Goyal Brothers Solutions for ICSE Class 9 Physics Chapter 9 Sound

Exercise

(A) Objective Questions

I. Multiple Choice Questions. Select the correct option:

Question. In case of longitudinal waves, the particles of the medium vibrate:
(a) in the direction of wave propagation
(b) opposite to the direction of wave propagation
(c) at right angles to the direction of wave propagation
(d) none of the above
Answer: (a) in the direction of wave propagation
In simple words: In this kind of wave, the particles of the air or water "shove" their neighbors back and forth in the same line that the sound is traveling, like people in a crowded line pushing forward and back.

📝 Teacher's Note: Use a Slinky spring to demonstrate this. Pushing it forward and pulling it back creates a longitudinal pulse where the coils move parallel to the wave.

🎯 Exam Tip: Remember: "Longitudinal" means "Along the length" (Parallel). Sound waves in air are always longitudinal.

Question. A longitudinal wave consists of:
(a) crest and trough in the medium
(b) compression and rarefaction in the medium
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer: (b) compression and rarefaction in the medium
In simple words: Sound travels by squashing air particles together (compression) and then letting them spread apart (rarefaction), repeating this pattern like an accordion.

📝 Teacher's Note: Draw a diagram of dots representing air particles. Areas with many dots are compressions; areas with few dots are rarefactions.

🎯 Exam Tip: Crests and troughs are for transverse waves (like water waves). Always associate compressions and rarefactions with sound.

Question. The longitudinal waves can propagate only in:
(a) solids
(b) liquids
(c) gases
(d) All of the options
Answer: (d) All of the options
In simple words: Sound can travel through almost anything that has "stuff" (matter) in it—whether it's a solid wall, a swimming pool, or the air around us.

📝 Teacher's Note: Ask students if they can hear underwater or through a closed door to prove sound travels through different states of matter.

🎯 Exam Tip: While sound travels through all three, its speed is fastest in solids and slowest in gases.

Question. A part of the longitudinal wave in which particles of the medium are closer than the normal particles is called:
(a) rarefaction
(b) crest
(c) trough
(d) compression
Answer: (d) compression
In simple words: Just like the word suggests, "compression" is when the particles are squeezed or compressed together into a small space.

📝 Teacher's Note: Relate this to high-pressure regions in the air. More particles in a smaller space = higher pressure.

🎯 Exam Tip: Associate "closeness" with "compression" and "high pressure."

Question. A part of longitudinal wave in which particles of the medium are farther away than the normal particles is called:
(a) rarefaction
(b) trough
(c) compression
(d) crest
Answer: (a) rarefaction
In simple words: Rarefaction is the opposite of a squeeze; it's when the particles have lots of room to spread out and be "rare" or thin.

📝 Teacher's Note: Contrast this with the previous answer. If compression is a "crowd," rarefaction is an "empty hallway."

🎯 Exam Tip: Rarefaction corresponds to a low-pressure region in the medium.

Question. In the region of compression or rarefaction, in a longitudinal wave, the physical quantity which does not change is:
(a) pressure
(b) mass
(c) density
(d) volume
Answer: (b) mass
In simple words: Even though the particles are getting squashed or spread out, the "total amount" of matter (the number of particles) doesn't change; they just move to different spots.

📝 Teacher's Note: Explain the Law of Conservation of Mass—you aren't adding or removing air, just rearranging where it is.

🎯 Exam Tip: Mass is an intrinsic property. While density (\( \text{mass/volume} \)) and pressure change, the actual mass of the particles involved remains constant.

Question. The wavelength is the linear distance between the:
(a) two consecutive compressions
(b) two consecutive rarefactions
(c) one compression and one rarefaction
(d) both (a) and (b)
Answer: (d) both (a) and (b)
In simple words: Wavelength is the length of one full "cycle" of the wave. You can measure it from one "squeeze" to the next "squeeze," or from one "spread-out part" to the next one.

📝 Teacher's Note: Use a wave diagram and mark the distance from center-to-center of compressions to help students visualize wavelength.

🎯 Exam Tip: The keyword is "consecutive." Measuring from the first compression to the third would be two wavelengths.

Question. The number of oscillations passing through a point in unit time is called:
(a) vibration
(b) frequency
(c) wavelength
(d) amplitude
Answer: (b) frequency
In simple words: Frequency is a "count" of how many waves zoom past a spot every second. High frequency sounds like a squeaky whistle; low frequency sounds like a deep drum.

📝 Teacher's Note: Use the "heartbeat" or "clock tick" analogy. How many times something happens per minute/second is its frequency.

🎯 Exam Tip: Frequency and time period are reciprocals (\( f = 1/T \)). Higher frequency means a shorter time for one wave.

Question. The SI unit of frequency is:
(a) hertz
(b) gauss
(c) decibel
(d) none of these
Answer: (a) hertz
In simple words: We measure frequency in Hertz (Hz). 1 Hertz means one wave happens every second.

📝 Teacher's Note: Mention that 1 Hz = 1 vibration per second. Mention Kilohertz (kHz) as 1,000 Hz, which they might see on radio tuners.

🎯 Exam Tip: Decibel is the unit for loudness, not frequency. Don't get them confused!

Question. If the frequency of a wave is 25 Hz, the total number of compressions and rarefactions passing through a point in 1 second is:
(a) 25
(b) 50
(c) 100
(d) none of these
Answer: (b) 50
In simple words: Frequency of 25 Hz means 25 full waves pass by. Since every single wave has one compression and one rarefaction, that's \( 25 + 25 = 50 \) total parts passing through.

📝 Teacher's Note: One oscillation = 1 compression + 1 rarefaction. Therefore, total "events" = \( 2 \times f \).

🎯 Exam Tip: Read carefully if the question asks for "waves" (25) or "total number of compressions AND rarefactions" (50).

Question. Which of the following is an elastic wave?
(a) light wave
(b) ratio wave
(c) sound wave
(d) microwave
Answer: (c) sound wave
In simple words: An elastic wave needs a physical material (like air or metal) to travel through. Light and radio waves can travel through empty space, but sound cannot.

📝 Teacher's Note: "Elastic" or "Mechanical" waves rely on the elasticity of the medium to return particles to their original position after they vibrate.

🎯 Exam Tip: "Sound needs a medium" is the most important rule to remember when identifying mechanical/elastic waves.

(B) Subjective Questions

Question. (a) What do you understand by the term sound energy? (b) State three conditions necessary for hearing sound.
Answer:
(a) Sound energy: “It is a form of energy that produces the sensation of hearing in our ears.” Sound is produced when a body vibrates.
(b) Necessary conditions for hearing sound:
1. There must be a vibrating body, capable of transferring its energy to its surroundings.
2. There must be a material medium to pick the energy and then propagate it in forward direction.
3. There must be a receiver, so as to receive the sound vibrations and then transmit them to the brain for final interpretation, such as human ear.
In simple words: Sound is energy you can hear. To hear it, you need something to make a noise, air for the noise to travel through, and an ear to catch it.

📝 Teacher's Note: Use the "clapping hands" example. Hands (source) vibrate air (medium), and the class hears it (receiver).

🎯 Exam Tip: Listing all three components—Source, Medium, and Receiver—is essential for full marks.

Question. Describe briefly an experiment to prove that vibrating bodies produce sound.
Answer:
Experiment: Take a tuning fork. It is a U-shaped fork made of steel provided with a handle.
Strike the tuning fork with a rubber hammer and hold it close to the ear. A sound is heard. Now take a freely suspended pith ball and touch one end of the tuning fork (which was hit) to it. It is observed that the pith ball repeatedly flies outward. This experiment proves that sound is produced by a vibrating body.
In simple words: If you hit a tuning fork, it hums. If you touch a tiny foam ball to the humming fork, the ball kicks away. This shows the fork is actually shaking very fast, even if you can't see it clearly.

📝 Teacher's Note: The pith ball acts as a "vibration detector." It magnifies the tiny movements of the fork tines so the whole class can see them.

🎯 Exam Tip: Always describe the observation (pith ball flies away) and the conclusion (vibrations produce sound).

Question. (a) What do you understand by the term infrasonic vibrations? (b) What do you understand by the term sonic vibrations? State the range of sonic vibrations for the human ear.
Answer:
(a) Infrasonic vibrations: Those vibrations whose frequency is less than 20 hertz are known as infrasonic vibrations.
(b) Sonic vibrations: Sonic vibrations are also known as audio vibrations. Those vibrations whose frequency is from 20 hertz to 20,000 hertz are known as sonic vibrations. The range of sonic vibrations is from 20 vibrations per second to 20,000 vibrations per second.
In simple words: Infrasonic sounds are too deep and slow for humans to hear (like a slow-moving earthquake). Sonic vibrations are the "sweet spot" that our ears can actually pick up.

📝 Teacher's Note: Use the piano keyboard analogy. Deep bass notes are closer to 20 Hz, while high-pitched tinkling notes are closer to the 20,000 Hz limit.

🎯 Exam Tip: The range "20 Hz to 20,000 Hz" is the most frequently asked fact in this chapter. Memorize it!

Question. (a) What do you understand by the term ultrasonic vibrations? (b) Name three animals which can hear ultrasonic vibrations.
Answer:
(a) Ultrasonic vibrations: Those vibrations whose frequency is more than 20,000 hertz and are not perceived by the human ear are known as ultrasonic vibrations.
(b) Dogs, bats, and dolphins can hear ultrasonic vibrations.
In simple words: Ultrasonic sounds are too high-pitched for us. It's like a whistle that is so "squeaky" it goes beyond what our ears can handle, but a dog can still hear it.

📝 Teacher's Note: Explain that "Ultra" means "beyond." So ultrasonic is beyond the hearing limit.

🎯 Exam Tip: Note that "Ultrasonic" is NOT the same as "Supersonic" (which refers to speed, not frequency).

Question. How do bats locate their prey during flight?
Answer: Bats produce ultra sound which returns after striking an obstacle in their way. By hearing the reflected sound, bats can judge the distance and direction of the obstacle or prey in their way, and hence bats can catch their prey during flight.
In simple words: Bats shout out high-pitched squeaks that we can't hear. Those squeaks bounce off bugs and fly back to the bat's ears, acting like a "sound-map" so they can "see" in the dark.

📝 Teacher's Note: This process is called Echolocation. It works exactly like a ball bouncing off a wall to tell you how far the wall is.

🎯 Exam Tip: Use the term "reflected sound" or "echo" to explain how the bat receives information.

Question. What is Galton’s whistle? To what use is it put?
Answer: A special whistle which can produce ultra sound, not heard by humans, is called Galton’s whistle. It is used to train dogs because they can hear ultrasounds up to a frequency of 40,000 hertz.
In simple words: It's a "silent" whistle. When you blow it, you hear nothing, but your dog will perk up its ears and come running because it can hear that high-pitched sound.

📝 Teacher's Note: Explain that different animals have different hearing "windows." Humans are actually quite limited compared to dogs or bats.

🎯 Exam Tip: Mention the specific frequency (up to 40,000 Hz) to make your answer more precise.

Question. State four practical uses of ultrasonic vibrations.
Answer: Uses of ultrasonic vibrations:
1. These are used for dissipating fogs on the runways at the airports.
2. These are used in the ultrasound scanning of internal organs of the human body.
3. These are used for making dishwashing machines. In these machines, water and detergents are vibrated with an ultrasonic vibrator. The vibrating particles rub against the plates and clean them.
4. These are used in SONAR (Sound Navigation and Ranging) to detect and find the distance of objects underwater.
In simple words: High-frequency sound can "shake" fog away, take pictures of a baby inside a mother, scrub dishes without using a brush, and help ships find hidden rocks or fish.

📝 Teacher's Note: SONAR is particularly interesting to students; relate it to how submarines find each other in movies.

🎯 Exam Tip: Ultrasound scanning (medical) and SONAR (navigation) are the two most important points to remember.

Question. Describe an experiment to prove that material medium is necessary for the propagation of sound.
Answer: A material medium is necessary for the propagation of sound. It can be proved with the help of the "Bell Jar Experiment":
An electric circuit consisting of a cell, a switch, and an electric bell is arranged inside a bell-jar, which stands on the platform of an evacuating pump.
The switch is pressed to close the circuit. Sound is heard when there is air within the bell-jar. Air is now gradually pumped out. The intensity of sound goes on decreasing, and no sound is heard when the air is completely removed. It is because the air, which acts as a medium for the propagation of sound energy, is removed. From this, it is clear that sound cannot be heard in the absence of air.
In simple words: If you put a ringing bell in a jar and suck all the air out, the bell keeps ringing, but you hear nothing! This proves sound can't travel through a "nothing" (vacuum); it needs air to carry it.

📝 Teacher's Note: Emphasize that the bell is still vibrating, but there are no air molecules to "pass the message" to your ears.

🎯 Exam Tip: Make sure to mention that sound "intensity decreases" as air is removed; it doesn't just cut off instantly.

Question. Why do astronauts talk to each other through radio telephone in space?
Answer: Sound cannot travel through vacuum. There is no material medium in space for the propagation of sound. Hence, astronauts talk to each other through radio telephone in space. In a radio telephone, the message reaches the other person with the help of radio waves, which can travel through a vacuum.
In simple words: Space is a silent vacuum. If an astronaut shouted, no sound would reach their friend. Radio waves are like light—they don't need air to move—so they are used to carry the voices instead.

📝 Teacher's Note: This is a perfect example to contrast Sound (mechanical wave) with Radio/Light (electromagnetic wave).

🎯 Exam Tip: The reason is two-fold: (1) Sound needs a medium and space has none, (2) Radio waves do not need a medium.

Question. Define the terms: 1. wavelength, 2. amplitude, 3. frequency.
Answer:
(i) Wavelength: The linear distance between two consecutive particles of a vibrating medium in the same phase is called its wavelength. It is denoted by the Greek letter lambda (\( \lambda \)). Or, the distance travelled by the wave in one time period of vibration.
(ii) Amplitude: The maximum displacement of a vibrating particle about its mean position is called amplitude.
(iii) Frequency: It is the number of complete vibrations executed by a vibrating particle of a medium about its mean position in one second. It is denoted by the letter (\( f \)).
In simple words: Wavelength is the length of one wave "unit." Amplitude is how "big" or tall the wave shakes. Frequency is how many waves happen per second.

📝 Teacher's Note: Use a rope to show amplitude. A small shake has low amplitude; a big wide swing has high amplitude. Use a swing to show frequency.

🎯 Exam Tip: Always use the word "consecutive" when defining wavelength to avoid ambiguity.

Question. State four differences between the sound wave and the light wave.
Answer:

In simple words: Light is a speed demon that can travel through empty space. Sound is much slower and gets "stuck" if there is no air or solid for it to ride on.

📝 Teacher's Note: Use the "Lightning and Thunder" example. You see the light instantly because it's fast, but wait for the sound because it's slow.

🎯 Exam Tip: The ability to travel through a vacuum is the most fundamental difference. Always include it.

Question. What is meant by the term wave motion?
Answer: Wave motion: The transference of energy when the particles of a medium move about their mean position is called wave motion.
In simple words: Wave motion is like "The Wave" at a sports stadium. Each person just stands up and sits down, but the "pulse" travels all the way around the stadium. Energy moves, but the people stay in their seats.

📝 Teacher's Note: Clarify that the particles do not travel from the source to the receiver; only the disturbance (energy) does.

🎯 Exam Tip: Use the word "transference of energy" to show a scientific understanding of waves.

Question. State the relation between the wavelength and the frequency.
Answer: Consider a wave is propagating through a medium.
Let \( f = \) Frequency of wave, \( \lambda = \) Wavelength, \( T = \) Time period.
In time \( T \), distance covered \( = \lambda \)
In 1 second, distance covered \( = \lambda / T \)
But distance covered in one second is velocity (\( v \)).

\( \implies v = \frac{\lambda}{T} \)
But we know \( f = \frac{1}{T} \)

\( \implies v = f \lambda \)
Which is the required relation between wavelength and frequency.
In simple words: The speed of a wave is just the length of one wave multiplied by how many waves happen every second.

📝 Teacher's Note: This formula is the "F=ma" of the sound chapter. It applies to all waves, including light.

🎯 Exam Tip: Practice rearranging this: \( f = v / \lambda \) and \( \lambda = v / f \). These are very common in numerical problems.

Question. What kinds of the waves are produced in solids, liquids and gases?
Answer: Elastic waves or material waves are produced in solids, liquids and gases.
In simple words: These are waves that need matter to exist. In air and water, we mostly see longitudinal waves, while solids can handle both longitudinal and transverse waves.

📝 Teacher's Note: Mention that gases only support longitudinal waves because they lack the "sideways" rigidity needed for transverse waves.

🎯 Exam Tip: Sound waves are the primary example of longitudinal elastic waves.

Question. The sound of an explosion on the surface of lake is heard by a boatman 100 m away and a diver 100 m below the point of explosion. 1. Of the two persons, who would hear the sound first? 2. Give reason. 3. If sound takes ‘t’ seconds to reach boatman, how much time for the diver?
Answer:
1. The diver would hear the sound first.
2. Because the velocity of sound in water (\( 1450 \text{ ms}^{-1} \)) is much more than the velocity of sound in air (\( 332 \text{ ms}^{-1} \)).
3. The time taken by sound to reach the diver would be less than \( t \). Specifically, it would be about 4.36 times less than the time taken for the boatman.
In simple words: Sound travels over 4 times faster through water than through air. So, even though they are the same distance away, the sound "swims" to the diver much faster than it "flies" to the boatman.

📝 Teacher's Note: This illustrates how the density and elasticity of a medium affect the speed of sound.

🎯 Exam Tip: Always compare the speeds (\( 1450 \) vs \( 332 \)) to justify your answer.

Question. What is approximate value of speed of sound in iron as compared to that in air? Illustrate with an experiment.
Answer: Speed of sound in iron is approximately sixteen times the speed of sound in air. [Iron: \( \approx 5100 \text{ ms}^{-1} \); Air: \( \approx 332 \text{ ms}^{-1} \)].
Experiment: If we put our ears to railway tracks, we can hear the sound of a distant train through the metal rails long before we can hear it through the air. This is because sound travels much faster through the solid iron rail than through the gaseous air.
In simple words: Metal is very stiff, so it passes sound vibrations along very quickly. It's like a high-speed train for sound, while air is more like a slow bus.

📝 Teacher's Note: This is why in old Western movies, people put their ears to the ground or the tracks—the ground (solid) carries the sound further and faster than air.

🎯 Exam Tip: The "16 times faster" is a good approximation to remember for comparative questions.

Question. How does a bat avoid obstacles in its way when in flight?
Answer: Bats make a series of twittering sounds, so high pitched that human ears cannot hear. These sound waves strike against the obstacles in their path and send back echoes to the bat’s ears. The echoes tell the bats how they must turn in the air to avoid colliding with the obstacles.
In simple words: A bat is like a pilot using radar. It listens to the "bounce" of its own voice to know where everything is, even in total darkness.

📝 Teacher's Note: Mention that the bat's brain is highly specialized to process these time delays into a 3D mental map.

🎯 Exam Tip: Use the keyword "echoes" or "reflection of ultrasound" to explain this phenomenon.

Question. A continuous disturbance is created on the surface of water with a small piece of cork floating on it. Describe the motion of the cork. What does it tell about the disturbance?
Answer: The cork will move up and down about its mean position but will stay at the same horizontal spot. It tells us that only the disturbance (energy) is transferred from one water particle to the next, but the particles (and the cork) do not travel along with the wave.
In simple words: If you drop a stone near a floating cork, the cork bobs up and down but it doesn't get pushed away to the shore. This shows the water is just "vibrating" in place, passing the energy along like a relay race.

📝 Teacher's Note: This is a crucial concept. Students often think waves "carry" matter. Correct them by using the cork example.

🎯 Exam Tip: The cork moves "perpendicular" to the wave direction, which is typical for a transverse wave on water.

Question. Draw a displacement-time graph for water wave.
Answer:
In simple words: This graph looks like a smooth "S" shape or a roller coaster. It shows how a particle moves up, then down, then back to the middle as time goes by.

📝 Teacher's Note: Ensure students label the Y-axis as "Displacement" and the X-axis as "Time." The distance between two top points is the Time Period (\( T \)).

🎯 Exam Tip: For a displacement-TIME graph, the distance between crests is the Time Period. For a displacement-DISTANCE graph, it's the wavelength.

Practice Problems 1

Question. 200 waves pass through a point in one second. Calculate the time period of wave.
Answer:
Waves passing through a point in 1 second \( = 200 \).
We know, frequency (\( f \)) \( = \) Number of waves passing through point in 1 second.
\( \therefore f = 200 \text{ Hz} \)
Also, \( T = \frac{1}{f} \)

\( \implies \text{Time period (T)} = \frac{1}{200} \)

\( \implies T = 0.005 \text{ s} \)
In simple words: If 200 waves happen in a second, each tiny wave only takes a small fraction of a second (\( 0.005 \)) to complete its trip.

📝 Teacher's Note: Time period is "seconds per wave." Frequency is "waves per second." They are inverses.

🎯 Exam Tip: Don't forget the units! Time period is measured in seconds (s).

Question. A bat emits an ultrasonic sound of frequency 0.25 MHz. Calculate the time in which one vibration is completed.
Answer:
Frequency of ultrasound \( = f = 0.25 \text{ MHz} \)
\( f = 0.25 \times 10^6 \text{ Hz} = 250,000 \text{ Hz} \)
Time in which one vibration is completed \( = T = \frac{1}{f} \)

\( \implies T = \frac{1}{0.25 \times 10^6} = \frac{100}{25} \times 10^{-6} \)

\( \implies T = 4 \times 10^{-6} \text{ s} \)
In simple words: This sound is vibrating so fast (quarter of a million times a second!) that one single vibration happens in just a few millionths of a second.

📝 Teacher's Note: 1 MHz (Megahertz) = \( 1,000,000 \) Hz. Help students handle scientific notation as it makes division much easier.

🎯 Exam Tip: Convert MHz to Hz first before using the \( T = 1/f \) formula.

Practice Problems 2

Question. The sonic boom of an aircraft has a time period of 0.00005s. Calculate the frequency of sound produced.
Answer:
Time period of sonic boom \( = T = 0.00005 \text{ s} \)
Frequency of sound \( = f = \frac{1}{T} \)

\( \implies f = \frac{1}{0.00005} \)

\( \implies f = \frac{100,000}{5} = 20,000 \text{ Hz} \)
\( \therefore f = 20 \text{ kHz} \)
In simple words: Since each wave takes such a tiny amount of time, we can fit 20,000 of them into one single second!

📝 Teacher's Note: Mention that 20,000 Hz is the very edge of human hearing. Older people often can't hear this frequency at all.

🎯 Exam Tip: Moving the decimal 5 places in the denominator is the same as adding 5 zeros in the numerator.

Question. An electromagnetic wave has a time period of \( 4 \times 10^{-8} \) s. Calculate its frequency in MHz.
Answer:
Time period \( = T = 4 \times 10^{-8} \text{ s} \)
Frequency of sound \( = f = \frac{1}{T} \)

\( \implies f = \frac{1}{4 \times 10^{-8}} = 0.25 \times 10^8 \text{ Hz} \)

\( \implies f = 25 \times 10^6 \text{ Hz} \)
\( \therefore f = 25 \text{ MHz} \)
In simple words: This wave vibrates 25 million times every second.

📝 Teacher's Note: To convert Hz to MHz, divide the final frequency by \( 10^6 \).

🎯 Exam Tip: Keep track of your powers of ten carefully during the inversion step.

Practice Problems 3

Question. An ultraviolet radiation has a wavelength of 300Å. If the velocity of electromagnetic wave is \( 3 \times 10^8 \text{ ms}^{-1} \). Calculate 1. frequency 2. time period.
Answer:
Wavelength \( = \lambda = 300 \text{ Å} = 300 \times 10^{-10} \text{ m} = 3 \times 10^{-8} \text{ m} \)
Velocity \( = v = 3 \times 10^8 \text{ ms}^{-1} \)
(i) We know \( v = f \lambda \)

\( \implies f = \frac{v}{\lambda} = \frac{3 \times 10^8}{3 \times 10^{-8}} \)

\( \implies f = \frac{3 \times 10^8}{3 \times 10^{-8}} = 10^{16} \text{ Hz} \)
(ii) Time period \( = T = \frac{1}{f} \)

\( \implies T = \frac{1}{10^{16}} = 10^{-16} \text{ s} \)
In simple words: Light waves are incredibly tiny and vibrate trillions of times a second. Because they are so fast, everything about them happens in a blink of an eye.

📝 Teacher's Note: 1 Angstrom (Å) = \( 10^{-10} \) m. This is a very common unit in radiation problems.

🎯 Exam Tip: Always convert all units to S.I. (meters and seconds) before starting your calculation.

Question. The wavelength of the vibrations produced on the surface of water is 2 cm. If the wave velocity is \( 16 \text{ ms}^{-1} \), calculate 1. no. of waves produced in one second 2. time required to produce one wave.
Answer:
Wavelength \( = \lambda = 2 \text{ cm} = 0.02 \text{ m} \)
Wave velocity \( = v = 16 \text{ ms}^{-1} \)
(i) Frequency \( = f = \frac{v}{\lambda} \)

\( \implies f = \frac{16}{0.02} = 800 \text{ Hz} \)
\( \therefore 800 \) waves are produced in one second.
(ii) Time period \( = T = \frac{1}{f} \)

\( \implies T = \frac{1}{800} = 0.00125 \text{ s} \)
In simple words: The wave is only 2 cm long but it's moving fast. To cover that much ground, 800 waves have to pass by every second.

📝 Teacher's Note: "Number of waves in one second" is just a wordy way of asking for the Frequency.

🎯 Exam Tip: Notice how we converted 2 cm to 0.02 m. Mixing cm and m/s will give the wrong answer!

Practice Problems 4

Question. A continuous progressive transverse wave of frequency 8 Hz moves across the surface of a ripple tank. (a) Describe the movement of water on the surface. (b) If the wavelength is 32 mm, calculate the speed.
Answer:
(a) Frequency \( = f = 8 \text{ Hz} \)
\( \implies \) 8 continuous progressive transverse waves move over the surface of water in one second. The water particles move up and down 8 times every second.
(b) Wavelength \( = \lambda = 32 \text{ mm} = 32 \times 10^{-3} \text{ m} \)
Speed \( = v = ? \)
\( v = f \lambda \)

\( \implies v = 8 \times 32 \times 10^{-3} = 256 \times 10^{-3} \text{ ms}^{-1} \)
\( v = 0.256 \text{ ms}^{-1} = 25.6 \text{ cms}^{-1} \)
In simple words: The water surface bobs up and down 8 times a second. The wave itself cruises along the tank at about 25 centimeters per second.

📝 Teacher's Note: In a ripple tank, waves are transverse because the water moves up and down while the energy moves across.

🎯 Exam Tip: If the question provides wavelength in mm, convert to meters (\( \div 1000 \)) first.

Question. A thin metal plate is placed against the teeth of a cog wheel. Cog wheel is rotated at a speed of 120 rotations per minute and has 160 teeth. Calculate: 1. frequency, 2. speed of sound if \( \lambda = 1.05 \) m, 3. doubling speed effect.
Answer:
Frequency of rotation \( = 120 \text{ rpm} = \frac{120}{60} = 2 \text{ rotations per second} \).
Number of teeth \( = n = 160 \).
(i) Frequency of sound (\( f \)) \( = n \times (\text{rotations per sec}) \)

\( \implies f = 160 \times 2 = 320 \text{ Hz} \)
(ii) Wavelength \( = \lambda = 1.05 \text{ m} \)
Speed of sound \( v = f \lambda \)

\( \implies v = 320 \times 1.05 = 336 \text{ ms}^{-1} \)
(iii) When speed of cog wheel is doubled, then frequency of sound increases and hence sound becomes shriller.
In simple words: Every time a tooth hits the plate, it makes a "click." With 160 teeth spinning twice a second, we get 320 clicks (vibrations) a second. If you spin it twice as fast, it clicks twice as fast and sounds higher (shrill).

📝 Teacher's Note: Shrillness is the sensory perception of frequency. High frequency = high pitch = shrill.

🎯 Exam Tip: For the third part, use the word "frequency" to explain "shrillness."

Practice Problems 5

Question. A sound wave of wavelength 1/3 m has a frequency 996 Hz. Keeping the medium same, if frequency changes to 1328 Hz. Calculate 1. velocity 2. new wavelength.
Answer:
Wavelength \( = \lambda = \frac{1}{3} \text{ m} \)
Frequency \( = f = 996 \text{ Hz} \)
(i) Velocity of sound \( = v = f \lambda \)

\( \implies v = 996 \times \frac{1}{3} = 332 \text{ ms}^{-1} \)
(ii) New frequency \( = f' = 1328 \text{ Hz} \). Since medium is same, \( v = 332 \text{ ms}^{-1} \).
New wavelength \( = \lambda' = \frac{v}{f'} \)

\( \implies \lambda' = \frac{332}{1328} = \frac{1}{4} \text{ m} \)
\( \therefore \lambda' = 0.25 \text{ m} \)
In simple words: The speed of sound doesn't change if you stay in the same room. If you make the vibrations faster (higher frequency), the waves must squeeze together and get shorter (smaller wavelength).

📝 Teacher's Note: This proves the inverse relationship: \( \lambda \propto 1/f \) when velocity is constant.

🎯 Exam Tip: "Keeping the medium same" is the hint that velocity remains constant. Don't recalculate it for the second part.

Question. Two tuning forks A and B of frequencies 256 Hz and 192 Hz respectively are vibrated in air. If the wavelength of A is 1.25 m, calculate the wavelength produced by B.
Answer:
Frequency of tuning fork A \( = f_A = 256 \text{ Hz} \)
Frequency of tuning fork B \( = f_B = 192 \text{ Hz} \)
Wavelength of A \( = \lambda_A = 1.25 \text{ m} \)
Case - I: Find velocity in air
\( v = f_A \lambda_A \)

\( \implies v = 256 \times 1.25 = 320 \text{ ms}^{-1} \)
Case - II: Find wavelength of B
\( v = f_B \lambda_B \)

\( \implies \lambda_B = \frac{v}{f_B} = \frac{320}{192} = 1.67 \text{ m} \)
In simple words: Tuning fork B has a lower frequency (lower pitch), so its waves are longer (\( 1.67 \text{ m} \)) compared to A's waves (\( 1.25 \text{ m} \)).

📝 Teacher's Note: Both forks are in air, so they must have the same wave velocity. This is the "bridge" between fork A and fork B.

🎯 Exam Tip: You can also use the ratio method: \( f_A \lambda_A = f_B \lambda_B \). This is often faster.

Practice Problems 6

Question. The distance between one crest and one trough of a sea wave is 4.5 m. If the waves are produced at the rate of 240/min, calculate 1. time period 2. wave velocity.
Answer:
Distance between one crest and one trough \( = \lambda/2 = 4.5 \text{ m} \).

\( \implies \text{Wavelength } \lambda = 4.5 \times 2 = 9 \text{ m} \).
Frequency \( = 240 \text{ min}^{-1} = \frac{240}{60} \text{ s}^{-1} \)

\( \implies f = 4 \text{ s}^{-1} \) or 4 Hz.
(i) Time period \( = T = \frac{1}{f} \)

\( \implies T = \frac{1}{4} = 0.25 \text{ s} \)
(ii) Wave velocity \( = v = f \lambda \)

\( \implies v = 4 \times 9 = 36 \text{ ms}^{-1} \)
In simple words: A full wave is two "half-waves" (crest to trough). So if crest to trough is 4.5m, the whole wave is 9m. Four waves pass every second, so the speed is 36 meters per second.

📝 Teacher's Note: Draw a wave. Label a crest and the very next trough. Show students that this is exactly half a wavelength.

🎯 Exam Tip: "Rate per minute" must always be divided by 60 to get frequency in Hertz.

Question. The distance between three consecutive crests of wave is 60 cm. If the waves are produced at the rate of 180/ min, calculate 1. wavelength 2. time period 3. wave velocity.
Answer:
(i) We know distance between two consecutive crests \( = \lambda \).
\( \therefore \) Distance between three consecutive crests \( = 2 \lambda = 60 \text{ cm} \).

\( \implies \lambda = \frac{60}{2} = 30 \text{ cm} \).
(ii) Frequency \( = 180 \text{ min}^{-1} \)

\( \implies f = \frac{180}{60} = 3 \text{ Hz} \).
\( \therefore \text{Time period } T = \frac{1}{f} = \frac{1}{3} = 0.33 \text{ s} \).
(iii) Wave velocity \( = v = f \lambda \)

\( \implies v = 3 \times 30 = 90 \text{ cms}^{-1} \)
In simple words: Three crests have two waves between them. If that distance is 60 cm, each wave is 30 cm long.

📝 Teacher's Note: Use the "fence post" analogy. Three posts have two gaps between them. Three crests have two wavelengths.

🎯 Exam Tip: For \( n \) consecutive crests, there are \( n-1 \) wavelengths.

Practice Problems 7

Question. The diagram given below shows a displacement distance graph of a wave. If the velocity of wave is \( 160 \text{ ms}^{-1} \), calculate 1. wavelength 2. frequency 3. amplitude.
Answer:
(i) Wavelength (\( \lambda \)) \( = \) Distance between two consecutive crests
From graph, crests are at \( 0.2 \text{ m} \) and \( 1.0 \text{ m} \).
\( \implies \lambda = 1.0 - 0.2 = 0.8 \text{ m} \).
(ii) Frequency (\( f \)) \( = v / \lambda \)
Velocity (\( v \)) \( = 160 \text{ ms}^{-1} \) (given).

\( \therefore f = \frac{160}{0.8} = 200 \text{ Hz} \)
(iii) Amplitude \( = \) Maximum displacement from mean position
From graph, height of peak is at 10 on the vertical axis.
\( \therefore \text{Amplitude} = 10 \text{ cm} \).
In simple words: We look at the graph "picture." The length of one wave is 0.8m. It is shaking 200 times a second and the waves are 10 cm tall.

📝 Teacher's Note: Teach students to find wavelength by picking any identifiable point (like a crest) and measuring to the very next identical point.

🎯 Exam Tip: Pay attention to the Y-axis units! Here displacement is in "cm" while distance is in "metres." Don't mix them up in the final answer.

Question. From diagram given below calculate 1. velocity of P and Q 2. frequency of P, when frequency of Q is 512 Hz. Assume both waves travel in same medium.
Answer:
For Q:
Wavelength (\( \lambda_Q \)) \( = \) Distance between two consecutive crests.
From graph, \( \lambda_Q = 0.5 - 0.1 = 0.4 \text{ m} \).
Frequency (\( f_Q \)) \( = 512 \text{ Hz} \) (given).
Velocity (\( v \)) \( = f \lambda = 0.4 \times 512 = 204.8 \text{ ms}^{-1} \).
Since both waves are in the same medium, velocity of P is also \( 204.8 \text{ ms}^{-1} \).
For P:
Wavelength (\( \lambda_P \)) \( = 1.0 - 0.2 = 0.8 \text{ m} \).
Frequency of wave P \( = f_P = ? \)
\( v = f_P \lambda_P \)

\( \implies f_P = \frac{v}{\lambda_P} = \frac{204.8}{0.8} = 256 \text{ Hz} \)
In simple words: Since they are in the same room, they move at the same speed. Wave P is twice as long as Q, so it must vibrate half as fast.

📝 Teacher's Note: This is a sophisticated graph-reading problem. It integrates the \( v = f \lambda \) formula with visual data analysis.

🎯 Exam Tip: Always state the assumption "Same medium \( \implies \) Same velocity" to earn marks for logical reasoning.