Sound Chapter 12 Class 9 Science Worksheets

Students should practice questions given in Sound Chapter 12 Class 9 Science Worksheets. These worksheets for Class 9 Science have a good collection of important questions and answers which are expected to come in your class tests and examinations. You should learn these solved worksheet questions for Science Class 9 as it will help you to understand all topics and give you more marks.

Class 9 Science Worksheets Chapter 12 Sound

Please refer to below questions and answers for Sound Chapter 12 Class 9 Science Worksheets. Prepared by expert teachers for Standard 9 Science

Question. What do you understand by the term infrasonic vibrations?

Question. Which of the following sound waves we can hear : 10Hz, 500 Hz, 1500 Hz, 12000 Hz, 25000 Hz?

Question. Where is density of air higher at compression or at rarefaction?

Question. Guess which sound has a higher pitch-guitar or car horn?

Question. Do the particles of the medium move from one place to another in a medium?

Question. What is intensity of sound?

Question. What is relation between time period and frequency?

Question. What is the range of frequencies associated with :
(a) Infrasound
(b) Ultrasound

Question. A Sitarist tries to adjust the tension and pluck the string suitably, before playing the orchestra in a musical concert. By doing so what is he adjusting?

Question. If the tension in the wire is increased four times, how will the velocity of wave in a string varies?

Question. Do waves transport matter?

Question. Does the velocity of wave motion depend on the nature of the medium?

Question. Does the velocity of wave motion depend on the nature or motion of the source?

Question. What is the other name of a long flexible spring?

Question. Explain, how is the principle of echo used by the dolphin to locate small fish as its prey?

Question. Give two practical applications of the reflection of sound waves.

Question. Why are longitudinal waves called pressure waves?

Question. Sound travels faster on a rainy day than on a dry day. Why?

Question. Define one hertz.

Question. Define wavelength.

Question. What is the audible range of the average human ear?

Question. What is sound and how is it produced?

Question. Why is sound wave called as longitudinal wave?

Question. Why do echoes produced in an empty auditorium usually decrease when it is full of audience?
Answer. When the hall is empty there are no obstacles in between to reflect the sound other than the walls.When the hall is full of audiences, the sound produced undergoes multiple reflections from the people and so it overlaps with the sound produced. Hence, the listener is not able to distinguish between the original sound and the echo.

Question. What are crests and troughs of a wave?
Answer. The elevation in a transverse wave is called crest. It is that part of transverse wave which s above the line of zero disturbance of the medium. The depression in a transverse wave is called trough. It is that part of the transverse wave which is below the line of zero disturbances.

Question. What is a stethoscope? Name the principle on which a stethoscope works.
Answer. Stethoscope is a medical instrument used for listening sounds produced within the body, chiefly in the heart or lungs. Stethoscope works on the principle of multiple reflection of sound.

Question. How moths of certain families are able to escape captures from bats? What is the range of frequencies associated with :
(a) infrasound?
(b) ultrasound?
Answer. They have very sensitive hearing equipment, that can hear the squeaks (ultrasound) of bat and know when a bat is flying nearby.
(a) Less than 20 Hz.
(b) More than 20,000 Hz.

Question. How does a stretched string on being set into vibration, produce the audible sound?
Answer. On being set into vibrations, the stretched string, forces the surrounding air to vibrate. This vibrating air, in turn, affects our eardrum and produces an audible sound.

Question. Will the sound be audible if the string is set into vibration on the surface of the Moon? Give reason for your answer.
Answer. No, we will not hear any audible sound on the surface of the Moon. This is because sound requires a medium to propagate, since there is no atmosphere on the surface of Moon, therefore, the sound will not be heard.

Question. What change, if any, would you expect in the characteristics of musical sound when we increase :
(i) its frequency, and
(ii) its amplitude?
Answer.
(i) Pitch of sound will increases,
(ii) Loudness of sound will increases.

Question. How do you account for the fact that two strings can be used to give notes of the same pitch and loudness but of different quality?
Answer. The ‘quality’ of a given note is determined by the overall effect of the harmonics present in it. The harmonics are multiples of the fundamental or basic frequency of the ‘note’. Depending on the conditions under which vibrations are taking place, sometimes we get one set of harmonics and sometimes another set.
The quality of the two notes will, therefore, different even though their fundamental frequencies may be the same.

Question. State any two characteristics of a wave motion.
Answer. The characteristics of wave motion are :
(i) It is a periodic disturbance.
(ii) Energy transfer takes place at a constant speed.

Question. A person fires a gun standing at a distance of 55 m from a wall. If the speed of sound is 330 ms^–1, find the time for an echo to be heard.
Answer.
Given d = 55 m, v = 330 ms^–1, t = ?
2d = v × t
t=2d/v=(2×55)/330=0.3s

Question. Give one example each of natural vibration, forced vibration and resonance.
Answer.
(i) Natural vibration : The vibrations of a simple pendulum about its mean position.
(ii) Forced vibration : A sonometer wire, under tension, vibrating under the influence of a vibrating tuning fork.
(iii) Resonance : A correctly adjusted length of a sonometer wire under proper tension, vibrating under the influence of a vibrating tuning fork.

Question. The pulse rate of a man is 80 beats in one minute. Calculate its frequency.
Answer.
No. of beats per minute = 80
No. of beats per second = 80/60 = 1.3
So, frequency = 1.3 Hz

Question. What is echo? Explain the conditions that have to be satisfied to hear an echo?
Answer. Reflection of sound wave from a large obstacle is called an echo. The most important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 second after the original sound is off and the obstacle is at least at a distance of 17 m.

Question. Which wave property determines :
(i) loudness (ii) pitch
Answer.
(i) The amplitude of the wave determines the loudness. More the amplitude of a wave, more is the loudness produced.
(ii) The pitch is determined by the frequency of the wave. Higher the frequency of a wave, more is its pitch and shriller is the sound.

Question. Explain, why can echoes not be heard in a small room?
Answer. For hearing echo, there should be at least a distance of 17 m between the source of sound and the body from which sound is reflected. In small rooms this is not the case, hence, echoes are not heard.

Question. The stem of a tuning fork is pressed against a table top. Answer the following questions :
(i) Would the above action produce any audible sound?
(ii) Does the above action cause the table to set into vibrations?
(iii) If the answer above is yes, what type of vibrations are they?
(iv) Under what conditions does the above action lead to resonance?
Answer.
(i) Yes, there is an audible sound produced.
(ii) Yes, the table top is set into ‘forced vibrations’ by this.
(iii) The vibrations are forced vibrations.
(iv) Pressing the stem of a vibrating tuning fork against a table top, would lead to resonance if the frequency of the tuning fork equals the natural frequency of oscillations of the table top.

Question. Give application of ultrasound (ultrasonic waves).
Answer. Ultrasonic waves have number of uses :
(1) Ultrasonic vibrations are used for homogenising milk. These vibrations break down the larger particles of the fat present in milk to smaller particles.
(2) Ultrasonic vibrations are used in dish washing machines. The vibrating detergent particles rub against the dirty utensils and thus, clean them.
(3) Ultrasonic vibrations produce a sort of depression in rats and cockroaches.
(4) Ultrasonic vibrations are used to study the growth of foetus in mother’s womb.
(5) Ultrasonic vibrations are used in relieving pain in joints and muscles.
(6) Ultrasonic vibrations are used in detecting flaws in articles made from metals. They are also used in finding the thickness of various parts of a metallic component.

Question. How does the sound produced by a vibrating object in a medium reach your ear?
Answer. When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure called compression. This compression starts to move away from the vibrating object. When vibrating object moves backwards, it creates a region of low pressure called refraction. As the object moves forth and back rapidly, a series of compressions and refractions are created in the air. These produce the sound wave that propagates through the medium.
This continues until the sound wave reaches to the ear of the listener.

Question. Write conditions for the production of an echo.
Answer. Conditions for the production of an echo are :
(i) Time gap between the original sound and the reflected sound.
The echo will be heard if the original sound reflected by an obstacle reaches our ears after 0.1 s.
(ii) Distance between the source of sound and obstacle.
Thus, the minimum distance (in air at 25°C) between the observer and the obstacle for the echo to be heard clearly should be 17.2 m.
(iii) Nature of the obstacle : For the formation of an echo, the reflecting surface or the obstacle must be rigid such as a building, hill or a cliff.
(iv) Size of the obstacle : Echoes can be produced if the size of the obstacle reflecting the sound is quite large.

Question. A particular transmitter of Aakashvani broadcasts at 420.5 m wavelength. (Given the speed of radio waves 3 × 108 ms^–1) Calculate the frequency at which the radio station broadcasts its program. What is the direction of oscillations of the medium particles through which a :
(i) Transverse wave is propagating?
(ii) Longitudinal wave is propagating?
Answer.
Given
l = 420.5 m, v = 3 × 10⁸ ms^–1, n = ?
Using the expression v = nλ

= 7 × 105 Hz
(i) The particles oscillate perpendicular to the direction of propagation of the wave.
(ii) The particles oscillate parallel to the direction of propagation of the wave.

Numerical Questions

Question. (a) A sound wave of wave length 0.33 m has a time period of 10^–3 s. If the time period is decreased to 10^–4 s, calculate the wave length and frequency of the new wave.
(b) Name the subjective property of sound related to its frequency and of light related to its wavelength.
Answer.
(a) λ = 0.33 m
Time taken to travel λ , t = 10^–3
velocity= λ/t=0.33/10^–3
= 300 m s^–1
Time period of 2^nd wave = 10^–4 s
Therefore, wavelength λ = v × t
= 330 × 10^–4 = 0.033 m
Frequency 1/t =1/10^–3
= 10³ Hz
(b) Pitch is related to the frequency of sound and colour is related to the wavelength of light.

Question. A human heart, on an average, is found to beat 75 times a minute. Calculate its frequency.
Answer. No. of beats of human heart = 75 min^–1
= 75/1 min
= 75/60s = 1.25 s^–1
So, average frequency of human heart beating
= 1.25 s^–1.

Question. A boat at anchor is rocked by waves whose consecutive crests are 100 m apart. The wave velocity of the moving crests is 20 m/s. What is the frequency of rocking of the boat?
Answer. Distance between two consecutive crests = 100 m
Wave velocity v = 20 m/s
The distance between two consecutive crests is equal
to the wavelength of the wave. So,
Frequency = Wave velocity/Wave length
20 m s^–1/100 m
= 0.2 s^–1
So, the frequency of rocking of the boat is 0.2 s^–1.

Question. A bat can hear sound at frequencies up to 120 kHz.Determine the wavelength of sound in the air at this frequency. Take the speed of sound in the air as 344
m/s.
Answer. Frequency, n = 120 kHz = 120 × 103 Hz
= 120 × 10³ s^–1 Velocity of sound in the air, v = 344 m/s
Wavelength of the sound wave = λ
We know,
Wavelength, λ = wave velocity/frequency
344 ms^–1/120 × 10³ s^–1
= 2.87 × 10³ m = 0.29 cm

Question. A gun is fired in the air at a distance of 660 m, from a person. He hears the sound of the gun after 2 s. What is the speed of sound?
Answer. Distance travelled by sound = 660 m,
Time taken by the sound = 2 s,
Speed of sound in air = ?
So, Speed of sound = Distance/Time
= 660 m/2 s = 330 m/s
Thus, the speed of sound in the air is 330 m/s.

Question. A child hears an echo from a cliff 4 seconds after the sound from a powerful cracker is produced. How far away is the cliff from the child? Velocity of sound in air at 20°C is 344 m/s.
Answer. Let the distance between the child and the cliff be d.
Then,
Total distance travelled by the sound = 2d
Total time taken by the sound = 4 s
Then, Velocity of sound = 344 m/s = d/2s
This gives, d = 344 m/s × 2 s
= 688 m
Thus, the cliff is at a distance of 688 m from the child.

Question. A ship sends on a high frequency sound wave and receives an echo after 1 second. What is the depth of the sea? Speed of sound in water is 1500 m/s.
Answer.
Let, Depth of the sea = d
So, total distance travelled by the sound wave = 2d
Time taken by sound to travel both ways = 1s
As per definition,
Speed of the sound = 1500 ms^–1 = 2d/1s
or d = 1500 ms^–1 × 1s/2
1 = 750 m
Thus, the depth of the sea is 750 metres.

Question. A sonar echo takes 2.2 s to return from a whale. How far away is the whale?
Answer.
Total time taken by the signal = 2.2 s
So, time taken the signal to reach the whale = 1.1 s
Distance of the whale = d (assume)
Speed of sound in sea water at 25°C = 1533 ms^–1
So, distance of the whale,
d = Speed of the signal × Time taken
or d = 1533 m s^–1 × 1.1 s = 1686.3 m

Question. A tuning fork produces 1024 waves in 4 seconds. Calculate the frequency to the tuning fork.
Answer. As the tuning fork produces 1024 waves in 4 seconds,hence
Frequency of tuning fork,
n = Number of vibration per second
= 1024/4 = 256 Hz

Question. Using the SONAR, sound pulses are emitted at the surface of water. These pulses after being reflected from the bottom are detected. If the time interval from the emission to the detection of the sound pulses is 2 seconds, find the depth of the water. Velocity of sound in water = 1498 m/s.
Answer. Let, depth of the water from the Earth’s surface be d. Then,
Total distance travelled by the pulse = 2d
Total time taken by the pulse = 2s
As per definition,
Velocity = Distance travelled/Time taken
So, Velocity of the sound =2d/2s =d/s
1498 m/s =d/s
This gives, d = 1498 m/s × 1s
= 1498 m
Thus, the depth of water is 1498 m.

Question. A wave moves a distance of 8 m in 0.05 s :
(a) Find the velocity of the wave.
(b) What is the wavelength of the wave if its frequency is 200 Hz?
Answer.
(a) Velocity 8/ 0.05= 160 ms^–1
(b)  λ= 160/200 = 0.8 m

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