1How does the sound produced by a vibrating object in a medium reach your ear?
Answer & Reference:
When an object vibrates, it sets the particles of the medium around it vibrating. The particles do not travel all the way from the vibrating object to the ear. A particle of the medium in contact with the vibrating object is first displaced from its equilibrium position. It then exerts a force on the adjacent particle. As a result, the adjacent particle gets displaced from its position of rest. After displacing the adjacent particle, the first particle comes back to its original position. This process continues in the medium till the sound reaches your ear. The disturbance created by a source of sound in the medium travels through the medium and not the particles of the medium. (Page 128, Section 11.2)
2Explain how sound is produced by your school bell.
Answer & Reference:
When the school bell is struck, it vibrates. These vibrations produce sound. Sound is produced due to vibration of different objects. The bell's vibrations create compressions and rarefactions in the surrounding air, which propagate as sound waves to our ears. (Page 128, Section 11.1)
3Why are sound waves called mechanical waves?
Answer & Reference:
Sound waves are called mechanical waves because they require a material medium (solid, liquid or gas) for their propagation. They are characterised by the motion of particles in the medium. The particles of the medium set neighbouring particles into motion, and they in turn produce similar motion in others. (Page 128, Section 11.2)
4Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer & Reference:
No, you will not be able to hear any sound produced by your friend on the moon. This is because sound requires a medium to travel, and the moon has no atmosphere (it's a vacuum). Since there is no medium for sound propagation, the sound cannot reach your ears. (Page 128, Section 11.2)
Answer & Reference:
(a) Loudness is determined by the amplitude of the sound wave.
(b) Pitch is determined by the frequency of the sound wave.
(Page 131, Section 11.2.2)
6Guess which sound has a higher pitch: guitar or car horn?
Answer & Reference:
A guitar has a higher pitch than a car horn. This is because the guitar produces sound with a higher frequency compared to a car horn. Higher frequency corresponds to higher pitch. (Page 131, Section 11.2.2)
7What are wavelength, frequency, time period and amplitude of a sound wave?
Answer & Reference:
- Wavelength (λ): The distance between two consecutive compressions or two consecutive rarefactions. Its SI unit is metre (m). (Page 130)
- Frequency (ν): The number of complete oscillations per unit time. Its SI unit is hertz (Hz). (Page 131)
- Time period (T): The time taken by two consecutive compressions or rarefactions to cross a fixed point, or the time taken for one complete oscillation. Its SI unit is second (s). (Page 131)
- Amplitude (A): The magnitude of the maximum disturbance in the medium on either side of the mean value. For sound, its unit will be that of density or pressure. (Page 131)
8How are the wavelength and frequency of a sound wave related to its speed?
Answer & Reference:
The speed (v), frequency (ν), and wavelength (λ) of a sound wave are related by the equation: v = λν. That is, speed = wavelength × frequency. (Page 132, Section 11.2.2)
9Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Answer & Reference:
Given:
Frequency (ν) = 220 Hz
Speed (v) = 440 m/s
Using the formula: v = λν
Therefore, λ = v/ν = 440/220 = 2 m
The wavelength of the sound wave is 2 metres. (Page 132, Section 11.2.2)
10A 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 & Reference:
The time interval between successive compressions is the time period (T) of the wave.
Given frequency (ν) = 500 Hz
Time period (T) = 1/ν = 1/500 = 0.002 seconds
Therefore, the time interval between successive compressions is 0.002 s. (Page 131, Section 11.2.2)
11Distinguish between loudness and intensity of sound.
Answer & Reference:
- Intensity: The amount of sound energy passing each second through unit area. It is an objective physical quantity.
- Loudness: A measure of the response of the ear to the sound. It is a physiological sensation and depends on the sensitivity of the ear.
Even when two sounds are of equal intensity, we may hear one as louder than the other simply because our ear detects it better. (Page 132, Section 11.2.2)
12In which of the three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer & Reference:
Sound travels fastest in iron (solid), then in water (liquid), and slowest in air (gas). This is because the speed of sound decreases when we go from solid to gaseous state. (Page 133, Section 11.2.3 and Table 11.1)
13An 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 & Reference:
Given:
Speed of sound (v) = 342 m/s
Time taken for echo (t) = 3 s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
Since sound travels to the reflecting surface and back, the distance to the reflecting surface is half of the total distance.
Therefore, distance = 1026/2 = 513 m
The reflecting surface is 513 m from the source. (Page 134, Example 11.2)
14Why are the ceilings of concert halls curved?
Answer & Reference:
The ceilings of concert halls are curved so that sound after reflection reaches all corners of the hall. This ensures that the sound is distributed evenly throughout the hall for better audibility. (Page 135, Section 11.3.3)
15What is the audible range of the average human ear?
Answer & Reference:
The audible range of sound for human beings extends from about 20 Hz to 20000 Hz (20 kHz). (Page 135, Section 11.4)
16What is the range of frequencies associated with (a) Infrasound? (b) Ultrasound?
Answer & Reference:
Sound is a form of energy which produces a sensation of hearing in our ears. Sound is produced due to vibration of different objects. When an object vibrates, it forces the neighbouring particles of the medium to vibrate, creating a disturbance that propagates as sound waves. (Page 127, Introduction and Page 137, "What you have learnt")
18Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Answer & Reference:
When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure called compression (C). When the vibrating object moves backwards, it creates a region of low pressure called rarefaction (R). As the object moves back and forth rapidly, a series of compressions and rarefactions is created in the air, which propagates as sound waves. (Page 128, Section 11.2 and Fig. 11.4)
19Why is sound wave called a longitudinal wave?
Answer & Reference:
Sound waves are called longitudinal waves because the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. In these waves, particles oscillate back and forth about their position of rest along the direction the wave travels. (Page 129, Section 11.2.1)
20Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer & Reference:
The characteristic that helps identify a friend's voice is the quality or timbre of sound. Quality is that characteristic which enables us to distinguish one sound from another having the same pitch and loudness. (Page 132, Section 11.2.2)
21Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Answer & Reference:
Thunder is heard after the flash is seen because light travels much faster than sound. The speed of light is about 3 × 10⁸ m/s, while the speed of sound in air is about 344 m/s. Therefore, the light from the lightning reaches us almost instantly, but the sound takes longer to travel the same distance. (Page 133, Section 11.2.3)
22A 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 & Reference:
Given:
Speed of sound (v) = 344 m/s
For frequency ν₁ = 20 Hz:
Wavelength λ₁ = v/ν₁ = 344/20 = 17.2 m
For frequency ν₂ = 20,000 Hz:
Wavelength λ₂ = v/ν₂ = 344/20000 = 0.0172 m = 1.72 cm
Therefore, the typical wavelengths range from 17.2 m to 1.72 cm. (Page 132, Section 11.2.2)
23Two 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 & Reference:
From Table 11.1 (Page 133):
Speed of sound in aluminium = 6420 m/s
Speed of sound in air (at 25°C) = 346 m/s
Let the length of the rod be L.
Time taken in aluminium, t₁ = L/6420
Time taken in air, t₂ = L/346
Ratio t₂/t₁ = (L/346) / (L/6420) = 6420/346 ≈ 18.55
Therefore, the ratio of times taken by sound in air to that in aluminium is approximately 18.55:1. (Page 133, Table 11.1)
24The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer & Reference:
Frequency = 100 Hz means 100 vibrations per second.
Number of vibrations in 1 minute (60 seconds) = 100 × 60 = 6000
Therefore, the source vibrates 6000 times in a minute. (Page 131, Section 11.2.2)
25Does sound follow the same laws of reflection as light does? Explain.
Answer & Reference:
Yes, sound follows the same laws of reflection as light. The directions in which the sound is incident and is reflected make equal angles with the normal to the reflecting surface at the point of incidence, and the three (incident ray, reflected ray, and normal) lie in the same plane. (Page 133, Section 11.3)
26When 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 & Reference:
On a hotter day, the speed of sound increases (as speed of sound increases with temperature). The time taken for the echo to return would be less than on a cooler day. However, as long as the time interval between the original sound and the reflected sound is at least 0.1 s, we will hear a distinct echo. Since the distance remains the same, and the speed increases, the time interval decreases. If this time interval becomes less than 0.1 s, we may not hear a distinct echo. (Page 134, Section 11.3.1)
27Give two practical applications of reflection of sound waves.
Answer & Reference:
Two practical applications of reflection of sound waves are:
1. Megaphones or loudhailers: Designed to send sound in a particular direction without spreading it in all directions by using multiple reflections.
2. Stethoscope: Used by doctors to listen to sounds produced within the body (like heartbeats) which reach the doctor's ears by multiple reflection of sound.
(Page 134-135, Section 11.3.3)
28A 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 & Reference:
First, calculate the time taken by the stone to reach the pond:
Using s = ut + ½gt²
500 = 0 + ½ × 10 × t₁²
t₁² = 100 → t₁ = 10 s
Now, calculate the time taken by sound to travel 500 m:
t₂ = distance/speed = 500/340 ≈ 1.47 s
Total time = t₁ + t₂ = 10 + 1.47 = 11.47 s
Therefore, the splash is heard at the top after approximately 11.47 seconds. (Page 133, Section 11.2.3)
29A 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 & Reference:
Given:
Speed (v) = 339 m/s
Wavelength (λ) = 1.5 cm = 0.015 m
Frequency (ν) = v/λ = 339/0.015 = 22,600 Hz = 22.6 kHz
The audible range for humans is 20 Hz to 20 kHz. Since 22.6 kHz is above 20 kHz, it is not audible to the average human ear. (Page 132, Section 11.2.2 and Page 135, Section 11.4)
30What is reverberation? How can it be reduced?
Answer & Reference:
Reverberation is the persistence of sound in an auditorium due to repeated reflection from the walls until it is reduced to a value where it is no longer audible. In an auditorium or big hall, excessive reverberation is highly undesirable.
To reduce reverberation, the roof and walls of the auditorium are generally covered with sound-absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are also selected on the basis of their sound absorbing properties. (Page 134, Section 11.3.2)
31What is loudness of sound? What factors does it depend on?
Answer & Reference:
Loudness is a physiological response of the ear to the intensity of sound. It depends on:
1. The amplitude of the sound wave - greater amplitude means louder sound.
2. The sensitivity of the ear - even sounds of equal intensity may be perceived as having different loudness based on how well the ear detects them.
(Page 132, Section 11.2.2 and Page 138, "What you have learnt")
32How is ultrasound used for cleaning?
Answer & Reference:
Ultrasound is used to clean parts located in hard-to-reach places (e.g., spiral tubes, odd-shaped parts, electronic components). Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out, resulting in thorough cleaning. (Page 136, Section 11.5)
33Explain how defects in a metal block can be detected using ultrasound.
Answer & Reference:
Ultrasound waves are allowed to pass through the metal block. Detectors are used to detect the transmitted waves. If there is even a small defect (crack or hole) inside the metal block, the ultrasound gets reflected back indicating the presence of the flaw or defect. Ordinary sound of longer wavelengths cannot be used for this purpose as it will bend around the corners of the defective location and enter the detector. (Page 136, Section 11.5 and Fig. 11.14)
34Provide the speed of sound in different media at 25°C as given in Table 11.1.