Ncert Solutions for Class 9 Science Chapter 11: Sound class 9 questions and answers
| Textbook | Ncert |
| Class | Class 9 |
| Subject | Science |
| Chapter | Chapter 11 |
| Chapter Name | Sound class 9 ncert solutions |
| Category | Ncert Solutions |
| Medium | English |
Are you looking for Class 9 Science Chapter 11 question answer ? Now you can download Sound class 9 questions and answers pdf from here.
In Text Questions Page No: 129
Question 1: How does the sound produced by a vibrating object in a medium reach your ear?
Answer 1: When an object vibrates, it causes the particles in the surrounding medium—such as air, water, or a solid—to vibrate as well. These vibrations travel in the form of sound waves, which are longitudinal waves made up of compressions and rarefactions of particles. As the object continues to vibrate, it creates a pattern of high-pressure and low-pressure regions that move outward from the source.
When these sound waves reach your ear, they enter the ear canal and cause the eardrum to vibrate. These vibrations are then transferred through the tiny bones in the middle ear and eventually reach the cochlea in the inner ear, where they are converted into electrical signals and sent to the brain. The brain interprets these signals as sound, allowing you to hear the original vibration.
Question 2: Explain how sound is produced by your school bell.
Answer 2: When the school bell is hit with a hammer, it moves forward and backwards, producing compression and rarefaction due to vibrations. This is how sound is produced by the school bell.
Question 3: Why are sound waves called mechanical waves?
Answer 3: Sound waves needs material medium to propagate therefore, they are called mechnical waves. Sound waves propagate through a medium because of theinteraction of the particles present in that medium.
Question 4: Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer 4: Sound waves are mechanical waves and hence need a medium to propagate. As the moon is devoid of any atmosphere, we cannot hear any sound on the moon.
In Text Questions Page No: 132
Question 1: Which wave property determines (a) loudness, (b) pitch?
Answer 1: (a) Amplitude: There is a direct correlation between a sound’s amplitude and loudness. The sound is louder when the amplitude is larger. (b) Frequency: There is a direct correlation between a sound’s pitch and frequency. A high pitch corresponds to a high frequency of sound.
Question 2: Guess which sound has a higher pitch: guitar or car horn?
Answer 2: Though car horn is louder than guitar, the latter has a higher pitch than car horn. Therefore, the frequency of guitar is higher than that of car horn.
In Text Questions Page No: 132
Question 1: What are wavelength, frequency, time period and amplitude of a sound wave?
Answer 1: Wavelength, frequency, time period, and amplitude are important characteristics of a sound wave:
- Wavelength is the distance between two consecutive compressions or rarefactions in a sound wave. It tells us how long one wave cycle is.
- Frequency is the number of wave cycles (vibrations) that occur in one second. It is measured in hertz (Hz) and determines the pitch of the sound.
- Time period is the time taken to complete one full wave cycle. It is the inverse of frequency and is measured in seconds.
- Amplitude is the maximum distance the particles of the medium move from their rest position due to the wave. It determines the loudness of the sound—the greater the amplitude, the louder the sound.
Question 2: How are the wavelength and frequency of a sound wave related to its speed?
Answer 2: The following equation relates a sound wave’s speed, wavelength, and frequency:
Speed (v) = Wavelength (λ) × Frequency (υ)
v = λ × υ
Question 3: Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer 3: To calculate the wavelength of a sound wave, we use the formula:
\(\text{Wavelength} (\lambda) = \frac{\text{Speed of sound} (v)}{\text{Frequency} (f)}\)
Given:
- Frequency, f = 220 Hz
- Speed, v = 440m/s
\(\lambda = \frac{440}{220} = 2 \, \text{meters}\)
The wavelength of the sound wave is 2 meters.
Question 4: 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 4: To find the time interval between successive compressions, we need to calculate the time period of the sound wave. The time period ( T ) is the inverse of the frequency:
\(T = \frac{1}{f}\)
Given:
- Frequency, f = 500Hz
\(T = \frac{1}{500} = 0.002 \, \text{seconds}\)
The time interval between successive compressions is 0.002 seconds.
In Text Questions Page No: 133
Question 1: Distinguish between loudness and intensity of sound.
Answer 1: Loudness and intensity of sound are related but different concepts:
| Loudness | Intensity |
|---|---|
| Loudness is how strong or weak a sound seems to a human ear. | Intensity is the amount of sound energy passing through a unit area per second. |
| It is a subjective quantity (depends on the listener). | It is an objective quantity (can be measured). |
| Measured in decibels (dB). | Measured in watts per square meter (W/m²). |
| Affected by both amplitude and sensitivity of the ear. | Depends only on the amplitude and distance from the source. |
In Text Questions Page No: 133
Question 1: In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer 1: Sound travels fastest in iron as compared to water and air.
An echo is returned in 3 s. mat is the distance of the reflecting surface from the source, given the speed of sound is 342 m s-1
In Text Questions Page No: 134
Question 1: 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 m s–1?
Answer 1: Speed of sound (v) = 342 ms-1
Echo returns in time (t) = 3 s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
In the given interval of time, sound must travel a distance which is twice the distance between the reflecting surface and the source.
Therefore, the distance of the reflecting surface from the source =1026/2 = 513 m
In Text Questions Page No: 135
Question 1: Why are the ceilings of concert halls curved?
Answer 1: Ceilings of concert halls are curved so that sound after reflection (from the walls) spreads uniformly in all directions.
In Text Questions Page No: 136
Question 1: What is the audible range of the average human ear?
Answer 1: The audible range of an average human ear is between 20 Hz to 20,000 Hz. Humans cannot hear sounds with frequency less than 20 Hzand greater than 20,000 Hz.
Question 2: What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
Answer 2: (a) 20 Hz (b) 20,000 Hz
Exercises
Question 1: What is sound and how is it produced?
Answer 1: Vibrations result in sound production. A body’s vibrations cause the medium’s nearby particles to vibrate as well. This disrupts the medium, which makes its way to the ear in the form of waves. As a result, sound is generated.
Question 2: Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer 2: When a vibrating body moves forward, it createsa region of high pressure in its vicinity. This region of high pressure is known as compressions. When it moves backward, it creates a region of low pressure in its vicinity. This region is known as a rarefaction. As the body continues to move forward and backwards, it produces a series of compressions and rarefactions. This is shown in below figure.
Question 3: Why is sound wave called a longitudinal wave?
Answer 3: Longitudinal wave motion is defined as a motion in which the individual particles of the medium vibrate back and forth along the direction of propagation of the wave. When sound moves in a medium, each particle of the medium vibrates about its mean position in the direction of wave propagation. That is why sound waves are called longitudinal waves.
Question 4: Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer 4: Quality of sound is a characteristic that helps us identify the voice of a particular person. Two people may have the same pitch and loudness, but their qualities will be different.
Question 5: Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer 5: Velocity of sound is 344 m/s and that of light is 3 × 108m/s . As the speed of light is greater than that of sound, the sound of thunder requires longer time than light to reach Earth. Therefore, before we hear thunder, a flash is seen.
Question 6: 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.
Answer 6: To find the wavelength of sound, we use the formula:
\(\lambda = \frac{v}{f}\)
Where:
- λ = wavelength
- v= 344m/s (speed of sound in air)
- f = frequency
For the lowest frequency (20 Hz):
\(\lambda = \frac{344}{20} = 17.2 \, \text{meters}\)
For the highest frequency (20,000 Hz or 20 kHz):
\(\lambda = \frac{344}{20000} = 0.0172 \, \text{meters}\)
- Wavelength at 20 Hz = 17.2 meters
- Wavelength at 20 kHz = 0.0172 meters
Question 7: 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.
Answer 7: Consider the length of the aluminium rod = d
Speed of sound wave at 25° C, V Al = 6420 ms-1
Time taken to reach the other end is,
T Al = d/ (V Al) = d/6420
Speed of sound in air, V air = 346 ms-1
Time taken by sound to each other end is,
T air = d/ (V air) = d/346
Therefore, the ratio of time taken by sound in aluminium and air is,
T air / t Al = 6420 / 346 = 18.55
Question 8: The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer 8: Vibrations made in 1 second = Frequency = 100
Vibrations made in 1 minute or in 60 s = 100 × 60 = 6,000
Hence, the source vibrates 6000 times in a minute, producing a frequency of 100 Hz.
Question 9: Does sound follow the same laws of reflection as light does? Explain
Answer 9: Yes, sound also follows the same laws of reflection as light waves. The laws of reflection of sound are as follows:
- The incident sound wave, the reflected sound wave and the normal drawn at the point of incidence. All three are in the same plane.
- The normal drawn at the point of incidence of the reflecting surface and the angle between the direction of incidence and the direction of reflection of the sound are equal.
Question 10: 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 10: An echo is heard when the time for the reflected sound is heard after 0.1 s
Time Taken= Total Distance / Velocity
On a hotter day, the velocity of sound is more. If the time taken by echo is less than 0.1 sec it will not be heard.
Question 11: Give two practical applications of reflection of sound waves.
Answer 11: Following are the two practical applications of reflection of sound waves:
(a) SONAR: SONAR is a technology where reflection of sound is used to measure the distance and speed of underwater objects.
(b) Stethoscope: A stethoscope is a device where the sound of the patient’s heartbeat reaches the doctor’s ear by multiple reflection of sound.
Question 12: 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.
Answer 12: Height (s) of tower = 500 m
Velocity (v) of sound = 340 m s−1
Acceleration (g) due to gravity = 10 m s−1
Initial velocity (u) of the stone = 0
Time (t1) taken by the stone to fall to the tower base:
As per the second equation of motion,
s= ut1 + (½) g (t1)2
500 = 0 x t1 + (½) 10 (t1)2
(t1)2 = 100
t1 = 10 s
Time (t2) taken by sound to reach the top from the tower base = 500/340 = 1.47 s
t = t1 + t2
t = 10 + 1.47
t = 11.47 s
Question 13: 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?
Answer 13: We are given:
- Speed of sound, v = 339 m/s
- Wavelength, λ = 1.5 cm = 0.015 m
Use the formula to find frequency
\(f = \frac{v}{\lambda} = \frac{339}{0.015} = 22,600 \, \text{Hz}\)
Check if it’s audible
The human hearing range is 20 Hz to 20,000 Hz.
Since 22,600 Hz is above 20,000 Hz, it is not audible to humans.
The frequency of the wave is 22,600 Hz, and it is not audible to humans.
Question 14: What is reverberation? How can it be reduced?
Answer 14: The persistence of sound in an auditorium is the result of repeated reflections of sound and is called reverberation. To reduce the undesirable effects due to reverberation, roofs and walls of the auditorium are generally covered with sound absorbent materials like compressed fiberboard, rough plaster or draperies. The seat materials are also selected having sound absorption properties.
Question 15: What is loudness of sound? What factors does it depend on?
Answer 15: Loudness is the physiological response of the human ear to the intensity of sound. It depends mainly on the amplitude of the vibrations that produce the sound—greater amplitude means louder sound. The force applied to make an object vibrate affects the amplitude, and therefore, the loudness. Loudness also depends on the sensitivity of the ear and the medium through which the sound travels. While the amount of vibrating air plays a role, it is essentially a result of the amplitude of the sound wave.
Question 16: How is ultrasound used for cleaning?
Answer 16: Ultrasounds are used to clean parts located in hard-to-reach places e.g., complicated electronic components, spiral or odd shaped parts. Appliances to be cleaned are placed in cleaning solutions and ultrasonic waves are sent through cleaning solution. Due to high frequency of ultrasounds, the dust, oil, grease and dirt get detached. The object, thus, gets thoroughly cleaned.
Question 17: Explain how defects in a metal block can be detected using ultrasound.
Answer 17: Defects in metal blocks do not allow ultrasound to pass through them and 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. The defective part of the metal block does not allow ultrasound to pass through it. As a result, it will not be detected by the detector. Hence, defects in metal blocks can be detected using ultrasound.
