CBSE Class 9 Science Sound VBQs

Question. Guess which sound has a higher pitch: guitar or car horn?
Answer : Guitar.

Question. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer : No, we will not be able to hear the sound because sound requires a medium for its propagation. On the moon there is no atmosphere, i.e., there is vacuum.

Question. How are the wavelength and frequency of a sound wave related to its speed?
Answer : Speed = Wavelength x Frequency.

Question. How does the sound produced by a vibrating object in a medium reach your ear?
Answer :  When a disturbance is created on an object, it starts vibrating and sets the particles of the medium to vibrate. These vibrating particles take the sound from object to our ear through the medium.

Question. Which wave property determines (a) loudness, (b) pitch?
Answer : (a) Loudness is determined by the amplitude of the sound wave which in turn depends on the force with which the object is made to vibrate.
(b) Pitch of a sound is determined by its frequency.

Question. Explain how sound is produced by your school bell.
Answer : When the school bell is struck with a hammer, it starts vibrating and as a result of these vibrations, sound waves are produced.

Question. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer : Frequency = f = 220 Hz
Speed of sound = v = 440 m/s
Wavelength = λ = v/f = 440/220 = 2m.

Question. Why are the ceilings of concert halls curved?
Answer : Ceilings of concert halls are curved to uniformly spread sound in all directions after reflecting from the walls.

Question. Why are sound waves called mechanical waves?
Answer : Because sound waves require a medium to propagate to interact with the particles present in it.

Question. 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?
Answer : Frequency = f = 500 Hz
Time between successive compressions = Time period of sound wave = T = 1/f = 1/500 =0.002 sec.

Question. What is the range of frequencies associated with (a) Infrasound? (b) Ultrasound?
Answer : (a) Infrasound: Less than 20 Hz ; (b) Ultrasound: More than 20000 Hz

Question. What are wavelength, frequency, time period and amplitude of a sound wave?
Answer : (a) Wavelength(λ) : The distance between two consecutive compressions (C) or two consecutive rarefactions (R) is called the wavelength, unit-metre.
(b) Frequency : The number of oscillations per unit time is called frequency, unit-Hz.
(c) Time period : The time taken by two consecutive compressions or rarefactions to cross a fixed point is called the time period.
(d) Amplitude : The magnitude of the maximum disturbance in the medium on either side of the mean value is called the amplitude of the wave.

Question. Distinguish between loudness and intensity of sound.
Answer :

Question. An echo is heard in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms-1 ?
Answer : Time = 3 sec
Speed of sound = v = 342 m/s
Let the distance of reflecting surface from the source is = S
Then, total distance travelled by sound is = 2S
Now, Speed = Distance/Time
342 = 25/3 ; 2S = 342 x 3 ; S = 1025/2 = 513 M.

Question. What is the audible range of the average human ear?
Answer : 20 Hz to 20,000 Hz.

Question. A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?
Answer : Time = t = 1.02 sec
Speed of sound = v = 1531 m/s
Let the distance of cliff from the submarine is = S
Then, total distance travelled by sound is = 2S
Now, Speed = Distance/Time
1531 = 2S/1.02 ; 2S = 342 x 1.02 : S = 1561.62/2 = 780.8 m.

Question. In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer : Iron (Solid).

Exercise Question-Answers

Question. Describe with the help of a diagram, how compressions and rarefactions are produced in the air near a source of sound.
Answer : (1) When a vibrating object moves forward, it pushes the air in front of it and compresses the air
creating a region of high pressure called compression (C).
(2) It starts moving away from the surface of the vibrating object.
(3) As this occurs the surface moves backward creating a region of low pressure called rarefaction (R)

Question. Cite an experiment to show that sound needs a material medium for its propagation.
Answer : Take an electric bell and an air tight glass bell jar. The electric bell is suspended inside an air tight
glass jar which is connected to a vacuum pump. Working:
1. When we press the switch, we will be able to hear the bell.
2. When the air in the jar is pumped out gradually, the sound becomes feeble although the same amount of current is flowing through the bell.
3. When the air is removed completely, are will not be able to hear the sound of the bell. Conclusion: This experiment shows that sound requires a medium for its propagation.

Question. Why sound wave is called a longitudinal wave?
Answer : A sound wave is called a longitudinal wave as it travels in a medium by the vibration of particles in a direction which is parallel to the direction of propagation of the sound wave.

Question. What is sound and how is it produced?
Answer : Sound is a form of energy and it is produced due to vibrations of different types of objects.

Question. 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 : Speed of sound = v = 344 m/s.
For Minimum audible frequency
Frequency = f = 20 Hz
Wavelength = λ = v/f = 344/20 = 17.2 m
For Maximum audible frequency
Frequency = f = 20000 Hz
Wavelength = λ = v/f = 344 / 20000 = 0.0172 m

Question. Two children are at opposite ends of an aluminum rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in the air and in aluminum to reach the second child.
Answer : Speed of sound in air = Vair = 344 m/s
Speed of sound in aluminum = Valuminum = 6420 m/s
As time is inversely proportional to the speed hence:
Tair / Taluminum = Valuminum / Vair = 6420 / 344 = 1605 : 86

Question. How is ultrasound used for cleaning?
Answer : Objects that need to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through the solution. The high frequency of ultrasound waves helps in detaching the dirt from the objects.

Question. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer : The speed of light is much higher than speed of sound. Due to this reason, the thunder takes more time
to reach the Earth as compared to the light. Hence, lightning is seen before whenever we hear the thunder.

Question. Does sound follow the same laws of reflection as light does? Explain.
Answer : Yes, sound follows the same laws of reflection as light does because,
(1) Angle of incidence of sound is always equal to angle of reflection of sound waves.
(2) The direction in which sound is incident, the direction in which it is reflected and normal all lie in the same plane.

 Question. Explain how defects in a metal block can be detected using ultrasound.
Answer : Ultrasounds can be used to detect cracks and flaws in metal blocks. Ultrasonic waves are allowed to pass through the metallic 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.

Question. 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 : An echo is heard when time interval between the reflected sound and the original sound is at least 0.1 second. As the temperature increases, the speed of sound in a medium also increases. On a hotter day, the time interval between the reflected and original sound will decrease and an echo is audible only if the time interval between the reflected sound and the original sound is greater than 0.1 s.

Question. What is reverberation? How can it be reduced?
Answer : The continuous multiple reflections of sound in a big enclosed space is reverberation. It can be reduced by covering walls and ceiling of enclosed space with the help of sound absorbing materials such as loose woolens, fibre boards.

Question. Explain how bats use ultrasound to catch prey.
Answer : The ultrasonic waves emitted by the bat are reflected from the prey (e.g., an insect) and are detected by its ears. The nature of reflected waves tells the bat about the location and the nature of its prey.

Question. 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 .
Answer : Height of tower = S = 500 m
Acceleration due to gravity = g = 10 m/s2
Initial speed of stone = u = 0
By equation of motion, S = ut + 1/2 at2
500 = 0 x t + 1/2 x 10 x t2
t2 = 500/5 = 100
Hence, t = 10 sec
Now time taken by sound to reach to the top of tower = T = distance/speed of sound = 500/340 = 1.47 sec
Total time takes to hear splash after dropping the stone = t + T = 10 + 1.47 = 11.47 sec.

Question. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer : Frequency = 100 Hz
Number of vibrations in 1 second = 100
Number of vibrations in 60 second = 100 x 60
Number of vibrations in 1 minute = 6000.

Question. 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 : Speed of sound = v = 339 m/s
Wavelength = λ = 1.5 cm = 15/100 = 0.015 m
Frequency = v/λ 339/0.015 = 22600 Hz
The sound is not audible as its frequency lies beyond the audible range (20 Hz to 20000 Hz).

Question. 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 : Time = t = 5 sec
Distance of object from submarine = d = 3625 m
Using, d = v X t / 2
Speed of sound in water = v = 2 xd/t = 2 X 3625/5 = 1450 m/s

Question. Give two practical applications of reflection of sound waves.
Answer : (i) Reflection of sound is used to measure the speed and distance of underwater objects. This method is called SONAR.
(ii) Working of a stethoscope - the sound of patient’s heartbeat reaches the doctor’s ear through multiple reflections of sound.

Question. What is loudness of sound? What factors does it depend on?
Answer : Loudness refers to how loud or soft a sound seems to a listener. Loudness of sound is determined of amplitude. The unit of intensity is the decibel (dB).

Question. Which characteristics of the sound help you to identify your friend by his voice while sitting with others in a dark room?
Answer : The quality (or timbre), pitch and loudness of sound.

Question. Explain the working and application of a SONAR.
Answer : SONAR;(-Sound Navigation and Ranging).It consists of a transmitter and a detector and is installed in a boat or a ship as shown in figure:
The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the sea bed, get reflected back and are sensed by the detector. SONAR calculates the distance (d) of the object by measuring time taken for transmission and reception of signal (t) by formula:
d = v X t / 2 : Where v is speed of sound in sea water. 
The SONAR technique is used to determine the depth of the sea and to locate the underwater hills, valleys, submarines, icebergs and sunken ships etc.

Question. Explain how the human ear works.
Answer : Various sounds produced by particles in our surroundings are collected by pinna that transfers these sounds to the ear drum through the ear canal. The eardrum begins to vibrate back and forth briskly as soon as the sound waves fall on it. The vibrating Eardrum initiates the small bone hammer to vibrate. These vibrations are passed from the hammer to the third bone stirrup via the second bone anvil. The stirrup strikes themembrane of the oval window to pass its vibration to the cochlea. The liquid in the cochlea produces electrical impulses in the nerve cells. These electrical impulses are carried to the brain by the auditory nerve. They are interpreted by the brain as sound and hence we get a sensation of hearing.