RBSE Solutions Class 9 Science Chapter 11 Sound

Objective Type Questions

 

Question 1. Wave that contains compression and rarefaction is called
(A) Transverse wave
(B) Longitudinal waves
(C) Light rays
(D) Ultraviolet ray
Answer: (B) Longitudinal waves
In simple words: A longitudinal wave makes the medium move back and forth in the same direction the wave travels. This creates squeezed parts (compressions) and stretched parts (rarefactions).

🎯 Exam Tip: Remember that sound waves are a common example of longitudinal waves, where particles vibrate parallel to the wave's direction.

 

Question 2. The relation between wave velocity (V), wave length and frequency (v) is:
(A) \( V = \nu\lambda \)
(B) \( \lambda = V\nu \)
(C) \( V =\lambda\nu \)
(D) \( V= \lambda/\nu \)
Answer: (A) \( V = \nu\lambda \)
In simple words: The speed of a wave is found by multiplying how often it wiggles (frequency) by the distance between its wiggles (wavelength). This formula helps us understand how waves move.

🎯 Exam Tip: Ensure you correctly identify the symbols: V for velocity, \( \nu \) (nu) for frequency, and \( \lambda \) (lambda) for wavelength. This is a fundamental wave equation.

 

Question 3. Longitudinal waves can be produced in:
(A) Solid and liquid
(B) Solids and gases
(C) Gases and liquids
(D) Solid, liquid and gases
Answer: (D) Solid, liquid and gases
In simple words: Longitudinal waves, like sound, can travel through all three main types of matter: solids (like a wall), liquids (like water), and gases (like air). This is because they rely on particles bumping into each other.

🎯 Exam Tip: The ability of longitudinal waves to travel through all states of matter distinguishes them from transverse waves, which often require a more rigid medium or boundary.

 

Question 4. In longitudinal waves, the vibration of the particles of the medium:
(A) is /are always in the direction of the wave
(B) is/ are always perpendicular to the direction of the wave
(C) molecules do not vibrate at all
(D) at an angle of 600 from the direction of wave
Answer: (A) is /are always in the direction of the wave
In simple words: When a longitudinal wave moves, the tiny bits of the material it passes through shake back and forth in the same way the wave is traveling. It's like a Slinky toy being pushed from one end.

🎯 Exam Tip: Visualizing a Slinky spring is a great way to remember how particles in a longitudinal wave move parallel to the wave's direction of propagation.

 

Question 6. In a clock, the time period of hour hand is:
(A) 1 hour
(B) 24 hours
(C) 12 hours
(D) None of the options
Answer: (C) 12 hours
In simple words: The hour hand on a clock takes 12 hours to complete one full circle and return to its starting position. This duration is its time period.

🎯 Exam Tip: Be careful not to confuse the hour hand's period with the minute hand's (1 hour) or the second hand's (1 minute).

 

Question 7. The speed of a wave is 350 m/s and its wavelength 50 cm, then the frequency of the wave will be:
(A) 13500 Hz
(B) 700 Hz
(C) 400 Hz
(D) 300 Hz
Answer: (B) 700 Hz
In simple words: If a wave travels at 350 meters per second and each wave is 50 centimeters long, it means 700 waves pass by every second. We first convert centimeters to meters, then divide speed by wavelength to get frequency.

🎯 Exam Tip: Always ensure all units are consistent (e.g., meters for wavelength if speed is in m/s) before performing calculations using the wave equation \( V = \nu\lambda \).

 

Question 8. A number of oscillation completed per second is called:
(A) Amplitude
(B) Speed
(C) Time period
(D) Frequency
Answer: (D) Frequency
In simple words: Frequency is how many times something wiggles back and forth in just one second. It tells us how fast the vibrations are.

🎯 Exam Tip: Frequency is measured in Hertz (Hz), where 1 Hz means one oscillation per second. It's the inverse of the time period.

 

Question 9. The time period of a vibrating body is 0.02s. The frequency of the vibrating body will be
(A) 100 Hz
(B) 20 Hz
(C) 50 Hz
(D) 1Hz
Answer: (C) 50 Hz
In simple words: If something takes 0.02 seconds to complete one shake, then it will shake 50 times in one second. Frequency is simply one divided by the time period.

🎯 Exam Tip: Remember the inverse relationship: frequency \( (\nu) = 1 / \text{Time Period (T)} \). If you know one, you can easily find the other.

 

Question 11. Human audible range is:
(A) 200 Hz to 20000 Hz
(B) 20Hz to 20000Hz
(C) 2 Hz to 20 Hz
(D) Above 20000Hz
Answer: (B) 20Hz to 20000Hz
In simple words: Humans can hear sounds that have a frequency between 20 shakes per second (very low sound) and 20,000 shakes per second (very high sound). Sounds outside this range are either infrasound or ultrasound.

🎯 Exam Tip: Frequencies below 20 Hz are infrasound, and those above 20,000 Hz are ultrasound. Animals like bats and dogs can hear frequencies outside the human range.

 

Question 12. The minimum time required to hear an echo,
(A) 0.1 S
(B) 0.5 S
(C) 1 S
(D) 2 S
Answer: (A) 0.1 S
In simple words: To clearly hear an echo, the reflected sound must reach your ears at least 0.1 seconds after you hear the original sound. This small time gap helps your brain tell the sounds apart.

🎯 Exam Tip: The minimum distance for an echo (at 20°C) is calculated using this time interval and the speed of sound, which is about 17.2 meters.

 

Question 13. The frequency of the waves used in ultrasonography is:
(A) 20 Hz
(B) Below 20 Hz
(C) 20 Hz to 20000 Hz
(D) Above 20000 Hz
Answer: (D) Above 20000 Hz
In simple words: Ultrasonography uses sound waves that are too high-pitched for humans to hear, meaning their frequency is greater than 20,000 Hertz. These high-frequency waves create images inside the body.

🎯 Exam Tip: Ultrasound is specifically sound with a frequency above the human hearing limit, making it useful for medical imaging and industrial applications.

 

Question 14. The unit of amplitude is:
(A) m
(B) m/s
(C) Hz
(D) None of the options
Answer: (A) m
In simple words: Amplitude measures how far a wave moves from its central resting position. Since it's a distance, its unit is meters (m). For a sound wave, amplitude relates to how loud it is.

🎯 Exam Tip: Amplitude is a displacement, so its unit is a unit of length (like meters, centimeters, or millimeters), not speed or frequency.

Sound Very Short Answer Type Questions

 

Question 16. Name the medium in which longitudinal waves can be produced.
Answer: Longitudinal waves can be made in all three types of mediums: solids, liquids, and gases. This is because they travel by pushing and pulling the particles of the medium. For example, sound waves travel through air, water, and even solid walls.
In simple words: Longitudinal waves can be made in solids, liquids, and gases.

🎯 Exam Tip: Remember that sound waves are a classic example of longitudinal waves, and they travel through almost any material.

 

Question 17. Name the type of sound waves produced in iron?
Answer: Longitudinal waves are produced in iron. Sound waves travel through solids, like iron, as longitudinal waves by compressing and expanding the material. Iron is an excellent conductor of sound.
In simple words: Sound waves in iron are longitudinal waves.

🎯 Exam Tip: In most solid materials, sound travels as a longitudinal wave, where particles vibrate in the same direction as the wave propagation.

 

Question 18. Name the sound wave produced in the air?
Answer: Longitudinal waves are produced in the air. Sound travels through the air by creating compressions (areas of high pressure) and rarefactions (areas of low pressure) as air particles vibrate back and forth. This is why sound needs a medium to travel.
In simple words: Sound waves in the air are longitudinal waves.

🎯 Exam Tip: Air is a gaseous medium, and sound always travels through gases as longitudinal waves due to the molecular interactions.

 

Question 19. A wire is tied between the two knobs and released perpendicular to its length. Name the wave produced in the wire.
Answer: Transverse waves are produced in the wire. When a wire is plucked or vibrated perpendicular to its length, the particles of the wire move up and down, while the wave itself travels along the length of the wire. This motion is characteristic of a transverse wave.
In simple words: Shaking a wire sideways creates transverse waves.

🎯 Exam Tip: In transverse waves, the particle motion is perpendicular to the direction the wave travels, like ripples on water or a wave on a string.

 

Question 20. What is the S.I unit of wavelength?
Answer: The S.I. unit of wavelength is the Metre (m). Wavelength is a measure of distance between two consecutive identical points on a wave, so it is expressed in units of length. It's a key property for describing wave size.
In simple words: Wavelength is measured in meters (m).

🎯 Exam Tip: Always use standard SI units in physics calculations unless specified otherwise, to avoid errors.

 

Question 21. What is the S.I unit of frequency?
Answer: The S.I. unit of frequency is Hertz (Hz). One Hertz means one oscillation or wave cycle per second. Frequency tells us how many times a wave repeats itself in a given period.
In simple words: Frequency is measured in Hertz (Hz).

🎯 Exam Tip: Hertz is a derived unit equal to \( \text{s}^{-1} \), meaning "per second."

 

Question 23. What type of motion do the hands of a clock exhibits?
Answer: The hands of a clock show periodic motion. Periodic motion is a type of motion that repeats itself at regular intervals of time. The clock hands move in a circle and return to the same position after a fixed period.
In simple words: Clock hands move in a repeating pattern, which is called periodic motion.

🎯 Exam Tip: Remember that any motion that repeats itself in a fixed time interval, like the Earth revolving around the Sun or a pendulum swinging, is periodic motion.

 

Question 24. Define wavelength?
Answer: Wavelength is the distance between two nearby points on a wave that are in the exact same phase. For example, in a transverse wave, it is the distance between two consecutive crests or two consecutive troughs. It is usually shown by the symbol \( \lambda \) (lambda) and its S.I. unit is meters (m).
In simple words: Wavelength is the length of one complete wave, measured from one peak to the next, or one valley to the next.

🎯 Exam Tip: Clearly define "in the same phase" by using examples like crest-to-crest or compression-to-compression. Always include the symbol and SI unit.

 

Question 25. Define frequency.
Answer: Frequency is the number of complete wave cycles or oscillations that pass a certain point in one second. It tells us how often a wave repeats. The S.I. unit of frequency is Hertz (Hz), where 1 Hertz means 1 wave cycle per second. It is the inverse of the time period.
In simple words: Frequency is how many wave cycles happen in one second.

🎯 Exam Tip: Frequency and time period (T) are inversely related: \( \nu = 1/T \). Understanding this relationship is crucial for solving wave problems.

 

Question 26. What is RADAR?
Answer: RADAR is an acronym that stands for RADIO DETECTION AND RANGING. It is a technique that uses radio waves to find out if objects are present in the air. RADAR works by sending out radio waves and listening for their echoes to detect objects like airplanes or storms.
In simple words: RADAR uses radio waves to find and locate objects far away.

🎯 Exam Tip: Know the full form and the basic principle of RADAR (radio wave reflection) for a complete answer.

Sound Short Answer type Questions

 

Question 28. What do you understand by wave motion?
Answer: Wave motion is a way for energy to move from one place to another without the actual material itself moving. It is a disturbance that travels through a physical medium, carrying energy but not matter. Think of ripples spreading in water.
In simple words: Wave motion is when energy moves through something, but the material itself stays in place.

🎯 Exam Tip: Emphasize that waves transfer energy, not matter, for a clear and concise definition.

 

Question 29. What are Longitudinal waves?
Answer: Longitudinal waves are waves where the particles of the medium vibrate back and forth (oscillate) in the same direction that the wave is traveling. They consist of compressions (squeezed parts) and rarefactions (stretched parts). A good example of this is a sound wave.
Examples:
(a) Sound waves are longitudinal waves.
(b) The waves produced in a spring when pushed and pulled are longitudinal.
In simple words: In longitudinal waves, particles wiggle in the same direction the wave moves, like sound traveling through air.

🎯 Exam Tip: Always remember the key characteristic: particle vibration is *parallel* to wave propagation. Providing examples helps illustrate the concept.

 

Question 30. What is necessary for wave propagation?
Answer: Waves generally need a medium to travel through. This medium is the substance that the wave moves in, such as a solid, liquid, or gas. Mechanical waves, like sound waves, require a material medium such as air, water, or steel to move. This means they cannot travel in a vacuum where there is no matter. Electromagnetic waves, however, can travel in a vacuum.
In simple words: Most waves need a material (like air or water) to travel through. They cannot move if there is nothing there, like in empty space.

🎯 Exam Tip: Distinguish between mechanical waves (which *require* a medium) and electromagnetic waves (which do *not* require a medium) when discussing wave propagation.

 

Question 31. In wave propagation, from one position to another position, what type of transfer takes place, 'Energy or matter".
Answer: In wave propagation, energy is transferred from one position to another. The matter in the medium only oscillates around its mean position and does not travel with the wave. This is a fundamental concept of wave motion.
In simple words: Waves move energy, not the material itself.

🎯 Exam Tip: This is a key distinction of wave motion: energy transfer without net mass transfer. Give a simple example like a ripple in water where the water moves up and down but not across the pond.

 

Question 32. Give two examples of Longitudinal waves?
Answer: Two examples of longitudinal waves are:
1. Sound waves: These waves travel through the air, water, or solids by making particles vibrate back and forth in the same direction as the sound moves.
2. Waves in a slinky spring: When you push and pull one end of a stretched Slinky spring, a wave of compressions and rarefactions travels along the spring.
In simple words: Sound and waves in a pushed Slinky are two examples of longitudinal waves.

🎯 Exam Tip: For examples, choose common and easily visualizable phenomena that clearly demonstrate the parallel particle vibration.

 

Question 33. Is the bullet fired from a gun and a stone from a slingshot follow wave motion?
Answer: Yes, the bullet fired from a gun and a stone from a slingshot will follow wave motion. While they appear as solid objects, their travel through the air creates shockwaves or sound waves, which are forms of wave motion. Also, in a broader sense, their trajectories can be analyzed using wave-like mathematical models in fluid dynamics, although typically we associate them with projectile motion.
In simple words: Yes, a bullet and a stone from a slingshot create waves as they fly, mainly sound or shockwaves.

🎯 Exam Tip: Think about the disturbance created by rapidly moving objects. While the object itself moves, it also generates waves in the medium it passes through.

 

Question 34. Describe the uses of RADAR and its system of locating position?
Answer: RADAR has several important uses and a specific system for locating positions:
**Uses of RADAR:**
1. It helps in weather forecasting, making predictions more accurate.
2. It is used to find and track surface-to-air missiles (SAMs).
3. It helps locate the position of enemy or unknown aircraft.
4. Scientists often use it to track how animals migrate.
5. It can determine the distance, height, direction, and speed of moving and fixed objects like aircraft, ships, and spacecraft.
**RADAR as a position locating device:**
Modern RADAR devices have a video monitor that shows where detected objects are. The computer in the RADAR measures the time it takes for the signal to go out, hit a target, and bounce back. Then, it calculates the distance to the target based on this time and the speed of radio waves. The basic idea is that a signal is sent, reflects off an object, and is then received by a detector. A cathode ray oscilloscope in the receiver then shows the object's exact position.
In simple words: RADAR helps predict weather, track missiles, find aircraft, and monitor animal migration. It works by sending radio waves and calculating distance from the time their echoes take to return.

🎯 Exam Tip: For descriptive answers like this, organize your points clearly, perhaps using subheadings, and always explain the underlying principle (reflection of radio waves).

 

Question 35. Distinguish Longitudinal waves and Transverse waves?
Answer: Here are the differences between longitudinal and transverse waves:

Longitudinal waves involve compressions and rarefactions, while transverse waves involve crests and troughs. Sound waves are a common example of longitudinal waves, and light waves are transverse.
In simple words: Longitudinal waves make particles move back and forth in the same direction as the wave, like sound. Transverse waves make particles move up and down, at right angles to the wave, like light or water ripples.

🎯 Exam Tip: When distinguishing between wave types, always focus on the direction of particle vibration relative to the wave's direction of travel. Use clear examples for each.

Sound Long Answer Type Questions

 

Question 36. Write the definition of the following:
(a) Amplitude
(b) Frequency
(c) Time period
(d) Wavelength
Answer:
(a) Amplitude: This is the maximum distance or displacement that a vibrating particle moves from its central resting position. It measures the strength or intensity of the wave. For example, for sound, a larger amplitude means a louder sound.
(b) Frequency: This is the number of complete oscillations or wave cycles that occur in one second. It is determined by the source of the wave. The S.I. unit for frequency is Hertz (Hz), where one Hz means one cycle per second.
(c) Time period: This is the time it takes for two consecutive compressions (or rarefactions) to pass a fixed point. In simpler terms, it's the time needed for one complete wave cycle to finish. It is shown by the symbol T, and its S.I. unit is second (s).
(d) Wavelength: This is the distance between two consecutive compressions (C) or two consecutive rarefactions (R) in a sound wave. It represents the length of one complete wave. Wavelength is shown by \( \lambda \) (lambda), and its S.I. unit is meters (m).
In simple words: Amplitude is how far a wave moves from its middle. Frequency is how many waves pass per second. Time period is how long one wave takes. Wavelength is the length of one wave.

🎯 Exam Tip: For definitions, provide a clear, concise explanation and, where applicable, include the symbol and SI unit. Understanding the interrelationship of these properties is also key.

 

Question 37. Find the relation between:
(a) The time period and frequency.
(b) Frequency, wave length and velocity.
Answer:
(a) **Relation between time period (T) and frequency \( (\nu) \):**
Let T be the time period of a vibrating body.
The number of oscillations completed in T seconds = 1
So, the number of oscillations completed in 1 second \( = \frac {1}{T} \).
The number of oscillations completed in 1 second is known as frequency \( (\nu) \).
Therefore, the relation is: \( \nu = \frac {1}{T} \)
\( \implies \) Frequency \( = \frac {1}{\text{Time period}} \). This means frequency and time period are inverses of each other.
(b) **Relation between frequency \( (\nu) \), wavelength \( (\lambda) \), and wave velocity (V):**
Wave velocity is defined as the distance traveled by a wave per unit time. When a wave travels a distance equal to its wavelength \( (\lambda) \), the time taken is its time period (T).
So, Wave velocity (V) \( = \frac {\text{Wave length}}{\text{Time period}} = \frac {\lambda}{T} \) ...(i)
We also know that frequency \( (\nu) = \frac{1}{T} \) ...(ii)
From equations (i) and (ii), we can substitute \( \frac{1}{T} \) with \( \nu \):
\( V = \lambda \times \frac{1}{T} \)
\( \implies V = \lambda \nu \)
So, Wave velocity \( = \) Wavelength \( \times \) Frequency. This equation is fundamental for all types of waves and shows that wave velocity depends on the medium.
In simple words: (a) Frequency is one divided by the time period, meaning they are opposite to each other. (b) Wave speed is found by multiplying the wave's frequency by its wavelength.

🎯 Exam Tip: Clearly state the derived formulas for each part and explain what each variable represents. The wave equation \( V = \nu\lambda \) is crucial for many wave-related calculations.

 

Question 38. Answer the following:
(a) Find the distance covered by any particle of the medium when it complete one oscillations.
(b) Sound waves travelling in air are longitudinal or transverse.
(c) Name the wave/waves produced in a long slinky.
(d) Give two examples each of transverse and longitudinal waves.
(e) Name the physical quantity whose unit is Hertz.
Answer:
(a) When a particle of the medium completes one oscillation, the distance covered by the wave it creates is equal to one wavelength. This means the wave travels one full 'cycle' during that oscillation.
(b) Sound waves traveling in air are longitudinal waves. This is because air particles vibrate parallel to the direction of the sound.
(c) When a long slinky is pushed and pulled along its length, longitudinal waves are produced. These waves consist of parts that are compressed and parts that are stretched out.
(d) **Examples of transverse waves are:**
1. All electromagnetic waves, such as light waves and radio waves, are transverse waves.
2. The water waves seen on the surface of water are also transverse waves, where water molecules move up and down from their normal position.
**Examples of longitudinal waves are:**
1. Sound waves are longitudinal waves that travel through various mediums.
2. Waves created by pushing and pulling a Slinky spring are longitudinal.
(e) The physical quantity whose unit is Hertz (Hz) is frequency. Hertz measures the number of cycles or oscillations per second.
In simple words: (a) A particle moves one wavelength during one full wiggle. (b) Sound in air is longitudinal. (c) A slinky creates longitudinal waves when pushed. (d) Light and water surface waves are transverse; sound and slinky waves are longitudinal. (e) Hertz is the unit for frequency.

🎯 Exam Tip: For multi-part questions, answer each part clearly and distinctly. For examples, ensure they are accurate and easy to understand.

 

Question 39. Give five uses of ultrasonic waves?
Answer: Here are five uses of ultrasonic waves:
1. Ultrasonic waves are used in echocardiography to create images of the heart, helping doctors see how it's working.
2. They are used to find cracks and flaws inside metal blocks without damaging the metal.
3. Ultrasonic waves can be used to kill bacteria in liquids, helping to clean and sterilize things.
4. They are used to determine the depth of the sea, a technique called sonar or echo ranging.
5. Ultrasonic waves are also used to observe the components inside nuclear reactors, helping monitor their condition safely.
In simple words: Ultrasonic waves are used to image the heart, find metal cracks, kill bacteria, measure sea depth, and check inside nuclear reactors.

🎯 Exam Tip: When listing uses, aim for distinct applications across different fields (medical, industrial, environmental) to show a broad understanding of the topic.

 

Question 40. Write a full form of SONAR. How do you determine the depth of the sea using echo ranging?
Answer: The full form of SONAR is **SOUND NAVIGATION AND RANGING**. This technique helps ships find submarines, icebergs, sunken objects, and determine sea depth. A SONAR device sends out ultrasonic waves to measure the distance, direction, and speed of underwater objects.
**Determination of the depth of a sea using SONAR:**
To find the sea depth using SONAR, a ship sends out an ultrasonic signal towards the seabed. This signal travels to the seabed and then reflects back up to the ship. The SONAR device measures the time interval (t) between sending the signal and receiving its echo. If the speed of sound through seawater is known (v), the total distance traveled by the ultrasound is \( 2d \) (down to the seabed and back up).
So, the total distance \( 2d = v \times t \).
Therefore, the depth of the sea (d) \( = \frac{v \times t}{2} \).
For example, if the sound takes 2 seconds to return and travels at 1500 m/s in water, the depth would be \( \frac{1500 \times 2}{2} = 1500 \) meters. This method provides a clear way to map the ocean floor.

In simple words: SONAR means Sound Navigation and Ranging. It finds sea depth by sending sound waves down and measuring how long it takes for the echo to return from the seabed. The time multiplied by speed, divided by two, gives the depth.

🎯 Exam Tip: When explaining SONAR, clearly state the full form, its working principle (echo ranging), and the formula used to calculate depth. A simple diagram can enhance your answer.

 

Question 41. Explain with figure how does human ear works?
Answer: The human ear is a complex organ that collects sound waves and converts them into electrical signals that the brain can understand. It has three main parts: the outer ear, middle ear, and inner ear. Each part plays a vital role in hearing.

Sound waves are collected by the **outer ear** (pinna) and channeled through the ear canal to the eardrum. When sound waves hit the **eardrum**, it starts to vibrate. These vibrations are then passed on to three tiny bones in the **middle ear**: the hammer (malleus), anvil (incus), and stirrup (stapes). These bones amplify the vibrations.
The amplified vibrations are then sent to the **inner ear**, specifically to a snail-shaped organ called the cochlea. The cochlea is filled with fluid and contains thousands of tiny hair cells. When the fluid vibrates, these hair cells move, converting the mechanical vibrations into electrical signals. These electrical signals are then sent to the brain via the auditory nerve, allowing us to interpret them as sound. This intricate process ensures we can hear a wide range of sounds.
In simple words: The outer ear catches sound, the middle ear makes it stronger, and the inner ear changes it into signals for the brain to understand.

🎯 Exam Tip: When explaining the ear's function, clearly describe the role of each main part (outer, middle, inner) and the process of converting sound vibrations into electrical signals for the brain.

 

Question 42. State the laws of Reflection of Sound. Arrange an activity to prove these laws?
Answer: The reflection of sound follows the same laws as the reflection of light. These laws govern how sound waves bounce off a surface.
**Laws of Reflection of Sound:**
1. The direction of the incident sound wave and the reflected sound wave both make equal angles with the normal (an imaginary line perpendicular to the surface) at the point of incidence. This means the angle of incidence is equal to the angle of reflection.
2. The incident sound wave, the reflected sound wave, and the normal to the reflecting surface, all lie in the same plane.
**Activity to prove these laws:**
1. Place a drawing board vertically on a table. This will act as our reflecting surface.
2. Take two metallic cardboard tubes, A and B. Position them at an angle to each other on the drawing board.
3. Place a ticking clock near one end of tube A (this is the sound source).
4. Put a solid screen between the two tubes to ensure the sound from the clock cannot be heard directly by an ear placed in front of tube B.
5. Now, place your ear at the end of tube B. Adjust the position of tube B until you can hear the clock's ticking clearly.
6. Observe the angles that the tubes make with the drawing board (representing the reflecting surface). You will notice that the angle of incidence (of tube A) is equal to the angle of reflection (of tube B). Also, the tubes and the normal at the point of reflection all lie on the same surface (the drawing board). This experiment clearly shows the laws of reflection of sound in action.
In simple words: Sound reflects like light: the incoming and outgoing sound waves make the same angle with the surface, and everything (incoming sound, outgoing sound, and the line straight up from the surface) stays in one flat space. We can show this by using tubes and a clock.

🎯 Exam Tip: When describing an activity, clearly state the aim, materials, procedure, and observation/conclusion. Relate the observations directly back to the laws being demonstrated.

 

Question 43. What is ultrasound? Explain how defects in a metal block can be detected, using ultrasound and explain how ultrasound is used to clean spiral tubes.
Answer:
**What is Ultrasound?**
Ultrasound refers to sound waves that have frequencies higher than 20,000 Hertz (Hz), which is beyond the range of human hearing. While humans cannot hear ultrasound, some animals like dogs, bats, and dolphins can. Due to its very high frequency, ultrasound has greater penetrating power compared to ordinary audible sound.
**Detection of defects in metal blocks:**
To detect flaws inside metal blocks, ultrasonic waves are sent through the block. Detectors are placed on the other side to pick up the transmitted waves. If there is a small defect, like a crack or an air bubble, inside the metal, the ultrasonic waves will reflect back from that defect instead of passing through. This reflected signal indicates the presence and location of the flaw, helping engineers find hidden problems without destroying the material.
**Cleaning of spiral tubes:**
Ultrasonic waves are also used for cleaning hard-to-reach parts, like spiral tubes, odd-shaped machinery, or electronic components. The object to be cleaned is placed in a cleaning solution. Then, ultrasonic waves are passed through the solution. The high-frequency waves create intense vibrations and cavitation (tiny bubbles forming and collapsing) in the liquid. These vibrations stir up and detach dust, dirt, and grease particles from the surfaces of the objects, thoroughly cleaning them. This is a very effective way to clean delicate or complex items.
In simple words: Ultrasound is sound too high for humans to hear. It can find hidden cracks in metal by bouncing off them. It also cleans tricky tubes by vibrating a liquid, shaking off dirt.

🎯 Exam Tip: Provide a clear definition of ultrasound. For applications, explain the mechanism for each use (e.g., reflection for defect detection, vibrations for cleaning) for a comprehensive answer.

 

Question 44. Enlist the Elements of RADAR and uses of RADAR?
Answer: RADAR (RADIO DETECTION AND RANGING) is a system used for detecting and locating objects. It consists of several key elements and has various important uses.
**Elements of RADAR are:**
1. **Modulator:** This component combines an electrical audio signal with high-frequency carrier waves, producing radio waves.
2. **Radio frequency oscillator:** It creates high-frequency power pulses, which form the actual RADAR signals.
3. **Antenna:** The antenna is used for both transmitting (sending out) the radar signals and receiving (picking up) the reflected signals.
4. **Receiver:** This part takes in the incoming radio signals that have bounced off objects.
5. **Indicator:** The indicator processes the information received by the radio waves and displays the position of the detected objects, often on a screen.
**Uses of RADAR:**
1. RADAR is widely used in the navigation of aeroplanes and ships, helping them find their way and avoid obstacles.
2. It is employed to guide rockets and missiles accurately to their targets.
3. It assists in weather forecasting by tracking storm systems and rainfall.
4. Air traffic control uses RADAR to monitor aircraft positions and movements, ensuring safe flights.
5. Law enforcement uses speed RADAR to measure vehicle speeds.
In simple words: RADAR uses a modulator, oscillator, antenna, receiver, and indicator to work. It helps airplanes and ships navigate, guides rockets, predicts weather, and tracks vehicle speeds.

🎯 Exam Tip: When listing elements of a system, briefly describe the function of each component. For uses, provide a diverse range of applications to demonstrate the versatility of RADAR technology.

Multiple Choice Questions (MCQs)

 

Question 1. The region of minimum density through which a wave propagates is called:
(A) Compression
(B) Rarefaction
(C) Crest
(D) Trough
Answer: (B) Rarefaction
In simple words: In a wave, a rarefaction is a stretched-out area where the particles are spread furthest apart, leading to the lowest density.

🎯 Exam Tip: Remember that longitudinal waves consist of alternating compressions (high density) and rarefactions (low density).

 

Question 2. Nature of sound wave is:
(A) Electromagnetic
(B) Transverse
(C) Longitudinal
(D) Seismic
Answer: (C) Longitudinal
In simple words: Sound waves are longitudinal because the particles in the medium (like air) vibrate back and forth in the same direction that the sound travels.

🎯 Exam Tip: Differentiate sound waves from light waves (electromagnetic and transverse) by their fundamental nature of particle vibration.

 

Question 3. The minimum distance between the source of the sound and the obstacle for an echo (at 20°C) to take place is:
(A) 17.2 m
(B) 1.72 m
(C) 17.8 m
(D) 34.4 m
Answer: (A) 17.2 m
In simple words: To hear a clear echo, the sound needs to travel to an obstacle and back. This round trip must take at least 0.1 seconds. Given the speed of sound at 20°C (approx. 344 m/s), the distance to the obstacle should be half of the total travel distance, which is 17.2 meters.

🎯 Exam Tip: This value (17.2 m) is derived from the minimum time for distinct hearing (0.1 s) and the speed of sound. Know both values to derive this distance.

 

Question 4. To change a loud sound into feeble sound, we decrease its:
(A) amplitude
(B) wavelength
(C) frequency
Answer: (A) amplitude
In simple words: To make a loud sound softer, you need to reduce its amplitude. Amplitude is how big the wave is, and a smaller wave means a quieter sound.

🎯 Exam Tip: Loudness is directly related to the amplitude of a sound wave. Pitch is related to frequency, and quality (timbre) to waveform.

Multiple Choice Questions (MCQs)

 

Question 5. The application of ultrasound is to
(A) get images of internal body parts
(B) detect cracks and flaws in metal blocks
(C) clean parts hard to reach
(D) All of the options
Answer: (D) All of the options
In simple words: Ultrasound waves are very useful for many things, like seeing inside the body, finding tiny cracks in metal, and even cleaning small, complicated objects. They help us in different ways because they are high-frequency sound waves.

🎯 Exam Tip: Remember the diverse applications of ultrasound beyond medical imaging, such as industrial inspection and cleaning, to score full marks.

 

Question 6. Before the main shock wave, an earthquake produces:
(A) audible sound
(B) ultrasound
(C) infrasound
(D) None of the options
Answer: (C) infrasound
In simple words: Before a big earthquake shock hits, the ground often makes very low-frequency sounds called infrasound. Humans usually cannot hear this type of sound. These infrasounds can travel long distances and are often detected by animals.

🎯 Exam Tip: Distinguish between audible, ultrasound, and infrasound based on their frequency ranges to correctly identify sounds associated with natural phenomena.

 

Question 7. Infrasound is the sound with frequencies:
(A) above 20 kHz
(B) below 20 Hz
(C) between 20 Hz and 20 kHz
(D) all the above
Answer: (B) below 20 Hz
In simple words: Infrasound is a type of sound that has a very low frequency, less than 20 Hertz. This is too low for human ears to hear. Animals like elephants, however, can hear and use infrasound for communication.

🎯 Exam Tip: Clearly define the human hearing range (20 Hz to 20 kHz) and use it as a reference point to explain infrasound and ultrasound.

 

Question 8. One tuning fork completes 1000 oscillations in 5 seconds, its frequency is:
(A) 1000 Hz
(B) 0.005 Hz
(C) 200 Hz
(D) 5000 Hz
Answer: (C) 200 Hz
In simple words: Frequency is how many times something vibrates in one second. If a tuning fork vibrates 1000 times in 5 seconds, you can find its frequency by dividing 1000 by 5, which gives you 200 vibrations per second, or 200 Hz.

🎯 Exam Tip: Remember the formula for frequency: Frequency = Number of oscillations / Time (in seconds). Always ensure the time is in seconds for calculations.

 

Question 9. In SONAR, we use:
(A) audible sound waves
(B) ultrasonic waves
Answer: (B) ultrasonic waves
In simple words: SONAR uses ultrasonic waves to detect things underwater. These are high-frequency sounds that travel well in water and bounce off objects, allowing ships to "see" what's around them. The high frequency allows for better resolution and less interference.

🎯 Exam Tip: Understand why high-frequency ultrasonic waves are preferred for SONAR (better resolution and less diffraction) compared to audible sound waves.

 

Question 10. Infrasound can be heard by:
(A) Elephant
(B) bat
(C) human beings
(D) dog
Answer: (A) Elephant
In simple words: Elephants can hear very low-frequency sounds called infrasound. This helps them talk to each other over long distances and detect distant storms or other large animal movements. These sounds are below what humans can hear.

🎯 Exam Tip: Be aware of examples of animals that use infrasound (like elephants) and ultrasound (like bats and dolphins) for communication and navigation.

 

Question 11. A speed greater than that of any type of sound waves is said to be:
(A) infrasonic speed
(B) sonic speed
(C) ultrasonic speed
(D) supersonic speed
Answer: (D) supersonic speed
In simple words: When something moves faster than the speed of sound, it is moving at supersonic speed. This creates a "sonic boom." This term is used for objects, not for the sound waves themselves.

🎯 Exam Tip: Differentiate between the types of sound waves (infrasonic, audible, ultrasonic) and the speed of an object relative to sound (subsonic, sonic, supersonic, hypersonic).

 

Question 12. Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting:
(A) the intensity of sound only
(B) the amplitude of sound only
(C) the frequency of sitar string with the frequency of another musical instrument
(D) the loudness of sound
Answer: (C) the frequency of sitar string with the frequency of another musical instrument
In simple words: A sitarist changes the tension of strings to match the pitch, or frequency, of their instrument with others. This makes all the instruments sound good together. When instruments are tuned, their frequencies are in harmony.

🎯 Exam Tip: Understand that adjusting string tension primarily changes the frequency (and thus the pitch) of the sound produced, which is crucial for musical harmony.

 

Question 13. Note is a sound:
(A) of a mixture of two frequency only
(B) of a mixture of several frequencies
(C) always unpleasant to listen
(D) of a single frequency
Answer: (B) of a mixture of several frequencies
In simple words: A musical "note" is usually made up of a main sound frequency along with other softer, related frequencies called overtones. These combined frequencies give the note its unique quality. This mix of sounds makes the note richer than a simple pure tone.

🎯 Exam Tip: Differentiate between a "tone" (single frequency) and a "note" (mixture of frequencies) in music, as this relates to sound quality or timbre.

 

Question 14. Human audible range is:
(A) 20 Hz to 20000 Hz
(B) 20Hz to 20000Hz
(C) 2 Hz to 20 Hz
(D) Above 20000Hz
Answer: (B) 20Hz to 20000Hz
In simple words: The normal range of sound frequencies that most humans can hear is from 20 Hertz, which is a very low rumble, up to 20,000 Hertz, which is a very high-pitched sound. Sounds outside this range are called infrasound (lower) or ultrasound (higher).

🎯 Exam Tip: Precisely recall the standard human audible frequency range to distinguish it from infrasound and ultrasound ranges.

 

Question 15. The amplitude of a sound is doubled and the frequency is reduced to one fourth the intensity of sound at the same point will be:
(A) increased by a factor of 2
(B) increased by a factor of 4
(C) decreased by a factor of 2
(D) decreased by a factor of 4
Answer: (D) decreased by a factor of 4
In simple words: The intensity of sound depends on the square of the amplitude and the square of the frequency. If the amplitude doubles, intensity increases by 4 times. If frequency reduces to one fourth, intensity decreases by 16 times. So, the total change is \( 4 \times \frac{1}{16} = \frac{1}{4} \), meaning the intensity decreases by a factor of 4.

🎯 Exam Tip: Remember that sound intensity is proportional to the square of both amplitude and frequency ( \( I \propto A^2 \nu^2 \) ). Apply this relationship carefully when calculating changes.

Sound Very Short Answer Type Questions

 

Question 1. Which mechanical waves are associated with
(a) compression and rarefaction
(b) crest and troughs
Answer:
(a) Longitudinal Waves: Waves like sound waves and those in a slinky have compressions (where particles are squished together) and rarefactions (where particles are spread apart). These waves transfer energy by particles vibrating parallel to the wave's direction. An example is sound traveling through the air.
(b) Transverse waves: Waves like light waves and ripples on water have crests (highest points) and troughs (lowest points). In these waves, particles move perpendicular to the direction the wave travels. For example, when you drop a stone in water, the water moves up and down while the wave spreads outwards.

🎯 Exam Tip: Clearly differentiate between longitudinal and transverse waves based on the particle motion relative to wave propagation, and provide appropriate examples for each.

 

Question 2. Which wave property determine
(a) Loudness
(b) Pitch
Answer:
(a) The amplitude of vibrations determines the loudness of sound. A bigger amplitude means a louder sound, like shouting instead of whispering. The energy carried by a sound wave is also related to its amplitude.
(b) A frequency of the sound determines its pitch. Higher frequency sounds have a higher pitch, like a whistle, while lower frequency sounds have a lower pitch, like a drum. This is why different musical notes sound distinct.

🎯 Exam Tip: Associate amplitude directly with loudness and energy, and frequency directly with pitch, as these are fundamental characteristics of sound waves.

 

Question 3. Explain how is a sound produced by your school bell?
Answer: When the school bell is hit with a hammer, it starts to vibrate very quickly. These vibrations then make the air molecules around the bell also vibrate, which creates sound waves that travel to our ears. All sounds are caused by vibrations. The ringing continues until the vibrations stop or become too small to hear.

🎯 Exam Tip: Always link sound production directly to vibrations and explain how these vibrations propagate through a medium to be heard.

 

Question 5. Name the waves used to break small stones formed in the kidneys into fine grains.
Answer: Ultrasound waves or ultrasonic waves are used for this purpose. These high-frequency sound waves can pass through the body and target the kidney stones. When the ultrasound waves hit the stones, they break them into tiny pieces that can then be passed out of the body naturally. This is a non-invasive medical procedure.

🎯 Exam Tip: Remember specific medical applications of ultrasound, such as lithotripsy (breaking kidney stones) and echocardiography, as these are common knowledge points.

 

Question 6. A girl hears an echo of her own voice from a distant tall building after 2s. What is the distance of the girl from the building?
Answer:
Given:
Time taken to hear the echo (\( t \)) = 2 s
Speed of sound in air (\( v \)) = 332 m/s
Let the distance of the girl from the building be \( s \).
The sound travels from the girl to the building and then back to the girl, so the total distance covered is \( 2s \).
Using the formula: Distance = Speed \( \times \) Time
\( 2s = v \times t \)
\( 2s = 332 \, \text{m/s} \times 2 \, \text{s} \)
\( 2s = 664 \, \text{m} \)
\( s = \frac{664 \, \text{m}}{2} \)
\( s = 332 \, \text{m} \)
Therefore, the distance of the girl from the building is 332 meters. This distance allows enough time for the sound to travel there and back, creating a clear echo.
In simple words: The sound goes to the building and comes back. This round trip takes 2 seconds. Since the sound travels at 332 meters per second, the total distance it covered is 332 times 2, which is 664 meters. The building is half of this distance away, so it's 332 meters from the girl.

🎯 Exam Tip: For echo problems, always remember to divide the total distance calculated by 2 because the sound travels to the obstacle and then returns.

 

Question 7. What are infrasonic and ultrasonic sounds?
Answer:
**Infrasonic sound:** A sound that has a frequency less than 20 Hz (Hertz) is called infrasonic sound. This is below the range that humans can hear. Animals like elephants and whales communicate using infrasound over long distances.
**Ultrasonic sound:** A sound that has a frequency greater than 20,000 Hz (20 kHz) is called ultrasonic sound. This is above the range that humans can hear. Animals like bats and dolphins use ultrasound for navigation and hunting (echolocation). These sounds are very high pitched.

🎯 Exam Tip: Memorize the specific frequency ranges for infrasound (below 20 Hz) and ultrasound (above 20 kHz) relative to the human hearing range (20 Hz - 20 kHz).

 

Question 8. Distinguish between loudness and intensity of sound.
Answer:
**Loudness:**
1. Loudness measures how our ears respond to sound. It's a subjective feeling.
2. Loudness depends on how sensitive a person's ear is. Two people might hear the same sound at different loudness levels.
**Intensity:**
1. Intensity is the amount of sound energy that passes through a specific area in one second. It's an objective measurement.
2. The intensity of sound does not depend on how sensitive the ear is. It's a physical property of the sound wave itself. A standard instrument can measure sound intensity.
In simple words: Loudness is what you feel in your ear, so it changes from person to person. Intensity is the actual power of the sound wave, which can be measured and is the same for everyone.

🎯 Exam Tip: The key distinction is that loudness is subjective (perception), while intensity is objective (measurable energy flow). This difference is crucial for understanding sound characteristics.

 

Question 9. Through which medium does sound travel fastest?
Answer: Sound travels fastest through iron. This is because sound waves travel quicker in denser and more rigid materials where particles are closer together and can transmit vibrations more efficiently. Solids, in general, conduct sound faster than liquids, and liquids conduct sound faster than gases.

🎯 Exam Tip: Remember the general rule: sound travels fastest in solids, then liquids, then gases. Provide a specific example like iron or steel for solids.

 

Question 11. Guess which sound has a higher pitch: guitar or car horn?
Answer: A guitar has a sound of higher pitch. Pitch is determined by the frequency of the sound wave. Generally, a guitar can produce much higher frequencies than a car horn. This is why guitar strings produce distinct musical notes. A car horn produces a lower, more singular frequency sound.

🎯 Exam Tip: Relate pitch directly to frequency; higher frequency means higher pitch. Think of common examples like musical instruments vs. heavy vehicle horns.

 

Question 12. Give two practical applications of reflection of sound waves?
Answer:
(a) In a stethoscope, the sound of a patient's heartbeat is guided along the tube to the doctor's ear through multiple reflections of sound. The sound travels efficiently from the chest piece through the narrow tubes.
(b) Soundboards are used in big halls to direct sound waves towards the audience. These curved boards reflect sound in a focused manner, making sure everyone in the audience can hear clearly. This improves the acoustics of the hall.

🎯 Exam Tip: For practical applications, choose examples that clearly demonstrate the principle of reflection, like stethoscopes (multiple reflections) and soundboards (focused reflection).

 

Question 13. What are longitudinal waves?
Answer: Longitudinal waves are waves in which the particles of the medium vibrate back and forth in the same direction that the wave is travelling. Think of a slinky: when you push one end, the compression travels along the slinky, but each coil just moves forward and backward. Sound waves are a common example of longitudinal waves because air particles vibrate parallel to the direction of sound propagation.

🎯 Exam Tip: Define longitudinal waves by describing the direction of particle vibration relative to wave propagation (parallel) and give a clear example like sound waves or slinky waves.

 

Question 14. What should be the minimum distance between the source of the sound and the reflector for the formation of echo?
Answer: For a clear echo to be heard, the minimum distance between the sound source and the reflector must be 17.2 meters. This distance ensures that the reflected sound reaches our ears at least 0.1 seconds after the original sound, which is the minimum time needed for the human ear to distinguish two separate sounds. Without this time gap, the sound would just seem prolonged. This value assumes the speed of sound in air is around 344 m/s.

🎯 Exam Tip: Remember the critical values: 0.1 seconds (persistence of hearing) and 17.2 meters (minimum distance for echo formation at standard temperature and pressure).

 

Question 15. Why are we able to hear the sound produced by the humming bees while the sound of vibrations of the pendulum is not heard?
Answer: We can hear the sound of humming bees because the frequency of the sound they produce falls within the human audible range (20 Hz to 20,000 Hz). In contrast, the frequency of a vibrating pendulum is very low, typically less than 20 Hz, placing it in the infrasonic region. Humans cannot hear infrasound. This means even though both are vibrating, only the bee's vibrations are within our hearing capabilities. The frequency determines whether we can perceive the sound.

🎯 Exam Tip: Always relate the ability to hear a sound to its frequency falling within the human audible range (20 Hz to 20,000 Hz).

 

Question 16. The frequency of a source of sound is 250 Hz. Calculate the number of times, the sound vibrates in one minute. Also, calculate the time period?
Answer:
Given frequency \( \nu \) = 250 Hz.
This means the source of sound vibrates 250 times in one second.
To find the number of vibrations in one minute:
Number of vibrations in one minute = Frequency \( \times \) 60 seconds
Number of vibrations = 250 Hz \( \times \) 60 s = 15,000 vibrations.
Now, to calculate the time period (\( T \)):
The time period is the inverse of frequency: \( T = \frac{1}{\nu} \)
\( T = \frac{1}{250 \, \text{Hz}} \)
\( T = 0.004 \, \text{seconds} \)
So, the sound vibrates 15,000 times in one minute, and the time period for one vibration is 0.004 seconds. The time period tells us how long each individual vibration takes.
In simple words: If a sound vibrates 250 times every second, then in one minute (60 seconds), it will vibrate 15,000 times. The time taken for just one vibration is very small, only 0.004 seconds.

🎯 Exam Tip: Remember the relationship between frequency and time period (they are reciprocals: \( T = 1/\nu \)). Ensure units are consistent, converting minutes to seconds for frequency calculations.

 

Question 17. If a man hears the thunder 5 seconds after lightning is seen, how far is the lightning from the man? Given, the speed of sound is 344 m/s.
Answer:
Given:
Time taken to hear thunder (\( t \)) = 5 seconds
Speed of sound in air (\( v \)) = 344 m/s
To find the distance of lightning from the man (\( S \)):
Distance = Speed \( \times \) Time
\( S = v \times t \)
\( S = 344 \, \text{m/s} \times 5 \, \text{s} \)
\( S = 1720 \, \text{meters} \)
Converting meters to kilometers:
\( S = 1.72 \, \text{km} \)
Therefore, the lightning is 1.72 kilometers away from the man. Light travels much faster than sound, so seeing lightning and then hearing thunder is a way to estimate distance.
In simple words: Light from lightning travels very fast, so you see it almost instantly. Sound travels slower. If it takes 5 seconds for the sound of thunder to reach you, and sound travels at 344 meters per second, then the lightning is 1720 meters, or 1.72 kilometers, away.

🎯 Exam Tip: In problems involving lightning and thunder, assume the speed of light is instantaneous and focus only on the speed of sound to calculate the distance.

 

Question 18. An echo returned in 3s. What is the distance of the reflecting surface from the source? Given that the speed of sound is 342 m/s?
Answer:
Given:
Time taken for echo to return (\( t \)) = 3 seconds
Speed of sound in air (\( v \)) = 342 m/s
Let the distance of the reflecting surface from the source be \( S \).
The sound travels to the reflecting surface and then back to the source, covering a total distance of \( 2S \).
Using the formula: Distance = Speed \( \times \) Time
\( 2S = v \times t \)
\( 2S = 342 \, \text{m/s} \times 3 \, \text{s} \)
\( 2S = 1026 \, \text{m} \)
\( S = \frac{1026 \, \text{m}}{2} \)
\( S = 513 \, \text{meters} \)
So, the reflecting surface is 513 meters away from the sound source. This distance ensures the sound travels there and back within the given time.
In simple words: The sound went to the wall and came back, taking 3 seconds. Since sound travels at 342 meters per second, the total distance it covered was 342 times 3, which is 1026 meters. The wall is half of that distance away, which is 513 meters.

🎯 Exam Tip: Always divide the total calculated distance by two when solving echo problems to find the one-way distance to the reflecting surface.

 

Question 19. A boy watches Dusshera celebration from a distance and sees the effigy of Ravana burn into flames and nears the explosion after 2 seconds. How far was he from the effigy if the speed of sound in air was 335 m/s.
Answer:
Given:
Time difference between seeing the fire and hearing the explosion (\( t \)) = 2 seconds
Speed of sound in air (\( v \)) = 335 m/s
Let the distance of the boy from the effigy be \( S \).
We assume that light from the burning effigy reaches the boy instantaneously.
The time delay is due to the sound taking time to travel the distance.
Using the formula: Distance = Speed \( \times \) Time
\( S = v \times t \)
\( S = 335 \, \text{m/s} \times 2 \, \text{s} \)
\( S = 670 \, \text{meters} \)
Therefore, the boy was 670 meters away from the effigy. This calculation relies on the large difference between the speed of light and the speed of sound.
In simple words: The boy saw the fire first and heard the explosion 2 seconds later. Since sound travels at 335 meters per second, the distance to the effigy is simply 335 multiplied by 2 seconds, which is 670 meters.

🎯 Exam Tip: This type of problem highlights the vast difference in speed between light and sound. The time delay is solely attributed to the sound's travel time.

 

Question 23. A slinky has a frequency of 20 Hz. Find the minimum separation distance between consecutive rarefactions of the slinky, if the speed of sound (waves) in it is 30 cm/s.
Answer:
Given:
Frequency (\( \nu \)) = 20 Hz
Speed of wave (\( V \)) = 30 cm/s = 0.30 m/s (converted to meters per second)
The minimum separation distance between consecutive rarefactions is equal to the wavelength (\( \lambda \)).
We use the wave equation: \( V = \nu \lambda \)
To find wavelength: \( \lambda = \frac{V}{\nu} \)
\( \lambda = \frac{0.30 \, \text{m/s}}{20 \, \text{Hz}} \)
\( \lambda = 0.015 \, \text{m} \)
So, the minimum separation distance between consecutive rarefactions is 0.015 meters. This value represents one complete wave cycle.
In simple words: The distance between two spread-out parts (rarefactions) of the slinky wave is its wavelength. We divide the wave's speed (0.30 meters per second) by its frequency (20 times per second) to get 0.015 meters.

🎯 Exam Tip: Remember that the distance between two consecutive compressions or rarefactions (or crests or troughs for transverse waves) represents one wavelength. Always ensure consistent units before calculation.

Sound Long Answer Type Questions

 

Question 1.
(a) What is reverberation? What will happen if the reverberation time in a big hall is too long? How can we reduce it?
(b) Which property of sound leads to the formation of echo? Briefly explain.
Answer:
(a) **Reverberation:** Reverberation is when sound waves repeatedly reflect off surfaces in a space, causing the sound to continue for some time even after the original sound source has stopped. It is like a prolonged echo.
If the reverberation time in a big hall is too long, the sound becomes distorted, blurred, and confusing. This makes it difficult to understand speech or appreciate music because sounds overlap with each other.
**Methods for reducing reverberation:**
1. To reduce too much reverberation, the roofs and walls of auditoriums are often covered with sound-absorbing materials. Examples include compressed fiberboard, rough plaster, or thick drapes. These materials soak up sound energy instead of reflecting it.
2. The materials for seats are also chosen for their sound-absorbing qualities. Carpets are used on floors, and thick curtains are hung on doors and windows to further absorb sound, helping to make the hall's acoustics clear.
(b) **Reflection of sound** is the property that leads to the formation of an echo. An echo happens when a sound wave hits a hard surface (like a wall or a hill) and bounces back to the listener. To hear a distinct echo, the reflecting surface must be far enough away (at least 17.2 meters) so that the reflected sound reaches the ear after the original sound has faded. If there are multiple reflecting surfaces at different distances, multiple echoes can be heard.
In simple words: (a) Reverberation is when sound bounces around too much in a room, making it unclear. If it lasts too long, sounds get muddled. We can fix this by using soft, sound-absorbing materials on walls, roofs, and even seats, and by adding carpets and curtains. (b) Echoes happen because sound waves bounce off things, like a ball bouncing off a wall. This bouncing back of sound is called reflection.

🎯 Exam Tip: Clearly differentiate between reverberation (persistence of sound due to multiple reflections) and echo (distinct, delayed reflection). Provide specific, practical methods for controlling reverberation, particularly in large spaces.

 

Question 3. A sonar echo takes 3 sec to return from a whale. How far away is the whale from the sonar? The speed of sound in seawater is 1440 m/s.
Answer:
Given:
Time taken for echo to return (\( t \)) = 3 seconds
Speed of sound in seawater (\( v \)) = 1440 m/s
Let the distance of the whale from the SONAR be \( d \).
The sound travels from the SONAR to the whale and then back to the SONAR, covering a total distance of \( 2d \).
Using the formula: Speed = Distance / Time
\( V = \frac{2d}{t} \)
Rearranging to find \( d \):
\( 2d = V \times t \)
\( 2d = 1440 \, \text{m/s} \times 3 \, \text{s} \)
\( 2d = 4320 \, \text{m} \)
\( d = \frac{4320 \, \text{m}}{2} \)
\( d = 2160 \, \text{m} \)
Therefore, the whale is 2160 meters (or 2.16 km) away from the SONAR. This calculation helps locate underwater objects accurately.
In simple words: The sonar sends a sound and it takes 3 seconds to reach the whale and come back. If sound travels at 1440 meters per second in water, the total distance covered is 1440 times 3, which is 4320 meters. So, the whale is half that distance away, which is 2160 meters.

🎯 Exam Tip: For SONAR calculations, remember that the time measured is for a round trip (to the object and back), so always divide the total distance by two to find the one-way distance.

 

Question 4. 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/s² and speed of sound = 340 m/s.
Answer:
Given:
Height of the tower (\( h \)) = 500 m
Acceleration due to gravity (\( g \)) = 10 m/s²
Speed of sound (\( v_{\text{sound}} \)) = 340 m/s
First, calculate the time it takes for the stone to fall from the top of the tower to the water surface. We can use the equation of motion:
\( h = ut + \frac{1}{2}gt^2 \)
Since the stone is dropped, initial velocity (\( u \)) = 0.
\( 500 = 0 \times t + \frac{1}{2} \times 10 \times t^2 \)
\( 500 = 5t^2 \)
\( t^2 = \frac{500}{5} \)
\( t^2 = 100 \)
\( t = \sqrt{100} \)
\( t_{\text{fall}} = 10 \, \text{s} \)
Next, calculate the time it takes for the sound of the splash to travel from the pond back to the top of the tower:
\( t_{\text{sound}} = \frac{\text{Distance}}{\text{Speed of sound}} \)
\( t_{\text{sound}} = \frac{500 \, \text{m}}{340 \, \text{m/s}} \)
\( t_{\text{sound}} \approx 1.47 \, \text{s} \)
The total time after which the splash is heard at the top is the sum of the time taken for the stone to fall and the time taken for the sound to travel back.
Total time = \( t_{\text{fall}} + t_{\text{sound}} \)
Total time = \( 10 \, \text{s} + 1.47 \, \text{s} \)
Total time = \( 11.47 \, \text{s} \)
So, the splash is heard at the top of the tower approximately 11.47 seconds after the stone is dropped. This problem combines concepts of free fall and sound propagation.
In simple words: First, the stone takes 10 seconds to fall 500 meters to the water. Then, the sound of the splash takes about 1.47 seconds to travel 500 meters back up to the top. So, the total time until the splash is heard is 10 seconds plus 1.47 seconds, which is about 11.47 seconds.

🎯 Exam Tip: This is a two-part problem: calculate the time for the object to fall first, then the time for the sound to travel back up. Sum these two times for the final answer. Pay attention to units and significant figures.

 

Question 4. 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/s² and speed of sound = 340 m/s.
Answer: First, we calculate the time it takes for the stone to fall into the water. The height of the tower (h) is 500 m, and the acceleration due to gravity (g) is 10 m/s². We use the formula \( h = ut + \frac{1}{2}gt^2 \), where initial velocity (u) is 0.
\( 500 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \)
\( 500 = 5t^2 \)
\( t^2 = 100 \)
\( t_{\text{stone}} = 10 \) seconds.
Next, we calculate the time it takes for the sound of the splash to travel from the pond's surface back to the top of the tower. The distance is 500 m, and the speed of sound is 340 m/s.
\( t_{\text{sound}} = \frac{\text{Distance}}{\text{Speed of sound}} = \frac{500}{340} \approx 1.47 \) seconds.
The total time after which the splash is heard at the top of the tower is the sum of these two times.
Total time \( = t_{\text{stone}} + t_{\text{sound}} = 10 + 1.47 = 11.47 \) seconds.
In simple words: The stone takes 10 seconds to fall. The sound of the splash then takes about 1.47 seconds to reach the top. So, the total time before hearing the splash is about 11.47 seconds.

🎯 Exam Tip: Remember to calculate both the time for the object to fall and the time for the sound to travel back. For falling objects, use kinematic equations, and for sound, use the simple distance/speed formula.

 

Question 5. With a diagram explaining the use of curved soundboard behind the speaker in conference halls?
Answer: In large halls, sound waves from the speaker often reflect off walls and ceilings, which can make the sound unclear or interfere with each other. To solve this, a curved soundboard is placed behind the speaker. This soundboard is shaped like a concave mirror.
When sound waves from the speaker hit the concave soundboard, they are reflected as parallel rays. This ensures that the sound travels a long distance and is spread evenly across the hall, making it clearer for the audience. This prevents the sound from spreading out too much in different directions.

In simple words: A curved board behind a speaker acts like a reflector. It takes the sound from the speaker and bounces it forward in straight lines, making it easier for everyone in a big room to hear clearly.

🎯 Exam Tip: When drawing, ensure the sound source is at the focal point of the concave soundboard, and the reflected rays are parallel to show effective sound projection.

 

Question 6. Is there any technique to detect cracks and flaws in a block of metal? If yes, then explain it with a labelled diagram. Or Explain how defects in a metal block can be detected using ultrasound.
Answer: Yes, there is a technique to detect cracks and flaws in metal blocks using **ultrasound**. This method is effective because ultrasonic waves cannot pass through a crack or a hole within the metal.
To detect a flaw, a source of ultrasonic sound is placed on one side of the metallic block. Detectors are placed on the opposite side. If there is a defect or flaw inside the metal, the ultrasonic waves will reflect back from it instead of passing through to the detectors. The detectors then pick up these reflected waves, indicating the presence and location of the flaw. This non-destructive testing method is crucial for ensuring the safety and integrity of structures.

In simple words: We can find hidden cracks in metal by sending high-frequency sound waves, called ultrasound, through it. If there's a crack, the sound waves will bounce back from it instead of going straight through. Special sensors then catch these bounced-back waves, showing us where the problem is.

🎯 Exam Tip: For this question, emphasize that ultrasound cannot pass through defects and that reflections are detected. A clear, labelled diagram showing the source, metal block, flaw, and detectors is essential for full marks.