain via auditory nerve. Brain interprets them as sound.