Frank Brothers Solutions for ICSE Class 10 Physics Chapter 3 Sound

Page 149

Question 1. Distinguish between longitudinal waves and transverse waves.
Answer:
The differences between these two wave types are as follows:

In simple words: Longitudinal waves wiggle back and forth in the direction they are traveling, like a Slinky. Transverse waves wiggle up and down, perpendicular to their travel, like ripples on water.

Exam Tip: Highlighting the direction of particle vibration relative to wave propagation is the most important point when distinguishing these waves.

Question 2. What do you understand by amplitude of an oscillation? State its SI unit.
Answer:
The maximum displacement of a vibrating particle from its central equilibrium position to either side (the height of a crest or depth of a trough) is defined as its amplitude. Its standard SI unit is the metre (m).
In simple words: Amplitude is how far a wave wiggles from its resting position.

Exam Tip: Be sure to mention both crest height and trough depth from the mean position for a complete definition.

Question 3. Write down the relation between wavelength, frequency and wave-velocity.
Answer:
The relationship between these wave properties is given by: \[ \text{Wave Velocity} = \text{Wavelength} \times \text{Frequency} \] Or, \( v = \lambda \times f \)
In simple words: To find out how fast a wave is moving, multiply its length by how many times it vibrates per second.

Exam Tip: Use standard symbols \( v \), \( \lambda \), and \( f \) when writing this equation in exams.

Question 4. State two factors on which the velocity of a wave depends.
Answer:
The propagation speed of a wave through a material depends on:
1. The elasticity of the medium.
2. The density of the medium.
In simple words: How fast sound travels depends on how stretchy and how heavy the material is.

Exam Tip: Name both properties explicitly as elasticity and density to secure full marks.

Question 5. How does the velocity of a wave vary with temperature?
Answer:
Wave speed is directly affected by temperature; it increases as the temperature of the medium rises.
In simple words: Sound waves travel faster when the air or material gets warmer.

Exam Tip: Mention that speed increases with temperature, often quantitatively by 0.61 m/s per degree Celsius in air.

Question 6. The speed of a wave is 350 \( \text{m s}^{-1} \). Find the wavelength of the wave whose frequency is 700 Hz.
Answer:
Given:
Wave speed (\( v \)) = \( 350 \text{ m s}^{-1} \)
Frequency (\( f \)) = \( 700 \text{ Hz} \)

Using the wave formula: \[ v = f \lambda \] Solving for wavelength (\( \lambda \)): \[ \lambda = \frac{v}{f} \] \[ \lambda = \frac{350}{700} = 0.5 \text{ m} \] Thus, the wavelength is \( 0.5 \text{ m} \).
In simple words: When we divide the speed of 350 by the frequency of 700, we find the wave length is exactly half a meter.

Exam Tip: Show the step-by-step division and always remember to write the unit as meters (m) for wavelength.

Page 150

Question 7. Explain how a longitudinal wave propagates in a medium.
Answer:
Longitudinal waves travel through a medium by forming alternating zones of high and low pressure, known as compressions and rarefactions:
- **Compressions:** When a vibrating source pushes forward, it squeezes the air directly in front of it, creating a high-density, high-pressure zone. The medium's particles are pushed close together.
- **Rarefactions:** When the source moves backward, it creates a sudden drop in pressure behind it. This creates a low-density, low-pressure zone where the particles spread out.
These disturbances are passed along neighboring particles, causing the wave to travel outward from the source.
In simple words: When something vibrates, it squashes the air in front of it (compression) and leaves a stretched space behind it (rarefaction). These pulses travel outwards as sound.

Exam Tip: Define compressions as regions of high density/pressure and rarefactions as regions of low density/pressure for a complete answer.

Question 8. How does a transverse wave propagate in a medium?
Answer:
A transverse wave moves forward through the creation of alternating crests and troughs. The uppermost positions of maximum positive displacement are termed crests, whereas the lowest positions of maximum negative displacement are termed troughs.
In simple words: Transverse waves move like ripples on water, with high hills called crests and low valleys called troughs.

Exam Tip: When defining crests and troughs, always state that they are the points of maximum positive and negative displacement, respectively.

Question 9. Define the term wave-velocity.
Answer:
Wave-velocity is defined as the distance traversed by a wave per unit time as it propagates through a given medium.
In simple words: Wave-velocity is simply how fast a wave is traveling through a material.

Exam Tip: Write the formula \( v = \frac{d}{t} \) or \( v = f \lambda \) along with the definition to show a thorough understanding.

Question 10. What is the nature of sound waves?
Answer:
By nature, sound waves are classified as mechanical (and specifically longitudinal) waves.
In simple words: Sound waves are mechanical waves because they need physical matter to move through.

Exam Tip: Do not just say "mechanical", specify "longitudinal mechanical waves" for complete accuracy.

Question 11. Which of the two, sound waves or electromagnetic waves, have a greater speed?
Answer:
Electromagnetic waves propagate at a vastly superior speed compared to sound waves (e.g., \( 3 \times 10^8 \text{ m/s} \) in vacuum).
In simple words: Light and other electromagnetic waves are much faster than sound waves.

Exam Tip: Mention the speed of light in vacuum (\( 3 \times 10^8 \text{ m/s} \)) versus the speed of sound in air (\( \sim 340 \text{ m/s} \)) to highlight the huge difference.

Question 12. Sound waves being mechanical waves need a material medium for their propagation. Can they travel in vacuum?
Answer:
Because sound waves are mechanical waves, they require a physical substance (such as a solid, liquid, or gas) to transmit their energy. Consequently, they are unable to propagate through a vacuum where no matter is present.
In simple words: Sound cannot travel in space or a vacuum because there are no particles to bounce into each other and pass the sound along.

Exam Tip: Clearly link the requirement of a material medium to the mechanical nature of sound waves.

Question 13. What are the key characteristics of wave motion?
Answer:
Key features of wave propagation include:
1. It requires a medium possessing both elasticity (to restore particles) and inertia (to store energy).
2. Energy supplied to any zone initiates periodic oscillations of particles around their baseline rest states.
3. Only energy is transported across space; the physical matter of the medium is not moved along with the wave.
4. The speed of the wave is determined entirely by the physical properties of the medium and is independent of the source's motion or characteristics.
5. The velocity of the wave itself is distinct from the instantaneous speed of the vibrating particles.
6. Transfer of energy from one particle to the next occurs over a regular, repeating time interval.
7. The total energy of the wave is a combination of the kinetic and potential energies of the medium's oscillating particles.
In simple words: A wave is just a disturbance that carries energy through a material without carrying the actual material along with it.

Exam Tip: Memorize at least three core characteristics, especially the fact that energy is transferred without any net transfer of matter.

Question 14. Is energy transferred during wave motion?
Answer:
Yes, the primary function of wave motion is the transfer of energy through a medium.
In simple words: Yes, waves are basically energy moving from one place to another.

Exam Tip: A simple "Yes" should be accompanied by a brief explanation that energy (not matter) is what propagates.

Question 15. Distinguish between electromagnetic waves and mechanical waves.
Answer:
The differences between electromagnetic and mechanical waves are summarized below:

In simple words: Electromagnetic waves (like light) can travel through empty space without any air or material, while mechanical waves (like sound) must have matter to travel through.

Exam Tip: The ability to travel through a vacuum is the most basic difference between electromagnetic and mechanical waves.

Question 16. Can transverse waves travel in air? Give a reason.
Answer:
No, transverse waves are unable to propagate through gases like air. They require a medium with shear rigidity (such as solids or liquid surfaces) to support perpendicular forces.
In simple words: Transverse waves cannot travel through air because air is a gas and isn't rigid enough to support side-to-side shear waves.

Exam Tip: Always explain that gases lack "rigidity", which is required for transverse waves to propagate.

Question 17. Define wavelength. What is its SI unit?
Answer:
Wavelength is the spatial length of one complete wave cycle.
- In a transverse wave, it is the distance between two consecutive crests or two consecutive troughs.
- In a longitudinal wave, it is the distance between two consecutive compressions or two consecutive rarefactions.
Its standard SI unit is the metre (m).
In simple words: Wavelength is the distance from the top of one wave to the top of the next one.

Exam Tip: Give definitions for both transverse and longitudinal wave types to write a comprehensive answer.

Question 18. What is the range of hearing for a normal human ear? What is this range called?
Answer:
The normal human ear can perceive sounds with frequencies ranging from \( 20 \text{ Hz} \) to \( 20,000 \text{ Hz} \). This is referred to as the audible range (or range of audibility).
In simple words: We can hear sounds that vibrate between 20 times a second and 20,000 times a second.

Exam Tip: Remember that the limits are 20 Hz and 20,000 Hz (or 20 kHz), and name the range as the range of audibility.

Question 19. What is an echo?
Answer:
An echo is a distinct sound heard due to the reflection of the original sound wave from a distant, solid barrier (such as a cliff, hillside, or building wall).
In simple words: An echo is when sound bounces off a wall far away and you hear it again.

Exam Tip: Mention that the reflecting surface must be a rigid obstacle for an echo to form.

Question 20. What is the full form of SONAR?
Answer:
The acronym SONAR stands for Sound Navigation and Ranging.
In simple words: SONAR means using sound waves to navigate and measure distances, especially underwater.

Exam Tip: Write the full form in lower or title case as requested, ensuring correct spelling of "Ranging".

Question 21. What is the full form of RADAR?
Answer:
The acronym RADAR stands for Radio Detection and Ranging.
In simple words: RADAR uses radio waves instead of sound to locate distant objects.

Exam Tip: Differentiate clearly between SONAR (sound waves) and RADAR (radio waves).

Question 22. How is the depth of a sea determined? Name the principle on which it is based.
Answer:
The depth of the ocean is calculated using a technique called echo depth sounding. This method relies on the principle of echo formation (reflection of ultrasonic waves).
In simple words: We can find out how deep the sea is by sending a sound down and timing how long it takes to bounce back.

Exam Tip: Name the method "echo depth sounding" and state that it is based on the reflection of sound.

Question 23. What are ultrasonic waves?
Answer:
Sound waves with a frequency exceeding the upper limit of human audibility (\( > 20,000 \text{ Hz} \)) are termed ultrasonic waves.
In simple words: Ultrasonic sounds are too high-pitched for humans to hear.

Exam Tip: Always mention the threshold of 20,000 Hz (or 20 kHz) in your definition of ultrasonic waves.

Question 24. What are infrasonic waves?
Answer:
Sound waves with frequencies below the lower threshold of human hearing (\( < 20 \text{ Hz} \)) are referred to as infrasonic waves.
In simple words: Infrasonic sounds are too low-pitched for humans to hear.

Exam Tip: Always specify that infrasonic waves are below 20 Hz.

Question 25. State two applications of supersonics.
Answer:
Supersonics (objects traveling faster than the speed of sound) are utilized in:
1. High-speed jet aircraft design.
2. Rockets and spacecraft propulsion systems.
In simple words: Supersonic flight is used by fast fighter jets and space rockets.

Exam Tip: Do not confuse "supersonic" (speed faster than sound) with "ultrasonic" (frequency higher than audibility).

Question 26. Define echo and reverberation. How are they different?
Answer:
While an echo is a single, distinct reflected sound heard from a distance, reverberation is the persistence or prolongation of sound caused by multiple overlapping reflections.
- **Echo:** Requires a minimum distance of \( 17 \text{ m} \) between the source and the reflector in air so that the reflection arrives at least \( 0.1 \text{ s} \) after the original sound.
- **Reverberation:** Occurs when the reflecting walls are closer than \( 17 \text{ m} \), causing the returning waves to blend with and prolong the original sound before it dies out.
In simple words: An echo is when you hear a clear copy of your voice a moment later, while reverberation is a prolonged roll of sound because of close walls.

Exam Tip: Highlight the distance of 17 m and the time interval of 0.1 s to explain why echoes are distinct while reverberations are not.

Question 27. State two conditions necessary for the formation of an echo.
Answer:
To successfully perceive an echo, two core criteria must be fulfilled:
1. The distance between the sound source and the reflecting obstacle must be at least \( 17 \text{ m} \) in air.
2. The reflected wave must arrive at the listener's ear at least \( 0.1 \text{ s} \) after the emission of the original sound, which corresponds to the persistence of hearing.
In simple words: To hear an echo, you must be at least 17 meters away from a wall, and the sound must take at least a tenth of a second to return.

Exam Tip: Connect the 0.1 s condition to the "persistence of hearing" of the human ear.

Question 28. What is the lowest and highest frequency audible to the human ear?
Answer:
The human auditory system can detect sound waves with a minimum frequency of \( 20 \text{ Hz} \) and a maximum frequency of \( 20,000 \text{ Hz} \).
In simple words: We can hear sounds as low as 20 Hz and as high as 20,000 Hz.

Exam Tip: Clearly state both the lower and upper limits of human hearing.

Question 29. Can sound waves be reflected?
Answer:
Yes, similar to light waves, sound waves undergo reflection when they strike a boundary.
In simple words: Yes, sound bounces off hard surfaces just like light bounces off a mirror.

Exam Tip: Mention that sound follows the laws of reflection just like light.

Question 30. Describe a simple experiment to demonstrate the reflection of sound waves.
Answer:
An experiment to verify the reflection of sound waves is conducted as follows:
1. Roll two identical long tubes out of thick chart paper.
2. Lay both tubes on a flat table pointing toward a smooth vertical wall at an angle.
3. Place a ticking mechanical clock near the outer opening of the first tube.
4. Put your ear near the outer opening of the second tube and adjust its angle until the ticking sound is loudest.
**Conclusion:** The sound waves from the clock pass down the first tube, bounce off the flat wall, and travel through the second tube to the ear. Measuring the angles reveals that the angle of incidence (\( i \)) equals the angle of reflection (\( r \)).
In simple words: Put two cardboard tubes on a table pointing at a wall. Put a clock at one end and listen at the other; you will hear the ticking best when the tubes are set at equal angles to the wall.

Exam Tip: Make sure to clearly state that the angle of incidence equals the angle of reflection in your conclusion.

Question 31. Explain a simple method to determine the velocity of sound in air.
Answer:
The speed of sound in air can be measured using the Echo Method:
1. A person stands at a measured distance (\( d \) meters) from a large flat wall or cliff.
2. The person fires a starting pistol to produce a loud sound and starts a stopwatch at the exact same instant.
3. The stopwatch is stopped the moment the person hears the returning echo.
4. Since the sound travels to the cliff and back, the total distance traveled is \( 2d \). The velocity (\( v \)) of sound is calculated as: \[ v = \frac{\text{Total Distance}}{\text{Time Taken}} = \frac{2d}{t} \] Repeating this process several times and averaging the results provides an accurate value.
In simple words: Stand a known distance from a cliff, make a loud noise, and time how long it takes to hear the echo. Since the sound goes back and forth, divide twice the distance by the time.

Exam Tip: Remember to double the distance (\( 2d \)) because the sound wave travels to the obstacle and back to the listener.

Question 32. Explain how (i) bats and (ii) dolphins make use of ultrasonic waves.
Answer:
(i) **Bats:** Bats emit and perceive ultra-high-frequency ultrasonic calls. As they fly, these signals bounce off nearby structures, insects, or obstacles. By listening to the returning echoes, bats build a mental map of their path in complete darkness, helping them navigate and capture prey without crashing. This is called sound ranging.
(ii) **Dolphins:** Dolphins send out ultrasonic pulses in all directions underwater. When these waves strike a fish, predator, or obstacle, they bounce back as echoes. By interpreting these echoes, dolphins determine the location, size, and movement of prey or threats.
In simple words: Bats and dolphins use ultrasonic squeaks that bounce off objects. By listening to these echoes, they can 'see' obstacles and prey in the dark or underwater.

Exam Tip: Use the term "echolocation" or "sound ranging" and specify that they utilize "ultrasonic waves" for this process.

Question 33. What is sound signaling?
Answer:
Sound signaling refers to the process of transmitting ultrasonic waves in various directions to navigate, communicate, or alert about obstacles based on returning echoes.
In simple words: Sound signaling is sending out high-pitched sounds to warn about objects or send messages.

Exam Tip: Define sound signaling as a method of communication or warning using ultrasonic waves.

Question 34. What is sound ranging?
Answer:
Sound ranging is the technique of locating or mapping distant objects and obstacles by transmitting ultrasonic pulses and measuring the time delay of their returning echoes.
In simple words: Sound ranging is figuring out where things are by measuring how long a high-pitched sound takes to bounce back.

Exam Tip: Emphasize that sound ranging specifically involves measuring distances based on the time delay of the echo.

Question 35. A man standing at a certain distance from a wall fires a gun and hears its echo after 5 s. He then moves 310 m towards the wall and on firing the gun again, hears the echo after 3 s. Calculate the velocity of sound in air.
Answer:
Let the initial distance of the person from the wall be \( d \) and the velocity of sound in air be \( v \).
Given:
Initial echo time (\( t \)) = \( 5 \text{ s} \)

Using the echo formula: \[ t = \frac{2d}{v} \] \[ 5 = \frac{2d}{v} \implies d = \frac{5v}{2} \quad \text{--- (i)} \]
When the person moves \( 310 \text{ m} \) closer to the wall:
New distance (\( d' \)) = \( d - 310 \)
New echo time (\( t' \)) = \( 3 \text{ s} \) \[ t' = \frac{2d'}{v} \] \[ 3 = \frac{2(d - 310)}{v} \implies d - 310 = \frac{3v}{2} \quad \text{--- (ii)} \]
Subtracting equation (ii) from equation (i): \[ d - (d - 310) = \frac{5v}{2} - \frac{3v}{2} \] \[ 310 = \frac{2v}{2} \] \[ v = 310 \text{ m/s} \] Thus, the velocity of sound in air is \( 310 \text{ m/s} \).
In simple words: We can find the speed of sound by comparing how much closer the person moved to how much sooner the echo returned. In this case, the speed of sound is 310 meters per second.

Exam Tip: Show both distance equations clearly and perform the subtraction step systematically to avoid calculation errors.

Question 36. How do bats fly safely in the dark and find their prey?
Answer:
Bats possess the ability to generate and detect high-frequency ultrasonic sounds. When flying, they emit these calls which strike any obstacle or prey ahead. The waves bounce back as echoes. By receiving and analyzing these echoes, bats determine the exact location and distance of obstacles even in complete darkness, allowing safe flight. This process is called sound ranging.
In simple words: Bats navigate in the dark by squeaking and listening to the echoes that bounce off objects around them.

Exam Tip: Mention "ultrasonic waves" and "sound ranging" as key terms in your answer.

Question 37. Give three examples of organisms or people who use the principle of echo to locate obstacles or prey, and explain how they do it.
Answer:
The principle of echo location is used by bats, dolphins, and fishermen:
- **Bats and Dolphins:** They emit ultrasonic pulses and listen to the returning reflections to locate prey and navigate around obstacles.
- **Fishermen:** They use sonar devices that send ultrasonic waves into the water. By detecting the echoes from shoals of fish or the sea bed, they can find fish and measure depth.
In simple words: Bats, dolphins, and fishermen all send out high-pitched sounds and listen for the echoes to spot objects, prey, or the sea floor.

Exam Tip: Clearly separate the examples of animals from humans (fishermen) to show a well-rounded response.

Question 38. What is SONAR? Explain its working principle and state the formula used to calculate the depth of an ocean.
Answer:
SONAR stands for Sound Navigation and Ranging. It is an instrument designed to calculate underwater distances using ultrasonic waves.
**Working Principle:** A transmitter on a ship sends a brief, high-energy ultrasonic pulse straight down toward the seabed. This signal travels through the water, strikes the sea bottom, and reflects back as an echo. A receiver detects the returning signal.
**Formula:** Knowing the speed of sound in water (\( v \)) and the time interval (\( t \)) between sending and receiving, the depth (\( d \)) is calculated using: \[ d = \frac{v \times t}{2} \]
In simple words: SONAR measures how deep the ocean is by timing how long an underwater beep takes to go down and bounce back up, then dividing by two.

Exam Tip: Remember to write the full form of SONAR and define all variables in the depth formula \( d = \frac{v \times t}{2} \).

Question 39. Which waves are sent in SONAR to find the depth of the sea?
Answer:
In SONAR systems, ultrasonic waves are utilized to measure the depth of the sea.
In simple words: SONAR uses high-pitched ultrasonic waves to measure sea depth.

Exam Tip: State clearly that ultrasonic waves are chosen because they can travel long distances in water without being absorbed much.

Question 40. A man standing between two parallel cliffs fires a gun. He hears the first echo after 4 s and the second echo after 6 s. If the speed of sound in air is 320 m/s, calculate the distance between the two cliffs.
Answer:
Let \( d_1 \) and \( d_2 \) be the distances from the person to the first and second cliff respectively.
Given:
Time for the first echo (\( t_1 \)) = \( 4 \text{ s} \)
Time for the second echo (\( t_2 \)) = \( 6 \text{ s} \)
Speed of sound (\( v \)) = \( 320 \text{ m/s} \)

Using the echo formula \( 2d = v \times t \):
For the first cliff: \[ 2d_1 = 320 \times 4 = 1280 \implies d_1 = 640 \text{ m} \] For the second cliff: \[ 2d_2 = 320 \times 6 = 1920 \implies d_2 = 960 \text{ m} \]
The total distance between the two cliffs is: \[ d_1 + d_2 = 640 + 960 = 1600 \text{ m} \] Thus, the cliffs are \( 1600 \text{ m} \) apart.
In simple words: The first cliff is 640 meters away, and the second cliff is 960 meters away. Adding them together gives a total distance of 1600 meters.

Exam Tip: Calculate the individual distances first, then add them together to find the total distance between the cliffs.

PAGE NO-159

Question 1. Explain how a wave propagates through a medium.
Answer:
Wave propagation is the process through which a disturbance travels through a medium. This occurs when the particles of the medium undergo repeated back-and-forth oscillations about their central rest positions, passing their momentum and energy sequentially to adjacent particles without shifting their own average positions.
In simple words: Waves travel when particles wiggle in place and pass the wiggle along to their neighbors, like a Mexican wave in a stadium.

Exam Tip: Explain that particles only oscillate about their mean positions and do not move along with the wave.

Question 2. Differentiate between transverse waves and longitudinal waves.
Answer:
The core differences between transverse and longitudinal waves are:

In simple words: Transverse waves wiggle up and down as they move, while longitudinal waves wiggle along the same direction they travel.

Exam Tip: Use tables to present comparative answers clearly to secure full marks.

Question 3. What are free vibrations?
Answer:
Free vibrations are the periodic oscillations of a body that continue with a constant amplitude and a constant natural frequency in the complete absence of any external dissipative forces.
In simple words: Free vibrations are when something wiggles at its own natural rate with a constant size, forever, without any air resistance slowing it down.

Exam Tip: Stress that constant amplitude only occurs in an ideal, resistance-free environment (like a vacuum).

Question 4. What are natural vibrations of a body? On what factors does their frequency depend?
Answer:
When a body clamped or fixed at a point is briefly disturbed from its resting position and left to vibrate, it oscillates at a specific rate called its natural frequency. These oscillations are known as natural vibrations. This natural frequency depends entirely on the size, shape, and elastic material properties of the body.
In simple words: Natural vibrations are the unique wiggles a clamped object makes when you tap it once. Its pitch depends on its size and shape.

Exam Tip: State clearly that the natural frequency depends on the shape, size, and material composition of the vibrating body.

Question 5. What are damped vibrations?
Answer:
Damped vibrations are periodic oscillations whose amplitude progressively decreases over time due to the presence of resistive forces (such as friction or air resistance) in the surrounding medium.
In simple words: Damped vibrations are oscillations that slowly die out and get smaller because of friction or air drag.

Exam Tip: Give a practical example, such as a swinging pendulum in air, to support your definition of damped vibrations.

Question 6. What is resonance? Give two examples of resonance.
Answer:
Resonance is a special state of forced vibration that occurs when the frequency of an externally applied periodic force matches the natural frequency of the vibrating system. This causes the system to oscillate with a dramatically increased amplitude.
**Examples:**
1. **Stringed Instruments:** Guitars and violins are equipped with a hollow sound box. When the strings vibrate, they force the air column inside the sound box to oscillate. Because the box is designed to match these frequencies, resonance occurs, producing a much louder sound.
2. **Radio Tuning:** When we tune a radio receiver to a specific station, we adjust the electrical components of the circuit to match its natural frequency to the frequency of the incoming broadcast waves. This produces resonance, amplifying the signal of that particular station.
In simple words: Resonance is when a sound or force matches the natural wiggle-rate of an object, making it vibrate with a massive amplitude.

Exam Tip: Mention that "large amplitude" is the key indicator of resonance when frequencies match.

Question 7. A string of length \( l \) can vibrate in different ways. (i) Which diagram shows the principal note? (ii) Which vibration has a frequency four times that of the first? (iii) What is the ratio of the frequency of the vibration in (a) and (b)?
Answer:
(i) Diagram (a) represents the fundamental note (or principal note) because the string oscillates as a single complete loop.
(ii) Diagram (c) illustrates a mode of vibration with a frequency four times that of the fundamental frequency (diagram a).
(iii) The ratio of the vibrational frequencies between diagram (a) and diagram (b) is \( 1:2 \).
In simple words: The first diagram shows the basic fundamental note with one loop. The third one vibrates four times faster, and the first two have a frequency ratio of one to two.

Exam Tip: Identify the number of loops (harmonics) to determine the relative frequency of each vibration mode.

Question 8. Compare music and noise.
Answer:
The comparison between music and noise is given below:

In simple words: Music is pleasant and periodic, while noise is irritating and random.

Exam Tip: Mention that music has a periodic waveform while noise has a completely irregular waveform.

Question 9. Give two examples of forced vibrations.
Answer:
Two common instances of forced vibrations are:
1. **Pressed Tuning Fork:** Pressing the stem of a vibrating tuning fork onto a wooden tabletop forces the tabletop to oscillate at the tuning fork's frequency, creating forced vibrations.
2. **Acoustic Guitar Strings:** Plucking guitar strings forces the wood of the bridge and the guitar body to vibrate at the same rate, amplifying the sound.
In simple words: Forced vibrations happen when a vibrating object touches another object and forces it to vibrate at the same speed, like a tuning fork pressed on a desk.

Exam Tip: Clearly identify both the forcing agent (the source) and the body experiencing the forced oscillations.

Question 10. What are forced vibrations? Under what condition does resonance occur?
Answer:
An oscillatory system executes forced vibrations when it is driven by a periodic external force, regardless of whether the external driving frequency matches the system's own natural frequency. Resonance is a specific case of these forced vibrations that occurs only when the frequency of the external force is exactly equal to (or is an integer multiple of) the system's natural frequency.
In simple words: Forced vibrations happen when a force makes something wiggle at any speed. Resonance only happens when that force matches the object's natural speed perfectly.

Exam Tip: Define forced vibrations first, then define resonance as the special case of frequency matching.

Question 11. State three factors on which the frequency of vibrations of a stretched string depends.
Answer:
The fundamental frequency (\( f \)) of a vibrating stretched string is determined by three physical factors:
1. **Length (\( l \)):** The frequency is inversely proportional to the vibrating length of the string (\( f \propto \frac{1}{l} \)).
2. **Tension (\( T \)):** The frequency is directly proportional to the square root of the mechanical tension in the string (\( f \propto \sqrt{T} \)).
3. **Linear Mass Density (\( m \)):** The frequency is inversely proportional to the square root of the mass per unit length (\( f \propto \frac{1}{\sqrt{m}} \)). Thus, thinner strings produce higher frequencies.
In simple words: A string vibrates faster if it is shorter, tighter, or thinner.

Exam Tip: Write down the proportional relations (\( f \propto \frac{1}{l} \), \( f \propto \sqrt{T} \), and \( f \propto \frac{1}{\sqrt{m}} \)) alongside the descriptions to show a complete mathematical understanding.

Question 12. State two ways by which the frequency of vibration of a stretched string can be increased.
Answer:
To increase the pitch or frequency of a stretched string:
1. Increase the tension by tightening the tuning pegs.
2. Decrease the active vibrating length of the string.
In simple words: You can make a guitar string sound higher by tightening it or by sliding your finger down to make it shorter.

Exam Tip: These two actions directly relate to the laws of transverse vibrations of a stretched string.

Question 13. Why are soldiers asked to break steps while crossing a suspension bridge?
Answer:
When marching across a suspension bridge, soldiers must break step to prevent disaster. If they march in unison, the periodic forces they exert are in phase, creating a strong vibration. If the frequency of these steps matches the bridge's natural frequency, resonance occurs. This causes the bridge to oscillate with a dangerously large amplitude, which could cause the structure to collapse.
In simple words: If soldiers march in perfect rhythm on a bridge, their collective steps might match the bridge's natural wiggle-rate. This creates resonance, making the bridge swing wildly and potentially collapse.

Exam Tip: Use the term "resonance" and explain how matching frequencies can cause dangerously high amplitude oscillations.

Question 14. Why does the amplitude of vibrations of a body in a medium decrease with time?
Answer:
Any real physical medium offers viscous drag or friction, which continually siphons off the mechanical energy of a vibrating body, causing its amplitude to decay over time.
In simple words: Air and other surroundings slow things down by friction, making the vibrations get smaller and smaller until they stop.

Exam Tip: Attribute the amplitude decay directly to the dissipative forces of the surrounding medium.

Question 15. Is pitch the same as frequency?
Answer:
No, pitch and frequency are not identical. Frequency is an objective, measurable physical property of a sound wave, whereas pitch is a subjective sensation perceived by the human ear.
In simple words: Frequency is a real physical measurement of sound speed, but pitch is how our ears and brain perceive that sound.

Exam Tip: Explain that frequency is objective (measured in Hertz) while pitch is subjective (perceived as high or low).

Question 16. In tuning stringed instruments, why is the tension changed and not the length of the string?
Answer:
For a given stringed instrument, the active length and mass density of each string are fixed parameters. Therefore, the tension must be adjusted to tune the instrument to the correct fundamental frequency.
In simple words: Since you cannot easily change the length or thickness of the strings permanently, you tighten or loosen them to tune the instrument.

Exam Tip: Relate this directly to the formula \( f \propto \sqrt{T} \) where length \( l \) and linear density \( m \) are constant.

Question 17. Why are stringed instruments provided with strings of different thicknesses?
Answer:
The fundamental frequency of a string is inversely proportional to its thickness (radius). In order to span a wide range of frequencies (from deep bass to high treble), stringed instruments are built with strings of varying thicknesses.
In simple words: Thicker strings make lower sounds, and thinner strings make higher sounds. That's why instruments have strings of different thicknesses.

Exam Tip: Mention the inverse relationship between fundamental frequency and string thickness (or mass per unit length).

Question 18. Four resonance tubes P, Q, R, and S are placed as shown. (i) With which tubes is no loud sound heard and with which is a loud sound heard? (ii) Explain why resonance occurs in Q and S but not in P and R. (iii) State the general condition for resonance in air columns.
Answer:
(i) A loud sound is heard only with tube S, while no loud sound is perceived from tubes P and R.
(ii) Resonance takes place in tubes Q and S, but fails to occur in tubes P and R. The air column in tube S has a natural frequency that is three times (an odd harmonic) the natural frequency of tube Q, leading to resonant reinforcement. Conversely, the air columns of P and R do not match the driving frequency.
(iii) In general, resonance occurs in an air column when its natural frequency matches either the fundamental frequency or any harmonic of the external vibrating tuning fork.
In simple words: Tube S resonates and makes a loud sound because its air column is the right size to match the vibration speed. Tubes P and R do not match, so they stay quiet.

Exam Tip: Explain that resonance requires the natural frequency of the air column to match the frequency of the tuning fork.

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Question 19. If a tuning fork of frequency 256 Hz is held over a resonance column, with which of the following tuning forks will the column resonate? (i) 256 Hz, (ii) 512 Hz, (iii) 1024 Hz.
Answer:
A tuning fork with a frequency of \( 512 \text{ Hz} \) will resonate with the column, as \( 512 \text{ Hz} \) is an exact integer multiple (the octave or second harmonic) of the fundamental frequency of \( 256 \text{ Hz} \).
In simple words: The column will vibrate in response to 512 Hz because it is exactly double the basic frequency of 256 Hz.

Exam Tip: State that resonance can occur at integer multiples (harmonics) of the fundamental natural frequency.

Question 20. (i) Describe an experiment to demonstrate the phenomenon of resonance using tuning forks. (ii) State the principle of resonance.
Answer:
(i) **Resonance Experiment with Tuning Forks:** Mount two identical tuning forks, A and B, of the same natural frequency, on separate hollow wooden soundboxes placed with their open ends facing each other. Set tuning fork A into vibration. After a brief moment, touch fork A to stop its vibration. You will hear a loud sound still coming from tuning fork B. The vibrations from A were carried through the air column to B, forcing B to vibrate at its resonant frequency.
(ii) **Principle of Resonance:** When a system is subjected to a periodic external force whose frequency matches the system's own natural frequency, the system begins to oscillate with a significantly larger amplitude.
In simple words: If you ring one tuning fork near a matching one, the second one will start ringing on its own. This happens because their natural frequencies match.

Exam Tip: Mention the role of the hollow soundboxes in carrying and amplifying the vibrations between the two forks.

Question 21. Describe an experiment using four pendulums to demonstrate resonance.
Answer:
**Pendulum Resonance Experiment:** Suspend four pendulums, P, Q, R, and S, from a common elastic string \( XX' \). Make pendulums P and S of equal length, while Q is shorter and R is longer.
**Observation:** Set pendulum P into oscillation. S soon starts swinging with a large amplitude, in phase with P. Pendulums Q and R also oscillate, but with very small, negligible amplitudes.
**Reason:** The vibrations from P are transmitted through the string. Since S has the same length as P, their natural time periods are identical, leading to resonance and large-amplitude swings. Q and R only undergo minor forced vibrations.
In simple words: If you hang four pendulums and make one swing, only the one with the exact same length will start swinging widely. This is because length determines a pendulum's natural swing-rate.

Exam Tip: State clearly that the pendulum of equal length (S) experiences resonance, while others of different lengths only show forced oscillations.

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Question 1. What is an echo? State the conditions necessary for its formation.
Answer:
An echo is the reproduction of a sound wave after it reflects off a distant obstacle.
**Necessary conditions:**
1. The distance to the reflector must be at least \( 17 \text{ m} \) in air.
2. The reflected wave must arrive at least \( 0.1 \text{ s} \) after the original sound.
In simple words: An echo is hearing your voice again after it bounces off a wall. You need to be at least 17 meters away to hear it.

Exam Tip: Make sure to list both the minimum distance (17 m) and the minimum time delay (0.1 s).

Question 2. Why are echoes usually decreased in an auditorium when the audience is present?
Answer:
The presence of an audience reduces echoes because human bodies and clothing act as highly effective sound-absorbing materials, preventing reflections.
In simple words: People and their clothes absorb sound, which stops echoes from bouncing around an auditorium.

Exam Tip: Use terms like "sound absorbers" or "acoustic damping" to explain why echoes decrease.

Question 3. A person fires a gun at a certain distance from a cliff and hears the echo after 2 s. If the speed of sound is 480 m/s, calculate the distance of the cliff.
Answer:
Given:
Time delay (\( t \)) = \( 2 \text{ s} \)
Speed of sound (\( v \)) = \( 480 \text{ m/s} \)

Using the echo distance relation: \[ d = \frac{v \times t}{2} \] \[ d = \frac{480 \times 2}{2} = 480 \text{ m} \] Thus, the cliff is \( 480 \text{ m} \) away.
In simple words: Since the echo took 2 seconds to travel there and back, the cliff is exactly 480 meters away.

Exam Tip: Always divide the total sound travel distance by 2 to get the actual distance to the reflecting surface.

Question 4. A girl stands at a distance of 660 m from a high wall. She claps her hands and hears the echo after some time. If the speed of sound in air is 330 m/s, after how long does she hear the echo?
Answer:
Given:
Distance to wall (\( d \)) = \( 660 \text{ m} \)
Speed of sound (\( v \)) = \( 330 \text{ m/s} \)

Using the echo time formula: \[ t = \frac{2d}{v} \] \[ t = \frac{2 \times 660}{330} \] \[ t = \frac{1320}{330} = 4 \text{ s} \] Thus, the girl hears the echo after \( 4 \text{ s} \).
In simple words: It takes the clap sound 4 seconds to travel the 1320 meters to the wall and back.

Exam Tip: Be sure to double the distance (\( 2 \times 660 = 1320 \text{ m} \)) before dividing by the speed of sound.

Question 5. A rifle is fired at a point between two parallel walls. The first echo is heard after 3 s and the second after 6 s. Find the width of the valley if the speed of sound at 0°C is 330 m/s and the temperature rises to 10°C. (Assume the velocity of sound increases by 0.61 m/s per °C rise in temperature).
Answer:
Given:
Speed of sound at \( 0^\circ\text{C} \) = \( 330 \text{ m/s} \)
Rate of increase in speed with temperature = \( 0.61 \text{ m/s per } ^\circ\text{C} \)
Temperature of air = \( 10^\circ\text{C} \)

First, calculate the speed of sound at \( 10^\circ\text{C} \): \[ v_{10} = 330 + (0.61 \times 10) = 330 + 6.1 = 336.1 \text{ m/s} \]
Let \( d_1 \) and \( d_2 \) be the distances to the two valley walls.
The time delays are \( t_1 = 3 \text{ s} \) and \( t_2 = 6 \text{ s} \).
Using the echo relation: \[ 2d_1 = v_{10} \times t_1 = 336.1 \times 3 = 1008.3 \text{ m} \] \[ 2d_2 = v_{10} \times t_2 = 336.1 \times 6 = 2016.6 \text{ m} \]
Adding these equations: \[ 2(d_1 + d_2) = 1008.3 + 2016.6 = 3024.9 \text{ m} \] \[ d_1 + d_2 = \frac{3024.9}{2} = 1512.45 \text{ m} \] Thus, the total width of the valley is \( 1512.45 \text{ m} \).
In simple words: At 10°C, sound travels at 336.1 meters per second. The total distance between the walls is 1512.45 meters.

Exam Tip: First adjust the speed of sound for the given temperature, then solve for individual distances and add them together.

Question 6. Explain how (a) bats and (b) dolphins use ultrasonic waves.
Answer:
(a) **Bats:** They produce and detect ultrasonic squeaks. When these waves encounter an obstacle, they reflect back. By listening to the echoes, bats can map out their surroundings and avoid crashing even in complete darkness. This is called sound ranging.
(b) **Dolphins:** They emit underwater ultrasonic signals in all directions. By analyzing the reflected waves, they identify threats and pinpoint small fish.
In simple words: Bats and dolphins use ultrasonic echoes as a sort of "sonar" to locate things around them.

Exam Tip: Emphasize "ultrasonic waves" and "echolocation" as core concepts.

Question 7. (a) Define frequency. Draw waveforms of high and low pitch. (b) Define loudness. Draw waveforms of soft and loud notes. (c) Define pitch and draw waveforms of waves of different pitch. (d) Define quality (timbre) and draw waveforms of different wave forms.
Answer:
(a) **Frequency:** The frequency of a vibrating object is the total number of complete wave cycles or oscillations executed per second. (b) **Loudness:** Loudness is the auditory attribute that allows a listener to distinguish between a strong (loud) sound and a weak (soft) sound of the same pitch and timbre. Loudness depends directly on the wave's amplitude. (c) **Pitch:** Pitch is the subjective interpretation of frequency. It is the quality that lets us categorize a note as high or low.
(d) **Quality (Timbre):** Quality is the characteristic of a sound that enables us to differentiate between two sounds of identical loudness and pitch produced by different sources. It depends on the waveform. In simple words: Frequency is how fast a wave vibrates, making it sound high or low. Loudness is how tall the wave is (amplitude), making it sound soft or loud.

Exam Tip: When drawing waveforms, show that high pitch has closely spaced waves (higher frequency) and loud notes have taller waves (higher amplitude).

Question 8. Define wave-velocity, wavelength and frequency. Write down the relation between them.
Answer:
- **Wave-velocity:** The distance covered by a wave per unit time as it moves through a medium.
- **Wavelength:** The distance between two consecutive identical points on a wave, such as two successive crests or troughs.
- **Frequency:** The number of complete wave cycles produced per second.
The relationship is given by: \[ \text{Wave-velocity} = \text{Wavelength} \times \text{Frequency} \]
In simple words: Wave speed is found by multiplying how long a wave is by how many waves pass by each second.

Exam Tip: Clearly define each term before stating the equation to secure full marks.

Question 9. A sound wave has a velocity of 330 m/s and frequency of 256 Hz. Calculate its wavelength.
Answer:
Given:
Velocity (\( v \)) = \( 330 \text{ m/s} \)
Frequency (\( f \)) = \( 256 \text{ Hz} \)

Using the wave equation: \[ \text{Wavelength} (\lambda) = \frac{\text{velocity}}{\text{frequency}} \] \[ \lambda = \frac{330}{256} \approx 1.3 \text{ m} \] Thus, the wavelength is \( 1.3 \text{ m} \).
In simple words: When we divide the speed of 330 by the frequency of 256, we get a wavelength of about 1.3 meters.

Exam Tip: Keep track of decimal points during division, and round to one decimal place where appropriate.

Question 10. On what factors do the pitch, loudness and quality of a musical note depend?
Answer:
The physical factors influencing these qualities are:
- **Pitch:** Depends primarily on the frequency of the sound wave.
- **Loudness:** Depends on the amplitude of the vibration.
- **Quality (Timbre):** Depends on the wave shape or waveform (the mix of fundamental and overtones).
In simple words: Pitch depends on vibration speed, loudness depends on wave height, and quality depends on the wave's shape.

Exam Tip: Always link pitch to frequency, loudness to amplitude, and quality to waveform.

Question 11. The displacement-distance graph of a wave is shown in the figure. (a) Find the wavelength of the wave. (b) Find the amplitude of the wave. (c) If the frequency of the wave is 50 Hz, find its velocity.
Answer:
(a) **Wavelength:** The distance between two consecutive crests of the wave is \( 10 \text{ cm} \) (or \( 0.1 \text{ m} \)).
(b) **Amplitude:** The peak displacement from the central axis is \( 4 \text{ cm} \).
(c) **Velocity:**
Given:
Frequency (\( f \)) = \( 50 \text{ Hz} \)
Wavelength (\( \lambda \)) = \( 10 \text{ cm} = 0.1 \text{ m} \)

Using the wave speed formula: \[ v = f \times \lambda = 50 \times 0.1 = 5 \text{ m/s} \]
In simple words: From the graph, the wavelength is 10 cm and the amplitude is 4 cm. At 50 Hz, the wave travels at 5 meters per second.

Exam Tip: Always convert wavelength from centimeters to meters before calculating wave velocity to keep units in SI.

Question 12. Distinguish between (a) light waves and sound waves, (b) free vibrations and forced vibrations, (c) radio waves and light waves.
Answer:
(a) **Light Waves vs. Sound Waves:**

(b) **Free Vibrations vs. Forced Vibrations:**

(c) **Radio Waves vs. Light Waves:**

In simple words: These tables compare light vs sound waves, free vs forced vibrations, and radio vs light waves.

Exam Tip: Memorize at least two distinct differences for each comparison to score full marks in descriptive sections.

Question 13. State three factors on which the frequency of vibrations of a stretched string depends.
Answer:
The natural frequency of a stretched string is determined by:
1. **String Length:** Frequency is inversely proportional to the active length of the string (\( f \propto \frac{1}{l} \)).
2. **String Tension:** Frequency is directly proportional to the square root of the string's tension (\( f \propto \sqrt{T} \)).
3. **Linear Mass Density:** Frequency is inversely proportional to the square root of the string's mass per unit length (\( f \propto \frac{1}{\sqrt{m}} \)).
In simple words: How fast a string vibrates depends on how long it is, how tightly it is stretched, and how thick it is.

Exam Tip: State the proportionalities clearly for a complete answer.

Question 14. What is resonance? Give two examples of resonance.
Answer:
Resonance is a specific form of forced vibration that occurs when the frequency of an external driving force matches the natural frequency of a nearby body, causing it to vibrate with a large amplitude.
**Examples:**
1. **Sound Box:** Stringed instruments are built with hollow soundboxes. The vibrating strings force the air column inside the box to vibrate. Since they match in frequency, resonance occurs, producing a much louder note.
2. **Tuning a Radio:** Turning the dial on a radio matching its internal natural frequency to the frequency of the incoming broadcast signal, resulting in resonance and amplification.
In simple words: Resonance is when a vibration matches an object's natural speed, causing it to vibrate loudly.

Exam Tip: Always mention that matching frequencies leads to a major increase in amplitude.

Question 15. Distinguish between music and noise.
Answer:
Sounds are divided into musical notes and noise. Music consists of regular, periodic oscillations that produce a pleasant sensation. Noise consists of random, discordant, and non-periodic vibrations that are irritating to the ear.
In simple words: Music is pleasant and periodic, while noise is irritating and random.

Exam Tip: Mention that music has a periodic waveform while noise has a completely irregular waveform.

Question 16. Explain how notes of different pitch and loudness are produced in (a) a guitar, (b) a flute, (c) a piano.
Answer:
(a) **Guitar:**
- **Pitch:** Adjusted by plucking strings of different thicknesses or changing their active length by pressing frets. Thinner strings vibrate faster, creating a higher pitch.
- **Loudness:** Increased by plucking the strings with greater force, which increases the amplitude. The hollow sound box undergoes forced vibrations, amplifying the sound over a large surface area.
(b) **Flute:**
- **Pitch:** Controlled by opening or closing key holes along the flute, which alters the effective length of the vibrating air column inside. Shorter air columns produce higher pitches.
- **Loudness:** Determined by how hard the player blows into the flute. Additionally, flutes with larger diameters enclose a larger volume of air, producing louder sounds.
(c) **Piano:**
- **Pitch:** Striking different keys strikes wires of different lengths and thicknesses. Thinner, shorter, and tighter wires produce higher pitch.
- **Loudness:** Controlled by how hard the keys are pressed. The vibrating wires are connected to a large soundboard, which vibrates to amplify the sound.
In simple words: Guitar and piano use strings (thinner or tighter strings sound higher), while flutes use air columns (shorter air columns sound higher). Blowing or plucking harder makes the sound louder.

Exam Tip: Explain how pitch is altered (by changing length/thickness) and how loudness is amplified (using sound boxes or soundboards) for each instrument.

Question 17. Explain the following observations: (a) When a nail is hammered into wood, its length outside wood gradually decreases. (b) When low notes are played on a pipe organ, the window panes sometimes rattle.
Answer:
(a) As a nail is driven deeper into wood, the length of the nail remaining outside the wood becomes shorter. Since the vibrating length decreases, its natural frequency increases, causing the pitch of the hammering sound to rise.
(b) The low-frequency notes played on a pipe organ can match the natural frequency of nearby window panes. This frequency match results in resonance, causing the windows to vibrate violently and rattle.
In simple words: Hammering a nail makes it shorter, so it vibrates faster and sounds higher. Window panes rattle because the deep organ notes match the windows' natural vibration speed, causing resonance.

Exam Tip: For the window panes, specifically mention the term "resonance" to get full marks.

PAGE NO-163

Question 18. What range of sound level is safe for human ears?
Answer:
Sound levels maintained below \( 120 \text{ dB} \) (ideally below \( 80 \text{ dB} \)) are considered safe and non-damaging for human hearing.
In simple words: Sounds under 120 decibels are generally safe for our ears.

Exam Tip: Cite \( 120 \text{ dB} \) as the threshold of pain, meaning anything below it is safe.

Question 19. What sound level causes noise pollution?
Answer:
Any sound level exceeding \( 120 \text{ dB} \) causes significant discomfort and is classified as noise pollution.
In simple words: Noise pollution happens when sounds get louder than 120 decibels.

Exam Tip: Mention that \( 120 \text{ dB} \) is the threshold of pain and cause of noise pollution.

Question 20. What determines the pitch of a musical note?
Answer:
The pitch of a note is determined primarily by its frequency. It is a subjective response of our hearing system to the frequency of the sound wave, following a nearly linear relationship.
In simple words: Pitch is determined by frequency; faster vibrations make us hear a higher pitch.

Exam Tip: State that pitch is the subjective counterpart of the objective physical frequency.

Question 21. What is the subjective property of sound related to frequency?
Answer:
The subjective quality of sound directly linked to its frequency is pitch.
In simple words: Pitch is the subjective way we perceive sound frequency.

Exam Tip: Keep the answer brief and use the term "pitch".

Question 22. Why is it possible to recognize a person by their voice without seeing them?
Answer:
We can identify a person by their voice because the vocal cords of each individual produce a unique waveform (timbre or quality) containing a specific mix of fundamental and subsidiary frequencies.
In simple words: Everyone's vocal cords create a slightly different wave shape, which is why we can recognize voices.

Exam Tip: Use terms like "waveform", "quality", or "timbre" to explain voice recognition.

Question 23. What factor determines the loudness of a sound wave?
Answer:
The loudness of a sound is determined by the amplitude of the wave; it is directly proportional to the square of the amplitude (\( L \propto a^2 \)).
In simple words: Loudness is decided by how tall the wave is (amplitude). Taller waves make louder sounds.

Exam Tip: Write down the proportional relationship \( L \propto a^2 \) to show a complete answer.

Question 24. Explain why stringed instruments are provided with a hollow sound box.
Answer:
Stringed instruments are built with a hollow sound box so that the vibrating strings force the large volume of air inside the box to vibrate. This large surface area increases the amplitude of the sound wave, making it much louder.
In simple words: The hollow box vibrates along with the strings, pushing a lot of air to make the sound louder.

Exam Tip: Explain that the sound box increases the vibrating surface area to increase loudness.

Question 25. Define intensity of a sound wave and state its unit.
Answer:
The intensity of a sound wave is defined as the amount of sound energy passing per second normally through a unit area. Its unit is the microwatt per square meter (\( \mu\text{W/m}^2 \)) or watt per square meter (\( \text{W/m}^2 \)).
In simple words: Intensity is the actual physical energy carried by the sound waves through a square meter every second.

Exam Tip: Clearly write down the SI unit of intensity as \( \text{W m}^{-2} \).

Question 26. State the relationship between loudness and intensity of sound.
Answer:
Loudness (\( L \)) and intensity (\( I \)) are related logarithmically by Weber-Fechner law: \[ L = k \log\left(\frac{I}{I_0}\right) \] where \( I_0 \) is the threshold intensity of hearing and \( k \) is a constant.
In simple words: Loudness is a logarithmic measure of sound intensity relative to the quietest sound we can hear.

Exam Tip: Write down the formula \( L = k \log\left(\frac{I}{I_0}\right) \) to secure full marks.

Question 27. (a) What happens to pitch if frequency is increased? (b) What happens to loudness if amplitude is increased?
Answer:
(a) If the frequency of a note is raised, its pitch will increase.
(b) If the amplitude of a note is increased, its loudness will increase.
In simple words: More frequency means a higher pitch, and more amplitude means a louder sound.

Exam Tip: Clearly separate parts (a) and (b) in your answer.

Question 28. What is the difference between loudness and intensity?
Answer:
Loudness is a subjective sensation that depends on the sensitivity of the listener's ear and cannot be directly measured physically. Intensity, on the other hand, is an objective, measurable physical quantity representing the sound energy passing per unit area per second.
In simple words: Loudness is how loud a sound feels to us, while intensity is the actual physical energy of the sound wave.

Exam Tip: Emphasize that loudness is subjective (depends on the observer) while intensity is objective (measurable with instruments).

Question 29. State three factors on which the loudness of a sound heard by a listener depends.
Answer:
The loudness of a sound perceived by a listener depends on:
1. The amplitude of the vibrating source (\( L \propto A^2 \)).
2. The distance of the listener from the source (\( L \propto \frac{1}{d^2} \)).
3. The surface area of the vibrating body.
In simple words: How loud a sound is depends on how hard the object is vibrating, how close you are to it, and how big the object is.

Exam Tip: List these three factors clearly to get full marks.

Question 30. Name the unit used to measure sound level.
Answer:
The unit used to measure sound level is the decibel (dB).
In simple words: Decibels (dB) are the units we use to measure how loud a sound is.

Exam Tip: Write "decibel" and its symbol "dB" correctly.

Question 31. Why do different musical instruments sound different even if they play notes of the same pitch and loudness?
Answer:
Different instruments playing notes of the same pitch and loudness sound distinct due to their different quality or timbre. This is because their waveforms differ, consisting of a different mixture of subsidiary vibrations (overtones) along with the fundamental note.
In simple words: Even if they play the same note at the same volume, different instruments have different wave shapes, which gives them a unique sound.

Exam Tip: Explain that quality or timbre is determined by the presence and relative amplitudes of overtones (harmonics).

Question 32. Why does the quality of sound of the same pitch differ when produced by different instruments?
Answer:
The quality of a note of the same pitch differs across instruments because each instrument produces a unique, complex waveform. When a note is played, it consists of a fundamental frequency combined with several overtones or subsidiary vibrations. The unique blend and strengths of these overtones give each instrument its distinct wave shape, allowing us to tell them apart easily.
In simple words: Instruments sound different because each one creates a different wave shape by mixing various overtones with the main note.

Exam Tip: Mention "overtones" or "subsidiary vibrations" as the cause of the complex waveform.

Question 33. What factors affect the quality of a musical sound?
Answer:
The quality of a musical sound is determined by the number, relative strengths, and frequencies of the harmonics and overtones present in the sound. Different combinations of these overtones produce the unique complex wave patterns characteristic of each instrument.
In simple words: The mix of extra frequencies (overtones) that accompany the main note is what determines the sound's quality.

Exam Tip: Explicitly mention both "harmonics" and "overtones" as the factors affecting sound quality.