Page 1 : Class- IX-CBSE-Science, , Sound, , CBSE NCERT Solutions for Class 9 Science Chapter 12, Back of Chapter Questions, 1., , What is sound and how is it produced?, , Solution:, Sound is the vibration of particles in a medium. It is produced by vibration. When a body vibrates,, it forces the neighbouring particles of the medium to vibrate. This disturbance, when reaches the, ear, we perceive it as sound., 2., , Describe with the help of a diagram, how compressions and rarefactions are produced in the air, near a source of sound, , Solution:, When a vibrating object moves forward, it pushes and compresses the air in front of it, creating, a region of high pressure. This region is called a compression (C), as shown in the figure. This, compression starts to move away from the vibrating object. When the vibrating object moves, backwards, it creates a region of low pressure called rarefaction (R), as shown in the figure. As, the object moves back and forth rapidly, a series of compressions and rarefactions are created in, the air. These make the sound wave that propagates through the medium., , 3., , Cite an experiment to show that sound needs a material medium for its propagation., , Solution:, Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the, airtight bell jar. The bell jar is connected to a vacuum pump, as shown in the figure., , Practice more on Sound, , Page - 1, , www.embibe.com

Page 2 : Class- IX-CBSE-Science, , Sound, , If you press the switch, you will be able to hear the bell. Now start the vacuum pump. When the, air in the jar is pumped out gradually, the sound becomes fainter, although the same current is, passing through the bell., If one keeps on pumping the air out of the bell-jar, then at one point, the glass-jar will be devoid, of any air. 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. When there is no air present inside, we can say that a, vacuum is produced. Sound cannot travel through a vacuum. This shows that sound needs a, material medium for its propagation., 4., , Why is sound wave called a longitudinal wave?, , Solution:, In a longitudinal wave, the vibration of the particles in the medium is along or parallel to the, direction of the wave. In a sound wave, the particles of the medium vibrate in the direction, parallel to the direction of the propagation of disturbance. Hence, a sound wave is called a, longitudinal wave., 5., , Which characteristic of the sound helps you to identify your friend by his voice while sitting with, others in a dark room?, , Solution:, The quality or timbre of the sound is that characteristic which enables us to distinguish one sound, from another having the same pitch and loudness. So, the sound produced by two persons may, have the same pitch and loudness, but the quality of the two sounds will be different., 6., , Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash, is seen, why?, , Solution:, , 7., , The speed of sound in air is approximately 331 m/s in the air which is quite less than the speed, of light which is 3 × 108 m/s. Hence, a flash is seen before we hear thunder., 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 ms–1 ., , Solution:, , Given: The speed of sound in air = 344 m/s, , For a sound wave, Speed = Wavelength × Frequency, , Practice more on Sound, , Page - 2, , www.embibe.com

Page 3 : Class- IX-CBSE-Science, , Sound, , v=λ×ν, , (i) For, ν1 = 20 Hz, λ1 =, , v 344, =, = 17.2 m, ν1, 20, , λ1 =, , v, 344, =, = 0.0172 m, ν1 20 × 103, , (ii) For, ν2 = 20 kHz, , 8., , Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m, , 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., , Solution:, Let the length of the aluminium rod be l., Speed of sound wave in aluminium at 25 ℃ = 6420 m s−1, , Therefore, time taken by the sound wave to reach the other end, 𝑡𝑡1 =, , 𝑙𝑙, 6420, , Therefore, time taken by the sound wave to reach the other end, 𝑡𝑡2 =, , 𝑙𝑙, 346, , Speed of sound wave in air at 25 ℃ = 346 m s−1, , The ratio of time taken by the sound wave in air and aluminium,, 9., , 𝑡𝑡2, 𝑡𝑡1, , =, , 𝑙𝑙, 346, 𝑙𝑙, 6420, , s, , s, , =, , 6420, 346, , = 18.55, , The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?, , Solution:, Given, Frequency of sound = 100 Hz, Total time =1 min = 60 s, , Number of vibrations = 100 × 60 = 6000, , 10., , Hence, the source vibrates 6000 times in a minute, producing a frequency of 100 Hz., , Does sound follow the same laws of reflection as light does? Explain., , Solution:, Yes, the sound wave follows the same laws of reflection as the light does. The laws of reflection, of sound are as follows:, The incident sound wave, the reflected sound wave and the normal at the point of the incident,, all lie in the same plane., The angle of incidence of the sound wave and angle of reflection of sound wave with the normal, to the reflecting surface are equal., , Practice more on Sound, , Page - 3, , www.embibe.com

Page 4 : Class- IX-CBSE-Science, , 11., , Sound, , 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?, , Solution:, On a hotter Day, the speed of sound increases with increase in temperature. Hence, the time after, which the echo is heard decreases. If the time by the reflected sound is less than 0.1 s After the, production of the original sound, then echo is not heard. However, If the time is greater than 0.1, s, then the echo will be heard., 12., , Give two practical applications of reflection of sound waves., , Solution:, The two practical application of reflection of sound waves are, The stethoscope is a medical instrument used for listening to sounds produced within the body,, chiefly in the heart or lungs. In stethoscopes, the sound of the patient’s heartbeat reaches the, doctor’s ears by multiple reflections of sound., Generally, the ceilings of concert halls, conference halls and cinema halls are curved so that, sound after reflection reaches all corners of the hall., 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., , Solution:, Given:, , The height of tower h = 500 m, , The initial velocity of the stone u = 0 m s −1, Gravitational acceleration g = 10 m s –2, , Speed of sound vsound = 340 m s −1, From the second equation of motion,, 1, 2, , h = ut + gt 2, , 1, 2, , ⇒ 500 = 0 + × (10) × 𝑡𝑡 2, ⇒ 𝑡𝑡 = 10 seconds, , Now, time taken by the sound to reach the top from the base of the tower, =, 14., , 500, 340, , = 1.47 seconds, , Therefore, the splash is heard at the top after time, 10 + 1.47 = 11.47 seconds, , 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?, , Solution:, Given: Speed of sound vsound = 339 m s−1, Practice more on Sound, , Page - 4, , www.embibe.com

Page 5 : Class- IX-CBSE-Science, , Sound, , Wavelength 𝜆𝜆 = 1.5 cm, , For wave, speed, wavelength, and frequency of a sound wave are related as, Speed (v) = Wavelength (λ) × Frequency(ν), Speed (v), , Frequency(ν) = Wavelength (λ) =, 15., , 339, 0.015, , = 22,600 Hz, , The frequency of the given sound is more than 20,000 Hz, it is not audible., , What is reverberation? How can it be reduced?, , Solution:, The repeated multiple reflections of sound in an 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, acoustic tiles, thick curtains etc., 16., , What is loudness of sound? What factors does it depend on?, , Solution:, Loudness is a measure of the response of the ear to the sound. Even when two sounds are of equal, intensity, we may hear one as louder than the other simply because our ear detects it better., Loudness depends upon the amplitude of vibration and sensitivity of the ear., 17., , Explain how bats use ultrasound to catch a prey., , Solution:, 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 location of his, prey., 18., , How is ultrasound used for cleaning?, , Solution:, Ultrasonic waves are used to clean the objects on hard to reach places like spiral tubes, etc. The, object is dipped in a cleaning solution and the ultrasonic waves are sent into the solution. These, ultrasonic waves create a scrubbing brush action within the fluid., 19., , Explain the working and application of a sonar., , Solution:, The acronym SONAR stands for Sound Navigation And Ranging. SONAR is a device that uses, ultrasonic waves to measure the distance, direction and speed of underwater objects., SONAR consists of a transmitter and a detector and is installed in a boat or a ship, The transmitter produces and transmits ultrasonic waves., These waves travel through water and after striking the object, get reflected and are sensed by, the detector., The detector converts the ultrasonic waves into electrical signals which are appropriately, interpreted., , Practice more on Sound, , Page - 5, , www.embibe.com

Page 6 : Class- IX-CBSE-Science, , Sound, , The distance of the object that reflected the sound wave can be calculated by knowing the speed, of sound in water and the time interval between transmission and reception of the ultrasound., 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., , Solution:, Given: Time taken to hear the echo, t = 5 s, , The 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, 21., , The velocity of sound in water, v =, , 2d, t, , =, , (2× 3625), 5, , = 1450 m s−1, , Explain how defects in a metal block can be detected using ultrasound., , Solution:, Ultrasounds can be used to detect cracks and flaws in metal blocks. The cracks or holes inside, the metal blocks, which are invisible from outside reduces the strength of the structure. Ultrasonic, waves are allowed to pass through the metal block and detectors are used to detect the transmitted, waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence, of the flaw or defect, as shown in the figure., , 22., , Explain how the human ear works., , Solution:, , The outer ear is called ‘pinna’. It collects the sound from the surroundings. The collected sound, passes through the auditory canal. At the end of the auditory canal, there is a thin membrane, called the eardrum or tympanic membrane. When compression of the medium reaches the, eardrum the pressure on the outside of the membrane increases and forces the eardrum inward., Practice more on Sound, , Page - 6, , www.embibe.com

Page 7 : Class- IX-CBSE-Science, , Sound, , Similarly, the eardrum moves outward when a rarefaction reaches it. In this way, the eardrum, vibrates. The vibrations are amplified several times by three bones (the hammer, anvil and stirrup), in the middle ear. The middle ear transmits the amplified pressure variations received from the, sound wave to the inner ear. In the inner ear, the pressure variations are turned into electrical, signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and, the brain interprets them as sound., ◆ ◆ ◆, , Practice more on Sound, , Page - 7, , www.embibe.com