1. Nature of Sound Waves
Sound is a longitudinal wave that requires a material medium (solid, liquid, or gas) to travel. It propagates as a series of compressions (high-pressure regions) and rarefactions (low-pressure regions).
2. Characteristics of a Sound Wave
Amplitude: The magnitude of maximum displacement of particles from their mean position. It determines the loudness of the sound.
Frequency: The number of vibrations per second. It determines the pitch of the sound (high or low). Its unit is Hertz (Hz).
Waveform: The shape of the wave. It determines the quality or timbre of the sound, which helps us distinguish between different sounds of the same pitch and loudness.
3. Speed of Sound
It is affected by the temperature, humidity, and density of the medium.
4. Reflection of Sound (Echo)
An echo is a reflected sound that we hear separately from the original sound. For an echo to be heard, the reflecting surface must be at least 17 meters away from the source (at 22°C).
5. Ultrasound
These are sound waves with a frequency above 20,000 Hz, which is beyond the range of human hearing. They are used in sonar (to find the depth of the sea or locate underwater objects), medical imaging, and cleaning.
6. Resonance
When the natural frequency of a vibrating object matches the frequency of the sound wave forced upon it, the object vibrates with a much larger amplitude. This phenomenon is called resonance.
7. Free and Forced Vibrations
Free Vibration: When a body vibrates with its own natural frequency without any external force.
Forced Vibration: When a body is made to vibrate by an external periodic force of a frequency other than its natural frequency.
This chapter explains how sound is produced, how it travels, and the principles behind common phenomena like echoes and the use of ultrasound.
EXERCISE. 7 (A)
Question 1: Define following terms in relation to a wave:
(a) amplitude
(b) frequency
(c) wavelength and
(d) wave velocity
Ans:(a) Amplitude
The amplitude of a wave is the maximum displacement of a particle from its rest position as the wave passes through. In simpler terms, it is a measure of the wave’s height or intensity. For a wave in water, it would be the height from the calm water level to the top of a crest. A larger amplitude means the wave carries more energy; for instance, a loud sound has a higher amplitude than a quiet one.
(b) Frequency
Frequency refers to the number of complete wave cycles that pass a fixed point in one second. It is measured in Hertz (Hz). If you watch waves hit a shore, a high frequency means many waves are arriving each second, while a low frequency means they are spaced further apart. In sound, a high frequency corresponds to a high-pitched note, and a low frequency to a low-pitched note.
(c) Wavelength
Wavelength is the distance between two successive, identical points on a wave. This is commonly measured from crest to crest or from trough to trough. It represents the length of one complete cycle of the wave. Imagine it as the distance between two neighbouring wave peaks. Long wavelengths are associated with low frequencies, while short wavelengths are associated with high frequencies.
(d) Wave Velocity
Wave velocity is the speed at which a wave propagates, or travels, through a medium. It defines how fast the wave’s energy is transferred from one point to another. For example, sound waves travel at about 343 meters per second in air, while light waves travel at nearly 300 million meters per second in a vacuum. The velocity of a wave is calculated by multiplying its frequency by its wavelength.
Question 2: A wave passes from one medium to another medium. Mention the property of the wave (i) which changes, (ii) which does not change.
Ans: (i) Property which changes:
Speed and wavelength.
(ii) Property which does not change:Frequency.
Question 3: State two factors on which the speed of a wave travelling in a medium depends.
Ans:The speed of a wave travelling in a medium primarily depends on two factors:
The elastic properties of the medium: A wave travels faster in media that are more rigid or have a higher elasticity. For instance, sound waves travel much faster in solid steel than in air because the particles in a solid are more tightly bound and can transmit vibrations more quickly.
The inertial properties (density) of the medium: The speed of a wave is slower in denser media if the elasticity is constant. This is because a denser material has greater inertia, making it harder for the particles to move and transfer energy. For example, sound travels slower in dense carbon dioxide compared to less dense air.
Question 4: State two differences between the light and sound waves.
Ans: Two differences between light and sound waves are:
First, light waves are electromagnetic and can travel through a vacuum, like the empty space from the sun to Earth. Sound waves, however, are mechanical vibrations and require a material medium, such as air or water, to travel through; they cannot propagate in a vacuum.
Second, light waves are transverse, meaning their vibrations are perpendicular to the direction of energy travel. In contrast, sound waves are longitudinal, meaning the particles vibrate back and forth parallel to the direction the wave is moving.
Question 5: What is meant by an echo? What is the condition necessary for an echo to be heard distinctly?
Ans:An echo is a distinct, reflected sound that we hear when the original sound wave bounces off a large and hard obstacle, like a cliff or a building, and returns to our ears after a short delay.
For an echo to be heard distinctly, two main conditions are necessary. First, the reflecting surface must be located at a sufficient distance from the sound source. This is because the human ear can only distinguish between two sounds if they are received at least 0.1 seconds apart. Since sound travels at approximately 342 meters per second in air, the minimum required distance between the observer and the obstacle is 17.1 meters. Second, the reflecting surface must be large and hard, such as a wall or a mountain, so that it can reflect the sound clearly and with enough intensity to be audible. Soft and porous surfaces tend to absorb sound waves, preventing a clear echo.
Question 6: A man is standing at a distance of 12 m from a cliff. Will he be able to hear a clear echo? Give a reason for your answer.
Ans:A clear echo requires a minimum distance of 17 meters from the reflecting surface. This is necessary for the sound to travel back to the listener after a delay of at least 0.1 seconds, which is the time needed for the human ear to perceive two separate sounds. Since sound travels at about 340 meters per second, it must cover a round trip of 34 meters. With the wall only 12 meters away, the sound returns too quickly, making a distinct echo inaudible.
Question 7: State two applications of echo.
Ans:Two common applications of echo are:
Firstly, echo is utilized in sonar technology. Ships use this principle by sending sound waves through water. When these waves hit an object like a submarine or the seabed, they reflect back as an echo. By measuring the time taken for the echo to return, the distance to that object can be accurately calculated, aiding in navigation and mapping.
Secondly, echocardiography is a vital medical application. In this diagnostic procedure, high-pitched sound waves (ultrasound) are sent into a patient’s body. The echoes produced when these waves bounce off the heart structures are used to create detailed moving images. This allows doctors to assess the health and function of the heart valves and chambers without any surgery.
Question 8: Explain how the speed of sound can be determined by the method of echo.
Ans: Measure the Speed of Sound with an Echo
You can calculate the speed of sound using a simple echo method in a large space with a flat wall.
What you need: A long tape measure, a stopwatch, and two blocks of wood to make a sharp clap.
The Method:
Measure: Find the exact distance (‘d’) from your clapping spot to the wall.
Time: Clap the wood together and start the stopwatch at the same moment. Stop it the instant you hear the echo. This time (‘t’) is for the sound to travel to the wall and back, a total distance of 2 × d.
Calculate: Use the formula: Speed of Sound = (2 × d) / t.
Example: If you are 50 meters from the wall and the echo takes 0.3 seconds, the calculation is (2 × 50) / 0.3 ≈ 333 meters per second.
Pro Tip for Accuracy:
Instead of timing one clap, clap in a steady rhythm that matches the returning echoes. Time 10 or 20 claps, then divide the total time by the number of claps to find a much more accurate average time for a single echo. This reduces errors caused by your reaction time.
This hands-on experiment is a fantastic way to discover a fundamental property of physics using everyday items.
Question 9: State the use of echo by a bat, dolphin and fisherman.
Ans: The use of echo, or echolocation, is a natural phenomenon used by animals and humans to navigate and locate objects by interpreting the reflection of sound waves. Here’s how it is specifically used by a bat, a dolphin, and a fisherman:
Bat:
Bats are nocturnal creatures that fly and hunt in complete darkness. They use echolocation brilliantly for navigation and hunting insects. A bat emits high-pitched ultrasonic squeaks that are inaudible to human ears. These sound waves travel forward and bounce off objects like trees, walls, or flying insects. By listening to the returning echo, the bat can determine the size, shape, distance, and even the direction of movement of the object. This allows it to weave through complex environments and catch its prey with incredible accuracy without relying on sight.
Dolphin:
Dolphins live in the ocean, where visibility is often very low. A dolphin produces a series of clicks, which are sound waves emitted from an organ in its head called the melon. These clicks travel through the water and bounce off objects like fish, rocks, or other dolphins. By analyzing the returning echo, the dolphin can not only find its prey but also judge its size, density, and internal structure. This ability is crucial for hunting, navigating murky waters, and communicating with their pod.
Fisherman:
Fishermen use a man-made electronic application of echolocation called a fishfinder or sonar. A device on their boat sends sound waves (pings) down into the water. When these waves hit objects like the seabed, a sunken log, or a school of fish, they reflect back to the device. The fisherman then sees this information displayed on a screen, showing the depth of the water and the location of fish. This helps them to find the best fishing spots efficiently, saving time and increasing their catch, instead of relying solely on guesswork.
Question 10: How do bats avoid obstacles in their way, when in flight?
Ans: To navigate the pitch-black world of night, bats rely on a remarkable natural sonar system known as echolocation. Since their eyes are of little use in total darkness, they have evolved this sophisticated auditory technique. They emit a series of high-frequency clicks and chirps, so shrill that they are inaudible to human ears. These emitted sound waves travel outward until they strike an object in the bat’s path, such as a branch or a building. The resulting echo then bounces back to the bat’s highly sensitive ears. By interpreting the subtle differences in the timing and quality of these returning signals, the bat can instantly calculate the distance to the object, determine its general shape and size, and even discern its surface texture. This incredible biological system allows bats to fly at high speeds through complex environments and capture tiny, fast-moving insects, all without the faintest glimmer of light.
Question 11: What is meant by sound ranging? Give one use of sound ranging.
Ans:Sound ranging locates distant sounds, like enemy artillery, by measuring the sound’s arrival time at several microphones in known positions. By comparing these times, the source of the sound can be precisely pinpointed. This method is a key military tool for detecting hidden enemy gun positions.
Question 12: Name the waves used for sound ranging. Why are the waves mentioned by you audible to us?
Ans: The statement is correct: the waves used for sound ranging are indeed ultrasonic waves.
Here’s a simple breakdown of why that is:
Ultrasonic waves are simply sound waves, but they vibrate at a frequency that is too high for our ears to detect. Human hearing has its limits; we can generally hear sounds within a range of about 20 vibrations per second (Hertz) up to 20,000 Hertz. Anything with a frequency higher than that upper limit of 20,000 Hz is called “ultrasound,” and it’s completely silent to us.
So, why use a sound we can’t hear for a task like “ranging” (which means finding the distance or location of an object)?
The key reasons are inaudibility and directionality.
They are Silent and Unobtrusive: Imagine a ship using sound to map the ocean floor or a doctor imaging a fetus. If they used audible sound waves, the process would create a loud, constant, and distracting noise. Using ultrasonic waves allows these technologies to operate quietly and without causing a disturbance.
They are Highly Directional and Accurate: High-frequency ultrasonic waves have a very short wavelength. This allows them to be focused into a tight, narrow beam, much like a flashlight beam compared to the glow of a bare lightbulb. This focused beam travels straight to a target, reflects off it, and returns clearly. By measuring the time it takes for this “ping” to return, the device can calculate the distance to the object with great precision. Lower frequency, audible sounds tend to spread out in all directions and are less effective for this kind of accurate targeting.
Question 13: What is sonar? State the principle on which it is based.
Ans:Sonar (which stands for Sound Navigation Ranging) is a technology that uses sound waves to navigate, communicate, and detect objects under the surface of water. It is often called echo-location and is used extensively in applications like mapping the ocean floor, locating underwater shipwrecks, detecting submarines, and by fishermen to find schools of fish.
More specifically, it works on the principle of sending out a burst of high-frequency sound waves (often called a “ping”) through the water and then listening for the returning echo. When these sound waves encounter an object or the seabed, they get reflected back. By carefully measuring the time taken for the sound wave to travel to the object and back as an echo, we can calculate the distance to that object.
This is calculated using a simple formula derived from the basic equation for speed:
Distance = (Speed of Sound in Water × Time) / 2
Where:
Speed of Sound in Water is approximately 1500 meters per second (it varies slightly with temperature and salinity).
Time is the total time taken for the sound pulse to go to the object and return.
We divide by 2 because the time measured is for the round trip (to the object and back), so we need to find the one-way distance.
Question 14: Name the waves which are used in sonar to find the depth of a sea. Give one reason for their use.
Ans: The statement is correct: the waves used in sonar to find the depth of the sea are indeed ultrasonic waves.
Why Ultrasonic Waves are Used
The primary reason for using ultrasonic waves (sound waves with a frequency higher than what the human ear can hear) in sonar comes down to their specific physical properties, which make them ideal for underwater exploration. Here’s a breakdown of the key reasons:
High Directivity and Resolution: Due to their very high frequency, ultrasonic waves have a much shorter wavelength compared to ordinary sound waves. This short wavelength allows them to be focused into a sharp, narrow beam, much like a flashlight beam. This focused beam can be directed precisely towards the seabed. When it reflects back, it provides a clear and accurate “echo,” allowing for precise measurement of the depth and detailed mapping of the ocean floor. Lower frequency sounds would spread out more, leading to fuzzy and less accurate readings.
Minimal Absorption and Long-Distance Travel: While water absorbs and scatters sound waves, the absorption is less severe for certain frequencies. Ultrasonic frequencies, particularly those in the range used for sonar (tens to hundreds of kilohertz), represent a good compromise. They can penetrate water effectively and travel for several kilometers without losing too much energy. This allows a single sonar pulse to probe great depths. If the frequency were too high, the water would absorb the sound too quickly.
Effective Reflection from Small Objects: The short wavelength of ultrasonic waves means they can be reflected effectively even from small objects and fine details on the sea bed. This is crucial for creating detailed images and for detecting underwater features, wrecks, or even schools of fish. A longer wavelength might simply pass over small obstructions without reflecting a strong enough signal.
Avoidance of Ambient Noise: The underwater environment is naturally filled with noise from waves, marine life (like whale songs and shrimp snaps), and ship traffic. Most of this background clutter exists at lower, audible frequencies. By operating at these high, ultrasonic frequencies, sonar systems can effectively “tune out” this ambient noise, ensuring that the returning echo is clear and distinguishable from the background racket.
Question 15: State the use of echo in medical science.
Ans:Use of Echo in Medical Science:
In medical science, echo or echocardiography is a diagnostic test. It uses ultrasound waves to create real-time images of the heart.
Its main uses are to:
Check the heart’s size, shape, and pumping function.
Assess the structure and movement of the heart valves.
Detect problems like heart muscle damage, birth defects, and fluid around the heart.
MUTIPLE CHOICE TYPE:
Question 1: The minimum distance between the source and the reflector in air. So that an echo is heard is approximately equal to:
(a) 10 m
(b) 17 m
(c) 34 m
(d) 50 m
Ans: (b) 17 m
Question 2: To detect the obstacles in their path, bats produce:
(a) infrasonic waves
(b) ultrasonic waves
(c) electromagnetic waves
(d) radio waves
Ans: (b) ultrasonic waves
NUMERICALS:
Question 1: The wavelength of waves produces on the surface of water is 20 cm. If the wave velocity is 24 m s-1, calculate: (i) the number of waves produces in one second , and (ii) the time in which one wave is produced.
Ans:Given:
Wavelength,
λ=20 cm=0.20 m
Wave velocity,
v=24 m/s
(i) Number of waves produced in one second
This is the frequency
v=fλ:
f=v/λ
=24/0.20
=120 Hz
So, 120 waves are produced in one second.
(ii) Time in which one wave is produced
This is the time period T
T=1/f
=1/120 s
s≈0.00833 s
Final Answer:
(i) 120 waves per second(ii) 0.00833 s
Question 2: Calculate the minimum distance in air required between the source of sound and the obstacle to hear an echo. Take speed of sound in air = 350 m s-1
Ans: Step 1: Understanding the echo condition
The human ear can distinguish between two sounds if the time gap between them is at least 0.1 seconds.
Step 2: Relationship between distance, speed, and time
Let the distance between the source and the obstacle be d
For an echo, sound travels to the obstacle and back, so total distance = 2d.
2d=v×t
where
v=350 m/s and
t=0.1 s.
Step 3: Calculation
2d=350×0.1
2d=35
d=17.5 m
Final answer:17.5
The minimum distance required to hear an echo in air is 17.5 m.
Question 3: What should be the minimum distance between source and reflector in water so that echo is heard distinctly? (The speed of sound in water = 1400 m s-1)
Ans: Step 1: Condition for hearing echo distinctly
The minimum time interval between the original sound and its echo must be at least 0.1 s for the human ear to distinguish them.
Step 2: Formula
Total distance travelled by sound = speed × time
Let d be the distance between source and reflector.
Sound goes to the reflector and comes back, so total distance = 2d.
2d=v×t
Step 3: Substitution
Given:
v=1400 m/s,
t=0.1 s
2d=1400×0.1
2d=140
d=70 m
Final answer:70 m
Question 4: A man standing 25 m away from a wall produces a sound and receives the reflected sound. (a) Calculate the time after which he receives the reflected sound if the speed of the sound in air is 350 m s-1. (b) will the man be able to hear a distinct echo? Explain the answer.
Ans: (a) Calculation of time
Distance from the man to the wall = 25 m.
Since the sound travels to the wall and returns, total distance = 25×2=50 m.
Speed of sound = 350 m/s.
Time=Total distance/Speed
=50/350
≈0.143 seconds.
(b) Will he hear a distinct echo?
Yes, because the time delay of 0.143 s is greater than 0.1 s, which is the minimum required for the human ear to perceive a distinct echo.
Question 5: A radar sends a signal to an aeroplane at a distance 45 km away with a speed of 3 × 108 m s -1. After how much time is the signal received back from the aeroplane?
Ans:Step 1: Understand the problem
The radar signal travels to the aeroplane and back.
Total distance covered by the signal = 2×45 km.
Step 2: Convert distance to metres
45 km=45×1000 m=45000 m
Total distance = 2×45000=90000 m.
Step 3: Use the formula
Time=Total distance/Speed
Time=90000/3×10/8 s
Time= 3×10 *8/90000s
Step 4: Calculate
90000/3×10/8
=9×10*4/3×10*8
=3×10*−4 s
Step 5: Convert to milliseconds (optional)
3×10*−4 s
s=0.0003 s=0.3 ms
Final Answer:0.0003 s
