Samacheer Kalvi Class 9 Science Solutions Chapter 8 Sound

I. Choose the Correct Answer:

 

Question 1. Which of the following vibrates when a musical note is produced by the cymbals in a orchestra?
(a) stretched strings
(b) stretched membranes
(c) air columns
(d) metal plates
Answer: (d) metal plates
In simple words: Cymbals are made of metal. When they are played, the metal plates themselves shake or vibrate to produce the musical sound.

🎯 Exam Tip: Remember that the part of an instrument that vibrates directly creates the initial sound waves.

 

Question 2. Sound travels in air:
(a) if there is no moisture in the atmosphere.
(b) if particles of medium travel from one place to another.
(c) if both particles as well as disturbance move from one place to another.
(d) if disturbance moves.
Answer: (d) if disturbance moves
In simple words: When sound moves through air, it's like a ripple effect. The sound itself, which is a disturbance, moves, but the air particles mostly stay in place and just shake back and forth.

🎯 Exam Tip: Always remember that sound is a wave that carries energy, not matter, meaning the medium particles oscillate but do not travel with the wave.

 

Question 3. A musical instrument is producing continuous note. This note cannot be heard by a person having a normal hearing range. This note must then be passing through
(a) wax
(b) vacuum
(c) water
(d) empty vessel
Answer: (b) vacuum
In simple words: Sound needs something to travel through, like air or water. If a sound cannot be heard, it means it is going through an empty space called a vacuum where there are no particles to carry the sound.

🎯 Exam Tip: Sound waves are mechanical waves, which means they always require a medium for propagation. A vacuum lacks any medium.

 

Question 4. The speed of vibrations which produces audible sound will be in
(a) seawater
(b) ground glass
(c) dry air
(d) Human blood
Answer: (b) ground glass
In simple words: Sound moves quickest through solid things. Out of the choices, ground glass is a solid, so sound will travel fastest in it.

🎯 Exam Tip: Remember the general rule that sound travels fastest in solids, then liquids, and slowest in gases, due to the density and elasticity of the particles.

 

Question 5. The sound waves travel faster
(a) in liquids
(b) in gases
(c) in solids
(d) in vacuum
Answer: (c) in solids
In simple words: Sound waves travel fastest in solids. This is because the particles in solids are very close, so they can pass on the sound vibrations quickly.

🎯 Exam Tip: The speed of sound depends on the medium's properties; solids have high density and elasticity, which are ideal for faster sound transmission.

II. Fill in the Blanks:

 

Question 1. Sound is a ............wave and needs a material medium to travel.
Answer: longitudinal mechanical
In simple words: Sound is a special kind of wave called a mechanical wave because it needs a substance to travel through. It is also a longitudinal wave, where the vibrations move in the same path as the sound.

🎯 Exam Tip: Remember the two key types of waves: mechanical waves (need a medium) and electromagnetic waves (do not need a medium).

 

Question 2. Number of vibrations produced in one second is.......
Answer: Frequency
In simple words: The count of how many times something shakes or moves back and forth in one second is called its frequency.

🎯 Exam Tip: Frequency is an important property of waves and is measured in Hertz (Hz), representing cycles per second.

 

Question 3. The velocity of sound in solid is ............... than the velocity of sound in air.
Answer: greater
In simple words: Sound moves faster in solids than in air. Solid materials have very close particles, so sound can pass through them very quickly.

🎯 Exam Tip: The speed of sound is directly related to the density and elasticity of the medium it travels through.

 

Question 4. Vibration of object produces ................
Answer: Sound
In simple words: When something shakes or vibrates, it makes sound.

🎯 Exam Tip: Remember that sound is always produced by vibrations, whether from a vocal cord, a drum, or a string.

 

Question 5. Loudness is proportional to the square of the ................
Answer: Amplitude of vibration of sound
In simple words: How loud a sound is depends on how big its vibration is. If the vibration is bigger, the sound is louder.

🎯 Exam Tip: Amplitude determines loudness (volume), while frequency determines pitch (how high or low a sound is).

 

Question 6. ..............is a medical instrument used for listening to sounds produced in the body.
Answer: Stethoscope
In simple words: A stethoscope is a tool doctors use to hear sounds inside our bodies, like heart sounds.

🎯 Exam Tip: Stethoscopes work by amplifying faint internal body sounds, aiding medical diagnosis.

 

Question 7. The repeated reflection that results in persistence of sound is called ................
Answer: Reverberation
In simple words: When sound keeps bouncing around and stays in a room for a bit longer, it is called reverberation.

🎯 Exam Tip: Reverberation is different from an echo; an echo is a distinct, separate sound, while reverberation is a prolonged sound due to many reflections that blend together.

III. Match the Following:

 

Answer: 1 - e, 2 - d, 3 - a, 4 - b, 5 - c
In simple words: A tuning fork makes sound. Sound travels as a longitudinal wave. Compressions are where air is most squeezed together. Amplitude is how far something moves from its resting place. Ultrasonics are sounds too high for humans to hear, with frequencies over 20,000 Hz.

🎯 Exam Tip: When matching, first identify the most direct relationships, like "Tuning fork" with "Production of sound", to simplify the remaining options.

IV. Answer in Brief:

 

Question 1. Through which medium sound travels faster, iron or water? Give reason.
Answer: Sound moves faster in iron than in water. The speed of sound changes depending on the material it travels through. Iron is a solid, and sound generally travels quicker through solids because their particles are very close and transmit vibrations more easily.
In simple words: Sound is faster in iron than in water. Iron is a solid, and sound moves faster in solids because the particles are packed tighter.

🎯 Exam Tip: Always remember that sound speed is highest in solids, followed by liquids, and slowest in gases, as it depends on the medium's elasticity and density.

 

Question 2. Name the physical quantity whose Sl unit is 'hertz'. Define.
Answer: The SI unit for frequency is Hertz. Frequency is defined as the total number of waves produced in one second, indicating how many cycles of a wave pass a point in that time.
In simple words: Hertz is the unit for frequency. Frequency is simply the number of waves made in one second.

🎯 Exam Tip: Clearly state the quantity first, then provide a simple and precise definition, avoiding complex terms.

 

Question 3. What is meant by supersonic speed?
Answer: Supersonic speed is when an object moves faster than the speed of sound in the surrounding air. The typical speed of sound in air is around 330 meters per second.
In simple words: Supersonic speed is when something moves faster than sound travels through the air.

🎯 Exam Tip: Mention the reference point (speed of sound in air) and its approximate value for a complete definition.

 

Question 4. How does the sound produced by a vibrating object in a medium reach your ears?
Answer: When an object vibrates, it makes the surrounding medium's particles vibrate too. These vibrating particles then push on their neighbors, passing the vibration along like a chain reaction. This movement of vibrations creates pressure changes that travel through the medium until they reach our ears, allowing us to hear the sound.
In simple words: An object vibrates and makes air particles around it shake. These shaking particles bump into the next ones, and this movement carries the sound energy all the way to our ears.

🎯 Exam Tip: Explain the process step-by-step: vibration of source, transfer to medium particles, propagation as a disturbance, and finally reception by the ear.

 

Question 5. You and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer: No, you would not be able to hear any sound on the moon. Sound waves need a medium, such as air, to travel. The moon does not have an atmosphere, so there is no medium for sound to be carried, making it impossible to hear anything.
In simple words: No, you cannot hear sound on the moon. Sound needs air to travel, and the moon has no air.

🎯 Exam Tip: The key concept here is that sound is a mechanical wave, strictly requiring a material medium for its propagation.

V. Answer in Detail:

 

Question 1. Describe with a diagram, how compressions and rarefactions are produced.
Answer: Sound waves are made when something vibrates, causing areas of high and low pressure to travel through a material like air.
**Compressions:** When a vibrating object moves forward, it pushes the particles of the medium together. This creates a region where the particles are crowded and the pressure is higher than normal. These regions are known as compressions.
**Rarefactions:** When the vibrating object moves backward, it pulls the particles of the medium apart. This results in a region where particles are spread out and the pressure is lower than normal. These regions are called rarefactions. Sound travels by these compressions and rarefactions moving through the medium. This continuous process of pushing and pulling creates the sound wave.
In simple words: When a vibrating thing moves forward, it squishes air particles together, making a tight spot called a compression. When it moves back, it pulls the air particles apart, making a loose spot called a rarefaction. These tight and loose spots travel to carry sound.

🎯 Exam Tip: Remember to clearly define both terms and how they relate to the vibrating source's movement to fully describe sound propagation.

 

Question 2. Verify experimentally the reflection of the laws of sound.
Answer: To verify the laws of reflection of sound, you can perform the following experiment:
1. **Prepare Pipes:** Make two long pipes that are exactly the same size using chart paper.
2. **Arrange Setup:** Place these pipes on a table, making sure they are near a smooth, hard wall.
3. **Place Sound Source:** Put a ticking clock at the open end of one of the pipes. This pipe will be your incident sound path.
4. **Listen:** Place your ear at the open end of the second pipe, which will be the reflected sound path.
5. **Adjust for Clarity:** Carefully adjust the position of the second pipe until you can hear the clock's sound most clearly through it.
Once the sound is clearest, you can confirm the laws of reflection:
* The angle at which the sound hits the wall (angle of incidence) will be the same as the angle at which it bounces off (angle of reflection).
* The path of the incoming sound, the path of the reflected sound, and an imaginary line drawn perpendicular to the wall (the normal) will all lie in the same flat plane.
This experiment clearly demonstrates the verification of the laws of reflection of sound, showing how sound behaves much like light when it bounces off surfaces.
In simple words: You use two paper tubes and a clock near a wall. You listen through one tube for the sound from the other. When you hear the clock clearest, it means the sound bounced off the wall following rules: the angle it hit the wall is the same as the angle it bounced away.

🎯 Exam Tip: When describing experiments, clearly outline the materials, setup, procedure, and expected observations to demonstrate a full understanding.

 

Question 3. List the applications of sound.
Answer: Ultrasound, a type of sound beyond human hearing, has several important applications:
* **Cleaning Technology:** Ultrasound can clean objects by removing tiny dirt particles. The objects are put in a liquid, and ultrasound waves are sent through it, shaking the dirt off.
* **Detecting Flaws:** It is used to find hidden cracks or defects inside large metal parts. The ultrasound waves reflect differently from flawed areas, revealing them.
* **Medical Imaging (Echocardiography):** Doctors use ultrasound to see inside the human body, especially to create pictures of the heart. This method helps check the heart's health without surgery.
* **Breaking Kidney Stones:** Small kidney stones can be broken into tiny grains using ultrasound. These small pieces can then easily leave the body.
In simple words: Ultrasound is used for cleaning things, finding cracks in metal, seeing inside the heart (like an ultrasound scan), and breaking kidney stones into small pieces.

🎯 Exam Tip: When listing applications, ensure each point is distinct and briefly explains the function or benefit of ultrasound in that context.

 

Question 4. Explain how do SONAR works?
Answer: SONAR, which means Sound Navigation And Ranging, is a system that uses ultrasonic waves to find and measure underwater objects.
1. **Sending Waves:** A SONAR device has a transmitter that sends out powerful ultrasonic sound waves into the water.
2. **Hitting Objects:** These sound waves travel through the water. When they hit an object, like a submarine or the seabed, they bounce back.
3. **Receiving Waves:** A detector on the ship then catches these bounced-back waves.
4. **Calculating Distance:** The system measures how long it took for the sound to go out and come back. Knowing the speed of sound in water, it can then figure out exactly how far away the object is.
This technology is very helpful for finding the depth of the ocean, locating things underwater like fish or shipwrecks, and mapping the sea floor. The use of sound waves allows detection over large distances underwater.
In simple words: SONAR sends out sound waves underwater. These waves hit objects and bounce back. By timing how long it takes for the waves to return, SONAR can tell us how far away and in what direction the underwater object is.

🎯 Exam Tip: When explaining SONAR, clearly define the acronym and describe the three main steps: transmission, reflection, and reception/calculation.

VI. Numerical Problems:

 

Question 1. The frequency of a source of sound is 600 Hz. Calculate the number of times it vibrates in a minute?
Answer:
**Given:**
Frequency \( (n) = 600 \, \text{Hz} \)
We know that 1 minute = 60 seconds.
**Calculation:**
Frequency represents the number of vibrations in one second.
To find the number of vibrations in one minute, we multiply the frequency by 60.
Number of vibrations in a minute \( = \text{Frequency} \times 60 \)
\( = 600 \, \text{Hz} \times 60 \)
\( = 36,000 \) vibrations
Thus, the source completes 36,000 vibrations in one minute.
In simple words: Since the sound source vibrates 600 times every second, to find out how many times it vibrates in a whole minute (60 seconds), we just multiply 600 by 60. This gives us 36,000 vibrations in one minute.

🎯 Exam Tip: Always pay attention to the units and the time frame given (seconds vs. minutes) when dealing with frequency and total vibrations.

 

Question 2. A stone is dropped from the top of a tower 750 m high into a pond of water at the base of the tower. Calculate the number of seconds for the splash to be heard? (Given \( g = 10 \, \text{m/s}^{-2} \) and speed of sound \( = 340 \, \text{m s}^{-1} \))
Answer:
**Given values:**
Height of tower, \( s = 750 \, \text{m} \)
Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
Speed of sound, \( v_{\text{sound}} = 340 \, \text{m/s} \)
Initial velocity of stone, \( u = 0 \, \text{m/s} \) (since the stone is dropped from rest)

**Part 1: Time for the stone to reach the pond (\( t_1 \))**
We use the second equation of motion: \( s = ut_1 + \frac{1}{2}gt_1^2 \)
Substitute the given values:
\( 750 = (0 \times t_1) + \frac{1}{2} \times 10 \times t_1^2 \)
\( 750 = 5t_1^2 \)
\( t_1^2 = \frac{750}{5} \)
\( t_1^2 = 150 \)
\( t_1 = \sqrt{150} \)
\( t_1 \approx 12.25 \, \text{s} \)

**Part 2: Time for the sound of the splash to travel back to the top of the tower (\( t_2 \))**
We use the formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
\( t_2 = \frac{s}{v_{\text{sound}}} \)
\( t_2 = \frac{750}{340} \)
\( t_2 \approx 2.21 \, \text{s} \)

**Part 3: Total time for the splash to be heard (\( T_{\text{total}} \))**
The total time is the sum of the time the stone falls and the time the sound travels up.
\( T_{\text{total}} = t_1 + t_2 \)
\( T_{\text{total}} = 12.25 \, \text{s} + 2.21 \, \text{s} \)
\( T_{\text{total}} \approx 14.46 \, \text{s} \)
Thus, the splash will be heard approximately 14.46 seconds after the stone is dropped.
In simple words: First, we figure out how long the stone takes to fall to the water. Then, we calculate how long it takes for the splash sound to travel back up. Adding these two times gives us the total time until the sound is heard.

🎯 Exam Tip: Break down complex problems into smaller, manageable steps, clearly identifying what each part calculates. This helps ensure accuracy and makes the solution easier to follow.

Intext Activities

 

Activity - 1
Aim: To show how a tuning fork vibrates and passes these vibrations to other things.
Materials Required:
1. Tuning fork
2. Rubber pad
Procedure:
1. Take a tuning fork and gently tap its prongs on a rubber pad.
2. Bring the vibrating tuning fork close to your ear. Listen for any sound.
3. Now, carefully touch the prongs of the vibrating tuning fork with your finger.
4. Try to feel any movement or sensation.
Observation:
When the tuning fork is struck on the rubber pad, it begins to vibrate. A sound is heard when it's near your ear, and you can also feel these vibrations with your finger.
Conclusion:
The tuning fork vibrates when hit, and these vibrations cause the nearby air particles to vibrate too, which produces sound. This movement from its normal position is what we call vibration. Sound needs these vibrations to travel.
In simple words: When you hit a tuning fork, it shakes. You can hear the sound and feel the shake. This shows that shaking (vibration) makes sound.

🎯 Exam Tip: When describing an activity, clearly list the aim, materials, step-by-step procedure, observations made, and the scientific conclusion drawn from them.

 

Activity - 2
Aim: To show how a longitudinal wave, like sound, travels.
Materials Required:
1. A coil (like a Slinky)
2. A spring
Procedure:
1. Take a long coil or spring.
2. Hold one end and move it quickly forward and backward in a straight line.
Observations:
1. You will notice that some parts of the coil get squeezed close together.
2. Other parts of the coil get stretched far apart.
These squeezed and stretched sections travel along the spring.
Conclusion:
1. The parts where the turns are closer represent compressions (high pressure/density).
2. The parts where the turns are far apart represent rarefactions (low pressure/density).
3. This demonstrates that longitudinal waves, like sound, move by creating these alternating areas of compressions and rarefactions. This constant movement helps sound travel through a medium.
In simple words: When you stretch and push a spring, you see tight parts (compressions) and loose parts (rarefactions) moving along. This shows how sound waves, which are longitudinal, travel.

🎯 Exam Tip: Use the Slinky analogy to help visualize compressions and rarefactions as regions of high and low particle density, respectively.

 

Activity - 3
Aim: To compare the different types of sounds that musical instruments make.
Materials Required:
1. Flute
2. Drums
3. Veena
(Note: Access to audio of these instruments would also be beneficial.)
Procedure:
1. Listen to music played by a flute, drums, and veena.
2. Make a table to note down how the sounds from each instrument are different.
Observation:

Conclusion:
The sounds produced by different musical instruments have unique characteristics. These include amplitude (which relates to loudness), wavelength, frequency (which determines pitch), and timbre (the unique quality that makes each instrument sound different). The part of the instrument that vibrates directly influences these sound qualities.
In simple words: Different musical instruments make different sounds because they have various vibrating parts. These parts create sounds with distinct loudness, pitch, and unique qualities.

🎯 Exam Tip: When comparing musical instruments, focus on how the vibrating component (e.g., string, air column, membrane) directly influences the fundamental properties of the sound produced, such as pitch and timbre.

9th Science Guide Sound Additional Important Questions and Answers

I. Choose the Correct Answer:

 

Question 1. Which of the following is a non-mechanical wave?
(a) sound
(b) water
(c) light
(d) None of the options
Answer: (c) light
In simple words: Light is a non-mechanical wave. This means it can travel without needing any material like air or water to move through.

🎯 Exam Tip: Differentiate between mechanical waves (require a medium, e.g., sound, water waves) and electromagnetic/non-mechanical waves (do not require a medium, e.g., light, radio waves).

 

Question 2. Sound cannot travel through
(a) darkness
(b) gases
(c) vacuum
(d) All of the options
Answer: (c) vacuum
In simple words: Sound needs something to travel through, like air. A vacuum is empty space, so sound cannot move there.

🎯 Exam Tip: Remember that sound is a mechanical wave, meaning it absolutely requires a medium (like solid, liquid, or gas) for its propagation and cannot travel in the absence of particles, such as in a vacuum.

 

Question 3. In longitudinal waves, the particle vibrates in a ................ direction of propagation.
(a) parallel
(b) perpendicular
(c) curved
(d) All of the options
Answer: (b) perpendicular
In simple words: In this type of wave, the particles of the medium move sideways, at a right angle, compared to the forward direction the wave is traveling.

🎯 Exam Tip: Understand how particles move relative to the direction of wave propagation for different wave types.

 

Question 4. Human beings cannot hear
(a) infrasonic sound
(b) sonic sound
(c) ultrasonic sound
(d) a & c
Answer: (d) a & c
In simple words: Humans cannot hear sounds that have very low frequencies (infrasonic) or very high frequencies (ultrasonic). Our ears are only designed to pick up sounds within a specific range.

🎯 Exam Tip: Remember that the human ear can typically hear sounds between 20 Hz and 20,000 Hz. Sounds outside this range are inaudible to us.

 

Question 5. The frequency greater than 20,000 Hz is ................ sound.
(a) infrasonic
(b) ultrasonic
(c) audible
(d) sonic
Answer: (b) ultrasonic
In simple words: Sounds that have a frequency higher than 20,000 Hz are called ultrasonic sounds. These are sounds with a pitch too high for human ears to detect.

🎯 Exam Tip: Know the definitions and frequency ranges for infrasonic, audible (sonic), and ultrasonic sounds, as these are common distinction points.

 

Question 6. of sound depends on the amplitude of the wave,
(a) Loudness
(b) Pitch
(c) Timbre
Answer: (a) Loudness
In simple words: The loudness of a sound is directly connected to the amplitude, or height, of its wave. A bigger wave means a louder sound that carries more energy.

🎯 Exam Tip: Distinguish between loudness (amplitude), pitch (frequency), and timbre (waveform/quality) as these are key characteristics of sound.

 

Question 7. ................ is a general term for the distinguishable characteristic of wave
(a) Pitch
(b) Amplitude
(c) Frequency
(d) Timbre
Answer: (d) Timbre
In simple words: Timbre is the special quality of a sound that lets you tell it apart from other sounds, even if they have the same loudness and pitch. It's like the unique "color" of a sound.

🎯 Exam Tip: Understanding timbre is important for describing why different instruments or voices sound distinct.

 

Question 8. Sound travels in air as ................ wave.
(a) longitudinal
(b) transverse
(c) electromagnetic
(d) all of these
Answer: (a) longitudinal
In simple words: When sound travels through the air, the air particles vibrate back and forth in the same direction that the sound is moving. This type of wave is called a longitudinal wave.

🎯 Exam Tip: Remember the key difference: longitudinal waves involve vibrations parallel to the wave direction, while transverse waves involve vibrations perpendicular to it.

 

Question 9. The distance between two adjacent crest is called ................
(a) amplitude
(b) wavelength
(c) time period
(d) frequency
Answer: (b) wavelength
In simple words: Wavelength is the measurement of one complete wave cycle, specifically the distance from the top of one wave (crest) to the top of the very next wave (crest).

🎯 Exam Tip: Clearly define wavelength, amplitude, frequency, and time period, as these terms are fundamental to understanding wave properties.

 

Question 10. When temperature increases the speed of sound
(a) increases
(b) decreases
(c) remains same
(d) none of these
Answer: (a) increases
In simple words: As the temperature of the air or any medium gets higher, the tiny particles inside it move faster. This makes them pass on sound vibrations more quickly, so the sound travels faster.

🎯 Exam Tip: The speed of sound is affected by the properties of the medium, with temperature being a key factor, usually increasing with higher temperatures in gases.

 

Question 11. The relation between speed, frequency and wavelength is ................
(a) \( V = \frac{v}{\lambda} \)
(b) \( v = \frac{\lambda}{\Upsilon} \)
(c) \( V = \gamma \lambda \)
(d) \( V = \frac{1}{\lambda} \)
Answer: (c) \( V = \gamma \lambda \)
In simple words: The formula that connects the speed of a wave, its frequency (how often it vibrates), and its wavelength (the length of one wave) is that speed equals frequency multiplied by wavelength.

🎯 Exam Tip: Remember the fundamental wave equation \( v = f \lambda \), where \( v \) is wave speed, \( f \) is frequency, and \( \lambda \) is wavelength. Be aware that different symbols may be used for frequency (like \( \gamma \) or \( \nu \)).

 

Question 12. SI unit of wavelength is ................
(a) Hertz
(b) metre per second
(c) decibel
(d) metre
Answer: (d) metre
In simple words: Wavelength is a measure of distance, specifically the length of one complete wave. The official unit for measuring length in science is the metre.

🎯 Exam Tip: Make sure to know the SI units for all basic wave properties like wavelength (metre), frequency (Hertz), time period (second), and speed (metre per second).

 

Question 13. The speed of sound in a gaseous medium depends on ................
(a) pressure
(b) temperature
(c) density
(d) all of these
Answer: (d) all of these
In simple words: How fast sound moves through a gas is affected by many things: the pressure of the gas, its temperature, and how dense it is. All these properties work together to determine the sound's speed.

🎯 Exam Tip: Remember that the properties of the medium (elasticity and density) are crucial for sound transmission. In gases, these are linked to pressure, temperature, and density.

 

Question 14. The elastic property of medium is a factor of ................ medium.
(a) gas
(b) liquid
(c) solid
(d) none of these
Answer: (c) solid
In simple words: The elastic property of a material, which is how well it springs back after being pushed, is a major reason why sound travels fastest through solids. Solids are typically very elastic.

🎯 Exam Tip: The speed of sound is generally highest in solids, followed by liquids, and slowest in gases, due to differences in their elastic properties and density.

 

Question 15. To hear an echo, the minimum distance of the obstacle from the source of sound must be ................
(a) 34 m
(b) 17 m
(c) 340 m
(d) 170 m
Answer: (b) 17 m
In simple words: To clearly hear an echo, the sound needs to travel to an object and bounce back. This requires the object to be at least 17 meters away from the sound source, allowing enough time for your brain to distinguish the original sound from its reflection.

🎯 Exam Tip: The minimum distance for an echo depends on the speed of sound and the persistence of hearing (about 0.1 seconds). Use the formula: Distance = (Speed of sound × Time) / 2.

 

Question 16. The part of ear that turns pressure variations into electrical signals is ................
(a) the hammer
(b) anvil
(c) cochlea
(d) stirrup
Answer: (c) cochlea
In simple words: Inside your inner ear, the cochlea is like a tiny processing center. It takes the vibrating signals from sound and changes them into electrical messages that your brain can understand, letting you hear.

🎯 Exam Tip: Be familiar with the main parts of the ear (outer, middle, inner) and their specific roles in the hearing process, especially the function of the cochlea.

 

Question 17. The frequency of source of sound is 60 Hz. Then the number of vibrations in a minute will be ................
(a) 36
(b) 360
(c) 3600
(d) 6000
Answer: (c) 3600
In simple words: If a sound source vibrates 60 times in one second (60 Hz), then in a full minute, which has 60 seconds, it will vibrate a total of 3600 times.

🎯 Exam Tip: Remember that 1 minute = 60 seconds. To convert from vibrations per second (Hz) to vibrations per minute, always multiply by 60.

 

Question 18. Sonar is a device that uses to measure the distance, direction and speed of the underwater objects.
(a) radio waves
(b) infrared waves
(c) water waves
(d) ultrasonic waves
Answer: (d) ultrasonic waves
In simple words: SONAR uses special, very high-pitched sound waves, called ultrasonic waves, to detect things under the water. These waves travel through water and bounce back, helping to find objects.

🎯 Exam Tip: Know the full form of SONAR (Sound Navigation And Ranging) and understand that it relies on the reflection of ultrasonic waves, which are ideal for underwater ranging due to their properties.

 

Question 19. The speeds of sound in four different media are given below. Which of the following is the most likely speed in ms-1 with which the two underwater whales in a sea can talk to each other when separated by a large distance?
(a) 5170
(b) 1280
(c) 340
(d) 1530
Answer: (d) 1530
In simple words: Whales live in the ocean, so their communication sounds travel through seawater. The typical speed of sound in seawater is around 1530 meters per second, which is the most realistic option for their underwater conversations.

🎯 Exam Tip: Remember that sound travels at different speeds in different mediums. For seawater, 1530 m/s is the characteristic speed.

 

II. Fill in the Blanks:

 

Question 1. The propagating disturbance that travels in a medium is called a ................
Answer: Wave
In simple words: A wave is like a moving ripple or shake that carries energy through something, without the material itself actually moving from one place to another.

🎯 Exam Tip: Recall that mechanical waves require a medium for propagation, unlike electromagnetic waves. Waves transmit energy, not matter.

 

Question 2. Rarefactions are the regions of ................ where particles are spread apart.
Answer: low pressure
In simple words: Rarefactions are the parts of a sound wave where the tiny particles in the air are spread out, making the air thinner and creating an area of low pressure.

🎯 Exam Tip: Rarefactions alternate with compressions in a longitudinal wave, creating the pressure variations that constitute sound.

 

Question 3. Sound travels faster in ................ and slower in ................
Answer: Solid, Gas
In simple words: Sound generally moves quickest through solid objects, like a metal bar, and slowest through gases, like the air we breathe.

🎯 Exam Tip: Sound travels fastest in solids, then liquids, and slowest in gases, due to the density and elasticity of the medium.

 

Question 4. The loudness of sound depends on the ................ of the sound wave.
Answer: Intensity
In simple words: How loud you perceive a sound to be is primarily determined by its intensity, which is a measure of the sound energy flowing per unit area.

🎯 Exam Tip: While loudness is subjective, intensity is an objective measure of sound energy. Loudness is closely related to a wave's amplitude.

 

Question 5. A sound of single frequency is called ................and a collection of tones is called ................
Answer: Tone, Note
In simple words: A single, pure sound with only one frequency is known as a tone. When you combine several different tones, you create what is called a musical note.

🎯 Exam Tip: Understand the difference between a simple tone (pure frequency) and a complex note (combination of frequencies/harmonics).

 

Question 6. The speed of sound in air at 0°C is ................
Answer: 330 ms\(^{-1}\)
In simple words: At a temperature of 0 degrees Celsius, sound travels through the air at a speed of 330 meters every second. This value is a common reference point.

🎯 Exam Tip: Memorize the approximate speed of sound in air at standard conditions (0°C) as it's a frequently used value in problems.

 

Question 7. ................is an unwanted sound.
Answer: Noise
In simple words: Noise is any sound that is unwanted, unpleasant, or disruptive to a person. It is often characterized by irregular vibrations.

🎯 Exam Tip: Noise is often characterized by irregular and non-periodic vibrations, making it distinct from musical sounds.

 

Question 8. Loudness equal to (or) greater than ................can be painful to the ear.
Answer: 120 dB
In simple words: Sounds that are as loud as or louder than 120 decibels (dB) can cause pain and potential damage to your ears. This is considered the pain threshold for hearing.

🎯 Exam Tip: Be aware of safe listening levels and the decibel (dB) scale for sound intensity, as prolonged exposure to high decibel levels can cause hearing damage.

 

Question 9. ................allows bats to navigate through dark caves and find insects for food.
Answer: Echolocation
In simple words: Bats use a clever method called echolocation, where they send out sounds and listen for the echoes. This helps them create a mental map of their surroundings and find prey in complete darkness.

🎯 Exam Tip: Echolocation is a fascinating biological application of sound reflection, similar to how SONAR works.

 

Question 10. ................is an image obtained by the use of reflected ultrasonic waves.
Answer: An echogram
In simple words: An echogram is a visual image or picture that is created by using echoes from ultrasonic sound waves, often used in medical scans like ultrasounds to see inside the body.

🎯 Exam Tip: Echograms are a medical application of ultrasound technology, useful for non-invasive imaging.

 

Question 11. Depth of an ocean can be determined by ................
Answer: SONAR
In simple words: The depth of the ocean can be accurately measured using SONAR, which sends sound pulses to the seabed and calculates the depth based on the time it takes for the echo to return.

🎯 Exam Tip: SONAR is critical for underwater mapping and navigation, as sound travels well in water where light cannot.

 

Question 12. The distance of underwater objects can be deduced by the formula ................
Answer: \( Id = v \times t \)
In simple words: To find the total distance sound travels to an underwater object and back, you multiply the speed of sound in water by the total time the sound takes. This helps determine how far away the object is.

🎯 Exam Tip: For objects where sound travels to and from, remember the formula \( \text{distance} = \frac{\text{speed} \times \text{time}}{2} \). The given formula \( Id = v \times t \) is for total distance traveled by sound (there and back).

 

Question 13. ................muscle contracts in response to electrical depolarisation of the muscle cells.
Answer: Cardiac
In simple words: The cardiac muscle, which is your heart muscle, is unique because it contracts automatically when it receives an electrical signal. This signal makes the muscle cells depolarize, causing the heart to beat.

🎯 Exam Tip: The cardiac muscle is unique for its involuntary, rhythmic contractions, driven by electrical impulses. This question is a bit out of context for a sound chapter, but good to know for biology.

 

Question 14. Velocity of sound by a tuning fork frequency of 480 Hz is 340 ms Its wavelength is ................
Answer: 0.7 m
In simple words: If a tuning fork creates sound waves at 480 vibrations per second, and those waves travel at 340 meters per second, then each wave is 0.7 meters long. We calculate this by dividing the speed by the frequency.

🎯 Exam Tip: Use the wave equation \( \lambda = \frac{v}{f} \) (wavelength = speed / frequency) to calculate wavelength. Ensure units are consistent.

 

Question 15. The part of ear that collects the sound from the surroundings is ................
Answer: Pinna
In simple words: The pinna is the visible, outer part of your ear that acts like a funnel. Its main job is to collect sound waves from the air around you and direct them deeper into your ear.

🎯 Exam Tip: The pinna acts like a funnel, efficiently collecting sound waves and channeling them towards the middle ear.

 

Question 16. Ultrasounds can also be used to detect cracks and flaws in ................
Answer: Metal blocks
In simple words: Ultrasonic waves, which are very high-frequency sounds, can be passed through metal objects. If there are any hidden cracks or defects, the way the waves reflect will change, allowing these flaws to be detected without damaging the metal.

🎯 Exam Tip: Industrial applications of ultrasound include non-destructive testing, where it's used to inspect materials for defects without damaging them.

 

Question 17. In the inner ear, the pressure variations are turned into electrical signals by the ................
Answer: Cochlea
In simple words: The cochlea in your inner ear is responsible for changing the physical pressure from sound vibrations into electrical signals. These electrical signals are then sent to your brain for interpretation as sound.

🎯 Exam Tip: This reinforces the importance of the cochlea as the main transducer in the auditory system, converting mechanical energy into neural impulses.

 

III. Match the Following:

Answer: 1-b, 2-c, 3-d, 4-a
In simple words: This match-up connects medical tools and sound phenomena with their correct descriptions or uses. For example, a stethoscope uses multiple sound reflections to amplify sounds, and an echogram is an image made with ultrasonic waves.

🎯 Exam Tip: For matching questions, understand the core function or definition of each term to correctly pair them. Think about how each item applies sound principles.

 

II.

Answer: 1-c, 2-d, 3-a, 4-b
In simple words: This section links various sound-related terms to their correct definitions. For example, a wave is a moving disturbance, and noise is sound we don't want to hear, while thunder is often heard as successive reflections.

🎯 Exam Tip: Pay close attention to precise definitions for fundamental concepts like 'wave' and 'noise' to avoid common errors in matching questions.

 

IV. Assertion and Reason Type Questions:

 

Mark the correct choice as :
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If assertion is false but reason is true.

 

Question 1. Assertion : Human beings Cannot hear infrasonic sound. Reason : frequency less than 20 Hz is called infrasonic sound.
Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion
In simple words: Both statements are correct: humans cannot hear infrasonic sounds, and infrasonic sounds are indeed defined as having a frequency below 20 Hz. The reason directly explains why humans can't hear them.

🎯 Exam Tip: For assertion-reason questions, first check if each statement is true individually. Then, evaluate if the reason correctly explains the assertion.

 

Question 2. Assertion : As the amplitude of sound increases, the loudness of sound decreases. Reason: The loudness of sound depends on the amplitude of the wave.
Answer: (d) Assertion is false but reason is true
In simple words: The first statement is false because a bigger sound wave (higher amplitude) means a louder sound, not a softer one. The second statement is true because how loud a sound is truly depends on its amplitude.

🎯 Exam Tip: Clearly differentiate between amplitude and loudness. A larger amplitude corresponds to a louder sound, so be careful with inverse relationships in assertions.

 

Question 3. Assertion : The sound of thunder is heard a little latter than the flash of light. Reason : The speed of sound is much less than the speed of light.
Answer: (a) Both Assertion and reason are true and Reason is the correct explanation of Assertion
In simple words: It is true that you see lightning before you hear thunder. This happens because light travels incredibly faster than sound, which is why the reason explains the assertion perfectly.

🎯 Exam Tip: This is a classic example demonstrating the significant difference in speeds between light and sound. Understand the magnitudes of these speeds.

 

Question 4. Assertion : The speed of sound in air is less than speed of sound in solid. Reason: If we increase the temperature of medium, the speed of sound increases in any medium.
Answer: (b) Both assertion and reason are true but reason is not the correct explanation of assertion
In simple words: Both statements are true: sound travels slower in air than in solids, and increasing temperature does make sound travel faster in any material. However, the reason about temperature doesn't explain why solids are faster than air in general.

🎯 Exam Tip: Remember to check if the reason directly explains the assertion, not just if both statements are individually true. Sound speed differences across states of matter relate to particle spacing and elasticity.

 

Question 5. Assertion : The sound which we hear again is called Echo. Reason: The sensation of sound (Echo) persists in our brain for about 1 s.
Answer: (c) Assertion is true but reason is false
In simple words: It is true that an echo is a sound heard again after reflection. But, the sensation of any sound lasts in our brain for only about 0.1 seconds, not a full 1 second, so the reason is incorrect.

🎯 Exam Tip: The persistence of hearing is critical for distinguishing echoes. Recall that the ear can distinguish two sounds if they arrive with a time interval greater than 0.1 seconds.

 

V. State Whether True or False. If False, Correct the Statement:

 

Question 1. Vibrating particles travel all the way from vibrating object to ear.
Answer: False.
Correct statement: Vibrating particles do not travel all the way from vibrating object to ear. They simply transfer energy by vibrating in place.
In simple words: This is false. The tiny particles in the air only vibrate back and forth to pass on sound energy; they do not travel all the way from the sound source to your ear.

🎯 Exam Tip: Understand that waves transmit energy, not matter. Particles in a medium oscillate around their equilibrium positions but do not travel with the wave.

 

Question 2. The small to and fro motion is called vibration.
Answer: True.
In simple words: Yes, this is true. Any small movement back and forth from a central point is defined as a vibration.

🎯 Exam Tip: Vibration is the fundamental cause of sound production. Any object that vibrates creates sound waves.

 

Question 3. Pitch is a general term for the distinguishable characteristic of a tone.
Answer: False.
Correct statement: Timbre is a general term for the distinguishable characteristic of a tone.
In simple words: This is false. Pitch tells us if a sound is high or low, but timbre is the overall quality that lets us tell different sounds apart, even if they have the same pitch.

🎯 Exam Tip: Be precise with sound characteristics: pitch relates to frequency, loudness to amplitude, and timbre (or quality) to the waveform or harmonic content.

 

Question 4. The speed of sound in a gaseous medium depends on the nature of gas.
Answer: True.
In simple words: Yes, this is a true statement. The speed at which sound travels through a gas is indeed affected by what kind of gas it is, due to its unique properties.

🎯 Exam Tip: The molecular weight and adiabatic index of a gas are key properties that influence the speed of sound within it.

 

Question 5. The direction of incident sound, direction of reflected sound and the normal are in different Planes.
Answer: False.
Correct statement: The direction of incident sound, direction of reflected sound and the normal are in the same plane.
In simple words: This is false. According to the law of reflection, the incoming sound, the outgoing sound, and an imaginary line straight out from the surface (the normal) all lie flat on the same surface.

🎯 Exam Tip: This statement relates to the law of reflection, which applies to both light and sound waves. The angle of incidence equals the angle of reflection, and all three lie in the same plane.

 

Question 6. Ultrasonic sounds cannot be heard by the Human beings.
Answer: True.
In simple words: This is true. Humans cannot hear ultrasonic sounds because their frequency is too high for the human ear to pick up, being above 20,000 Hz.

🎯 Exam Tip: Recall the human audible range (20 Hz to 20,000 Hz) and categorize sounds as infrasonic (below 20 Hz) or ultrasonic (above 20,000 Hz).

 

Question 7. The Eardrum moves inward when a rarefaction reaches it.
Answer: False.
Correct statement: The Eardrum moves outward when a rarefaction reaches it.
In simple words: This statement is false. When a rarefaction (an area of low pressure in a sound wave) hits your eardrum, the eardrum actually bulges outwards because the pressure inside is now higher than outside.

🎯 Exam Tip: Understand how compressions (high pressure) and rarefactions (low pressure) in a sound wave interact with the eardrum, causing it to vibrate. Compressions push the eardrum inward, while rarefactions pull it outward.

 

VI. Very Short Answer Questions:

 

Question 1. What is a wave?
Answer: A wave is a moving disturbance that carries energy from one place to another through a medium (or sometimes through empty space), without the medium's particles themselves moving along with the energy. It's essentially a transfer of energy.
In simple words: A wave is like a wobble or shake that passes energy from one spot to another, but the material it travels through just wiggles in place, it doesn't move with the wave.

🎯 Exam Tip: Emphasize that waves transfer energy, not matter, and are a disturbance that propagates through a medium (or space, for electromagnetic waves).

 

Question 2. Define compression and rarefaction.
Answer:
Compression: In a sound wave, a compression is a region where the particles of the medium are squeezed together, causing the density and pressure to be higher than normal. Think of it as the "squished" part of a wave.
Rarefaction: A rarefaction is the opposite; it's a region where the particles of the medium are spread out, resulting in lower density and pressure. This is the "stretched" part of a wave.
In simple words: Compressions are the tight, crowded spots in a sound wave where particles are packed close. Rarefactions are the open, spread-out spots where particles are far apart.

🎯 Exam Tip: Clearly define both terms and remember that they alternate in a longitudinal wave to propagate sound.

 

Question 3. Name the five characteristics of a sound wave.
Answer: The five main characteristics that describe a sound wave are: 1. Amplitude (how big the vibration is), 2. Frequency (how many vibrations per second), 3. Time period (how long one vibration takes), 4. Wavelength (the length of one complete wave), and 5. Velocity or Speed (how fast the wave travels).
In simple words: The five key things describing a sound wave are its size (amplitude), how often it vibrates (frequency), how long one full shake takes (time period), the length of one wave (wavelength), and how fast it moves (speed).

🎯 Exam Tip: Memorizing these five key characteristics and their definitions is fundamental to understanding wave mechanics.

 

Question 4. Name the distinguishing factors of sound.
Answer: The factors that help us tell one sound from another are: 1. Loudness (how strong the sound is), 2. Pitch (how high or low the sound is), and 3. Timbre (the unique quality or "color" of the sound). These allow us to recognize different voices or instruments.
In simple words: We can tell sounds apart by how loud they are, how high or low their pitch is, and their special unique quality called timbre.

🎯 Exam Tip: Understand how each of these factors (loudness, pitch, timbre) relates to the physical properties of a sound wave (amplitude, frequency, waveform).

 

Question 5. Define Intensity.
Answer: Intensity is defined as the amount of sound energy that passes through a specific area (perpendicular to the sound's direction) in one unit of time. It tells us how powerful the sound wave is at that particular point.
In simple words: Intensity measures how much sound energy hits a certain spot in one second. It's like checking how strong the sound is at that exact place.

🎯 Exam Tip: Intensity is an objective measure (power per unit area) closely related to the subjective experience of loudness, but not the same. It is proportional to the square of the amplitude.

 

Question 6. Differentiate tone from note.
Answer:
Tone: A tone is a very simple sound that has only one pure frequency. It sounds clean and unmixed, like the sound from a tuning fork.
Note: A note is a more complex sound that is made up of several different tones combined. It usually includes a main frequency and other related frequencies, which give it a richer sound, like a musical instrument playing.
In simple words: A tone is a single, pure sound, like a simple beep. A note is a richer sound made from several tones put together, like a sound from a musical instrument.

🎯 Exam Tip: Think of a tuning fork producing a pure tone, while a musical instrument playing a note produces a rich sound with various frequencies (harmonics).

 

Question 7. What is Reverberation?
Answer: Reverberation is when sound repeatedly reflects off surfaces, causing it to persist or echo in a space after the original sound has stopped. This is common in large, empty rooms where sounds linger.
In simple words: When sound bounces back many times and stays for a while, it's called reverberation.

🎯 Exam Tip: Reverberation is distinct from a discrete echo; it refers to the prolonged persistence of sound due to multiple reflections, whereas an echo is a single, clearly distinguishable reflection.

 

Question 8. Expand the abbreviation of the term SONAR.
Answer: SONAR stands for Sound Navigation And Ranging. This technology is widely used in underwater exploration and by marine animals.
In simple words: SONAR means Sound Navigation And Ranging. It's a way to find things using sound waves.

🎯 Exam Tip: Remember that SONAR uses sound waves, not light waves, because light cannot penetrate water effectively over long distances.

 

Question 9. What type of waves are illustrated by the movement of a rope whose one end is moved up and down?
Answer: When you move one end of a rope up and down, it creates a transverse wave. In this type of wave, the particles of the rope move perpendicular (at a right angle) to the direction the wave travels. Light waves are also a type of transverse wave, where the electric and magnetic fields oscillate perpendicular to the direction of propagation.
In simple words: Moving a rope up and down makes a transverse wave. The bits of rope move up and down, but the wave goes forward.

🎯 Exam Tip: Focus on the direction of particle movement relative to the wave's propagation to correctly identify wave types. Transverse is perpendicular, longitudinal is parallel.

 

Question 10. Name the type of waves produced when a tuning fork is struck in air.
Answer: When a tuning fork is hit in the air, it creates longitudinal waves. These waves cause air particles to vibrate back and forth in the same direction the sound travels.
In simple words: Hitting a tuning fork makes longitudinal waves in the air.

🎯 Exam Tip: Understand that sound waves are typically longitudinal because they involve compressions and rarefactions in the medium.

 

Question 11. A stone is dropped on the surface of water in a pond. Name the types of waves produced.
Answer: When a stone falls into a pond, it makes transverse waves on the water surface. While water waves on the surface are primarily transverse, sound waves within the water itself are longitudinal.
In simple words: Dropping a stone in water creates transverse waves that spread out.

🎯 Exam Tip: Surface water waves are a good example of transverse waves because the water particles move up and down, perpendicular to the wave's horizontal travel.

 

Question 12. State one observation from every day life which shows that sound travels much more slower than light.
Answer: We see a flash of lightning before we hear the thunder. This happens because the speed of sound (about \( 344 \text{ m/s} \)) is much slower than the speed of light (about \( 3 \times 10^8 \text{ m/s} \)). So, light reaches our eyes almost instantly, but sound takes time. This time difference can even be used to estimate how far away a storm is.
In simple words: We see lightning first, then hear thunder. Light is super fast, but sound is much slower.

🎯 Exam Tip: Always use the lightning and thunder example to explain the difference in speeds between light and sound, as it's a common and relatable observation.

 

Question 13. What name is given to those air-crafts which fly at speeds greater than the speed of sound?
Answer: Aircraft that fly faster than the speed of sound are called supersonic aircraft. When these aircraft break the sound barrier, they create a 'sonic boom'.
In simple words: Very fast planes that go quicker than sound are called supersonic aircraft.

🎯 Exam Tip: Relate supersonic speed to the phenomenon of a 'sonic boom' for a more complete answer.

 

Question 14. Name the device which is used to produce sound in laboratory experiments.
Answer: A tuning fork is used to create sound in science experiments. Tuning forks produce a very pure tone, meaning a single frequency, which is useful for precise experiments.
In simple words: A tuning fork makes sound for lab tests.

🎯 Exam Tip: Mention the purity of the tone (single frequency) when discussing tuning forks, as it's a key characteristic for experiments.

 

Question 15. What should an object do to produce sound?
Answer: For an object to make sound, it needs to vibrate (move back and forth) in a medium like air or water. These vibrations then travel as disturbances, reaching our ears and letting us hear the sound. Without a medium, like in space, even strong vibrations won't produce audible sound.
In simple words: An object needs to shake quickly in air or water to make sound. These shakes travel and we hear them.

🎯 Exam Tip: The core concept for sound production is vibration; ensure this is always included in your answer.

 

Question 16. Can sound travel through vacuum?
Answer: Sound cannot travel through a vacuum because it needs a material medium (like air, water, or steel) for its vibrations to pass through. In a vacuum, there are no particles to carry the sound energy. This is why there is no sound in outer space.
In simple words: Sound cannot go through empty space. It needs air, water, or something else to move.

🎯 Exam Tip: Emphasize that sound is a mechanical wave, which by definition requires a medium for propagation.

 

Question 1. How do we hear sound?
Answer: We hear sound because mechanical energy causes an object to vibrate. These vibrations then travel through the air as waves, enter our ears, and are interpreted as sound by our brain. The tiny bones in our middle ear play a crucial role in amplifying these vibrations.
In simple words: An object vibrates, sending waves to our ears, and our brain understands them as sound.

🎯 Exam Tip: Key elements to include are "vibrations," "travels through a medium," and "interpreted by the brain."

 

Question 2. Why are sound waves called mechanical waves?
Answer: Sound waves are called mechanical waves because they require a material medium (like air, water, or solids) to travel. They move by causing the particles of this medium to vibrate and bump into each other, transferring energy. Unlike electromagnetic waves such as light, mechanical waves cannot travel through a vacuum.
In simple words: Sound waves are mechanical waves because they need something solid, liquid, or gas to travel through, moving particles along.

🎯 Exam Tip: Always state that mechanical waves require a medium and contrast them with electromagnetic waves to highlight this property.

 

Question 3. Define: (i) Amplitude (ii) Frequency
Answer:
(i) Amplitude (A): This is the largest distance a particle of a medium moves from its calm, resting spot when a wave passes through it. The unit for amplitude is meters (m).
(ii) Frequency (n): This is the count of how many vibrations or waves are made in just one second. The unit for frequency is Hertz (Hz) or \( \text{s}^{-1} \). Amplitude is related to the loudness of a sound, while frequency determines its pitch.
In simple words: (i) Amplitude: How far a particle swings from its middle position. (ii) Frequency: How many swings happen in one second.

🎯 Exam Tip: Clearly distinguish between amplitude (related to loudness) and frequency (related to pitch), and remember their respective SI units.

 

Question 4. Define : Wavelength \( (\lambda) \) and Timbre
Answer:
(i) Wavelength \( (\lambda) \): This is the distance over which a sound wave completely repeats itself. More simply, it's the space between the centers of two compressions (where particles are close) or two rarefactions (where particles are spread out) next to each other. The unit for wavelength is meters (m).
(ii) Timbre: This is the special quality of a sound that lets us tell the difference between two sounds that have the same loudness and pitch but come from different sources, like a guitar and a piano playing the same note. Wavelength and frequency are inversely related; a longer wavelength means a lower frequency.
In simple words: (i) Wavelength: The length of one complete wave, from one peak to the next. (ii) Timbre: The special sound color that makes a trumpet sound different from a flute, even if they play the same note.

🎯 Exam Tip: For wavelength, define it using both repeating distance and distance between compressions/rarefactions. For timbre, emphasize distinguishing sounds from different sources at the same pitch and loudness.

 

Question 5. How are the Wavelength and frequency of a Sound wave related to its speed?
Answer: The speed of a sound wave \( (v) \) is connected to its wavelength \( (\lambda) \) and frequency \( (\nu) \) by a simple formula. To find the speed, you multiply the wavelength by the frequency. This shows how quickly the waves travel based on their length and how often they occur. This fundamental wave equation applies to all types of waves, not just sound.
\( v = \lambda \times \nu \)
In simple words: The speed of sound is found by multiplying its wavelength (how long one wave is) by its frequency (how many waves pass per second).

🎯 Exam Tip: Remember the universal wave equation \( v = \lambda \nu \). Clearly state that speed is directly proportional to both wavelength and frequency.

 

Question 6. What are the factors that intensity of Sound depends upon?
Answer: The loudness or intensity of a sound depends on these factors:

• The strength of the sound source's vibration (its amplitude).
• How far away the person listening is from the sound source.
• The size of the vibrating surface that creates the sound.
• The density (how tightly packed the particles are) of the material the sound travels through.
• The frequency (how fast it vibrates) of the sound source.

Sound intensity decreases rapidly with distance because the energy spreads out over a larger area.
In simple words: Sound intensity depends on how strong the vibration is, how far you are, the size of what makes the sound, how dense the air is, and the sound's pitch.

🎯 Exam Tip: List multiple factors for comprehensive answers. Amplitude and distance are usually the most critical for intensity.

 

Question 7. On which factors the speed of sound in gaseous medium depends?
Answer: The speed of sound as it travels through a gas is affected by these factors:

• The pressure of the gas.
• The temperature of the gas.
• The density of the gas (how much mass is in a given volume).
• The type or nature of the gas itself.

Generally, sound travels faster in warmer gases and slower in colder ones.
In simple words: How fast sound moves in gas depends on the gas's pressure, temperature, how thick it is, and what kind of gas it is.

🎯 Exam Tip: Remember that temperature has a significant effect on the speed of sound in gases, as it influences molecular motion.

 

Question 8. On which factors the speed of sound in solid medium depends?
Answer: The speed of sound moving through a solid material relies on these factors:

• The elastic properties of the solid (how well it returns to its original shape after being stretched or compressed).
• The temperature of the solid.
• The density of the solid (how much mass it has in a certain space).

Sound travels fastest in solids because their particles are closely packed and strongly linked, allowing vibrations to transfer quickly.
In simple words: How fast sound moves in a solid depends on how stretchy the solid is, its temperature, and how heavy it is for its size.

🎯 Exam Tip: Elasticity is a key factor for sound speed in solids; emphasize that denser, more rigid solids generally transmit sound faster.

 

Question 9. Mention the fequency ranges of infrasoncis and ultrasonics.
Answer: We categorize sound waves based on their frequency.

• Infrasonics are sound waves with frequencies that are too low for humans to hear, specifically less than \( 20 \text{ Hz} \).
• Ultrasonics are sound waves with frequencies that are too high for humans to hear, specifically greater than \( 20,000 \text{ Hz} \).

Many animals, like elephants and whales, can communicate using infrasound, while bats and dolphins use ultrasound for navigation and hunting.
In simple words: Infrasonics are sounds too low for us to hear (below \( 20 \text{ Hz} \)). Ultrasonics are sounds too high for us to hear (above \( 20,000 \text{ Hz} \)).

🎯 Exam Tip: Clearly state the exact frequency thresholds (\( 20 \text{ Hz} \) and \( 20,000 \text{ Hz} \)) for audible, infrasonic, and ultrasonic ranges.

 

Question 10. Why is a sound wave called a longitudinal wave?
Answer: A sound wave is called a longitudinal wave because it creates regions of compression (where particles are crowded) and rarefaction (where particles are spread out) in the air. In this wave, the air particles vibrate back and forth in the same direction that the sound energy is traveling. Think of a Slinky toy; when you push one end, the compression travels along the spring in the same direction as the push.
In simple words: Sound is a longitudinal wave because it squishes and stretches the air. The air particles move in the same path as the sound.

🎯 Exam Tip: The key point for longitudinal waves is that particle vibration is parallel to the direction of wave propagation; ensure this is clearly explained.

 

Question 11. Which characteristics of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer: The timbre, or unique quality, of a sound is what allows you to recognize your friend's voice even when you can't see them in a dark room and other people are also talking. Timbre is determined by the unique mix of different frequencies and overtones present in a sound.
In simple words: The special 'color' or quality of your friend's voice, called timbre, helps you know it's them, even in the dark.

🎯 Exam Tip: Timbre is often confused with pitch or loudness; emphasize that it's the unique "voice" or "color" of a sound source.

 

Question 12. List the causes of noise pollution.
Answer: The main causes of noise pollution include:

• Issues that lead to hearing problems.
• Damage to the eardrum and the resulting loss of hearing.
• Mental and emotional harm, also known as psychological damages.

Common sources of noise pollution include traffic, construction, industrial activities, and loud music.
In simple words: Noise pollution causes hearing problems, eardrum damage, and can harm your mind.

🎯 Exam Tip: Provide diverse examples of the negative impacts of noise pollution, covering both physical and psychological effects.

 

Question 13. Why are roots and walls of the auditorium covered with rough plaster?
Answer: The rough plaster on the ceilings and walls of an auditorium is used to reduce reverberation. Reverberation is when sound bounces around too much, making it hard to hear clearly. The rough surfaces absorb and scatter sound waves, preventing them from reflecting too many times. Sound-absorbing materials like curtains, carpets, and acoustic panels are also used for the same purpose.
In simple words: Rough plaster on auditorium walls helps stop sound from bouncing too much. This makes sounds clearer and not echo for too long.

🎯 Exam Tip: Connect rough surfaces directly to their ability to absorb and scatter sound, thereby reducing unwanted reflections and improving sound clarity.

 

Question 14. Give two practical applications of reflection of Sound?
Answer: Here are two practical uses of sound reflection:

• SONAR: This system uses sound waves to find things underwater. It sends out sound, and when the sound bounces back (reflects) from an object, SONAR measures the time and direction to figure out the object's distance, direction, and speed.
• Stethoscope: Doctors use stethoscopes to hear heartbeats and breathing sounds. The sound of a patient's heartbeat travels up the tube of the stethoscope to the doctor's ears through many reflections inside the tube.

Echos, which are also a form of sound reflection, are used in medical imaging and by bats for navigation.
In simple words: SONAR uses sound reflections to find things underwater. Stethoscopes use reflections to let doctors hear heartbeats.

🎯 Exam Tip: When providing applications, briefly explain how the reflection principle works in each example for better understanding.

 

Question 15. What is acoustics?
Answer: Acoustics is a field in physics that focuses on studying sound. It looks at how sound is made, how it travels, and how it affects things. Acousticians design concert halls and recording studios to optimize sound quality.
In simple words: Acoustics is the science of sound. It studies how sound works.

🎯 Exam Tip: Define acoustics as a branch of physics and mention its broad scope, from production to perception of sound.

 

Question 1. The frequency of a tuning fork is \( 484 \text{ Hz} \). What is the time period?
Answer: To find the time period, we use the formula that connects frequency and time period.
Given: Frequency of the tuning fork, \( n = 484 \text{ Hz} \).
The relationship between frequency \( (n) \) and time period \( (T) \) is:
\( n = \frac{1}{T} \)
This means the time period is the inverse of the frequency:
\( T = \frac{1}{n} \)
Now, we put in the given frequency value:
\( T = \frac{1}{484} \text{ s} \)
\( \implies T \approx 0.00207 \text{ s} \)
So, the time period is approximately \( 0.00207 \) seconds. A key concept here is that time period is the inverse of frequency.
In simple words: The tuning fork vibrates \( 484 \) times in one second. To find the time for one vibration (time period), we divide \( 1 \) by \( 484 \). This gives about \( 0.00207 \) seconds.

🎯 Exam Tip: Always show the formula \( T=1/n \) or \( n=1/T \) and include units in your calculations and final answer.

 

Question 2. Distinguish between Loudness and Intensity of the Sound?
Answer:
Here are the differences between loudness and intensity of sound:
Loudness
1. Loudness is how loud or soft a sound appears to us. It helps us tell different sounds apart.
2. The loudness of a sound is directly related to its amplitude. A bigger amplitude means a louder sound. So, loudness is proportional to the amplitude of the wave.
Intensity
1. The intensity of a sound wave describes the amount of sound energy that passes through a specific area in one second. It's a measure of the power of the sound.
2. Similar to loudness, the amplitude of the sound wave also determines its intensity. A larger amplitude leads to higher intensity. This means a more powerful sound.
In simple words: Loudness is how we feel sound, linked to its height (amplitude). Intensity is the actual power of the sound passing through an area, also linked to amplitude.

🎯 Exam Tip: Remember that loudness is a subjective perception, while intensity is an objective, measurable physical quantity related to energy flow.

 

Question 3. What is an echo? Write the conditions to hear an echo.
Answer: An echo is a sound that we hear again because sound waves have bounced off a surface and returned to us.
Conditions to hear an echo:

• To clearly hear an echo, there must be a time gap of at least \( 0.1 \) seconds between the original sound and its reflection. This means the sound must travel to an obstacle and come back to the listener in more than \( 0.1 \) seconds.
• The total distance the sound travels from the source to the obstacle and back to the listener must be at least \( 34 \) meters (calculated as speed of sound \( 340 \text{ m/s} \times 0.1 \text{ s} \)).
• Therefore, the minimum distance required for the obstacle (like a wall or mountain) from the sound source is half of this total distance, which is \( \frac{34 \text{ m}}{2} = 17 \text{ m} \).

This principle is used in sonar systems to measure depths and locate objects underwater.
In simple words: An echo is a sound that repeats because it bounces off something. To hear it, the sound needs to travel at least \( 17 \) meters to a wall and come back, taking more than \( 0.1 \) seconds.

🎯 Exam Tip: Crucial conditions are the time interval (\( 0.1 \) s) and the minimum distance (\( 17 \) m, based on the speed of sound in air), as these are frequently tested.

 

Question 4. Explain the working of human ear with diagram.
Answer: The human ear is made up of three main parts: the outer ear, the middle ear, and the inner ear.

• Outer Ear (Pinna): This is the visible part of the ear. It gathers sound waves from the surroundings and guides them into the ear canal.
• Middle Ear: When sound travels through the ear canal, it hits and vibrates the eardrum, a thin membrane. Three tiny bones-the hammer, anvil, and stirrup-are connected to the eardrum. These bones amplify (make stronger) the vibrations many times and pass them on to the inner ear.
• Inner Ear (Cochlea): The inner ear receives these amplified vibrations. Inside the cochlea, these vibrations are changed into electrical signals. These signals are then sent to the brain through the auditory nerve.

The brain then understands these electrical signals as the sounds we hear. The structure of the cochlea, a snail-shaped organ, is filled with fluid and tiny hair cells that detect different frequencies.
In simple words: The outer ear catches sound. The middle ear (with eardrum and tiny bones) makes the sound stronger. The inner ear (cochlea) changes it to electric signals for the brain to hear.

🎯 Exam Tip: When describing the ear, break it down into outer, middle, and inner sections, explaining the function of key parts like the eardrum, ossicles (hammer, anvil, stirrup), and cochlea in sequence.

 

Question 1. The frequency of a tuning fork is \( 484 \text{ Hz} \). What is the time period?
Answer: To find the time period, we use the formula that connects frequency and time period.
Given: Frequency of the tuning fork, \( n = 484 \text{ Hz} \).
The relationship between frequency \( (n) \) and time period \( (T) \) is:
\( n = \frac{1}{T} \)
This means the time period is the inverse of the frequency:
\( T = \frac{1}{n} \)
Now, we put in the given frequency value:
\( T = \frac{1}{484} \text{ s} \)
\( \implies T \approx 0.00207 \text{ s} \)
So, the time period is approximately \( 0.00207 \) seconds. A key concept here is that time period is the inverse of frequency.
In simple words: The tuning fork vibrates \( 484 \) times in one second. To find the time for one vibration (time period), we divide \( 1 \) by \( 484 \). This gives about \( 0.00207 \) seconds.

🎯 Exam Tip: Always show the formula \( T=1/n \) or \( n=1/T \) and include units in your calculations and final answer.

 

Question 2. Velocity of sound in air is \( 344 \text{ ms}^{-1} \). What is the range of wavelength of audible sound?
Answer: To find the range of wavelengths for sounds that humans can hear, we use the formula linking speed, frequency, and wavelength.
Given:
Speed of sound in air, \( v = 344 \text{ m/s} \).
The frequency range for audible sound (what humans can hear) is from \( 20 \text{ Hz} \) to \( 20,000 \text{ Hz} \).
The formula is: Speed \( (v) = \) Frequency \( (\nu) \times \) Wavelength \( (\lambda) \).
This can be rearranged to find wavelength: \( \lambda = \frac{v}{\nu} \).
Now, let's calculate the wavelength for the lowest and highest audible frequencies:

• For the lowest frequency (\( \nu = 20 \text{ Hz} \)):
  \( \lambda = \frac{344 \text{ m/s}}{20 \text{ Hz}} = 17.2 \text{ m} \)
• For the highest frequency (\( \nu = 20,000 \text{ Hz} \)):
  \( \lambda = \frac{344 \text{ m/s}}{20,000 \text{ Hz}} = 0.0172 \text{ m} \)

Therefore, the range of wavelengths for audible sound is from \( 0.0172 \text{ meters} \) to \( 17.2 \text{ meters} \). This means very low sounds have long wavelengths and very high sounds have short wavelengths.
In simple words: Sound travels at \( 344 \text{ m/s} \). Humans hear sounds from \( 20 \text{ Hz} \) to \( 20,000 \text{ Hz} \). We use speed divided by frequency to get wavelength. So, the longest sound wave we hear is \( 17.2 \text{ m} \) and the shortest is \( 0.0172 \text{ m} \).

🎯 Exam Tip: Remember to convert all given values to consistent SI units (e.g., cm to m) before performing calculations and clearly state the audible frequency range.

 

Question 3. A body vibrates \( 3000 \) times in one minute. If the velocity of sound in air \( 330 \text{ ms}^{-1} \), find (i) Frequency of vibration (ii) Wavelength.
Answer: We need to find the frequency and wavelength of a vibrating body.
Given:
The body vibrates \( 3000 \) times in one minute.
The velocity of sound in air, \( v = 330 \text{ m/s} \).
(i) Frequency of vibration:
Frequency is the number of vibrations per second. Since there are \( 60 \) seconds in a minute:
\( \nu = \frac{\text{Number of vibrations}}{\text{Time in seconds}} = \frac{3000}{60} \text{ Hz} \)
\( \implies \nu = 50 \text{ Hz} \)
(ii) Wavelength:
We use the wave equation: Speed \( (v) = \) Frequency \( (\nu) \times \) Wavelength \( (\lambda) \).
Rearranging to find wavelength: \( \lambda = \frac{v}{\nu} \).
Substitute the given velocity and calculated frequency:
\( \lambda = \frac{330 \text{ m/s}}{50 \text{ Hz}} \)
\( \implies \lambda = 6.6 \text{ m} \)
So, the frequency is \( 50 \text{ Hz} \) and the wavelength is \( 6.6 \text{ meters} \). This calculation is essential for understanding the properties of specific sound waves.
In simple words: The body vibrates \( 3000 \) times in \( 60 \) seconds, so its frequency is \( 50 \) times per second. With sound speed at \( 330 \text{ m/s} \), the wavelength is \( 330 \) divided by \( 50 \), which is \( 6.6 \) meters.

🎯 Exam Tip: Always convert time to seconds when calculating frequency from vibrations per minute to ensure correct units.

 

Question 4. A person is listening to a tone of \( 400 \text{ Hz} \) from a distance of \( 500 \text{ m} \) from the source of sound. What is the time interval between successive compressions from source?
Answer: We need to find the time between two compressions of a sound wave. This time is actually the wave's time period.
Given:
Frequency of the sound tone, \( \nu = 400 \text{ Hz} \).
The time interval between two consecutive compressions (or rarefactions) from the source is the definition of the wave's time period \( (T) \).
The formula for time period is: \( T = \frac{1}{\nu} \) (inverse of frequency).
Substitute the given frequency:
\( T = \frac{1}{400 \text{ Hz}} \)
\( \implies T = 0.0025 \text{ s} \)
So, the time interval between successive compressions from the source is \( 0.0025 \) seconds. The distance of \( 500 \text{ m} \) is extra information not needed for this specific part of the question. This short time period indicates a high-pitched sound.
In simple words: The sound has a frequency of \( 400 \) times per second. The time between each squeeze (compression) of the sound wave is the time period. We get this by dividing \( 1 \) by \( 400 \), which is \( 0.0025 \) seconds.

🎯 Exam Tip: Recognize that the time interval between successive compressions (or rarefactions) is simply the time period of the wave.

 

Question 5. An echo returned in \( 3\text{s} \). What is the distance of reflecting surface from the source given that the speed of sound is \( 330 \text{ ms}^{-1} \).
Answer: We need to find the distance to the surface that created the echo.
Given:
Speed of sound, \( v = 330 \text{ m/s} \).
Time taken for the echo to return, \( t = 3 \text{ s} \).
The sound travels from the source to the reflecting surface and then back to the listener. So, the total distance covered by the sound is twice the distance to the reflecting surface (\( 2d \)).
Using the formula: Distance = Speed \( \times \) Time.
Total distance \( (2d) = v \times t \)
\( 2d = 330 \text{ m/s} \times 3 \text{ s} \)
\( \implies 2d = 990 \text{ m} \)
To find the distance \( (d) \) from the source to the reflecting surface, we divide the total distance by \( 2 \):
\( d = \frac{990 \text{ m}}{2} \)
\( \implies d = 495 \text{ m} \)
So, the reflecting surface is \( 495 \) meters away from the sound source.
In simple words: Sound travels \( 330 \text{ m/s} \). An echo came back in \( 3 \) seconds. So, the sound traveled a total of \( 330 \times 3 = 990 \) meters. Since this is a round trip, the wall is half that distance away, which is \( 495 \) meters.

🎯 Exam Tip: For echo problems, always remember that the calculated total distance is for a round trip (to the obstacle and back), so you must divide by two to get the distance to the obstacle.

 

Question 6. Two persons are at opposite ends of a steel rod. One strikes the end of rod with a stone. Find the ratio of time taken by the sound wave in air and in steel to reach the second person. (Sound of speed in steel = \( 5960 \text{ ms}^{-1} \), Sound of speed in air = \( 344 \text{ ms}^{-1} \))
Answer: We need to find the ratio of how long sound takes to travel through air versus steel over the same distance. Let the length of the steel rod be \( L \).
Given:
Speed of sound in steel, \( v_{\text{steel}} = 5960 \text{ m/s} \).
Speed of sound in air, \( v_{\text{air}} = 344 \text{ m/s} \).
The formula for time is: Time = Distance / Speed.
Let's assume the distance \( L \) is the same for both.
Time taken for sound in air, \( T_{\text{air}} = \frac{L}{v_{\text{air}}} = \frac{L}{344} \)
Time taken for sound in steel, \( T_{\text{steel}} = \frac{L}{v_{\text{steel}}} = \frac{L}{5960} \)
Now, we find the ratio of the time taken in air to the time taken in steel:
\( \frac{T_{\text{air}}}{T_{\text{steel}}} = \frac{L/344}{L/5960} \)
The \( L \) terms cancel out:
\( \frac{T_{\text{air}}}{T_{\text{steel}}} = \frac{5960}{344} \)
\( \implies \frac{T_{\text{air}}}{T_{\text{steel}}} \approx 17.33 \)
So, the ratio of time taken by sound in air to sound in steel is approximately \( 17.33:1 \). This means sound travels much faster through steel than through air. This difference in speed is due to steel being much denser and more elastic than air, allowing vibrations to propagate more efficiently.
In simple words: Sound travels \( 5960 \text{ m/s} \) in steel and \( 344 \text{ m/s} \) in air. To find how much longer it takes in air, we divide the speed in steel by the speed in air. The sound takes about \( 17.33 \) times longer to travel through air than through steel.

🎯 Exam Tip: When dealing with ratios of time for constant distance, remember that time is inversely proportional to speed, so the ratio of times is the inverse of the ratio of speeds.

 

Question 7. A Sound wave travels at a speed of \( 344 \text{ ms}^{-1} \). If the wavelength is \( 1.5 \text{ cm} \), what is the frequency of Wave? Will it be audible?
Answer: We need to calculate the frequency of a sound wave and determine if it's audible to humans.
Given:
Speed of sound, \( v = 344 \text{ m/s} \).
Wavelength of sound, \( \lambda = 1.5 \text{ cm} \). We must convert this to meters: \( \lambda = 1.5 \text{ cm} = 0.015 \text{ m} \).
The formula relating speed, frequency, and wavelength is: Speed \( (v) = \) Frequency \( (\nu) \times \) Wavelength \( (\lambda) \).
Rearranging to find frequency: \( \nu = \frac{v}{\lambda} \).
Substitute the given values:
\( \nu = \frac{344 \text{ m/s}}{0.015 \text{ m}} \)
\( \implies \nu \approx 22933.33 \text{ Hz} \)
Now, let's check if this frequency is audible. The human audible frequency range is typically from \( 20 \text{ Hz} \) to \( 20,000 \text{ Hz} \).
Since \( 22,933.33 \text{ Hz} \) is greater than \( 20,000 \text{ Hz} \), this frequency is not audible to humans. It falls into the ultrasonic range. Sounds above \( 20,000 \text{ Hz} \) are called ultrasound and are used in many medical and industrial applications.
In simple words: Sound travels at \( 344 \text{ m/s} \). The wave is \( 1.5 \text{ cm} \) long (or \( 0.015 \text{ m} \)). Frequency is speed divided by wavelength, which is about \( 22,933 \text{ Hz} \). Since humans can only hear up to \( 20,000 \text{ Hz} \), this sound is too high-pitched for us to hear.

🎯 Exam Tip: Always remember to convert units (e.g., cm to m) before calculation. Also, compare the calculated frequency with the human audible range (\( 20 \text{ Hz} \) to \( 20,000 \text{ Hz} \)) to answer the "audible" part of the question.