CBSE Class 9 Physics Sound Worksheet Set B

Question. If the amplitude of a wave is doubled, what will be the effect on its loudness?
Answer. Loudness depends upon the square of the amplitude of the wave, therefore, when the amplitude of wave is doubled, the loudness becomes four times.


Question. How do the frequency and amplitudes affect a musical sound?
Answer. The ‘frequency’ of a musical sound affects its ‘pitch’. The more the frequency of a (musical) sound, the ‘sharper’ and ‘shriller’ the sound becomes.
The ‘amplitude’ of a musical sound affects its loudness, or intensity. The more the amplitude of the sound, the louder (or more intense) the sound is.


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. Mention one practical use of echoes.
Answer. Echoes are used in radars to estimate the distance of flying objects.


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. 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 longitudinal wave is produced on a toy slinky. The wave travels at a speed of 30 cm/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions of the slinky?
Answer. Wave speed,                 v = 30 cm/s
Frequency of the wave,          v = 20 Hz = 20 s^–1
The minimum separation between the consecutive
compressions is equal to the wavelength. Therefore,
                             Wavelength=30 cm s^–1/20s-1= 1.5 cm


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
                   =344ms-1/120x103s^–1
                   = 2.87 × 10–3 m = 0.29 cm


Question. Give uses of multiple reflection of sound.
Answer. There are several uses of multiple reflection of sound :
(i) Megaphone is a device used to address public meetings. It is horn-shaped. When we speak through megaphone, sound waves are reflected by the megaphone. These reflected sound waves are directed towards the people (or audience) without much spreading.
(ii) The ceilings of concert halls and auditoriums are made curved. This is done so that the sound reaches all the parts of the hall after reflecting from the ceiling. Moreover, these ceilings are made up of sound absorbing materials to reduce the reverberation.
(iii) Stethoscope is a device used by doctors to listen the sound produced by heart and lungs. The sound produced by heart beat and lungs of a patient reaches the ears of a doctor due to multiple reflection of sound.
(iv) Sound boards are curved surfaces (concave) which are used in a big hall to direct the sound waves towards the people sitting in a hall. The speaker is (i.e., source of sound) placed at the focus of the sound board.
(v) Sound waves from the speaker are reflected by die sound board and these reflected waves are directed towards the people (or audience).
(vi) Hearing aid is used by a person who is hard of hearing. The sound waves falling on hearing aid are concentrated into a narrow beam of sound waves by reflection. This narrow beam of sound waves is made to fall on the diaphragm of the ear. Thus, diaphragm of the ear vibrates with large amplitude. Hence, the hearing power of the person is improved.


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. 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. 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 /60 = 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 ms-1/100 m= 0.2 s^–1
So, the frequency of rocking of the boat is 0.2 s^–1.

Question. Do waves transport matter?
Ans : No.

Question. Sound is produced due to a vibratory motion, then why a vibrating pendulum does not produce sound?
Ans : The frequency of the vibrating pendulum does not lie within the audible range (20 Hz to 20,000 Hz) and hence, it does not produce audible sound.

Question. Why do echoes produced in an empty auditorium usually decrease when it is full of audience?
Ans : 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 is a wave?
Ans : A wave is a disturbance that travels in a medium due to repeated periodic motion of particles about their mean position – such that the disturbance is handed over from one particle to the other without the actual motion of the medium.

Question. What is a transverse wave?
Ans : It is a wave in which the particles of the medium vibrate perpendicular to the direction of propagation of the wave.

Question. What is a longitudinal wave?
Ans : It is a wave in which the particles of the medium vibrate in the direction of propagation of the wave.

Question. What do you understand by the term ultrasonic vibrations?
Ans : Sounds of frequency higher than 20,000 Hz are called the ultrasonics.

Question. What do you understand by the term echo?
Ans : The sound heard after reflection from a rigid obstacle is called an echo.

Question. Do the particles of the medium move from one place to another in a medium?
Ans : No.

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

Question. Does the velocity of wave motion depend on the nature or motion of the source?
Ans : No.

Question. What are transverse waves? Give two examples.
Ans : A wave in which the particles of the medium vibrate up and down at right angle to the direction in which the wave is moving.
Example : (i) The waves produced by moving one end of a long spring up and down rapidly.
(ii) Ripples formed on the surface of water in a pond.

Question. What are crests and troughs of a wave?
Ans : 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. 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?
Ans : He is adjusting frequency if the sitar string with the frequency of the other musical instrument.

Question. If the tension in the wire is increased four times, how will the velocity of wave in a string varies?
Ans : Velocity of the wave in string is directly proportional to the square root of the tension thus if tension is increased 4 times, the velocity will be doubled.

Question. Explain, how is the principle of echo used by the dolphin to locate small fish as its prey?
Ans : Dolphins are aquatic animals which send out ultrasonic sound to communicate with each other. They have a sound sensing system which enables them to find animals underwater with great accuracy due to the echo of the ultrasonic sound produced by them.

Question. Give two practical applications of the reflection of sound waves.
Ans : (i) In stethoscope the sound of patient’s heartbeat reaches the doctor’s ears by multiple reflections in the tubes.
(ii) Megaphones are designed to send sound waves in particular direction are based on the reflection of sound

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

Question. Distinguish between tone and note.
Ans : A pitch is a particular frequency of sound, for example : 440 Hz.
A note is a named pitch. For example : Western music generally refers to the 440 Hz pitch as A, specifically A4.

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?
Ans : 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. Can you produce both types of waves (i.e., longitudinal and transverse) on a slinky?
Ans : Yes, we can produce both types of waves (i.e., longitudinal and transverse) on a slinky.

CHARACTERISTICS OF SOUND WAVES:

Sound waves have 4 characteristics
1. Amplitude 2. Wavelength 3. Frequency 4. Speed

• As the sound wave propagates in a medium, the density as well as the pressure of the medium at a given time varies with distance above and below the average value.
• Increase in density is not the same throughout compression. Maximum increase in density is seen at the centre of compression.

• i) Amplitude- The amplitude of sound wave is the height of the crest or tough. The amplitude is how high the crests are. In a sound wave, the maximum displacement associated with the particle constituting a wave is called its amplitude. It is represented by “A’. SI unit is metre.

Amplitude depends upon the force with which an object vibrates . E.g., When we hit a table hard, a loud sound is produced due to its larger amplitude.
Similarly, if we hit the table slowly, the sound produced is low as its amplitude is small. Thus, loudness as well as soft sound is determined by its amplitude.

• ii) Wavelength- 

The wavelength is the distance between 2 consecutive compressions or 2 consecutive rarefaction is called wavelength n is represented as λ (lambda). Its SI unit is metre. Wavelength can also be considered as the distance over which graph/wave is repeated.

• iii) Frequency- The number of vibrations completed by a particle in one second is the frequency of the sound wave.
Frequency = Number of Oscillations / Total Time. =1/T

We can calculate the frequency of sound by calculating the number of compressions or rarefaction in one second. It is represented by a Greek letter (Greek letter nu). SI unit is Hertz. SI unit of frequency is named after Heinrich Rudolph Hertz who laid foundation 4 future development of radio, telephone, telegraph and TV.

• iv) Time period- The time taken by the particle of the medium for completing one oscillation/vibration is called the time period. It is represented by the symbol “T”. SI unit is second. Time period of a sound wave is the time between 2 successive compressions or 2 successive rarefactions

v) Velocity of sound wave- It is the distance travelled by a wave in one second. Speed is represented by V. Speed with which compression and rarefactions move ahead is called velocity. SI unit is meter per second (m/s).

The speed of sound is more in solids, less in liquids and least in gases

RELATIONSHIP BETWEEN SPEED V, FREQUENCY AND WAVELENGTH OF SOUND

Wave Velocity= Distance covered/ time taken = Wavelength/time taken
= λ /T......................(1)
As =1/T , eq(1), connecting V and in terms of frequency can be written as
V= λ×f......................................(2)
(Or)
Wave velocity= wavelength × Frequency

The velocity of sound remains almost same for all frequencies in a given medium under the same physical conditions.

Questions based on above topics:

Question. What is the formula of time period?
Solution: T= 1/f. Here f stands for frequency.

Question. A wave covers 25 oscillation with its crest and through from point A to B what is the time period of the wave from A to B?
Solution: As we know T= 1/f
Given f = 25 Hz.
( no. Of oscillation of wave = frequency and the SI unit of f is Hz)
T= 1/25 = 0.4Hz.

Question. What affects amplitude of a wave?
Solution: The amount of energy carried by a wave is related to the amplitude of the wave. A high energy wave is characterized by a high amplitude; a low energy wave is characterized by a low amplitude. The energy imparted to a pulse will only affect the amplitude of that pulse.

Question. Does amplitude decrease with distance?
Solution: The energy spreads out in a spherical shell, the energy density decreases as the square of the distance from the source. That's your amplitude decrease. However, the frequency and wavelength stay the same as long as it keeps traveling at the speed of light.

Question. Compute the amplitude of the wave if a wave travels a distance of 0.5 m and has a frequency of 5 Hz..
Solution: Given: Distance D = 0.5 m,
Frequency f = 5 Hz
The amplitude is given by
A= Distance /Frequency
A= 0.5/5 = 0.1 meter.

Question. How are the wavelength and frequency of a sound wave related to its speed?
Solution: Wavelength, speed, and frequency are related in the following way:
Speed = Wavelength x Frequency
v ( speed) = λ × f

Question. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Solution: Frequency = (Number of oscillations) / Total time
Number of oscillations = Frequency × Total time
Given, Frequency of sound = 100 Hz
Total time = 1 min (1 min = 60 s)
Number of oscillations or vibrations = 100 × 60 = 6000
The source vibrates 6000 times in a minute and produces a frequency of 100 Hz.

Question. Calculate the wavelength of a wave whose frequency is 220hz and speed is 440m/sec in medium?
Solution: frequency=velocity/wavelength
So, 220=440/wavelength
wavelength =2m.

Question. Frequency of sound is 100Hz. How many times does it vibrates in a minute?
Solution: Given f= 100 Hz
Time = 1 min
As we know f=1/T
1min = 60 second
F= 1/60
vibration in one minute =100 x 60=6000 times.

Question. A person is listening to a tone of 500Hz sitting at a distance of 450m from the source of sound. What is the time interval between successive compressions from the source?
Solution: Given f= 500 Hz.
T=1/f T=1/500
T=0.002 sec.