Selina Concise Solutions for ICSE Class 9 Physics Chapter 8 Propagation Of Sound Waves

Exercise 8(A)

Question 1S. How is sound caused?
Answer: Sound is caused due to vibrations of a body.
In simple words: Sound happens when something jiggles back and forth very quickly—like a guitar string or your vocal cords. This jiggling is called vibration.

📝 Teacher's Note: You can show this to students by having them touch their throats while humming. They will feel the vibrations causing the sound.

🎯 Exam Tip: "Vibration" is the single most important keyword for any question asking about the origin of sound.

Question 2S. What is sound? How is it produced?
Answer: Sound is a form of energy that produces the sensation of hearing in our ears. Sound is produced by a vibrating body.
In simple words: Sound is a type of energy that our ears can detect. It always starts with something moving back and forth (vibrating).

📝 Teacher's Note: Remind students that energy can change forms. In a drum, mechanical energy of the hit changes into sound energy.

🎯 Exam Tip: Define sound as a "form of energy" to get full credit.

Question 3S. Complete the following sentence: Sound is produced by a ______ body.
Answer: Vibrating
In simple words: Only things that move back and forth can create sound.

📝 Teacher's Note: This is a fundamental concept. If vibrations stop, sound stops.

🎯 Exam Tip: This is a common fill-in-the-blank or one-word answer question.

Question 4S. Describe an experiment to show that a vibrating body produces sound.
Answer: Experiment: A tuning fork is taken and its one arm is struck on a rubber pad and it is brought near a tennis ball suspended by a thread as shown in figure.
It is noticed that as the arm of the vibrating fork is brought close to the ball, it jumps back and forth and sound of the vibrating tuning fork is heard. When its arm stop vibrating, the ball becomes stationary and no sound is heard.
In simple words: When you hit a tuning fork, it hums. If you touch it to a hanging ball, the ball bounces away. This proves the fork is jiggling even if you can't see it clearly.

📝 Teacher's Note: The "suspended ball" acts as an indicator. Since vibrations are often too fast for the eye to see, the ball magnifies the motion.

🎯 Exam Tip: When drawing this, ensure the ball is shown in two positions (stationary and being pushed) to indicate motion.

Question 5S. Describe an experiment to demonstrate that a material medium is necessary for the propagation of sound.
Answer: An electric bell is suspended inside an airtight glass bell jar. The bell jar is connected to a vacuum pump as shown in figure. As the circuit of electric bell is completed by pressing the key, the hammer of the electric bell begins to strike the gong repeatedly due to which sound is heard.
Keeping the key pressed, air is gradually withdrawn from jar by starting the vacuum pump. It is noticed that the loudness of sound goes on decreasing as the air is taken out from the bell jar and finally no sound is heard when all the air from the jar has been drawn out. The hammer of the electric bell is still seen striking the gong repeatedly which means that sound is still produced but it is not heard.
When the jar is filled with air, the vibrations produced by the gong are carried by the air to the walls of jar which in turn set the air outside the jar in vibration and sound is heard by us but in absence of air, sound produced by bell could not travel to the wall of the jar and thus no sound is heard. It proves that material medium is necessary for the propagation of sound waves.
In simple words: If you put a ringing bell in a jar and suck out all the air, the bell goes silent even though you can see it still hitting. This shows that without air or something similar, sound has no way to travel to your ears.

📝 Teacher's Note: This is the famous "Bell Jar Experiment." It helps students realize that sound is a mechanical wave that needs physical particles to travel.

🎯 Exam Tip: Make sure to mention that the hammer is *seen* moving but the sound is *not heard*. This confirms sound is being produced but not transmitted.

Question 6S. Why can we not hear each other on the moon's surface?
Answer: We cannot hear each other on moon’s surface because there is no air on moon and for sound to be heard, a material medium is necessary.
In simple words: The Moon is a giant vacuum with no air. Since there's no air to carry the jiggles from my mouth to your ears, it's perfectly silent there.

📝 Teacher's Note: Use the analogy of a bridge: if the bridge (air) is missing, you cannot cross the river (reach the ear).

🎯 Exam Tip: The scientific reason is "absence of a material medium."

Question 7S. State the requisites of the medium for the propagation of sound.
Answer: Requisites of the medium for propagation of sound:
1. The medium must be elastic.
2. The medium must have inertia.
3. The medium should be frictionless.
In simple words: For sound to travel, the stuff it's moving through must be able to spring back (elastic), have some weight (inertia), and not swallow the energy as heat (frictionless).

📝 Teacher's Note: Elasticity allows particles to return to their original position after being disturbed, which is essential for wave motion.

🎯 Exam Tip: List these three properties precisely to score full marks in short-answer questions.

Question 8S. Explain how sound travels in air as a longitudinal wave using the example of a vibrating metal strip.
Answer: Take a vertical metal strip with its lower end fixed and upper end being free to vibrate as shown in fig (a).
As the strip is moved to right from a to b as shown in Fig (b), the air in that layer is compressed (compression is formed at C). The particles of this layer compress the layer next to it, which then compresses the next layer and so on. Thus, the disturbance moves forward in form of compression without the particles themselves being displaced from their mean positions.
As the metal strip returns from b to a as shown in Fig (c) after pushing the particles in front, the compression C moves forward and particles of air near the strip return to their normal positions.
When the strip moves from a to c as shown in Fig (d), it pushes back the layer of air near it towards left and thus produces a low pressure space on its right side i.e. layers of air get rarefied. This region is called rarefaction (rarefaction is formed at R).
When the strip returns from C to its mean position A in Fig (e), the rarefaction R travels forward and air near the strip return to their normal positions.
Thus, one complete to and fro motion of the strip forms one compression and one rarefaction, which together form one wave. This wave through which sound travels in air is called longitudinal wave.
In simple words: Sound travels like a slinky. When the metal strip pushes out, it squashes air particles together (compression). When it pulls back, it leaves extra space (rarefaction). This pattern of "squash and stretch" moves forward to your ear.

📝 Teacher's Note: Emphasize that the particles don't actually travel to the ear; they just bump into their neighbors and stay in their own small area.

🎯 Exam Tip: Define "Compression" as a high-pressure region and "Rarefaction" as a low-pressure region.

Question 10S. In what forms does sound travel in a medium?
Answer: Sound travels in a medium in form of longitudinal and transverse waves.
In simple words: Sound can move as a back-and-forth pulse (longitudinal) or a side-to-side wiggle (transverse).

📝 Teacher's Note: Clarify that sound travels *only* as longitudinal waves in gases (like air), but can be transverse in some solids.

🎯 Exam Tip: Note that transverse sound waves only occur in solids and on the surface of liquids.

Question 11S. What is a longitudinal wave?
Answer: A type of wave motion in which the particle displacement is parallel to the direction of wave propagation is called a longitudinal wave. It can be produced in solids, liquids as well as gases.
In simple words: It's a wave where the particles move the same way the wave is going—like people in a crowded line pushing forward and back.

📝 Teacher's Note: The word "parallel" is the technical requirement for this definition.

🎯 Exam Tip: This is the most common type of sound wave. Remember it works in all three states of matter.

Question 12S. What is a transverse wave?
Answer: A type of wave motion in which the particle displacement is perpendicular to the direction of wave propagation is called a transverse wave. It can be produced in solids and on the surface of liquids.
In simple words: It's a wave where the particles move up and down while the wave moves forward—like a jump rope or a wave in the ocean.

📝 Teacher's Note: Transverse waves require a medium with shear strength, which is why they don't happen inside gases or most liquids.

🎯 Exam Tip: The word "perpendicular" is the key term here.

Question 13S. How does a longitudinal wave propagate?
Answer: A longitudinal wave propagates by means of compressions and rarefactions.
When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called a compression (C), as shown in Fig. This compression starts to move away from the vibrating object. When the vibrating object moves backwards, it creates a region of low pressure called rarefaction (R), as shown in Fig.
Compressions are the regions of high density where the particles of the medium come very close to each other and rarefactions are the regions of low density where the particles of the medium move away from each other.In simple words: The wave moves by squeezing and spreading the medium. High pressure parts are compressions, and low pressure parts are rarefactions.

📝 Teacher's Note: Use the density of dots in a drawing to represent particles. More dots = compression.

🎯 Exam Tip: Define compressions as "regions of high density/pressure" and rarefactions as "regions of low density/pressure."

Question 14S. What are crests and troughs in a transverse wave?
Answer: A crest is a point on the transverse wave where the displacement of the medium is at a maximum.
A point on the transverse wave is a trough if the displacement of the medium at that point is at a minimum.
In simple words: The "mountain peaks" of the wave are crests, and the "valleys" are troughs.

📝 Teacher's Note: In a transverse wave, the "zero line" is the equilibrium position where particles stay when there is no wave.

🎯 Exam Tip: Crest = maximum positive displacement; Trough = maximum negative displacement.

Question 15S. Describe an experiment to show that in a wave motion, only energy is transferred, but particles of the medium do not move.
Answer: If we drop a piece of stone in the still water of pond, we hear a sound of stone striking the water surface. Actually a disturbance is produced in water at the point where the stone strikes it. This disturbance spreads in all directions radially outwards in form of circular waves on the surface of water.
If we place a piece of cork on water surface at some distance away from the point where the stone strikes it, we notice that cork does not move ahead, but it vibrates up and down, while the wave moves ahead. The reason is that particles of water (or medium) start vibrating up and down at the point where the stone strikes. These particles then transfer their energy to the neighboring particles and they themselves come back to their mean positions. Thus only energy is transferred but the particles of the medium do not move.
In simple words: If you put a cork in a wavy pond, the cork bobbles up and down but stays in the same spot. It doesn't get carried away by the wave. This proves that the water stays put, and only the energy "ghost" moves through it.

📝 Teacher's Note: This is a very common student mistake—thinking waves push matter. Use the "Mexican Wave" in a stadium as an analogy: people stay in their seats, but the "wave" travels around the stadium.

🎯 Exam Tip: Use the phrase "particles vibrate about their mean positions" to explain why they don't move with the wave.

Question 16S. Define amplitude of a wave and state its S.I. unit.
Answer: The maximum displacement of the particle of medium on either side of its mean position is called the amplitude of wave. Its SI unit is metre.
In simple words: Amplitude is the "height" of the wave jiggle. It tells you how far a particle moves from its home base.

📝 Teacher's Note: Amplitude determines the loudness of a sound. Bigger amplitude = Louder sound.

🎯 Exam Tip: Measured from the "mean position" to the "peak," not from trough to crest.

Question 17S. Define frequency of a wave and state its S.I. unit.
Answer: The number of vibrations made by the particle of the medium in one second is called the frequency of the wave. It can also be defined as the number of waves passing through a point in one second. Its SI unit is hertz (Hz).
In simple words: Frequency is how fast the jiggling is happening. If it jiggles 10 times in a second, the frequency is 10 Hertz.

📝 Teacher's Note: Frequency determines the pitch. High frequency = High pitch (like a whistle).

🎯 Exam Tip: 1 Hz = 1 vibration per second.

Question 18S. State the relationship between frequency and time period.
Answer: Frequency of a wave is the reciprocal of the time period.
\( \upsilon = \frac{1}{T} \)
In simple words: If it takes a long time for one jiggle (big T), the jiggles aren't very frequent (small frequency). They are opposites.

📝 Teacher's Note: Greek letter \( \nu \) (nu) is often used for frequency, though some books use 'f'.

🎯 Exam Tip: Frequency (\( \nu \)) \( \times \) Time period (\( T \)) = 1.

Question 19S. Define wave velocity and state its S.I. unit.
Answer: The distance travelled by a wave in one second is called its wave velocity. Its SI unit is metre per second (\( \text{ms}^{-1} \)).
In simple words: Wave velocity is simply the speed of the wave. It tells you how many meters the sound travels in one second.

📝 Teacher's Note: In air, sound travels at about 330 meters per second.

🎯 Exam Tip: Always use the exponent notation \( \text{ms}^{-1} \) for velocity in senior physics exams.

Question 22S. Derive the relationship between wave velocity (V), frequency (\( \nu \)), and wavelength (\( \lambda \)).
Answer: Let the velocity of a wave be V, time period T, frequency \( \nu \) and wavelength \( \lambda \).
By the definition of wavelength,
Wavelength = Distance travelled by the wave in one time period i.e., in T second
Or, wavelength = Wave velocity \( \times \) Time period
Or, \( \lambda = V \times T \)
Or, \( \lambda = V \times \frac{1}{\nu} \) [As, \( T = \frac{1}{\nu} \)]
Therefore, \( V = \nu \lambda \)
Therefore, Wave velocity = Frequency \( \times \) wavelength
In simple words: The speed of a wave can be found by multiplying how many times it jiggles (frequency) by how long each jiggle is (wavelength).

📝 Teacher's Note: This is the most important formula in the chapter. It links the speed of the wave to its physical characteristics.

🎯 Exam Tip: This derivation is often asked as a 2 or 3-mark question. Ensure you start by defining each symbol.

Question 23S. Upon what factors does the speed of sound in a medium depend?
Answer: The speed of sound in a medium depends upon its elasticity and density.
In simple words: Sound moves faster in materials that are "springy" (elastic) and not too heavy (less dense).

📝 Teacher's Note: The formula is \( v = \sqrt{\frac{E}{\rho}} \) where E is elasticity and \( \rho \) is density.

🎯 Exam Tip: Mention both "elasticity" and "density" for a complete answer.

Question 24S. Compare the speed of sound in solids (\( V_s \)), liquids (\( V_l \)), and gases (\( V_g \)).
Answer: \( V_g < V_l < V_s \)
In simple words: Sound travels slowest in air, faster in water, and fastest in solid metal or wood.

📝 Teacher's Note: Sound is fastest in solids because the atoms are packed tightly and have high elasticity.

🎯 Exam Tip: Remember the order: Solids > Liquids > Gases.

Question 25S. State the speed of light and sound in air.
Answer: (i) Speed of light in air = \( 3 \times 10^8 \text{ m s}^{-1} \)
(ii) Speed of sound in air = \( 330 \text{ m s}^{-1} \).
In simple words: Light is a million times faster than sound! Light could circle the Earth 7 times in one second; sound can barely cross a few football fields.

📝 Teacher's Note: Use this to explain lightning and thunder: you see the flash instantly, but the sound takes time to arrive.

🎯 Exam Tip: Memorize these two values exactly.

Question 27S. (i) Can sound travel in a vacuum? (ii) How does speed vary in different states of matter?
Answer: (i) No, sound cannot travel in vacuum as it requires a material medium for its propagation.
(ii) Speed of sound is maximum in solids, less in liquids and least in gases.
In simple words: Sound needs "stuff" to move through. Without stuff, it's silence. It likes moving through solid stuff best.

📝 Teacher's Note: This reinforces the mechanical nature of sound waves.

🎯 Exam Tip: "No" for vacuum; "Solids > Liquids > Gases" for speed.

Question 28S. Why is lightning seen before thunder is heard?
Answer: This happens because the light travels much faster than sound.
In simple words: Both happen at once, but light is like a super-fast race car, and sound is like a slow walker. The flash wins the race to your eyes first.

📝 Teacher's Note: The time gap can tell you how far away the storm is.

🎯 Exam Tip: Mention the specific speed values (3 \( \times \) 10⁸ vs 330) to make your reasoning stronger.

Question 30S. If a gun is fired at a distance, who hears it first: a person on land or a diver underwater? Why?
Answer: (i) The diver would hear the sound first.
(ii) This is because sound travels faster in water than in air.
(iii) It would take 0.25t to reach the diver because sound travels almost four times faster in water.
In simple words: Water is a better "highway" for sound than air. It lets the sound speed through much more quickly.

📝 Teacher's Note: Sound travels at about 1400-1500 m/s in water, compared to 330 m/s in air.

🎯 Exam Tip: Always justify the "who first" question with a "because sound travels faster in [medium]" statement.

Question 31S. Describe the effect of the following on the speed of sound in air: (i) Frequency, (ii) Temperature, (iii) Pressure, (iv) Moisture.
Answer: (i) Frequency of sound has no effect on the speed of sound.
(ii) Speed of sound increases with the increase in the temperature of sound.
(iii) Pressure of sound has no effect on the speed of sound.
(iv) Speed of sound increases with the increase in presence of moisture in air.
In simple words: Sound moves faster on hot, humid days. How high the note is (frequency) or how heavy the air is pushing (pressure) doesn't change the speed.

📝 Teacher's Note: This is a common trap. Students often think higher pressure or higher frequency makes sound faster. Only temperature and humidity really matter in air.

🎯 Exam Tip: Remember: Temperature and Moisture = INCREASE speed. Pressure and Frequency = NO CHANGE.

Question 33S. Why is the speed of sound more in humid air?
Answer: Speed of sound is more in humid air because in presence of moisture, the density of air decreases and sound travels with greater speed.
\( V \propto \frac{1}{\sqrt{\rho}} \)
In simple words: Moist air is actually "lighter" (less dense) than dry air. Since lighter air is easier to move, sound can travel through it faster.

📝 Teacher's Note: Explain that water vapor molecules are lighter than Nitrogen and Oxygen molecules, which is why humid air is less dense.

🎯 Exam Tip: Mention the inverse relationship between speed and square root of density.

Question 34S. By how much does the speed of sound increase with temperature?
Answer: The speed of sound increases by \( 0.61 \text{ m s}^{-1} \) for each \( 1^\circ\text{C} \) rise in temperature.
In simple words: For every degree warmer it gets, sound gains about 60 centimeters of speed every second.

📝 Teacher's Note: This is a linear approximation used for near-room temperature calculations.

🎯 Exam Tip: Memorize the value \( 0.61 \text{ ms}^{-1} \) for numerical reasoning.

Question. Describe an experiment to calculate the speed of sound in air.
Answer: The simple experiment that a person can do to calculate the speed of sound in air is that a person stands at a known distance (d meter) from the cliff and fires a pistol and simultaneously start the stopwatch. He stops the stopwatch as soon as he hears an echo. The distance travelled by the sound during the time (t) seconds is 2d. So, speed of sound = distance travelled / time taken = \( \frac{2d}{t} \).
The approximation made is that speed of sound remains same for the time when the experiment is taking place.
In simple words: You fire a gun toward a mountain and time how long it takes for the "Bang" to bounce back. Since the sound went to the mountain and back, it covered twice the distance. Divide that total distance by the time to get the speed.

📝 Teacher's Note: This is the basis of sonar and radar. Distance is doubled because sound makes a round trip.

🎯 Exam Tip: Don't forget the '2' in \( 2d/t \). It's the most common mistake in these numericals.

Question 36S. Fill in the blanks:
(a) Sound cannot travel in ______, it requires a material ______.
(b) While sound travels, the particles of the medium ______, but only the disturbance ______.
(c) A longitudinal wave consists of compression and ______.
(d) The minimum displacement in a transverse wave is called ______.
Answer: (a) Vacuum, medium (b) do not move, moves (c) rarefaction (d) trough.
In simple words: Basic wave vocabulary: Sound needs a medium, particles just jiggle, waves have troughs and rarefactions.

📝 Teacher's Note: These words form the "ABC" of acoustics.

🎯 Exam Tip: Re-read the chapter notes to ensure you know the definitions of these fill-in terms.

Question 1M. Differentiate light and sound based on their need for a medium.
Answer: Sound needs medium, but light does not need medium for its propagation.
In simple words: Sound is like a car (needs a road/air); light is like a ghost (can walk through walls/empty space).

📝 Teacher's Note: This is why we see the sun (light) but cannot hear its massive explosions (sound).

🎯 Exam Tip: Light is an electromagnetic wave; sound is a mechanical wave.

Question 2M. What type of wave is a sound wave in air?
Answer: Longitudinal wave
In simple words: Sound in air moves by pushing and pulling air in a straight line.

📝 Teacher's Note: Always specify "in air" as sound can behave differently in solids.

🎯 Exam Tip: This is the most common sound wave classification.

Question 3M. What is the approximate speed of sound in air?
Answer: 330 m s^-1
In simple words: In one second, sound covers roughly the length of 3.3 football fields.

📝 Teacher's Note: This value changes with temperature, but 330 is the standard "room temp" value used in schools.

🎯 Exam Tip: Note the units \( \text{m s}^{-1} \).

Question 4M. What is the approximate speed of light?
Answer: 3 x 10⁸ m s^-1
In simple words: Light travels at 300,000 kilometers every single second.

📝 Teacher's Note: This is the universal speed limit.

🎯 Exam Tip: 10⁸ means 1 followed by 8 zeros.

Question 1N. A heart beats 75 times a minute. Calculate (a) Frequency and (b) Time period.
Answer: Given, heart beats 75 times a minute
(a) Frequency = No. of times heart beats in 1 s
Or, \( \nu = 75/60 = 1.25 \text{ s}^{-1} \)
(b) Time period, \( T = \frac{1}{\nu} \)
Or, \( T = \frac{1}{1.25} = 0.8 \text{ s} \)
In simple words: The heart beats 1.25 times every second. The time between each beat is exactly 0.8 seconds.

📝 Teacher's Note: Always convert minutes to seconds first (60 s). Frequency is "how many in 1 second."

🎯 Exam Tip: Frequency unit is Hz or \( \text{s}^{-1} \); Time period unit is seconds.

Question 2N. If the time period of a wave is 2 seconds, find its frequency.
Answer: Frequency, \( \nu = \frac{1}{T} \)
Or, \( \nu = 1/2 = 0.5 \text{ Hz} \)
In simple words: If it takes 2 seconds for one jiggle, that means only half a jiggle (0.5) happens in one second.

📝 Teacher's Note: Use this to reinforce the reciprocal relationship.

🎯 Exam Tip: Simple division: \( 1/T \).

Question 3N. Wavelength of a wave is 100m and speed is 20 m/s. Find frequency.
Answer: Given, wavelength = 100m
Wave velocity = 20 m/s
We know that,
Wave velocity = Frequency \( \times \) Wavelength
Or, Frequency = Wave velocity / wavelength
Or, \( \nu = 20/100 = 0.2 \text{ Hz} \)
In simple words: If the wave moves 20 meters every second, and each wave is 100 meters long, only 0.2 (one-fifth) of a wave passes by every second.

📝 Teacher's Note: Rearrange \( V = \nu \lambda \) to solve for the missing variable.

🎯 Exam Tip: Always write the formula first before putting in the numbers.

Question 4N. Wave velocity = 0.3 m/s, Frequency = 20 Hz. Find wavelength.
Answer: Wave velocity = 0.3 m/s
Frequency = 20 Hz
Separation between two consecutive compressions is the wavelength of a wave.
We know that,
Wave velocity = Frequency \( \times \) Wavelength
Or, wavelength = Wave velocity / frequency
Or, \( \lambda = 0.3 / 20 = 1.5 \times 10^{-2} \text{ m} \)
In simple words: The distance from one squash of air to the next is 0.015 meters (or 1.5 cm).

📝 Teacher's Note: "Separation between consecutive compressions" is just a long name for wavelength.

🎯 Exam Tip: Scientific notation like \( 10^{-2} \) is standard for small measurements.

Question 5N. A frequency of wave is 40 and time is 0.4s. Find frequency.
Answer: Frequency of wave = number of waves per second
Or, \( \nu = 40/0.4 = 100 \text{ Hz} \)
In simple words: If 40 waves pass in 0.4 seconds, we divide to find that 100 waves pass in 1 full second.

📝 Teacher's Note: This is a variation of the basic frequency definition.

🎯 Exam Tip: \( \text{Frequency} = \frac{\text{Total Waves}}{\text{Total Time}} \).

Question 6N. Distance between two observers is 1650 m. Speed of sound is 330 m/s. How long after a gun is fired will the second observer hear it?
Answer: Distance between the two observers = 1650 m
Speed of sound = 330 m/s
Time in which B hears the sound = Distance / speed = \( \frac{1650}{330} = 5\text{s} \)
Thus, B will hear the sound 5s after the gun is shot.
In simple words: To find the time, divide the total distance (1650) by the speed (330). It takes 5 seconds for the sound to travel that far.

📝 Teacher's Note: This is a standard \( \text{Time} = \text{Distance} / \text{Speed} \) problem.

🎯 Exam Tip: Unit check: meters and m/s result in seconds.

Question 7N. Thunder is heard 5s after lightning is seen. Find distance. (V = 330 m/s)
Answer: Speed of sound in air (V) = 330 m/s
Time in which thunder is heard after lighting is seen (t) = 5s
Thus, distance between flash and observer = \( V \times t = (330 \times 5) = 1650 \text{ m} \)
In simple words: If you count 5 seconds after a flash, the storm is 1650 meters (about 1.6 km) away.

📝 Teacher's Note: We ignore the time light takes to travel because it is virtually instantaneous over this distance.

🎯 Exam Tip: Use the formula \( \text{Distance} = \text{Speed} \times \text{Time} \).

Question 10N. A sound travels through a 3.3 km iron rail. Speed in iron = 5280 m/s, Speed in air = 330 m/s. (a) Find time in rail. (b) Find time in air.
Answer: (a) Length of iron rail (D) = 3.3 km = 3300 m
Speed of sound in iron (V) = 5280 m/s
Time taken by sound to travel in iron rod (t) = D/V
Or, \( t = (3300 / 5280) \text{ s} = 0.625 \text{ s} \)
(b) Length of iron rail (D) = 3.3 km = 3300 m
Speed of sound in air (V) = 330 m/s
Time taken by sound to travel in iron rod (t) = D/V
Or, \( t = (3300/330) \text{ s} = 10 \text{ s} \)
In simple words: Through the metal rail, sound arrives in less than a second! Through the air, it takes a long 10 seconds. This is why you hear the train "buzz" in the rails long before you hear the engine.

📝 Teacher's Note: This problem compares transmission through different states of matter. Metals are much faster conductors of sound.

🎯 Exam Tip: Convert kilometers to meters (3.3 km = 3300 m) before starting.

Question 11N. Speed in air = 340 m/s, Speed in water = 1360 m/s. Find time to travel 1700 m in both.
Answer: (i) Distance travelled (D) = 1700
Speed of sound in air (V) = 340 m/s
Time taken (t) = D/V = \( (1700 / 340) \text{ s} = 5 \text{ s} \)
(ii) Distance travelled (D) = 1700
Speed of sound in water (V’) = 1360 m/s
Time taken (t) = D/V = \( (1700 / 1360) \text{ s} = 1.25 \text{ s} \)
In simple words: The same sound trip takes 5 seconds in air but only 1.25 seconds in water.

📝 Teacher's Note: This is a direct proof that sound is about 4 times faster in water than in air.

🎯 Exam Tip: Label part (i) and (ii) clearly to distinguish between the two media.

Exercise 8(B)

Question 1S. Define audible range of frequency.
Answer: The range of frequency within which the sound can be heard by a human being is called the audible range of frequency.
In simple words: This is the "hearing window"—any sound too deep or too squeaky for our ears is outside this range.

📝 Teacher's Note: Human hearing is limited. Other animals, like dogs, have a much wider window.

🎯 Exam Tip: Use the term "audible" to describe human hearing.

Question 2S. What is the audible range for humans?
Answer: The audible range of frequency for humans is 20 Hz to 20 kHz.
In simple words: Humans can hear anything that jiggles between 20 times a second and 20,000 times a second.

📝 Teacher's Note: 20 kHz is the same as 20,000 Hz.

🎯 Exam Tip: This range is standard. Know both 20-20,000 Hz and 20 Hz - 20 kHz.

Question 3S. In which range are human ears most sensitive?
Answer: Human ears are most sensitive for the range 2000 Hz to 3000 Hz.
In simple words: Our ears are "tuned" best to hear middle-pitch sounds, like human speech.

📝 Teacher's Note: This is where we can hear even the softest whispers.

🎯 Exam Tip: Sensitive range is 2k to 3k Hz.

Question 4S. Which has higher frequency: sound or ultrasonic?
Answer: Ultrasonic has higher frequency.
In simple words: "Ultra" means beyond. Ultrasonic is sound that is too high-pitched for us to hear.

📝 Teacher's Note: "Sound" usually refers to the audible range.

🎯 Exam Tip: Ultrasonic > 20,000 Hz.

Question 5S. Fill in the blanks: (a) Human hearing is ____ to ____ (b) Ultrasonic is ____ (c) Infrasonic is ____ (d) ____ is high frequency (e) ____ is low frequency.
Answer: (a) 20 Hz, 20 kHz (b) above 20 kHz (c) below 20 Hz (d) ultrasonic (e) infrasonic.
In simple words: High = Ultra, Low = Infra. Hearing = 20 to 20k.

📝 Teacher's Note: Help students with the prefixes: Infra (below), Ultra (above).

🎯 Exam Tip: Be consistent with units (Hz vs kHz).

Question 7S. Why can we not hear a seconds pendulum?
Answer: No, we cannot hear the sound produced due to vibrations of a seconds pendulum because the frequency of sound produced due to vibrations of seconds pendulum is 0.5 Hz which is infrasonic.
In simple words: A clock's swinging pendulum only moves back and forth once every two seconds. That's way too slow for our ears to catch.

📝 Teacher's Note: The human ear needs at least 20 vibrations per second to start "hearing." 0.5 is far below that.

🎯 Exam Tip: State that 0.5 Hz is "below the lower limit of audible frequency."

Question 8S. What is ultrasound?
Answer: Sounds of frequency above 20 kHz are called ultrasound.
In simple words: Ultrasound is sound energy that vibrates so fast (over 20,000 times a second) that we can't hear it, but machines and some animals can.

📝 Teacher's Note: We use ultrasound for baby scans and cleaning jewelry.

🎯 Exam Tip: Frequency > 20,000 Hz.

Question 9S.
Answer: The disturbance

Question 10S. Name two properties of ultrasound that make it useful.
Answer: Two properties of ultrasound which make it useful to us are:
1. High energy contents
2. High directivity
In simple words: Ultrasound carries a lot of "punch" (energy) and travels in a very straight, narrow beam without spreading out.

📝 Teacher's Note: Because it stays in a beam, we can use it to "see" inside the body precisely.

🎯 Exam Tip: Mention "directivity" as it explains why ultrasound gives clear images.

Question 11S. How do bats use ultrasound?
Answer: Bats locate the obstacles and prey in their path by producing and hearing the ultrasound. They emit an ultrasound which returns after striking an obstacle in their way. By hearing the reflected sound and from the time interval (when they produce ultrasound and they receive them back), they can judge the direction and the distance of the obstacle in their way.
In simple words: Bats shout invisible screams that bounce off bugs. By timing the echo, the bat knows exactly where the bug is, even in total darkness.

📝 Teacher's Note: This is called Echolocation. It works exactly like a ship's sonar.

🎯 Exam Tip: Mention "emission" and "reflection" of ultrasound.

Question 12S. State two applications of ultrasound.
Answer: Two applications of ultrasound:
1. Ultrasound is used for drilling holes or making cuts of desired shape in materials like glass.
2. Ultrasound is used in surgery to remove cataract and in kidneys to break the small stones into fine grains.
In simple words: We use high-speed sound jiggles to cut glass perfectly and to crumble painful kidney stones without having to cut the patient open.

📝 Teacher's Note: These applications rely on the high energy density of ultrasonic waves.

🎯 Exam Tip: Mentioning "medical use" (kidney stones) and "industrial use" (drilling) provides a well-rounded answer.

Question 13S. How does a longitudinal wave propagate?
Answer: A longitudinal wave propagates by means of compressions and rarefactions.
When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure. This region is called a compression (C). This compression starts to move away from the vibrating object. When the vibrating object moves backwards, it creates a region of low pressure called rarefaction (R).
Compressions are the regions of high density where the particles of the medium come very close to each other and rarefactions are the regions of low density where the particles of the medium move away from each other.
In simple words: The wave moves by squeezing air particles together into high-pressure "packets" and then letting them spread out into low-pressure "gaps."

📝 Teacher's Note: Use the density of vertical lines in a drawing to help students visualize pressure differences. Denser lines = higher pressure.

🎯 Exam Tip: Define "Compression" as a high-density/high-pressure region and "Rarefaction" as a low-density/low-pressure region.

Question 14S. What are crests and troughs in a transverse wave?
Answer: A crest is a point on the transverse wave where the displacement of the medium is at a maximum.
A point on the transverse wave is a trough if the displacement of the medium at that point is at a minimum.
In simple words: The "mountain peaks" of the wave are crests, and the "valleys" are troughs.

📝 Teacher's Note: The "displacement" is measured from the equilibrium position (the flat center line).

🎯 Exam Tip: Use the diagram to show that a wave is made of one full crest and one full trough cycle.

Question 15S. Describe an experiment to show that in a wave motion, only energy is transferred, but particles of the medium do not move.
Answer: If we drop a piece of stone in the still water of pond, we hear a sound of stone striking the water surface. Actually a disturbance is produced in water at the point where the stone strikes it. This disturbance spreads in all directions radially outwards in form of circular waves on the surface of water.
If we place a piece of cork on water surface at some distance away from the point where the stone strikes it, we notice that cork does not move ahead, but it vibrates up and down, while the wave moves ahead. The reason is that particles of water (or medium) start vibrating up and down at the point where the stone strikes. These particles then transfer their energy to the neighboring particles and they themselves come back to their mean positions. Thus only energy is transferred but the particles of the medium do not move.
In simple words: If you put a cork in a wavy pond, the cork bobbles up and down but stays in the same spot. It doesn't get carried away by the wave. This proves that the water stays put, and only the "wave energy" travels across the pond.

📝 Teacher's Note: Use the "Mexican Wave" in a stadium as an analogy: people stay in their seats (particles stay in place), but the "wave" travels all the way around the stadium (energy transfer).

🎯 Exam Tip: Focus your explanation on the fact that "particles vibrate about their mean positions."

Question 16S. Define amplitude and state its S.I. unit.
Answer: The maximum displacement of the particle of medium on either side of its mean position is called the amplitude of wave. Its SI unit is metre.
In simple words: Amplitude is the "height" of the wave's jiggle. It tells you how far a particle moves from its home base.

📝 Teacher's Note: Higher amplitude sounds carry more energy and sound louder to our ears.

🎯 Exam Tip: Amplitude is measured in meters (m).

Question 17S. Define frequency and state its S.I. unit.
Answer: The number of vibrations made by the particle of the medium in one second is called the frequency of the wave. It can also be defined as the number of waves passing through a point in one second. Its SI unit is hertz (Hz).
In simple words: Frequency is just how fast the wave is jiggling. If it jiggles 100 times in one second, the frequency is 100 Hertz.

📝 Teacher's Note: High frequency sounds have a high "pitch" (like a bird), and low frequency sounds have a low "pitch" (like a drum).

🎯 Exam Tip: 1 Hz = 1 vibration per second.

Question 18S. State the relationship between frequency and time period.
Answer: Frequency of a wave is the reciprocal of the time period.
\( \upsilon = \frac{1}{T} \)
In simple words: If it takes a long time for one wave to pass (big T), then the waves aren't very frequent (small frequency). They are opposites!

📝 Teacher's Note: Use the Greek letter \( \nu \) (nu) for frequency as is standard in many physics curricula.

🎯 Exam Tip: Frequency (\( \nu \)) \( \times \) Time Period (\( T \)) = 1.

Question 19S. Define wave velocity and state its S.I. unit.
Answer: The distance travelled by a wave in one second is called its wave velocity. Its SI unit is metre per second (\( \text{ms}^{-1} \)).
In simple words: Wave velocity is simply the speed of the wave. It tells you how fast the sound is zipping through the air.

📝 Teacher's Note: In air, sound travels at approximately 330 to 340 meters per second depending on temperature.

🎯 Exam Tip: Do not confuse wave velocity (how fast the wave moves) with particle velocity (how fast an individual molecule jiggles).

Question 20S. Draw a displacement-time graph for a wave.
Answer:
In simple words: This graph shows how a single particle moves up and down over time as the wave passes by.

📝 Teacher's Note: A displacement-time graph is used to identify the Time Period (\( T \)) and Amplitude (\( a \)).

🎯 Exam Tip: Ensure the axis labels ("Displacement" and "Time") are correctly placed.

Question 21S. Draw a displacement-distance graph for a wave.
Answer:
In simple words: This graph is like a snapshot photo of the whole wave frozen in time, showing all the crests and valleys at once.

📝 Teacher's Note: A displacement-distance graph is used to identify the Wavelength (\( \lambda \)) and Amplitude (\( a \)).

🎯 Exam Tip: The distance between two consecutive crests on this graph is the wavelength.

Question 22S. State the relationship between wave velocity, frequency, and wavelength.
Answer: Let the velocity of a wave be V, time period T, frequency \( \nu \) and wavelength \( \lambda \). By the definition of wavelength,
Wavelength = Distance travelled by the wave in one time period i.e., in T second
Or, wavelength = Wave velocity \( \times \) Time period
Or, \( \lambda = V \times T \)
Or, \( \lambda = V \times \frac{1}{\nu} \) [As, \( T = \frac{1}{\nu} \)]
Therefore, \( V = \nu \lambda \)
Therefore, Wave velocity = Frequency \( \times \) wavelength
In simple words: The speed of a wave can be found by multiplying how many times it jiggles (frequency) by how long each jiggle is (wavelength).

📝 Teacher's Note: This is the most fundamental equation in wave physics. It applies to sound, light, and any other type of wave.

🎯 Exam Tip: This derivation is often asked as a 3-mark question. Be sure to show the step where \( T \) is replaced by \( 1/\nu \).

Question 23S. Upon what factors does the speed of sound in a medium depend?
Answer: The speed of sound in a medium depends upon its elasticity and density.
In simple words: Sound moves faster in materials that are "springy" (elastic) and not too heavy (less dense).

📝 Teacher's Note: The mathematical relationship is \( v = \sqrt{\frac{E}{\rho}} \) where E is elasticity and \( \rho \) is density.

🎯 Exam Tip: Mentioning "Elasticity" and "Density" together is required for full marks.

Question 24S. Compare the speed of sound in different states of matter.
Answer: \( V_g < V_l < V_s \)
In simple words: Sound moves slowest in gases (air), faster in liquids (water), and fastest in solids (iron or wood).

📝 Teacher's Note: This is because solid particles are closer together and transmit vibrations more efficiently than loose gas particles.

🎯 Exam Tip: Sound travels about 15 times faster in iron than in air.

Question 25S. State the speed of light and sound in air.
Answer: (i) Speed of light in air = \( 3 \times 10^8 \text{ m s}^{-1} \)
(ii) Speed of sound in air = \( 330 \text{ m s}^{-1} \).
In simple words: Light is nearly a million times faster than sound! This is why you see a firework explode way before you hear the "bang."

📝 Teacher's Note: Use these standard values for any numerical problems in this chapter.

🎯 Exam Tip: Be sure to write the units correctly: \( \text{m s}^{-1} \).

Question 26S. What is the ratio of speed of sound in gas, liquid, and solid?
Answer: 1 : 4 : 15
In simple words: If sound moves 1 unit of distance in air, it moves about 4 units in water and 15 units in metal in the same amount of time.

📝 Teacher's Note: This is an approximate ratio that helps students compare the relative speeds across media.

🎯 Exam Tip: This ratio is often asked in competitive exams.

Question 27S. (i) Can sound travel in a vacuum? (ii) How does speed vary?
Answer: (i) No, sound cannot travel in vacuum as it requires a material medium for its propagation.
(ii) Speed of sound is maximum in solids, less in liquids and least in gases.
In simple words: Sound needs "stuff" to travel. In empty space (vacuum), there's no stuff, so there's no sound. It travels fastest through solid metal.

📝 Teacher's Note: This reinforces the concept that sound is a mechanical wave.

🎯 Exam Tip: Use the speed comparison (\( \text{Solids} > \text{Liquids} > \text{Gases} \)) to answer reasoning questions about hearing distant trains through rails.

Question 28S. Why is lightning seen before thunder is heard?
Answer: This happens because the light travels much faster than sound.
In simple words: The flash and the boom happen at the same time. But light is like a super-fast race car that arrives instantly, while sound is like a slow walker that takes a few seconds to reach you.

📝 Teacher's Note: You can calculate the distance of a storm by counting the seconds between the flash and the sound (every 3 seconds is roughly 1 km).

🎯 Exam Tip: Always compare the magnitudes of speed: \( 3 \times 10^8 \text{ ms}^{-1} \) vs \( 330 \text{ ms}^{-1} \).

Question 29S. Why is sound in an iron rail heard before sound in air?
Answer: Sound travels in iron faster than in air so first the sound travelled in iron rail is heard and then the sound travelled in air is heard.
In simple words: Vibrations zip through solid metal much faster than through the air.

📝 Teacher's Note: Native American trackers and old-time railway workers used this fact to listen for approaching horses or trains by putting their ears to the ground or the rails.

🎯 Exam Tip: Mention that iron is a solid and air is a gas to justify the speed difference.

Question 30S. If a diver is underwater and a gun is fired at a distance, who hears it first: the diver or someone on land?
Answer: (i) The diver would hear the sound first.
(ii) This is because sound travels faster in water than in air.
(iii) It would take 0.25t to reach the diver because sound travels almost four times faster in water.
In simple words: Water is a better conductor for sound than air, so the sound zips through it more quickly to reach the diver.

📝 Teacher's Note: Sound travels at about 1450 m/s in water compared to 330 m/s in air.

🎯 Exam Tip: State clearly that "Speed of sound in water \( > \) Speed of sound in air."

Question 31S. Describe the effect of the following on the speed of sound: (i) Frequency, (ii) Temperature, (iii) Pressure, (iv) Moisture.
Answer: (i) Frequency of sound has no effect on the speed of sound.
(ii) Speed of sound increases with the increase in the temperature of sound.
(iii) Pressure of sound has no effect on the speed of sound.
(iv) Speed of sound increases with the increase in presence of moisture in air.
In simple words: Sound moves faster on hot, humid days. How high the note is (frequency) or how heavy the air is pushing (pressure) doesn't change the speed.

📝 Teacher's Note: This is a common trap. Many students think higher pressure makes sound faster, but it actually doesn't because density increases equally with pressure, cancelling the effect.

🎯 Exam Tip: Remember: Temperature and Moisture INCREASE speed. Pressure and Frequency have NO EFFECT.

Question 32S. Does the speed of sound change with amplitude or wavelength?
Answer: (i) Speed of sound does not change with a change in amplitude.
(ii) Speed of sound does not change with a change in wavelength.
In simple words: Making a sound louder (amplitude) or changing the note (wavelength) won't make it travel faster across the room.

📝 Teacher's Note: In a specific medium at a specific temperature, the speed is constant for all sounds.

🎯 Exam Tip: If frequency changes, wavelength changes automatically to keep the speed constant (\( V = \nu \lambda \)).

Question 33S. Why does sound travel faster in humid air than dry air?
Answer: Speed of sound is more in humid air because in presence of moisture, the density of air decreases and sound travels with greater speed.
\( V \propto \frac{1}{\sqrt{\rho}} \)
In simple words: Humid air is actually "lighter" (less dense) than dry air. Since it's lighter, it's easier for the sound vibrations to move through it quickly.

📝 Teacher's Note: Explain that water vapor (\( \text{H}_2\text{O} \)) molecules are lighter than Nitrogen (\( \text{N}_2 \)) and Oxygen (\( \text{O}_2 \)) molecules which make up most of the air.

🎯 Exam Tip: Use the inverse relationship between speed and square root of density to justify your answer.

Question 34S. By how much does the speed of sound increase for every \( 1^\circ\text{C} \) rise in temperature?
Answer: The speed of sound increases by \( 0.61 \text{ m s}^{-1} \) for each \( 1^\circ\text{C} \) rise in temperature.
In simple words: For every degree warmer it gets, sound zips about 60 centimeters faster every second.

📝 Teacher's Note: This is a linear approximation used for temperatures near room temperature.

🎯 Exam Tip: Memorize the value \( 0.61 \text{ ms}^{-1} \) for numerical calculations.

Question 35. Describe a simple experiment to calculate the speed of sound in air.
Answer: The simple experiment that a person can do to calculate the speed of sound in air is that a person stands at a known distance (d meter) from a cliff and fires a pistol and simultaneously starts the stopwatch. He stops the stopwatch as soon as he hears an echo. The distance travelled by the sound during the time (t) seconds is 2d. So, speed of sound = distance travelled / time taken = \( \frac{2d}{t} \).
The approximation made is that speed of sound remains same for the time when the experiment is taking place.
In simple words: You fire a gun toward a mountain and time how long it takes for the "Bang" to bounce back to you. Since the sound went to the mountain and back, it covered twice the distance. Divide that total distance by the time to find the speed.

📝 Teacher's Note: Remind students to stand at least 17 meters away from the cliff to hear a distinct echo.

🎯 Exam Tip: Do not forget the '2' in the distance. The sound travels to the cliff AND back to your ear.

Question 36S. Fill in the blanks:
(a) Sound cannot travel in ______, it requires a material ______.
(b) While sound travels, the particles of the medium ______, but only the disturbance ______.
(c) A longitudinal wave consists of compression and ______.
(d) The minimum displacement in a transverse wave is called ______.
Answer: (a) Vacuum, medium (b) do not move, moves (c) rarefaction (d) trough.
In simple words: These are the basic vocabulary words of sound science!

📝 Teacher's Note: This summarizes the entire chapter's key terminology.

🎯 Exam Tip: Review these definitions regularly as they form the basis of more complex theory.

Question 1M. How do light and sound differ in their propagation?
Answer: Sound needs medium, but light does not need medium for its propagation.
In simple words: Sound is like a car that needs a road (air) to move. Light is like a ghost that can walk through empty space.

📝 Teacher's Note: This is why we can see the light from distant stars but cannot hear the massive explosions happening on them.

🎯 Exam Tip: Light is an electromagnetic wave; sound is a mechanical wave.

Question 2M. What type of wave is a sound wave?
Answer: Longitudinal wave
In simple words: Sound moves as a series of back-and-forth pushes in the air.

📝 Teacher's Note: Always specify "in air" if the question gets more specific, as sound can be transverse in some solids.

🎯 Exam Tip: Longitudinal is the most common answer for sound waves.

Question 3M. What is the speed of sound in air at normal room temperature?
Answer: 330 m s^-1
In simple words: Sound travels about 330 meters in one single second.

📝 Teacher's Note: Standard value used in school textbooks; actual speed at 20°C is about 343 m/s.

🎯 Exam Tip: Note the units: \( \text{m s}^{-1} \).

Question 4M. What is the speed of light?
Answer: 3 x 10⁸ m s^-1
In simple words: Light travels at 300,000 kilometers every second!

📝 Teacher's Note: This is the universal speed limit.

🎯 Exam Tip: Memorize this value as it is used in both Sound and Light chapters.

Question 1N. A heart beats 75 times a minute. Calculate (a) Frequency and (b) Time period.
Answer: Given, heart beats 75 times a minute
(a) Frequency = No. of times heart beats in 1 s

Or, \( \nu = 75/60 = 1.25 \text{ s}^{-1} \)
(b) Time period, \( T = 1/ \nu \)

Or, \( T = 1 / 1.25 = 0.8 \text{ s} \)
In simple words: The heart beats 1.25 times every second. Each single beat takes exactly 0.8 seconds to happen.

📝 Teacher's Note: Always convert "per minute" to "per second" by dividing by 60 to find the frequency in Hertz.

🎯 Exam Tip: Frequency and time period are reciprocals of each other.

Question 2N. If the time period is 2 seconds, find frequency.
Answer: Frequency, \( \nu = 1/T \)

Or, \( \nu = 1 / 2 = 0.5 \text{ Hz} \)
In simple words: If it takes 2 seconds for one beat, only half a beat (0.5) happens in one second.

📝 Teacher's Note: Hertz (Hz) is the standard unit for frequency.

🎯 Exam Tip: Don't forget to include the unit 'Hz' in your final answer.

Question 3N. Find the frequency of a wave if its wavelength is 100 m and its wave velocity is 20 m/s.
Given, wavelength = 100m
Wave velocity = 20 m/s
Answer:
We know that,
Wave velocity = Frequency \( \times \) Wavelength
Or, Frequency = Wave velocity / wavelength
Or, \( \nu = 20/100 = 0.2 \text{ Hz} \)

In simple words: To find how many waves pass by every second (frequency), you take the speed of the wave and divide it by the length of one single wave. Here, sound travels 20 meters in a second, and each wave is 100 meters long, so only 0.2 (or one-fifth) of a wave passes per second.

📝 Teacher's Note: Use the analogy of a train: if a train is moving at a certain speed and each carriage has a specific length, you can calculate how many carriages pass a point in one second. This helps students visualize the abstract concept of frequency.

🎯 Exam Tip: Always state the formula \( v = \nu \lambda \) clearly. Ensure all units are in SI (meters and seconds) before performing the division to avoid common calculation errors.

Question 4N. Wave velocity = 0.3 m/s, Frequency = 20 Hz. Separation between two consecutive compressions is the wavelength of a wave. Calculate the wavelength.
Answer: Wave velocity = 0.3 m/s
Frequency = 20 Hz
Separation between two consecutive compressions is the wavelength of a wave.
We know that,
Wave velocity = Frequency \( \times \) Wavelength
Or, wavelength = Wave velocity / frequency
Or, \( \lambda = 0.3 / 20 = 1.5 \times 10^{-2} \text{ m} \)
In simple words: Wavelength is the length of one full wave cycle. By dividing how fast the wave moves by how often it jiggles, we find that each wave is 0.015 meters long.

📝 Teacher's Note: Help students understand that for longitudinal waves, the distance between the center of one compression and the next is the most accurate way to measure wavelength.

🎯 Exam Tip: Always show the division step clearly; in this case, \( \frac{0.3}{20} \) is a common place where students make decimal errors.

Question 5N. Calculate the frequency of a wave if 40 waves pass through a point in 0.4 seconds.
Answer: Frequency of wave = number of waves per second
Or, \( \nu = 40 / 0.4 = 100 \text{ Hz} \)
In simple words: Frequency is a measure of how many waves pass a spot in one full second. If 40 pass in less than a second, we calculate that 100 would pass in one whole second.

📝 Teacher's Note: Use the heartbeat analogy to explain frequency: if your heart beats 72 times in 60 seconds, the frequency is \( 72/60 \) beats per second.

🎯 Exam Tip: The unit "Hz" (Hertz) is mandatory for frequency. Writing the number without the unit may result in a loss of marks.

Question 6N. The distance between two observers is 1650 m. If the speed of sound is 330 m/s, find the time in which the second observer hears the sound after a gun is shot.
Answer: Distance between the two observers = 1650 m
Speed of sound = 330 m/s
Time in which B hears the sound = Distance / speed = 1650 / 330 = 5s
Thus, B will hear the sound 5s after the gun is shot.
In simple words: Since sound travels at a steady speed, we divide the total distance it needs to cover by its speed to find out how long the journey takes.

📝 Teacher's Note: This is a standard application of the \( \text{Speed} = \text{Distance} / \text{Time} \) formula. Ensure students are comfortable rearranging it to \( \text{Time} = \text{Distance} / \text{Speed} \).

🎯 Exam Tip: Always verify that the units for distance (m) and speed (m/s) match so that the final answer is correctly calculated in seconds.

Question 7N. A person sees a flash of lightning and hears the thunder 5 seconds later. If the speed of sound in air is 330 m/s, find the distance of the flash from the observer.
Answer: Speed of sound in air (V) = 330 m/s
Time in which thunder is heard after lighting is seen (t) = 5s
Thus, distance between flash and observer = \( V \times t = (330 \times 5) = 1650 \text{ m} \)
In simple words: Light travels so fast it's almost instant. By counting the 5 seconds it took for the sound to reach us, we find the storm is about 1.6 kilometers away.

📝 Teacher's Note: Explain that we ignore the time light takes to travel because it is roughly a million times faster than sound, making the visual flash a perfect "start" signal for the timer.

🎯 Exam Tip: Use the formula \( S = v \times t \). This is a classic "real-world" physics application often found in exams.

Question 8N. An observer standing at a distance hears the sound of a fire 2.5 seconds after the flash is seen. If the speed of sound is 340 m/s, calculate the distance.
Answer: Speed of sound in air (V) = 340 m/s
Time in which sound of fire is heard after flash is seen (t) = 2.5s
Thus, distance between flash and observer = \( V \times t = (340 \times 2.5) = 850 \text{ m} \)
In simple words: To find the distance, we multiply the speed of sound by the seconds it took to arrive. Here, the source was 850 meters away.

📝 Teacher's Note: Note that the speed of sound (340 m/s) varies slightly from the previous problem (330 m/s); explain that this is usually due to temperature changes in the air.

🎯 Exam Tip: Always use the specific speed value given in the question rather than assuming a standard 330 m/s.

Question 9N. An observer hears the sound of two tanks firing. The sound from the first tank A arrives at 3.5s and from the second tank B at 2s. If the distance between the two tanks is 510m, calculate the speed of sound.
Answer: Time taken by the observer to hear the sound of the first tank A = 3.5s
Time taken by the observer to hear the sound of the second tank B = 2s
Time taken by the sound to travel from the position of tank A to tank B = (3.5 – 2)s = 1.5s
Distance between the two tanks = 510m
Speed = 510 / 1.5 = 340 m/s
In simple words: Since we know the gap between the tanks is 510 meters and the sound took 1.5 seconds longer to reach us from the farther tank, we divide that distance by the extra time to find the speed.

📝 Teacher's Note: This problem tests the relative distance/time logic. The observer, Tank B, and Tank A must be in a straight line for this calculation to be perfectly accurate.

🎯 Exam Tip: The key first step is finding the time difference (\( \Delta t \)). Without this, the problem cannot be solved.

Question 10N. A sound travels through an iron rail of length 3.3 km.
(a) Find the time taken by sound to travel through the iron rod if the speed in iron is 5280 m/s.
(b) Find the time taken through air if the speed in air is 330 m/s.
Answer: (a) Length of iron rail (D) = 3.3 km = 3300 m
Speed of sound in iron (V) = 5280 m/s
Time taken by sound to travel in iron rod (t) = D/V
Or, \( t = (3300 / 5280) \text{ s} = 0.625 \text{ s} \)

(b) Length of iron rail (D) = 3.3 km = 3300 m
Speed of sound in air (V) = 330 m/s
Time taken by sound to travel in air (t) = D/V
Or, \( t = (3300 / 330) \text{ s} = 10 \text{ s} \)
In simple words: Sound travels much faster through metal than through air. It takes less than a second to move through the metal rail, but a full 10 seconds to move through the air above it.

📝 Teacher's Note: This numerical illustrates why track workers can hear a train coming from several kilometers away by putting their ear to the rail, while it remains silent in the air.

🎯 Exam Tip: Always convert kilometers to meters (3.3 km = 3300 m) before starting calculations involving meters per second.

Question 11N. Sound travels a distance of 1700 m.
(i) Calculate the time taken in air if the speed is 340 m/s.
(ii) Calculate the time taken in water if the speed is 1360 m/s.
Answer: (i) Distance travelled (D) = 1700
Speed of sound in air (V) = 340 m/s
Time taken (t) = D/V = (1700 / 340) s = 5 s

(ii) Distance travelled (D) = 1700
Speed of sound in water (V’) = 1360 m/s
Time taken (t) = D/V = (1700 / 1360) s = 1.25 s
In simple words: Sound moves faster in water than in air. The same trip takes 5 seconds in air but only about 1 second in water.

📝 Teacher's Note: Use this to contrast with light, which slows down in water. Sound, being a mechanical wave, travels better in denser, more elastic media like water.

🎯 Exam Tip: Label the two different speeds (V and V') clearly in your working to avoid confusion between the two parts of the question.

Exercise 8(B)

Question 1S. What is the audible range of frequency?
Answer: The range of frequency within which the sound can be heard by a human being is called the audible range of frequency.
In simple words: This is the "hearing window"—any sound vibrating too slowly or too fast for our ears to catch is outside this range.

📝 Teacher's Note: Remind students that different animals have different audible ranges; for example, dogs can hear frequencies higher than 20 kHz.

🎯 Exam Tip: Use the term "human being" specifically in your definition to show clarity.

Question 2S. State the audible range of frequency for humans.
Answer: The audible range of frequency for humans is 20 Hz to 20 kHz.
In simple words: Humans can hear any sound that jiggles between 20 times a second and 20,000 times a second.

📝 Teacher's Note: 20 kHz is equal to 20,000 Hz. Both notations are commonly used in textbooks.

🎯 Exam Tip: This is a very common one-mark objective question. Memorize the range exactly.

Question 3S. For which frequency range are human ears most sensitive?
Answer: Human ears are most sensitive for the range 2000 Hz to 3000 Hz.
In simple words: Our ears are "tuned" best to hear middle-pitch sounds, which is where most human talking happens.

📝 Teacher's Note: This sensitivity range explains why most alarms and sirens use frequencies in this bracket—it's what we notice most easily.

🎯 Exam Tip: Don't confuse "audible range" with "sensitive range." The sensitive range is a much smaller window inside the audible range.

Question 4S. Does ultrasonic sound have a higher or lower frequency than audible sound?
Answer: Ultrasonic has higher frequency.
In simple words: Ultrasonic is the "squeaky" sound that is so high-pitched it goes beyond what our ears can detect.

📝 Teacher's Note: "Ultra" means beyond. So ultrasonic literally means "beyond the sound range."

🎯 Exam Tip: Ultrasonic frequency is always defined as being greater than 20,000 Hz.

Question 5S. Fill in the blanks:
(a) The audible range is ______ to ______.
(b) Ultrasonic is ______ 20 kHz.
(c) Infrasonic is ______ 20 Hz.
(d) ______ is high frequency.
(e) ______ is low frequency.
Answer: (a) 20 Hz, 20 kHz (b) above 20 kHz (c) below 20 Hz (d) ultrasonic (e) infrasonic.
In simple words: These are the standard names for the different speed limits of sound our ears can or cannot catch.

📝 Teacher's Note: Use the prefixes "infra" (below) and "ultra" (above) to help students remember which category is which.

🎯 Exam Tip: Ensure you include the units (Hz or kHz) correctly for each blank.

Question 6S. Categorize the following frequencies:
(a) 10 Hz
(b) 5000 Hz
(c) 15 kHz
(d) 40 kHz
Answer: (a) Infrasonic (b) Audible (c) Audible (d) Ultrasonic.
In simple words: 10 Hz is too slow to hear, while 40 kHz is too fast. The middle ones are just right for our ears.

📝 Teacher's Note: Have students convert 15 kHz to 15,000 Hz to show them it falls comfortably within the 20 to 20,000 Hz audible range.

🎯 Exam Tip: Carefully check if the frequency is given in Hz or kHz before categorizing it.

Question 7S. Why can we not hear the sound produced by a seconds pendulum?
Answer: No, we cannot hear the sound produced due to vibrations of a seconds pendulum because the frequency of sound produced due to vibrations of seconds pendulum is 0.5 Hz which is infrasonic.
In simple words: A clock's swinging arm only jiggles once every two seconds. That is way too slow for our ears to recognize as sound.

📝 Teacher's Note: A "seconds pendulum" has a time period of 2 seconds (1 second for each swing), so its frequency is \( 1 / 2 = 0.5 \text{ Hz} \).

🎯 Exam Tip: State the specific frequency (0.5 Hz) and mention it is below the human hearing limit (20 Hz).

Question 8S. What is ultrasound?
Answer: Sounds of frequency above 20 kHz are called ultrasound.
In simple words: Ultrasound is sound energy that jiggles so fast (over 20,000 times a second) that it becomes invisible to our ears.

📝 Teacher's Note: Mention that ultrasound has the same speed as regular sound in air, even though the frequency is much higher.

🎯 Exam Tip: 20 kHz (or 20,000 Hz) is the standard cutoff point for ultrasound.

Question 9S. What is the approximate speed of ultrasound in air?
Answer: The approximate speed of ultrasound in air is 330 m/s.
In simple words: Even though it's super high-pitched, ultrasound travels at the same speed as the sounds we can hear.

📝 Teacher's Note: This is a common misconception; students think high frequency means high speed. Speed depends on the material (air), not the pitch.

🎯 Exam Tip: Remember: speed of sound is independent of frequency.

Question 10S. List two properties of ultrasound that make it useful to us.
Answer: Two properties of ultrasound which make it useful to us are:
1. High energy contents
2. High directivity
In simple words: Ultrasound has a lot of "punch" (energy) and travels in a very straight line (directivity) without spreading out, which makes it perfect for seeing inside things.

📝 Teacher's Note: Directivity means it can be sent in narrow beams over long distances without scattering.

🎯 Exam Tip: "High directivity" is the key property for medical imaging (scans).

Question 11S. How do bats locate obstacles and prey using ultrasound?
Answer: Bats locate the obstacles and prey in their path by producing and hearing the ultrasound. They emit an ultrasound which returns after striking an obstacle in their way. By hearing the reflected sound and from the time interval (when they produce ultrasound and they receive them back), they can judge the direction and the distance of the obstacle in their way.
In simple words: Bats shout invisible screams that bounce off bugs. By listening for the echo, the bat knows exactly where the bug is, even in pitch blackness.

📝 Teacher's Note: This biological process is called "echolocation." It is the same principle used by SONAR in ships.

🎯 Exam Tip: Mention both the "emission" and the "reflection" (echo) to explain the full process.

Question 12S. State two applications of ultrasound.
Answer: Two applications of ultrasound:
1. Ultrasound is used for drilling holes or making cuts of desired shape in materials like glass.
2. Ultrasound is used in surgery to remove cataract and in kidneys to break the small stones into fine grains.
In simple words: We use powerful jiggles to cut glass perfectly and to crumble painful kidney stones without needing surgery.

📝 Teacher's Note: In medical use, the high energy of ultrasound is focused on a specific point to vibrate and break down tissues or stones.

🎯 Exam Tip: Providing one industrial use and one medical use provides a well-balanced answer.

Question 1M. Frequency of 1 kHz is equal to:
(a) 1 Hz
(b) 10 Hz
(c) 100 Hz
(d) 1000 Hz
Answer: (d) 1000 Hz
In simple words: Just like 1 kilometer is 1000 meters, 1 kiloHertz is 1000 Hertz.

📝 Teacher's Note: 'k' stands for 'kilo', which is the metric prefix for one thousand.

🎯 Exam Tip: Basic unit conversion question; often used as a trap in multi-part numericals.

Question 2M. Properties of ultrasound that make it useful are:
(a) High power and good directivity
(b) Low power and high speed
(c) High amplitude and low frequency
(d) None of the options
Answer: (a) High power and good directivity
In simple words: Ultrasound is useful because it is strong and travels in a very straight line.

📝 Teacher's Note: Directivity allows for sharp, clear imaging in sonar and ultrasound scans.

🎯 Exam Tip: These are the standard "merits" of ultrasound discussed in textbooks.

Question 3M. Sound vibrations above 20,000 Hz are called:
(a) Audible
(b) Infrasonic
(c) Ultrasound
(d) Low frequency
Answer: (c) Ultrasound
In simple words: Anything above the human hearing limit is categorized as ultrasound.

📝 Teacher's Note: Terminology check: Audible (20-20,000), Infrasonic (< 20), Ultrasound (> 20,000).

🎯 Exam Tip: Ensure you memorize the three main frequency categories.