Goyal Brothers Solutions for ICSE Class 10 Physics Chapter 7 Sound

Exercise - 1

Question 1. (a) State the laws of reflection of sound.
(b) How will you verify laws of reflection of sound experimentally ?
Answer:
(a) The laws governing the reflection of sound are:
1. The angle at which the sound wave strikes the reflecting surface (angle of incidence, \( \angle i \)) is exactly equal to the angle at which it bounces off (angle of reflection, \( \angle r \)). Thus, \( \angle i = \angle r \).
2. The incident sound wave, the reflected sound wave, and the perpendicular line (normal) drawn at the point of incidence on the reflecting surface all lie within the same flat plane.

(b) **Experimental Verification:**
1. Set up a large, smooth, polished wooden board vertically on a flat table.
2. Place a wooden partition screen perpendicular to the center of this board.
3. Position a long, narrow, internally smooth metal or plastic tube on either side of the partition.
4. Put a ticking mechanical watch at the outer end of tube A.
5. Place your ear at the outer end of tube B and adjust its angle by rotating it slightly to the left or right until the ticking sound of the watch is heard most clearly.
6. Measure the angle of incidence \( \angle \text{PKN} \) (made by tube A with the screen) and the angle of reflection \( \angle \text{BKN} \) (made by tube B with the screen).
7. It is observed that \( \angle \text{PKN} = \angle \text{BKN} \), which confirms that the angle of incidence is equal to the angle of reflection, thus verifying the laws of reflection of sound.

In simple words: Sound bounces off walls just like a rubber ball bounces off the ground. The angle at which sound hits a flat board is exactly equal to the angle at which it bounces away from it.

Exam Tip: Remember to write the main equations like \( \angle i = \angle r \) and state clearly that all three lines lie in the 'same plane' to score full marks.

Question 2. Describe any two applications of reflection of sound.
Answer: Two common applications of the reflection of sound are:
1. **Megaphone (or Speaking Tube):** This is a funnel-shaped device used to address large crowds. As sound waves travel through the tube, they undergo consecutive reflections along the widening walls, which prevents the sound energy from spreading in all directions and forces it forward in a concentrated beam.
2. **Hearing Aid:** Designed for individuals with hearing difficulties, this trumpet-shaped device has a wide opening that gathers sound waves from the speaker and channels them through multiple internal reflections into a narrow tube that enters the ear canal, significantly boosting the sound intensity.
In simple words: Reflection of sound is used in megaphones to direct sound forward in a single direction and in hearing aids to collect and concentrate sound waves so they sound louder.

Exam Tip: For questions asking for applications, use well-structured bullet points and define how reflection helps in each device (e.g., preventing the spreading of sound energy or focusing sound waves).

Question 3. (a) What is an echo ?
(b) State two conditions necessary for the formation of an echo.
Answer:
(a) An **echo** is the distinct repetition of a sound heard when the original sound wave is reflected from a distant, rigid surface (such as a hillside, building wall, or cliff) after the initial sound has completely stopped.
(b) Two essential conditions required to hear a clear echo are:
1. The minimum distance between the source of the sound and the reflecting obstacle must be at least \( 17\ \text{m} \) (calculated based on a sound speed of \( 340\ \text{m/s} \) and the human ear's persistence of hearing of \( 0.1\ \text{s} \)).
2. The sound must be sufficiently loud and the reflecting surface large enough so that the reflected wave retains enough energy to be heard.
In simple words: An echo is when a sound hits a far-away wall and bounces back to your ears so you hear it again. For this to happen, the wall must be at least 17 meters away, and the sound has to be loud enough.

Exam Tip: Always mention the persistence of hearing of the human ear (\( 0.1\ \text{s} \)) when explaining the minimum distance condition, as this shows the exact scientific basis for the \( 17\ \text{m} \) rule.

Question 4. What are reverberation ? Give two examples.
Answer: **Reverberation** is the persistent prolongation of sound caused by multiple, rapid reflections from nearby surfaces when the reflecting barrier is located less than \( 17\ \text{m} \) from the sound source.
Two examples of reverberation are:
1. The prolonged, rolling sound produced when speaking inside a large, completely empty hall.
2. The echoing, lingering sound when clapping inside enclosed historical tombs like the Taj Mahal.
In simple words: Reverberation is the lingering hum or echo-like sound you hear in an empty room because the sound keeps bouncing off the close walls, making the sound stretch out.

Exam Tip: Distinguish clearly between an echo (which requires a distance \( \ge 17\ \text{m} \)) and reverberation (where distance is \( < 17\ \text{m} \)), as examiners look specifically for this distinction.

Question 5. How will you determine speed of sound by the method of echos ?
Answer: To measure the speed of sound using the echo method:
1. Stand at a precisely measured distance \( d \) (which should be at least \( 50\ \text{m} \)) facing a large vertical reflecting surface, like a cliff or a high building wall.
2. Produce a sharp sound (such as firing a starting pistol or clapping wooden boards) and start a highly accurate stopwatch at the same instant.
3. Stop the stopwatch the exact moment you hear the returned echo. Let this total elapsed time interval be \( t \).
4. Since the sound wave travels to the obstacle and back, the total distance covered is \( 2d \). The speed of sound \( v \) is then calculated using the formula:
\( v = \frac{2d}{t}\ \text{m/s} \)
In simple words: To find the speed of sound, stand a known distance from a wall, make a sound, and time how long it takes to bounce back. Since the sound travels to the wall and back, divide twice the distance by the measured time.

Exam Tip: Be sure to write the formula \( v = \frac{2d}{t} \) clearly, and emphasize that the distance \( d \) is doubled because the sound travels back and forth.

Question 6. What is sonar ? State its principle. How is it used to find the depth of sea ?
Answer: **SONAR** stands for **So**und **Na**vigation **a**nd **R**anging. It is an active acoustic device used to detect underwater objects and measure water depth.
- **Principle:** It is based on the principle of the reflection of sound (echoes).
- **Method to find sea depth:** High-frequency ultrasonic waves are emitted by a transmitter mounted on the bottom of a ship. These waves travel down through the water, strike the seabed or any underwater object (like a submarine or shipwreck), and bounce back. A receiver on the ship detects the reflected waves and records the time interval \( t \) between transmission and reception.
If the speed of sound in seawater is \( V \), the depth of the sea \( d \) is calculated using:
\( d = \frac{V \times t}{2} \)

In simple words: SONAR uses high-frequency ultrasound waves to measure how deep the ocean is. A ship sends a sound wave to the ocean floor and measures the time it takes to bounce back, calculating the depth from the speed of the wave.

Exam Tip: Remember to state that ultrasonic waves are used in SONAR because they can travel long distances in water without spreading out or being easily absorbed.

Question 7. How do bats locate their prey ? Explain in detail.
Answer: Bats navigate and locate their prey using a highly developed biosonar system called **echolocation**:
1. While flying at night, bats emit extremely high-frequency ultrasonic squeaks (often up to and beyond \( 20\ \text{kHz} \)).
2. These silent, high-frequency sound waves travel through the air, strike nearby objects or insects (prey), and bounce back.
3. The bat's highly sensitive ears detect these returning echoes. By analyzing the time delay and the change in pitch of the returning sound, the bat's brain instantly calculates the exact distance, size, direction, and speed of the prey.
4. This detailed audio map allows bats to hunt flying insects and avoid obstacles in absolute darkness with incredible precision.
In simple words: Bats locate insects in the dark by making high-pitched squeaks that humans cannot hear. These sounds hit the insects and bounce back as echoes, which the bat's sensitive ears use to locate the prey.

Exam Tip: In your explanation, highlight that bats use 'ultrasonic waves' (waves with frequency greater than \( 20,000\ \text{Hz} \)) and detail the process of emission, reflection, and reception.

Question 8. How do the following use echoes ?
1. army,
2. geologists,
3. fishermen.
Answer:
1. **Army:** The military uses echoes via RADAR (which works on the same reflection principle using radio waves instead of sound) to track enemy aircraft. Additionally, sound-ranging techniques are used to locate the positions of enemy artillery.
2. **Geologists:** Geologists use echoes produced by controlled underground explosions (seismic reflection) to map different rock strata and locate valuable mineral deposits or oil reserves.
3. **Fishermen:** Commercial fishermen use sonar devices to send ultrasonic waves down into the sea. When these waves hit a school of fish, they bounce back to the receiver, letting fishermen calculate the exact depth and location of the fish using:
\( d = \frac{v \times t}{2} \) (where \( v = 1450\ \text{m/s} \) is the speed of sound in seawater).
In simple words: The army uses echoes to find enemy positions, geologists use underground sound reflections to search for minerals and oil, and fishermen use sonar to locate schools of fish under the water.

Exam Tip: When discussing fishermen, make sure to state that they use ultrasonic waves specifically, and include the calculation formula \( d = \frac{v \times t}{2} \) to show a thorough understanding.

Multiple Choice Questions

Question 1. The practical application based on the reflection of sound is:
(a) megaphone
(b) sounding board
(c) sonometer
(d) both (a) and (b)
Answer: (d) both (a) and (b)
In simple words: Both megaphones and sounding boards are designed specifically to use the reflection of sound waves to make them louder and guide them in a certain direction.

Exam Tip: Both of these devices are standard applications of sound reflection. A sonometer, by contrast, is used to study the frequency of vibrating strings.

Question 2. Which is not the condition for the formation of echoes ?
(a) Minimum distance between the source of sound and reflecting body should be 17 m.
(b) The temperature of air should be above 20°C.
(c) The wavelength of sound should be less than the height of the reflecting body.
(d) The intensity of sound should be sufficient so that it could be heard after reflection.
Answer: (a) Minimum distance between the source of sound and reflecting body should be 17 m.
In simple words: To hear a distinct echo, a minimum distance is required, but it changes slightly with temperature. Therefore, a fixed 17 m is not a rigid condition under all air temperatures.

Exam Tip: Note that since the speed of sound fluctuates with temperature, the minimum distance required for an echo is not always exactly 17 m.

Question 3. For hearing an echo, the minimum distance between the source of sound and reflecting body should be
(a) 12 m
(b) 24 m
(c) 17 m
(d) 51 m
Answer: (c) 17 m
In simple words: You need to stand at least 17 meters away from a wall to hear your own voice echo clearly.

Exam Tip: This value is based on the calculation \( d = \frac{v \times t}{2} \) using the speed of sound in air as approximately \( 340\ \text{m/s} \) and the persistence of hearing of \( 0.1\ \text{s} \).

Question 4. To locate its prey in the darkness the owl or the bat emits:
(a) infrasonic waves
(b) ultrasonic waves
(c) sonic waves
(d) infrared waves
Answer: (b) ultrasonic waves
In simple words: Bats use high-frequency ultrasound waves to navigate and find insects in the dark.

Exam Tip: Always remember that bats produce and detect 'ultrasonic waves', which lie beyond the human audible range of \( 20\ \text{Hz} - 20,000\ \text{Hz} \).

Numerical Problems on Echoes

Practice Problems 1

Question 1. A person fires a gun in front of a building 167 m away. If the speed of sound is 334 ms-1, calculate time in which he hears an echo.
Answer: Given:
- Distance to the building, \( d = 167\ \text{m} \)
- Speed of sound in air, \( v = 334\ \text{m/s} \)
The total distance traveled by the sound wave to the building and back is:
\( 2d = 2 \times 167\ \text{m} = 334\ \text{m} \)
Let the time taken to hear the echo be \( t \). We know the relation:
\( t = \frac{2d}{v} \)
Substituting the given values:
\( t = \frac{334\ \text{m}}{334\ \text{m/s}} = 1.0\ \text{s} \)
Thus, the person will hear the echo after \( 1.0\ \text{second} \).

In simple words: Since the sound has to travel 167 meters to the wall and another 167 meters back (334 meters total), and sound travels at 334 meters per second, it takes exactly 1 second to hear the echo.

Exam Tip: Always write down the 'Given' values first with their units, and show the formula \( t = \frac{2d}{v} \) before substituting the values to secure full steps marks.

Question 2. An echo is heard after 0.8 s, when a person fires a cracker, 132.8 m from a high building. Calculate the speed of sound.
Answer: Given:
- Time taken to hear the echo, \( t = 0.8\ \text{s} \)
- Distance from the building, \( d = 132.8\ \text{m} \)
We need to find the speed of sound, \( v \). The relationship is:
\( v = \frac{2d}{t} \)
Substituting the given values:
\( v = \frac{2 \times 132.8}{0.8} \)
\( v = \frac{265.6}{0.8} = 332\ \text{m/s} \)
Therefore, the speed of sound in air is \( 332\ \text{m/s} \).
In simple words: The sound travels 132.8 meters to the wall and 132.8 meters back, covering a total distance of 265.6 meters. Since this takes 0.8 seconds, dividing the distance by the time gives a speed of 332 meters per second.

Exam Tip: When performing decimal division like \( \frac{265.6}{0.8} \), simplify it by writing it as \( \frac{2656}{8} = 332 \) to prevent calculation mistakes.

Question 3. The speed of sound is 310 ms-1. A person fires a gun. An echo is heard after 1.5 s. Calculate the distance of person from the cliff from which echo is heard.
Answer: Given:
- Speed of sound, \( v = 310\ \text{m/s} \)
- Time interval for the echo, \( t = 1.5\ \text{s} \)
Let the distance of the person from the cliff be \( d \). We know the formula:
\( v = \frac{2d}{t} \implies 2d = v \times t \)
\( d = \frac{v \times t}{2} \)
Substituting the given values:
\( d = \frac{310 \times 1.5}{2} \)
\( d = 155 \times 1.5 = 232.5\ \text{m} \)
Thus, the distance of the person from the cliff is \( 232.5\ \text{m} \).
In simple words: In 1.5 seconds, sound travels a total of 465 meters. Since it goes to the cliff and back, the one-way distance to the cliff is half of that, which is 232.5 meters.

Exam Tip: Always double-check if the question asks for 'total distance traveled' (\( 2d \)) or 'distance from the obstacle' (\( d \)), as confusing the two is a very common mistake.

Practice Problems 2

Question 1. An echo is heard by a radar in 0.08 s. If velocity of radio waves is 3 × 108 ms-1, how far is the enemy plane ?
Answer: Given:
- Time interval of returned signal, \( t = 0.08\ \text{s} \)
- Speed of radio waves, \( v = 3 \times 10^8\ \text{m/s} \)
Let the distance to the enemy aircraft be \( d \). Using the reflection formula:
\( d = \frac{v \times t}{2} \)
Substituting the values:
\( d = \frac{3 \times 10^8 \times 0.08}{2} \)
\( d = 1.5 \times 10^8 \times 0.08 \)
\( d = 0.12 \times 10^8\ \text{m} = 1.2 \times 10^7\ \text{m} \)
Converting meters to kilometers:
\( d = \frac{1.2 \times 10^7}{1000} = 12000\ \text{km} \)
Thus, the enemy plane is at a distance of \( 12000\ \text{km} \).
In simple words: Radio waves travel incredibly fast. In 0.08 seconds, they travel a round-trip of 24,000 kilometers, meaning the enemy aircraft is exactly 12,000 kilometers away.

Exam Tip: Remember to convert large distances into kilometers (\( \text{km} \)) if required or requested, and write intermediate values in standard scientific notation.

Question 2. An enemy plane is at a distance of 300 km from a radar. In how much lime the radar will be able to detect the plane ? Take velocity of radiowaves as 3 × 108 ms-1.
Answer: Given:
- Distance to the aircraft, \( d = 300\ \text{km} = 300 \times 1000\ \text{m} = 3 \times 10^5\ \text{m} \)
- Speed of radio waves, \( v = 3 \times 10^8\ \text{m/s} \)
Let \( t \) be the total round-trip time for the radar signal to detect the plane. Using the relation:
\( t = \frac{2d}{v} \)
Substituting the given values:
\( t = \frac{2 \times 3 \times 10^5}{3 \times 10^8} \)
\( t = \frac{2 \times 10^5}{10^8} = 2 \times 10^{-3}\ \text{s} = 0.002\ \text{s} \)
Therefore, the radar will detect the plane in \( 0.002\ \text{seconds} \) (or \( 2\ \text{milliseconds} \)).
In simple words: Since the radar wave has to travel to the plane and back (600 kilometers total) at the speed of light, it takes only 0.002 seconds for the radar to detect the plane.

Exam Tip: Be careful with unit conversions. Always convert distance in kilometers (\( \text{km} \)) to meters (\( \text{m} \)) first before using the wave speed in \( \text{m/s} \).

Practice Problems 3

Question 1. A man stands between two buildings A and B. He fires a gun and hears the first echo from building A after 0.4 s and the second echo from building B after 1.6 s. If the speed of sound in air is 332 m/s, calculate the distance between the two buildings.
Answer: Given:
- Speed of sound, \( v = 332\ \text{m/s} \)
- Time of echo from building A, \( t_1 = 0.4\ \text{s} \)
- Time of echo from building B, \( t_2 = 1.6\ \text{s} \)

1. **Calculate the distance to building A (\( d_1 \)):**
\( d_1 = \frac{v \times t_1}{2} = \frac{332 \times 0.4}{2} = 166 \times 0.4 = 66.4\ \text{m} \)

2. **Calculate the distance to building B (\( d_2 \)):**
\( d_2 = \frac{v \times t_2}{2} = \frac{332 \times 1.6}{2} = 166 \times 1.6 = 265.6\ \text{m} \)

3. **Calculate the total distance between the two buildings (\( D \)):**
\( D = d_1 + d_2 = 66.4\ \text{m} + 265.6\ \text{m} = 332.0\ \text{m} \)
Thus, the distance between the two buildings is \( 332\ \text{m} \).
In simple words: The man is between the buildings. The nearest building is 66.4 meters away, and the farther building is 265.6 meters away. Adding these distances together, the buildings are exactly 332 meters apart.

Exam Tip: Be sure to divide the round-trip echo times by 2 when calculating each individual distance to find the true spacing between objects.

Question 2. A man stands in between two parallel cliffs and explodes a cracker. He hears the first echo after 0.6 s and second echo after 2.4 s. Calculate the distance between the cliffs. [Speed of sound is 336 ms-1]
Answer: Let \( d_1 \) be the distance from the man to the nearer cliff, and \( d_2 \) be the distance to the farther cliff.
Given:
- Speed of sound, \( v = 336\ \text{m/s} \)
- Time for the first echo, \( t_1 = 0.6\ \text{s} \)
- Time for the second echo, \( t_2 = 2.4\ \text{s} \)

1. **Calculate the distance to the nearer cliff (\( d_1 \)):**
\( d_1 = \frac{v \times t_1}{2} = \frac{336 \times 0.6}{2} = 168 \times 0.6 = 100.8\ \text{m} \)

2. **Calculate the distance to the farther cliff (\( d_2 \)):**
\( d_2 = \frac{v \times t_2}{2} = \frac{336 \times 2.4}{2} = 168 \times 2.4 = 403.2\ \text{m} \)

3. **Calculate the total distance between the cliffs (\( D \)):**
\( D = d_1 + d_2 = 100.8\ \text{m} + 403.2\ \text{m} = 504.0\ \text{m} \)
Thus, the total distance between the two parallel cliffs is \( 504\ \text{m} \).

In simple words: The man is standing closer to one cliff and farther from the other. The distance to the closer cliff is 100.8 meters, and the distance to the farther cliff is 403.2 meters. Adding these together, the total distance between the cliffs is 504 meters.

Exam Tip: In multiple cliff problems, calculate the individual distances \( d_1 \) and \( d_2 \) separately using the respective echo times before summing them to get the total distance.

Practice Problems : 4

Question 1. A man stands in between two cliffs, such that he is at a distance of 133.6 m from nearer cliff. He fires a gun and hears first echo after 0.8 s and second echo after 1.8 s. Calculate :
1. speed of sound
2. distance between two cliffs.
Answer: Given:
- Distance to the nearer cliff A, \( d_1 = 133.6\ \text{m} \)
- Time for the first echo (from cliff A), \( t_1 = 0.8\ \text{s} \)
- Time for the second echo (from cliff B), \( t_2 = 1.8\ \text{s} \)

1. **Calculate the speed of sound (\( v \)):**
Using the echo formula for the nearer cliff A:
\( v = \frac{2d_1}{t_1} \)
\( v = \frac{2 \times 133.6}{0.8} = \frac{267.2}{0.8} = 334\ \text{m/s} \)

2. **Calculate the distance to the farther cliff B (\( d_2 \)):**
Using the calculated speed of sound and the time for the second echo:
\( d_2 = \frac{v \times t_2}{2} \)
\( d_2 = \frac{334 \times 1.8}{2} = 167 \times 1.8 = 300.6\ \text{m} \)

3. **Calculate the total distance between the two cliffs (\( D \)):**
\( D = d_1 + d_2 = 133.6\ \text{m} + 300.6\ \text{m} = 434.2\ \text{m} \)

Thus:
1. The speed of sound is \( 334\ \text{m/s} \).
2. The total distance between the two cliffs is \( 434.2\ \text{m} \).
In simple words: Using the closer cliff, we find the speed of sound is 334 meters per second. With this speed, the farther cliff is calculated to be 300.6 meters away. Adding both distances, the two cliffs are 434.2 meters apart.

Exam Tip: Remember to use the calculated speed from the first part of the question to solve for the second part, as these multi-step questions rely on accurate intermediate values.

Question 2. A person stands in between two parallel cliffs which are 99 m apart. He fires a gun and hears two successive echoes after 0.2 s and 0.4 s. Calculate :
1. the distance of the person from the nearer cliff
2. speed of sound.
Answer: Let \( d_1 \) be the distance of the person from the nearer cliff A, and \( d_2 \) be the distance from the farther cliff B.
Given:
- Total distance between the cliffs, \( D = d_1 + d_2 = 99\ \text{m} \)
- Time for the first echo, \( t_1 = 0.2\ \text{s} \)
- Time for the second echo, \( t_2 = 0.4\ \text{s} \)
- Let the speed of sound be \( v \).

We can write:
\( d_1 = \frac{v \times t_1}{2} = \frac{v \times 0.2}{2} = \frac{v}{10}\ \text{m} \)
\( d_2 = \frac{v \times t_2}{2} = \frac{v \times 0.4}{2} = \frac{v}{5}\ \text{m} \)

Since the sum of these distances is \( 99\ \text{m} \):
\( d_1 + d_2 = 99 \)
\( \frac{v}{10} + \frac{v}{5} = 99 \)
Taking the LCM (10):
\( \frac{v + 2v}{10} = 99 \)
\( \frac{3v}{10} = 99 \)
\( 3v = 990 \)
\( v = 330\ \text{m/s} \)

Now, we find the individual distances:
1. **Distance from the nearer cliff (\( d_1 \multi-line \)):**
\( d_1 = \frac{v}{10} = \frac{330}{10} = 33\ \text{m} \)

Thus:
1. The distance of the person from the closer cliff is \( 33\ \text{m} \).
2. The speed of sound in air is \( 330\ \text{m/s} \).
In simple words: Since the second echo takes twice as long as the first, the farther cliff is twice as far as the nearer cliff. Splitting the total 99 meters into this 1:2 ratio gives 33 meters for the closer cliff and 66 meters for the farther one, with the speed of sound being 330 m/s.

Exam Tip: For algebraic problems where both speed and individual distances are unknown, set up a linear equation in terms of \( v \) using the total distance, which resolves both variables easily.

Practice Problems : 5

Question 1. A man stands in front of a vertical cliff and fires a gun. He hears an echo after 2.5 s. On moving 80 m closer to the cliff he again fires the gun and he hears an echo after 2 s. Calculate:
(a) distance of man from cliff to his initial position
(b) speed of sound.
Answer: Let \( d_1 \) be the initial distance of the man from the cliff, and \( v \) be the speed of sound.
- First scenario (initial position):
Time taken for echo, \( t_1 = 2.5\ \text{s} \)
Using the echo equation:
\( v = \frac{2d_1}{2.5} = \frac{4d_1}{5} \) --- (i)

- Second scenario (moving \( 80\ \text{m} \) closer):
New distance, \( d_2 = d_1 - 80\ \text{m} \)
Time taken for echo, \( t_2 = 2\ \text{s} \)
Using the echo equation:
\( v = \frac{2(d_1 - 80)}{2} = d_1 - 80 \) --- (ii)

**(a) Calculate the initial distance (\( d_1 \)):**
Equating equations (i) and (ii):
\( \frac{4d_1}{5} = d_1 - 80 \)
\( 4d_1 = 5(d_1 - 80) \)
\( 4d_1 = 5d_1 - 400 \)
\( 5d_1 - 4d_1 = 400 \)
\( d_1 = 400\ \text{m} \)

**(b) Calculate the speed of sound (\( v \)):**
Substitute \( d_1 = 400\ \text{m} \) into equation (ii):
\( v = 400 - 80 = 320\ \text{m/s} \)

Thus:
(a) The initial distance of the man from the cliff is \( 400\ \text{m} \).
(b) The speed of sound is \( 320\ \text{m/s} \).
In simple words: Moving 80 meters closer reduced the echo round-trip time by 0.5 seconds, meaning sound takes 0.5 seconds to travel 160 meters. This gives a speed of 320 m/s, and a starting distance of 400 meters.

Exam Tip: In movement-based echo problems, set up two equations for the speed of sound \( v \) and solve them simultaneously. This is a very common type of exam question.

Question 2. A boy stands in front of a cliff, on the other side of a river. He fires a gun and hears an echo after 6 seconds. The boy then moves 170 m backwards and again fires the gun. He hears an echo after 7 seconds. Calculate :
(a) width of river
(b) speed of sound.
Answer: Let the width of the river (initial distance from the cliff) be \( d_1 \), and the speed of sound be \( v \).
- First position (width of the river):
Time taken for echo, \( t_1 = 6\ \text{s} \)
\( v = \frac{2d_1}{6} = \frac{d_1}{3} \) --- (i)

- Second position (moving \( 170\ \text{m} \) backwards):
New distance from the cliff, \( d_2 = d_1 + 170\ \text{m} \)
Time taken for echo, \( t_2 = 7\ \text{s} \)
\( v = \frac{2(d_1 + 170)}{7} \) --- (ii)

**(a) Calculate the width of the river (\( d_1 \)):**
Equating equations (i) and (ii):
\( \frac{d_1}{3} = \frac{2(d_1 + 170)}{7} \)
\( 7d_1 = 6(d_1 + 170) \)
\( 7d_1 = 6d_1 + 1020 \)
\( d_1 = 1020\ \text{m} \)

**(b) Calculate the speed of sound (\( v \)):**
Substitute \( d_1 = 1020\ \text{m} \) into equation (i):
\( v = \frac{1020}{3} = 340\ \text{m/s} \)

Thus:
(a) The width of the river is \( 1020\ \text{m} \).
(b) The speed of sound is \( 340\ \text{m/s} \).
In simple words: Moving 170 meters further back added 1 second to the echo round-trip time, meaning sound takes 1 second to travel 340 meters (to the cliff and back). This gives a speed of 340 m/s and a river width of 1020 meters.

Exam Tip: Make sure you write 'moving backwards' as an addition to distance (\( d_1 + 170 \)) and 'moving closer' as a subtraction (\( d_1 - 80 \)) to avoid setting up wrong algebraic equations.

Exercise - 2

Question 1. Define the following :
1. natural vibrations
2. forced vibrations
3. damped vibrations
4. natural frequency
Answer:
1. **Natural vibrations:** The periodic oscillations executed by a body when it is slightly disturbed from its equilibrium position and allowed to vibrate freely in the absence of any external resistive forces or friction.
2. **Forced vibrations:** The oscillations produced in a body when it is driven to vibrate under the continuous influence of an external periodic force, vibrating with the frequency of that external force rather than its own natural frequency.
3. **Damped vibrations:** Periodic oscillations whose amplitude continuously decreases over time due to the presence of external resistive forces (like friction or air resistance) until the motion eventually stops.
4. **Natural frequency:** The specific, constant frequency with which a body naturally oscillates when it is allowed to vibrate freely without any external forces acting upon it.
In simple words: Natural vibrations happen when an object shakes freely on its own, and its speed of shaking is its natural frequency. Forced vibrations happen when an outside force makes it shake, and damped vibrations are when friction slowly stops the shaking.

Exam Tip: In definitions, use keywords like 'absence of external forces' for natural vibrations, 'influence of external periodic force' for forced vibrations, and 'decreasing amplitude' for damped vibrations.

Question 2. State four characteristics of forced vibrations.
Answer: Four key characteristics of forced vibrations are:
1. The vibrating body adopts the frequency of the external periodic driving force rather than its own natural frequency.
2. If the driving frequency is significantly different from the natural frequency of the body, the resulting amplitude of the vibration remains very small.
3. The amplitude of the forced vibrations stays constant with time, but its size is highly dependent on the frequency of the external force.
4. When the frequency of the external driving force becomes exactly equal to, or an integral multiple of, the natural frequency of the body, resonance occurs, causing the amplitude of the oscillations to become extremely large.
In simple words: In forced vibrations, the object shakes at the speed of the outside force. The shaking has a steady but small amplitude unless the outside force matches the object's natural speed, which makes it shake violently.

Exam Tip: When listing characteristics, number each point clearly. Be sure to mention how the amplitude depends on the difference between the driving frequency and natural frequency.

Question 3. Give two examples of forced vibrations.
Answer: Two common examples of forced vibrations are:
1. When playing an acoustic guitar, plucking a string forces the wooden sounding board and the air inside the hollow body to vibrate at the string's frequency.
2. The vibrations produced in a hollow sound box or a table top when a vibrating tuning fork's stem is pressed firmly against it.
In simple words: Plucking a guitar string forces the guitar's wooden body to vibrate, and placing a vibrating tuning fork on a wooden table forces the table top to vibrate.

Exam Tip: Choose clear, physical examples involving musical instruments or tuning forks on sounding boards, as these are highly recognized in marking schemes.

Question 4. Why are the stringed instruments provided with large wind box?
Answer: Stringed instruments are equipped with a large hollow sound box (wind box) so that when a string is plucked, its vibrations are transmitted to the large volume of air enclosed within the box. This forces the air column to execute forced vibrations. Because of the large surface area of the box and the large mass of vibrating air, a significantly greater amount of energy is transmitted to the surrounding air, producing a much louder and richer sound.
In simple words: Guitar strings are thin and cannot move much air on their own. The large hollow body captures the string's vibrations and forces a large amount of air to shake together, making the sound much louder.

Exam Tip: Examiners look for the term 'forced vibrations of the air column' and 'large surface area' in this explanation. Ensure both are included.

Question 5. What do you understand by the term resonance ? Give two conditions for producing resonance.
Answer: **Resonance** is a special category of forced vibrations that occurs when the frequency of an externally applied periodic force on a body is exactly equal to its natural frequency, causing the body to vibrate with a dramatically increased amplitude.
Two necessary conditions for producing resonance are:
1. The natural frequency of the driven body must be equal to (or an integer multiple of) the frequency of the external vibrating source.
2. There must be a physical connection or medium to transfer energy from the vibrating source to the driven body with sufficient force.
In simple words: Resonance is when an outside vibration matches an object's natural vibration speed, causing it to shake with a huge, amplified motion. This only happens if the two frequencies match perfectly.

Exam Tip: Clearly define resonance as a 'special case of forced vibrations' and explicitly state that the driving frequency must equal the natural frequency.

Question 6. Explain the following :
(a) Why does the frame of a motorbike vibrate violently at some particular speed ?
(b) Why does an odd piece of cutlery start vibrating violently when a note of some particular frequency is played ?
(c) Why are the soldiers instructed to march out of step while crossing a bridge ?
Answer:
(a) As a motorbike is ridden, the periodic motion of the engine's piston creates mechanical vibrations. At a specific speed, the frequency of these piston movements matches the natural frequency of the motorbike's metal frame. This triggers **resonance**, causing the frame to vibrate violently.
(b) When a musical note of a specific frequency is played, it creates sound waves in the air. If the frequency of this note matches the natural frequency of a loose piece of cutlery, the cutlery absorbs energy from the sound waves and undergoes **resonance**, beginning to rattle and vibrate violently.
(c) When soldiers march in unison, their synchronized footsteps exert a powerful, periodic downward force on the bridge. If the frequency of their steps happens to match the natural frequency of the bridge's structural span, the bridge will begin to execute forced vibrations at that frequency. This leads to **resonance**, causing the bridge to sway with a very large amplitude, which could easily cause it to fracture and collapse. Therefore, soldiers are ordered to break their step while crossing.
In simple words: All three situations are caused by resonance. When engine parts, sound notes, or marching feet match the natural shaking speed of a bike frame, a spoon, or a bridge, it makes them shake with a huge, dangerous motion.

Exam Tip: In all three parts, the core keyword to mention is 'resonance'. Explain how the frequency of the external force (piston, sound note, or footsteps) matches the natural frequency of the object.

Multiple Choice Questions

Question 1. A string is stretched between two nails fixed in the opposite walls and plucked from middle. The vibrations produced by the string are : –
(a) forced vibrations
(b) free vibrations
(c) damped vibrations
(d) resonant vibrations
Answer: (b) free vibrations
In simple words: Plucking a fixed string and letting it go causes it to vibrate naturally on its own, which is a free vibration.

Exam Tip: Since the string is only plucked once and allowed to oscillate without any periodic external force, its vibrations are free (or damped free) vibrations.

Question 2. Water from a tap is allowed to fall in a vessel with a thin neck. The pitch of sound produced by falling water with the volume of water in the vessel.
(a) decreases
(b) increases
(c) remains same
(d) none of these
Answer: (b) increases
In simple words: As water fills the bottle, the empty air column inside gets shorter, which makes the air vibrate faster and produces a higher-pitched sound.

Exam Tip: The frequency of a vibrating air column is inversely proportional to its length (\( f \propto \frac{1}{l} \)). As the water level rises, the air column shortens, increasing the frequency and pitch.

Question 3. The amplitude of forced vibrations is generally than the amplitude of applied external force.
(a) more
(b) less
(c) equal to
(d) none of these
Answer: (b) less
In simple words: Usually, an object driven into forced vibrations does not shake as strongly as the actual force pushing it.

Exam Tip: The amplitude of forced vibrations is smaller than that of the external force because of energy losses and the mismatch between natural and driving frequencies.

Question 4. A tuning fork has a frequency of 212 Hz. It will produce resonance in a wooden board of frequency
(a) 106 Hz
(b) 318 Hz
(c) 212 Hz
(d) 448 Hz
Answer: (c) 212 Hz
In simple words: Resonance can only happen if the frequency of the board matches the 212 Hz frequency of the tuning fork exactly.

Exam Tip: Resonance requires the natural frequency of the body to be exactly equal to (or an integer multiple of) the driving frequency of the tuning fork.

Exercise - 3

Question 1. Define the following
1. musical sound,
2. noise
Answer:
1. **Musical sound:** A pleasant, harmonious, and uniform sound produced by regular, periodic, and continuous vibrations of a source, which is pleasing to the human ear.
2. **Noise:** An unpleasant, harsh, and discordant sound produced by irregular, non-periodic, and discontinuous vibrations, which causes irritation and discomfort to the listener.
In simple words: Musical sounds are smooth and pleasant because their vibrations are regular. Noise is a chaotic and annoying sound because its vibrations are messy and irregular.

Exam Tip: Differentiate between the two based on their physical cause: musical sound is caused by 'regular, periodic vibrations', while noise is caused by 'irregular, non-periodic vibrations'.

Question 2. Give three differences between musical sound and noise.
Answer: Three key differences between musical sound and noise are:

In simple words: Musical sounds have a regular, repeating pattern that sounds pleasant. Noise has a completely random, jagged pattern that sounds irritating and harsh.

Exam Tip: Drawing a quick sketch of a smooth wave versus a jagged wave is a highly effective way to earn extra credit in descriptive answers.

Question 3. State three characteristics of musical sound.
Answer: The three fundamental characteristics of a musical sound are:
1. **Pitch:** The characteristic that allows a listener to distinguish between a shrill (acute) sound and a flat (grave) sound, which is determined by frequency.
2. **Loudness (or Intensity):** The characteristic that distinguishes a strong sound from a faint one, determined primarily by the amplitude of the vibration.
3. **Quality (or Timbre):** The characteristic that enables us to differentiate between two sounds of the same pitch and loudness produced by different musical instruments, determined by the waveform.
In simple words: The three features of music are pitch (how high or low it is), loudness (how strong it is), and quality (the unique tone of different instruments).

Exam Tip: Do not confuse loudness (subjective sensation) with intensity (objective physical quantity). State clearly what physical factor governs each characteristic.

Question 4. (a) What do you understand by the term pitch of sound ?
(b) How is the pitch of sound related to the frequency of a vibrating body ?
Answer:
(a) **Pitch** is the characteristic of musical sound that enables us to distinguish between a sharp, shrill note and a heavy, grave note, even if both sounds have the same loudness. For instance, it allows us to differentiate between different notes on a keyboard.
(b) Pitch is directly related to the frequency of the vibrating source: the higher the frequency of the vibration, the higher (shriller) is the pitch of the sound produced.
In simple words: Pitch is what makes a sound feel high or low, like a bird's chirp versus a lion's roar. A faster vibration (higher frequency) creates a higher pitch.

Exam Tip: Always write that pitch is 'directly proportional to frequency'. Use a real-life comparison like a female voice (high pitch/frequency) versus a male voice (low pitch/frequency) to illustrate.

Question 5. (a) What do you understand by the term loudness of sound ?
(b) How is the loudness of sound related to :
1. amplitude of the vibrating body,
2. distance of the observer from the vibrating body,
3. density of the medium producing sound,
4. frequency of sound.
Answer:
(a) **Loudness** is the physiological sensation perceived by the ear that distinguishes a strong sound from a faint one. It is related to the rate at which sound energy flows through a unit area perpendicular to the direction of propagation.
(b) Loudness depends on these factors as follows:
1. **Amplitude:** Loudness is directly proportional to the square of the amplitude of vibration (\( L \propto A^2 \)). A larger amplitude produces a much louder sound.
2. **Distance:** Loudness decreases as the distance between the listener and the source increases. Specifically, intensity is inversely proportional to the square of the distance (\( I \propto \frac{1}{r^2} \)).
3. **Density of the medium:** Loudness is directly proportional to the density of the medium through which the sound travels. Denser media transmit sound energy more effectively.
4. **Frequency:** Although loudness is governed by amplitude, our ears are more sensitive to certain frequencies. Thus, two sounds of equal amplitude but different frequencies may be perceived with different levels of loudness.
In simple words: Loudness is how strong a sound feels to your ears. It gets louder if the vibration is bigger (amplitude), if you are closer to the source, or if the air is dense.

Exam Tip: Make sure to state the mathematical relationship \( L \propto A^2 \) for amplitude and \( I \propto \frac{1}{r^2} \) for distance, as these specific ratios are highly graded.

Question 6. (a) What do you understand by the term intensity of sound?
(b) Name the unit in which intensity of sound (loudness) is measured.
(c) What is the normal range of loudness ?
(d) What is the range of loudness when sound becomes painful?
Answer:
(a) **Intensity of sound** is the amount of sound energy passing per second through a unit area held perpendicular to the direction of propagation of the sound waves.
(b) The level of loudness is measured in **decibels (dB)**.
(c) The comfortable, normal range of loudness for human hearing is between **\( 50\ \text{dB} \) and \( 80\ \text{dB} \)**.
(d) Sound becomes painful to human ears at levels **above \( 80\ \text{dB} \)**.
In simple words: Intensity is the physical energy of a sound wave, measured in decibels. Normal comfortable sounds are between 50 and 80 decibels, and anything louder than 80 decibels can hurt your ears.

Exam Tip: Differentiate between 'intensity' (which is measured in \( \text{W/m}^2 \)) and 'loudness level' (measured in \( \text{dB} \)), as they represent objective versus subjective properties.

Question 7. List one source of noise in :
(a) transportation,
(b) homes,
(c) factories,
(d) surroundings.
Answer: One common source of noise in each category is:
(a) **Transportation:** Heavy engine sounds and continuous honking from petrol and diesel vehicles.
(b) **Homes:** Loud music players, vacuum cleaners, or desert air coolers.
(c) **Factories:** Continuous clatter of heavy machinery and grinders.
(d) **Surroundings:** Loudspeakers used at high volumes during festivals, marriages, or religious assemblies.
In simple words: Noise comes from traffic honking, loud home appliances, clanging factory machines, and outdoor loudspeakers.

Exam Tip: Keep your examples simple and realistic, matching the categories given in the question.

Question 8. List four harmful effects of sound pollution.
Answer: Four harmful physiological and psychological effects of noise pollution are:
1. Constant exposure to noise can trigger severe headaches, irritability, and elevated nervous tension.
2. Prolonged exposure to high-volume sound can cause temporary or permanent hearing loss, leading to deafness.
3. Excessive noise interferes with normal face-to-face communication, causing misunderstandings and vocal strain.
4. It disrupts sleep patterns, which can lead to chronic fatigue, anger, and concentration issues.
In simple words: Noise pollution causes headaches and stress, can damage your ears and make you deaf, makes it hard to talk to others, and ruins your sleep.

Exam Tip: Use terms like 'nervous tension', 'permanent hearing loss', and 'disruption of sleep patterns' to write a high-scoring, scientifically precise answer.

Question 9. List four ways of reducing noise pollution.
Answer: Four effective measures to control and reduce noise pollution are:
1. Industrial units and factories should be located far away from residential neighborhoods.
2. Heavy transport vehicles and trucks should be restricted from entering quiet residential zones.
3. At home, televisions, radios, and music systems should always be played at a moderate, low volume.
4. Machinery and vehicle engines should be regularly serviced and fitted with high-quality silencers to minimize noise.
In simple words: We can reduce noise by placing factories far from homes, keeping trucks out of neighborhoods, playing music softly at home, and keeping machinery well-serviced.

Exam Tip: Group your suggestions into logical, actionable points (e.g., zoning laws, personal habits, and engineering solutions) for a well-structured answer.

Question 10. What do you understand by the term quality of sound ?
Answer: **Quality of sound** (or **timbre**) is the characteristic of musical sound that allows a listener to distinguish between two sounds of the exact same pitch and loudness produced by different sources or instruments. This unique tonal quality arises because different instruments produce different waveforms and mixtures of overtones.
In simple words: Quality or timbre is why a flute and a piano sound completely different, even when they play the exact same note at the exact same loudness.

Exam Tip: Always explain that quality is determined by the 'waveform' or 'presence of overtones/harmonics' to score full marks.

Multiple Choice Questions

Question 1. The amplitude of a sound wave is increased from 1 mm to 2 mm. The loudness of the sound will:
(a) increase two time
(b) increase four times
(c) same
(d) decrease
Answer: (b) increase four times
In simple words: Since loudness depends on the square of the amplitude, doubling the amplitude (\( 2^2 \)) makes the sound four times louder.

Exam Tip: Always write out the formula \( L \propto A^2 \) to justify why doubling the amplitude results in a fourfold increase in loudness level.

Question 2. By decreasing the amplitude of a pure note its :
(a) speed decreases
(b) wavelength decreases
(c) quality changes
(d) loudness decreases
Answer: (d) loudness decreases
In simple words: Making the physical vibration smaller simply makes the sound quieter (decreases loudness).

Exam Tip: Amplitude is directly tied to the energy and loudness of the sound wave. Decreasing it does not affect speed or wavelength.

Question 3. Two notes are produced from a flute and piano, such that they have same loudness and same pitch. The notes so produced differ in their :
(a) waveform
(b) wavelength
(c) frequency
(d) speed
Answer: (a) waveform
In simple words: Even with the same pitch and volume, different instruments create uniquely shaped sound waves, called waveforms.

Exam Tip: The 'waveform' determines the quality or timbre, which is the only characteristic that differs when pitch and loudness are identical.

Question 4. The voice of women is shrill as compared to men because of the difference in their :
(a) speed
(b) loudness
(c) frequency
(d) all these
Answer: (c) frequency
In simple words: A female voice has a higher frequency of vibration, which makes it sound higher-pitched or shriller.

Exam Tip: High frequency corresponds to high pitch or shrillness. Female vocal cords are shorter and thinner, vibrating faster than male vocal cords.

Question 5. The sound produced by two tuning forks A and B have same amplitude and same waveform, but the frequency of A is three times more than B. In such a case :
(a) quality of sound of A differs from B
(b) the note produced by A is shriller than B
(c) the note produced by B is shriller than A
(d) the note produced by A has more speed than B.
Answer: (b) the note produced by A is shriller than B
In simple words: Since tuning fork A vibrates three times faster than B, it creates a much higher-pitched, shriller sound.

Exam Tip: Shrillness is a direct function of pitch, which depends entirely on frequency. Fork A has a higher frequency and is therefore shriller.

Questions from ICSE Examination Papers

2006

Question 1. Explain why musical instruments like a guitar are provided with a hollow box.
Answer: A guitar is equipped with a large, hollow wooden sounding box so that when its strings are plucked, the vibrations are transferred to the enclosed air column. The hollow box is designed such that its natural frequency matches the frequency of the vibrating strings. This triggers **resonance**, setting the air column into vigorous forced vibrations of large amplitude, which dramatically amplifies the sound produced.
In simple words: The hollow box inside a guitar is filled with air. When you pluck a string, it forces this air to vibrate at the same speed, creating a rich, resonant sound that is much louder.

Exam Tip: Ensure you mention the key terms 'resonance' and 'forced vibrations of the enclosed air column' to satisfy the grading parameters.

Question 2. A tuning fork, struck by a rubber pad, is held over a length of air column in a tube. It produces a loud sound for a fixed length of the air column.
(a) Name the above phenomenon.
(b) How does the frequency of the loud sound compare with that of the tuning fork ?
(c) State the unit for measuring loudness.
Answer:
(a) The phenomenon responsible for this sudden surge in volume is called **acoustic resonance**.
(b) At resonance, the frequency of the loud sound is **exactly equal** to the natural frequency of the vibrating tuning fork.
(c) The unit used to measure the level of loudness is the **decibel (dB)**.
In simple words: This is called resonance. The loud sound has the exact same frequency as the tuning fork because they are vibrating in perfect sync, and loudness is measured in decibels.

Exam Tip: Make sure you clearly state that the frequencies are equal, as resonance only occurs when the driving frequency matches the natural frequency of the system.

2007

Question 3. Define the terms :
(a) Amplitude
(b) Frequency (as applied to sound waves).
Answer:
(a) **Amplitude:** The maximum displacement of a vibrating particle of the medium from its central equilibrium (mean) position on either side as the sound wave passes through.
(b) **Frequency:** The total number of complete waves or vibrational cycles completed by a particle of the medium in one second, measured in Hertz (Hz).
In simple words: Amplitude is how far a particle swings from its resting spot, and frequency is how many times it shakes back and forth in one second.

Exam Tip: Always mention 'from its mean position' when defining amplitude, and state that frequency is measured in 'Hertz (Hz)' to provide a complete, technical definition.

Question 4. A man standing in front of a vertical cliff fires a gun. He hears the echo after 3 seconds. On moving closer to the cliff by 82.5 m, he fires again. This time, he hears the echo after 2.5 seconds. Calculate :
(a) the distance of the cliff from the initial position of the man.
(b) the velocity of sound.
Answer: Let the initial distance of the man from the cliff be \( d_1 \), and the speed of sound be \( v \).
- At the initial position:
Time taken for echo, \( t_1 = 3\ \text{s} \)
\( v = \frac{2d_1}{3} \implies d_1 = \frac{3v}{2} \) --- (i)

- At the second position (moving \( 82.5\ \text{m} \) closer):
New distance from the cliff, \( d_2 = d_1 - 82.5\ \text{m} \)
Time taken for echo, \( t_2 = 2.5\ \text{s} \)
\( v = \frac{2(d_1 - 82.5)}{2.5} = \frac{4(d_1 - 82.5)}{5} \) --- (ii)

**(a) Calculate the starting distance (\( d_1 \)):**
Substitute equation (i) into equation (ii):
\( v = \frac{4(\frac{3v}{2} - 82.5)}{5} \)
\( 5v = 4 \left(\frac{3v}{2}\right) - 4 \times 82.5 \)
\( 5v = 6v - 330 \)
\( 6v - 5v = 330 \)
\( v = 330\ \text{m/s} \)

**(b) Calculate the velocity of sound (\( v \)) and distance (\( d_1 \)):**
Now substitute \( v = 330\ \text{m/s} \) into equation (i):
\( d_1 = \frac{3 \times 330}{2} = 3 \times 165 = 495\ \text{m} \)

Thus:
(a) The initial distance of the man from the cliff is \( 495\ \text{m} \).
(b) The velocity of sound in air is \( 330\ \text{m/s} \).

In simple words: By moving 82.5 meters closer, the echo round-trip path is shortened by 165 meters. Since this saves 0.5 seconds of travel time, the sound's speed is 330 m/s, making the starting distance to the cliff 495 meters.

Exam Tip: In multi-step algebraic problems, keep track of variable names. Substituting \( d_1 \) in terms of \( v \) simplifies the calculations significantly.

2008

Question 5. Point a radar sends a signal to an aeroplane at a distance 45 km away with a speed of 3 × 108 ms-1. After how long is the signal received back from the aeroplane ?
Answer: Given:
- One-way distance to the plane, \( d = 45\ \text{km} = 45 \times 1000\ \text{m} = 45,000\ \text{m} \)
- Velocity of radar waves, \( v = 3 \times 10^8\ \text{m/s} \)
Since the signal travels to the aeroplane and reflects back, the total distance is \( 2d \). Let \( t \) be the round-trip time. Using the relation:
\( t = \frac{2d}{v} \)
Substituting the given values:
\( t = \frac{2 \times 45,000}{3 \times 10^8} \)
\( t = \frac{90,000}{3 \times 10^8} = \frac{9 \times 10^4}{3 \times 10^8} \)
\( t = 3 \times 10^{-4}\ \text{s} = 0.0003\ \text{s} \)
Thus, the signal is received back after \( 3 \times 10^{-4}\ \text{seconds} \) (or \( 0.3\ \text{milliseconds} \)).
In simple words: The radar wave has to travel a total round-trip of 90 kilometers. Traveling at the speed of light, it takes only 0.0003 seconds for the signal to bounce back.

Exam Tip: Make sure you write your final answer clearly in scientific notation (\( 3 \times 10^{-4}\ \text{s} \)) or standard decimals to avoid leaving room for ambiguity.

Question 6. (a) :
1. What is meant by an echo ? Mention one important condition that is necessary for an echo to be heard distinctly.
2. Mention one important use of echo.
(b) :
1. Sometimes when a vehicle is driven at a particular speed, a rattling sound is heard. Explain briefly, why this happens and give the name of the phenomenon taking place.
2. Suggest one way by which the rattling sound can be stopped.
Answer:
**(a)**
1. An **echo** is the clear repetition of a sound heard when the original sound waves bounce back after striking a distant, rigid surface. A vital condition to hear a distinct echo is that the reflecting obstacle must be positioned at least \( 17\ \text{m} \) away from the sound source in air.
2. **Use of echo:** Echoes are used in sonar devices for underwater distance measurements (depth sounding) and by geologists for mapping underground mineral deposits.
**(b)**
1. This rattling sound is caused by **resonance**. Every physical part of a vehicle's body has its own natural frequency. When the car reaches a particular speed, the periodic movements of the engine's pistons match this natural frequency. This drives that specific part of the frame to vibrate with a very large, loud amplitude, creating a rattle.
2. To stop the rattling sound, simply change the vehicle's speed. This shifts the engine piston's frequency away from the body's natural frequency, breaking the resonance condition.
In simple words: An echo is a sound bounce, which requires a 17-meter gap. Rattling in a car happens when the engine vibrations match the natural shaking speed of a loose car part, causing resonance. Changing speed stops it immediately.

Exam Tip: Explicitly use the term 'resonance' when explaining the rattling sound, and state that changing the engine speed disrupts the matching frequency condition.

2009

Question 7. An ultrasonic wave is sent from a ship towards the bottom of the sea. It is found that the time interval between the sending and the receiving of the wave, is 1.5 second. Calculate the depth of the sea if the velocity of sound in sea water is 1400 ms’1.
Answer: Given:
- Round-trip time interval, \( t = 1.5\ \text{s} \)
- Velocity of sound in seawater, \( v = 1400\ \text{m/s} \)
Let the depth of the sea be \( d \). Using the reflection formula:
\( d = \frac{v \times t}{2} \)
Substituting the given values:
\( d = \frac{1400 \times 1.5}{2} \)
\( d = 700 \times 1.5 = 1050\ \text{m} \)
Thus, the depth of the sea is \( 1050\ \text{m} \).
In simple words: Since the sound wave travels to the bottom and back in 1.5 seconds, the one-way trip takes 0.75 seconds. At 1400 meters per second, this means the sea is 1050 meters deep.

Exam Tip: Always write the formula \( d = \frac{v \times t}{2} \) clearly before calculating depth from echo times to demonstrate a solid understanding.

Question 8. (a) A stringed musical instrument, such as the Sitar, is provided with a number of wires of different thicknesses. Explain the reason for this.
(b) What is meant by noise pollution ? Write the name of one source of sound that causes noise pollution.
Answer:
**(a)** In stringed musical instruments like the sitar, the frequency of vibration \( f \) of a stretched string is inversely proportional to its thickness (radius \( r \)), given by the relation:
\( f \propto \frac{1}{r} \)
By providing strings of different thicknesses, musicians can produce a wide range of different fundamental frequencies and notes. Thinner strings vibrate faster to produce high-pitched notes, while thicker strings vibrate slower to produce deep, low-pitched notes.
**(b)** **Noise pollution** refers to the presence of unwanted, harsh, and loud sounds in our environment that cause psychological discomfort, stress, or physiological damage to human health. One prominent source of noise pollution is the honking of horns from road traffic.
In simple words: Thicker strings vibrate slower and make low notes, while thinner strings vibrate faster and make high notes. Noise pollution is annoying, loud sound in our surroundings, like heavy traffic noise.

Exam Tip: Cite the relation \( f \propto \frac{1}{r} \) or mention that frequency depends on thickness to provide a mathematically complete answer for part (a).

Question 9. (a) (i) What is the principle on which sonar is based?
(ii) Calculate the minimum distance at which a person should stand in front of a reflecting surface so that he can hear a distinct echo. (Take speed of sound in air = 350 ms-1.)
(b) :
1. Name the characteristic of sound which enables a person to differentiate between two sounds with equal loudness but having different frequencies.
2. Define the characteristic named by you in (i).
3. Name the characteristic of sound which enables a person to differentiate between two sounds of the same loudness and frequency but produced by different instruments.
(c) :
1. A person is tuning his radio set to a particular station. What is the person trying to do to tune it ?
2. Name the phenomenon involved, in tuning the ratio set.
3. Define the phenomenon named by you in part (ii).
Answer:
**(a)**
1. SONAR is based on the principle of the **reflection of sound waves (echo)**.
2. The human ear retains a sensation of sound for about \( 0.1\ \text{s} \) (persistence of hearing). For an echo to be heard distinctly, the reflected sound wave must return after this time interval has elapsed.
Let \( d \) be the minimum distance. Using the formula:
\( d = \frac{v

Question 9(a). (i) What is the principle on which sonar is based?
(ii) Calculate the minimum distance at which a person should stand in front of a reflecting surface so that he can hear a distinct echo. (Take speed of sound in air = 350 ms-1.)
Answer: (i) SONAR operates based on the reflection of acoustic waves, commonly known as the echo principle.
(ii) The auditory sensation of any sound lingers in human ears for approximately \( 0.1\ \text{s} \) (persistence of hearing) after the source stops. Let \( d \) represent the separation between the observer and the reflecting barrier, and \( v \) represent the speed of sound. The acoustic wave travels to the barrier and back, covering a total path of \( 2d \). The time \( t \) required to perceive the echo is:
\( t = \frac{2d}{v} \)
For a clear echo, this duration must be at least \( 0.1\ \text{s} \). Given the speed of sound \( v = 350\ \text{m/s} \):
\( 0.1 = \frac{2d}{350} \)
\( \implies 2d = 35 \)
\( \implies d = 17.5\ \text{m} \)
Therefore, the observer must stand at least \( 17.5\ \text{m} \) away from the barrier to hear a distinct echo.
In simple words: SONAR works on the reflection of sound (echoes). To hear an echo clearly when sound travels at 350 m/s, you must stand at least 17.5 meters from the reflecting wall so that the returning sound doesn't overlap with the original one.

Exam Tip: Ensure you use the specified speed from the prompt (350 m/s) instead of the standard 340 m/s. Write down the persistence of hearing time of 0.1 s clearly in your steps.

Question 9(b). 1. Name the characteristic of sound which enables a person to differentiate between two sounds with equal loudness but having different frequencies.
2. Define the characteristic named by you in (i).
3. Name the characteristic of sound which enables a person to differentiate between two sounds of the same loudness and frequency but produced by different instruments.
Answer: 1. The attribute that helps distinguish between two sounds of identical volume but varying frequencies is called **Pitch**.
2. **Pitch** is defined as the characteristic of a sound that allows a listener to distinguish a sharp, high-pitched (shrill) sound from a deep, low-pitched (grave) one.
3. The property that helps differentiate two sounds having identical volume and frequency but generated by different sources is called **Quality** (or **Timbre**).
In simple words: Pitch is the feature of sound that makes it sound high or low. Quality is the feature that lets us tell a violin apart from a flute playing the exact same note at the same loudness.

Exam Tip: Clearly state that pitch is determined by frequency, and quality is determined by the waveform or overtones of the sound wave.

Question 9(c). 1. A person is tuning his radio set to a particular station. What is the person trying to do to tune it ?
2. Name the phenomenon involved, in tuning the ratio set.
3. Define the phenomenon named by you in part (ii).
Answer: 1. The listener is adjusting the internal frequency of the radio components to align with the broadcasting station's transmitting frequency, which creates resonance to make the sound clear and audible.
2. This occurrence is known as **Resonance**.
3. **Resonance** is a specific category of forced vibrations. It occurs when the frequency of an external periodic force matches the natural frequency of an object, causing it to vibrate with a significantly increased amplitude.
In simple words: Tuning a radio means matching the frequency of your radio to the station's frequency. This causes resonance, making the sound signal clear and loud.

Exam Tip: Define resonance clearly as a special case of forced vibrations. Use keywords like 'frequency matching' and 'increased amplitude' to score full marks.

2010

Question 10. (a) State two differences between light waves and sound waves.
(b) The waves of the same pitch have their amplitudes in the ratio 2 : 3.
1. What will be the ratio of their loudness ?
2. What will be the ratio of their frequencies ?
(c) Name the subjective property
1. on sound related to its frequency.
2. of light related to its wavelength
Answer: (a) Two fundamental differences between light waves and sound waves are:

(b)
1. Loudness (\( L \)) is directly proportional to the square of the wave's amplitude (\( A \)), i.e., \( L \propto A^2 \). Given the amplitude ratio is \( 2:3 \), the ratio of their loudness is:
\( (2)^2 : (3)^2 = 4 : 9 \).
2. Because both waves share the same pitch, their frequencies must be identical. Thus, the ratio of their frequencies is \( 1 : 1 \).

(c)
1. The subjective property of sound related to its frequency is **Pitch**.
2. The subjective property of light related to its wavelength is **Colour**.
In simple words: Light waves don't need a medium and travel very fast, while sound waves need air or a medium and travel much slower. Loudness depends on amplitude squared, so an amplitude ratio of 2:3 gives a volume ratio of 4:9. Shaking speed determines pitch for sound and color for light.

Exam Tip: Always remember that loudness is proportional to the square of amplitude (\( L \propto A^2 \)). Make sure to construct a neat table when asked to differentiate between two wave types.

Question 11. (a) A man stands at a distance of 68 m from a cliff and fires a gun. After what time interval will he hear the echo, if the speed of sound in air is 340 ms-1.
(b) If the man had been standing at a distance of 12 m from the cliff would he have hear the clear echo ?
Answer: (a) Let the distance to the cliff be \( d = 68\ \text{m} \) and the velocity of sound be \( v = 340\ \text{m/s} \). The sound must travel to the cliff and back, covering a total distance of \( 2d \). The time interval \( t \) after which the echo is perceived is:
\( t = \frac{2d}{v} \)
\( t = \frac{2 \times 68}{340} = \frac{136}{340} = 0.4\ \text{s} \)
So, he will perceive the echo after \( 0.4\ \text{s} \).

(b) If the distance is \( d = 12\ \text{m} \nolinebreak \), the time interval is:
\( t = \frac{2 \times 12}{340} = \frac{24}{340} \approx 0.07\ \text{s} \)
No, the listener will not be able to perceive a distinct echo. The human ear has a persistence of hearing of \( 0.1\ \text{s} \). Since \( 0.07\ \text{s} \) is less than \( 0.1\ \text{s} \), the reflected sound will merge with the original sound instead of being heard as a separate echo.
In simple words: At 68 meters, the sound takes 0.4 seconds to return, which is slow enough for us to hear it as a separate sound. At 12 meters, the sound returns in only 0.07 seconds, which is too fast for our brain to separate from the original clap, so no echo is heard.

Exam Tip: Always remember that the human ear cannot distinguish two sounds if they arrive less than 0.1 seconds apart. Show this calculation clearly whenever you are asked if an echo can be heard.

2011

Question 12. (a) When acoustic resonance takes place, a loud sound is heard. Why does this happen ? Explain.
(b) :
1. Three musical instruments give notes of the frequencies listed below. Flute : 400 Hz; Guitar : 200 Hz; Trumpet: 500 Hz. Which of these has the highest pitch?
2. Which of the following frequencies does a tuning fork of 256 Hz resonate ? 288 Hz, 333 Hz, 512 Hz.
Answer: (a) During acoustic resonance, the external driving frequency perfectly matches the natural frequency of the body. This causes the body to vibrate with a highly amplified displacement. Because loudness depends directly on the square of the vibration amplitude, this large amplitude generates a noticeably louder sound.
(b)
1. The **Trumpet** (at \( 500\ \text{Hz} \)) has the highest pitch because pitch increases with higher frequency.
2. The \( 256\ \text{Hz} \) tuning fork will resonate with the **\( 512\ \text{Hz} \)** sound because it is an exact integer multiple (harmonic) of the fundamental frequency, leading to secondary resonance.
In simple words: When waves sync up in resonance, the vibration gets much bigger, which makes the sound way louder. Higher frequency means higher pitch, so the trumpet is highest. The 256 Hz fork matches with 512 Hz because it's a perfect multiple.

Exam Tip: For pitch comparisons, always state the frequency of the instrument to justify your answer. For resonance, note that it can happen at integer multiples of the fundamental frequency.

Question 13. (a) :
1. Name the type of waves which are used for sound ranging.
2. Why are these sound waves mentioned in (i) above are not audible to us ?
3. Give one use of sound ranging.
(b) A man is standing 25 m away from a wall produces a sound and receives the reflected sound.
1. Calculate the time after which he receives reflected sound ¡f the speed of sound is 350 ms1
2. Will the man be able to hear a distinct echo ? Give a reason for your answer.
Answer: (a)
1. The waves used for this application are **ultrasonic waves**.
2. These waves are silent to humans because their high frequencies exceed the upper threshold of human hearing, which is limited to the range of \( 20\ \text{Hz} \) to \( 20,000\ \text{Hz} \).
3. Sound ranging is widely used to map the depth of seabeds or locate submerged obstructions.

(b)
1. Given distance \( d = 25\ \text{m} \) and velocity \( v = 350\ \text{m/s} \). The time taken for the sound to travel back and forth is:
\( t = \frac{2d}{v} = \frac{2 \times 25}{350} = \frac{50}{350} \approx 0.14\ \text{s} \)
2. Yes, the listener can clearly distinguish the echo because the calculated time interval (\( 0.14\ \text{s} \)) is greater than the \( 0.1\ \text{s} \) limit needed for human hearing persistence, and the distance exceeds the minimum threshold.

In simple words: Ultrasonic waves are used in sonar because they can travel long distances underwater without scattering. We cannot hear them because their frequencies are too high. At 25 meters, sound takes 0.14 seconds to return, which is slow enough for us to hear as an echo.

Exam Tip: Always remember to calculate the time using \( t = \frac{2d}{v} \) and compare it to the persistence of hearing of 0.1 seconds to prove whether an echo can be heard.

2012

Question 14. (a) Which characteristic of sound will change, if there is a change in
1. its amplitude
2. its waveform.
(b) :
1. Name one factor which affects the frequency of sound emitted due to vibrations in an air column.
2. Name the unit used for measuring the sound level.
Answer: (a)
1. If there is a modification in the amplitude, the **loudness** of the sound wave will change.
2. If the waveform is modified, the **quality** (or **timbre**) of the sound will change.
(b)
1. The **length of the vibrating air column** determines the pitch or frequency of the emitted sound. Specifically, a longer air column produces a lower frequency, and a shorter air column produces a higher frequency.
2. The **decibel (dB)** is the standard unit of measurement used to quantify the intensity or level of sound.
In simple words: Changing the wave's height (amplitude) changes how loud a sound is, while changing its wave shape changes its musical quality. For air tubes, shorter columns vibrate faster, and sound level is measured in decibels.

Exam Tip: Remember that loudness depends on amplitude, pitch depends on frequency, and quality depends on the waveform. These direct links are highly evaluated in the marking scheme.

Question 15. (a):
1. What is meant by Resonance ?
2. State two ways in which Resonance differs from Forced vibrations.
(b):
1. A man standing between two cliffs produces a sound and hears two successive echoes at intervals of 3 s and 4 s respectively. Calculate the distance between the two cliffs. The speed of sound in the air is 330 ms-1.
2. Why will an echo not be heard when the distance between the source of sound and the reflecting surface is 10 m?
(c) The diagram below shows the displacement-time graph for a vibrating body.
1. Name the type of vibrations produced by the vibrating body.
2. Give one example of a body producing such vibrations.
3. Why is the amplitude of the wave gradually decreasing?
4. What will happen to the vibrations of the body after some time ?
Answer: (a)
1. **Resonance** is a unique type of forced vibration that takes place when the frequency of an externally applied periodic force matches the natural frequency of a body, causing it to vibrate with a significantly increased amplitude.
2. Two differences between resonance and forced vibrations are:
- Resonance occurs only when the frequency of the external force is exactly equal to the natural frequency of the body, whereas during forced vibrations, the body is driven to oscillate at any arbitrary frequency applied by the external force.
- A highly amplified, loud sound is produced during resonance, while the sound amplitude remains comparatively small in ordinary forced vibrations.

(b)
1. Let the distance of the man from the first cliff be \( x \), and the distance from the second cliff be \( y \).
Given:
- Speed of sound in air, \( v = 330\ \text{m/s} \)
- Time for the first echo, \( t_1 = 3\ \text{s} \)
- Time for the second echo, \( t_2 = 4\ \text{s} \)
Using the echo formula for the first cliff:
\( \frac{2x}{t_1} = v \implies \frac{2x}{3} = 330 \)
\( \implies x = \frac{330 \times 3}{2} = 495\ \text{m} \)
Using the echo formula for the second cliff:
\( \frac{2y}{t_2} = v \implies \frac{2y}{4} = 330 \)
\( \implies y = \frac{330 \times 4}{2} = 660\ \text{m} \)
The total distance between the two cliffs \( AB \) is:
\( AB = x + y = 495\ \text{m} + 660\ \text{m} = 1155\ \text{m} \)

2. The human ear has a persistence of hearing of \( 0.1\ \text{s} \) (\( \frac{1}{10} \) of a second). For a distinct echo to be heard, the reflected sound wave must return to the ear at least \( 0.1\ \text{s} \) after the original sound. At a distance of \( 10\ \text{m} \), the sound travels a total distance of \( 20\ \text{m} \), returning in:
\( t = \frac{2d}{v} = \frac{20}{330} \approx 0.06\ \text{s} \)
Since this time is less than \( 0.1\ \text{s} \), the reflected sound overlaps with the original sound, meaning no distinct echo can be perceived.

(c)
1. The body is executing **damped transverse vibrations**.
2. A plucked guitar string vibrating in air is a classic example of this phenomenon.
3. The amplitude decreases because the wave continuously loses its energy to the surrounding medium due to air resistance and internal friction.
4. As energy continues to dissipate, the oscillations will eventually stop completely, bringing the body to rest.

In simple words: Resonance happens when an outside force shakes something at its natural rhythm, making it vibrate with massive energy. In the cliff problem, the cliffs are 1155 meters apart. Sound dies out in air because of friction, which is why the waves get smaller over time until they stop.

Exam Tip: In echo calculation problems with two cliffs, remember to solve for the distance to each cliff separately using \( d = \frac{v \times t}{2} \) and then add them together. For dampening, focus on keywords like 'frictional resistance of medium' and 'dissipation of energy'.

2013

Question 16. (a) : A bucket kept under a running tap is getting filled with water. A person sitting at a distance is able to get an idea when the bucket is about to be filled.
1. What changes take place in the sound to give this idea?
2. What causes the change in the sound ?
(b): A sound made on the surface of a lake takes 3 s to reach a boatman. How much time will it take to reach a diver inside the water at the same depth ? [Velocity of sound in air = 330 ms-1 ; Velocity of sound in water = 1450 ms-1]
Answer: (a)
1. As the bucket fills up, the sound produced changes from a low-pitched, deep sound to a high-pitched, shrill sound. The sound fades away completely once the water reaches the top.
2. As water fills the bucket, the length of the empty air column above the water decreases. Since the frequency (pitch) of sound is inversely proportional to the length of the vibrating air column, the frequency increases, causing the pitch to rise.
(b)
The distance between the boatman and the source of the sound can be calculated as:
\( \text{Distance} = \text{Speed in air} \times \text{Time} = 330\ \text{m/s} \times 3\ \text{s} = 990\ \text{m} \)
Since the diver is at the same distance under the water, the time taken for the sound to reach the diver through water is:
\( \text{Time} = \frac{\text{Distance}}{\text{Speed in water}} = \frac{990\ \text{m}}{1450\ \text{m/s}} \approx 0.68\ \text{s} \)
In simple words: Splashing water sounds higher and higher in pitch as the bucket fills because the column of air inside gets shorter and vibrates faster. Sound travels much faster in water than in air, taking only 0.68 seconds to reach a diver compared to 3 seconds for the boatman.

Exam Tip: When explaining the bucket sound, always mention the inverse relationship between the length of the vibrating air column and the frequency/pitch of the sound.

Question 17. (a) :
1. What is the principle on which SONAR is based.
2. An observer stands at a certain distance away from a cliff and produces a loud sound. He hears the echo of the sound after 1.8 s. Calculate the distance between the cliff and the observer if the velocity of sound in air is 340 ms-1.
(b) : A vibrating tuning fork is placed over the mouth of a burette filled with water: The tap of the burette is opened and the water level gradually starts falling. It is found that the sound from the tuning fork becomes very loud for a particular length of the water column.
1. Name the phenomenon taking place when this happens.
2. Why does the sound become very loud for this length of the water column ?
(c) :
1. What is meant by the terms (a) amplitude (b) frequency of a wave ?
2. Explain why stringed musical instruments, like the guitar, are provided with a hollow box.
Answer: **(a)**
1. SONAR is based on the **reflection of sound waves** (the echo principle).
2. Given:
- Velocity of sound, \( v = 340\ \text{m/s} \)
- Time taken to perceive the echo, \( t = 1.8\ \text{s} \)
Let the distance to the cliff be \( d \). The sound travels to the cliff and back, covering a total distance of \( 2d \).
\( d = \frac{v \times t}{2} \)
\( d = \frac{340 \times 1.8}{2} = 306\ \text{m} \)
Therefore, the distance between the observer and the cliff is \( 306\ \text{m} \).

**(b)**
1. This occurrence is known as the **resonance of sound**.
2. At a specific length of the air column, its natural frequency of vibration matches the frequency of the tuning fork. The two wave systems reinforce each other constructively, causing the air column to vibrate with maximum amplitude and produce a loud sound.
**(c)**
1. **Amplitude:** The maximum displacement of a vibrating particle from its mean or equilibrium position.
**Frequency:** The total number of waves or complete cycles passing through a point in one second.
2. Stringed instruments like guitars are built with a large, hollow sounding box because strings have a very small surface area and cannot set much air into vibration on their own. The vibrations of the strings are transmitted to the hollow box, forcing the large volume of air inside to undergo forced vibrations of high amplitude, creating a loud and rich sound.
In simple words: Sonar works on echoes. In the cliff problem, the cliff is 306 meters away. The sound in the water tube gets loud because of resonance - when the air column's natural shaking speed matches the tuning fork. Guitars have hollow boxes to force a larger volume of air to vibrate, making the sound much louder.

Exam Tip: In (c)2, make sure to use the term 'forced vibrations of the enclosed air column' and mention that the large surface area of the sounding box increases the volume of vibrating air, which is the key to amplifying the sound.

2014

Question 18. (a) :
1. What are mechanical waves ?
2. Name one property of waves that do not change when the wave passes from one medium to another.
(b) :
The diagram above shows three different modes of vibrations P, Q and R of the same string.
1. Which vibrations will produce a louder sound and why?
2. The sound of which vibration will have maximum shrillness?
3. State the ratio of wavelengths P and R.
Answer: **(a)**
1. **Mechanical waves** are periodic disturbances that require a material medium (like a solid, liquid, or gas) to propagate and cannot travel through a vacuum.
2. The **frequency** of a wave remains constant and does not alter when the wave transitions from one medium to another.
**(b)**
1. Vibration **R** will produce the loudest sound because its wave has a larger amplitude than P and Q.
2. Vibration **P** will produce the sound with the maximum shrillness because it has the highest frequency of vibration (having the shortest loops and most oscillations).
3. Let the length of the string be \( L \).
For mode P (three loops):
\( L = \frac{3 \lambda_P}{2} \implies \lambda_P = \frac{2L}{3} \)
For mode R (one loop):
\( L = \frac{\lambda_R}{2} \implies \lambda_R = 2L \)
The ratio of their wavelengths is:
\( \frac{\lambda_P}{\lambda_R} = \frac{2L/3}{2L} = \frac{1}{3} \)
Therefore, the ratio of wavelengths \( \lambda_P : \lambda_R = 1 : 3 \).

In simple words: Mechanical waves must have a physical medium like air or water to travel. When a wave enters a new medium, its frequency never changes. R makes the loudest sound because its waves are the tallest (biggest amplitude), and P is the shrillest because it has the fastest vibrations (highest frequency).

Exam Tip: Remember that frequency depends only on the source of the wave, which is why it remains completely unchanged across medium boundaries. For the string ratio, show the simple algebraic steps connecting string length \( L \) and wavelength \( \lambda \) to secure full marks.

2014

Question 19. (a) : A type of electromagnetic wave has wavelength 50A.
1. Name the wave.
2. What is the speed of this wave in vacuum?
3. State one use of this type of wave.
(b) :
1. State one important property of waves used for echo depth sounding.
2. A radar sends a signal to an aircraft at a distance of 30 km away and receives it back after 2 x 10-4 second. What is the speed of the signal?
Answer:
**(a)**
1. An electromagnetic wave with a wavelength of \( 50\ \text{Å} \) (or \( 5\ \text{nm} \)) is an **X-ray**.
2. In a vacuum, this wave propagates at the speed of light, which is \( 3 \times 10^8\ \text{m/s} \).
3. X-rays are widely utilized in medical imaging to detect fractures in bones or locate foreign objects inside the body.
**(b)**
1. The waves employed for echo depth sounding (ultrasonic waves) must have high energy and short wavelengths, enabling them to travel long distances in a medium without significant deviation, be focused into a narrow beam, and resist absorption.
2. Given:
Distance of the aircraft, \( d = 30\ \text{km} = 30,000\ \text{m} \)
The total distance traveled by the radar signal to the aircraft and back is:
\( 2d = 2 \times 30,000\ \text{m} = 60,000\ \text{m} \)
Time taken, \( t = 2 \times 10^{-4}\ \text{s} \)
The speed of the signal \( V \) is calculated as:
\( V = \frac{2d}{t} \)
\( V = \frac{60,000}{2 \times 10^{-4}} \)
\( V = 30,000 \times 10^4\ \text{m/s} \)
\( \implies V = 3 \times 10^8\ \text{m/s} \)
Thus, the speed of the signal is \( 3 \times 10^8\ \text{m/s} \).
In simple words: An electromagnetic wave with a wavelength of 50 Å is an X-ray, which travels at the speed of light and is used to inspect bones. Ultrasonic waves are used in oceans because they stay focused in a narrow beam. Using the radar's echo time, the speed of the signal is calculated to be 300,000 km/s.

Exam Tip: When working with electromagnetic wave equations, remember that their speed in vacuum is always \( 3 \times 10^8\ \text{m/s} \). Use this constant as a double-check for your numerical answers.

2015

Question 20. 1. Draw a graph between displacement and the time for a body executing free vibrations.
2. Where can a body execute free vibrations?
Answer:
1. The displacement-time graph representing a body undergoing free vibrations (maintaining a constant amplitude over time) is plotted below:

2. True free vibrations can only take place in a perfect vacuum. In any physical medium (like air or water), resistive forces like friction and drag gradually drain the system's energy, causing the amplitude of the vibrations to decay over time (damped vibrations).
In simple words: Free vibrations keep the exact same size (amplitude) forever. This is only possible in a vacuum because any air or gas would create friction, slowing the vibrations down.

Exam Tip: In the graph for free vibrations, ensure the wave peaks and troughs touch the exact same maximum height (+a) and minimum depth (-a) boundaries to show constant amplitude.

Question 21. (a) : A person standing between two vertical cliffs and 480 m from the nearest cliff shouts. He hears the first echo after 3s and the second echo 2s later. Calculate:
1. The speed of sound.
2. The distance of the other cliff from the person.
(b) : In the diagram below, A, B, C, D are four pendulums suspended from the same elastic string PQ. The length of A and C are equal to each other while the length of pendulum B is smaller than that of D. Pendulum A is set in to a mode of vibrations.
1. Name the type of vibrations taking place in pendulums B and D?
2. What is the state of pendulum C?
3. State the reason for the type of vibrations in pendulums B and C.
Answer:
**(a)**
1. Let the distance to the closer cliff be \( d_1 = 480\ \text{m} \), and the time for the first echo be \( t_1 = 3\ \text{s} \). The speed of sound \( v \) is given by:
\( v = \frac{2d_1}{t_1} \)
\( v = \frac{2 \times 480}{3} = \frac{960}{3} = 320\ \text{m/s} \)
Thus, the speed of sound is \( 320\ \text{m/s} \).
2. The second echo is perceived \( 2\ \text{s} \) after the first, making the total round-trip time \( t_2 = 3 + 2 = 5\ \text{s} \). Let the distance to the farther cliff be \( d_2 \).
\( d_2 = \frac{v \times t_2}{2} \)
\( d_2 = \frac{320 \times 5}{2} = 160 \times 5 = 800\ \text{m} \)
So, the distance of the farther cliff from the person is \( 800\ \text{m} \).

**(b)**
1. Pendulums B and D execute **forced vibrations** under the influence of the periodic driving force transmitted from pendulum A.
2. Pendulum C undergoes **resonance** with pendulum A.
3. Vibrations are passed from A to B and C through the elastic string PQ. Since pendulum B is of a different length than A, its natural frequency does not match that of A; hence, it oscillates with a smaller amplitude in a forced state. On the other hand, pendulum C has the exact same length as A, meaning their natural frequencies are identical. Consequently, C oscillates in phase with A with a very large amplitude because it achieves resonance.

In simple words: If you stand between two cliffs and hear echoes at 3 seconds and 5 seconds, you can calculate the distance to each cliff using the speed of sound, which adds up to 1280 m total. In the pendulum system, since C has the same length as A, it shakes very easily when A is set into motion because their natural frequencies match.

Exam Tip: In double cliff echo questions, the second echo is heard '2s later', meaning its total round-trip time is \( t_1 + t_{\text{delay}} = 3 + 2 = 5\ \text{s} \). Failing to add the times is a very common student mistake.

2016

Question 22. (a) : The ratio of amplitude of two waves is 3:4. What is the ratio of their :
1. loudness
2. frequencies?
(b) : State two ways by which the frequency of transverse vibrations of a stretch string can be increases.
(c) : What is meant by noise pollution? Name one source of sound causing noise pollution.
Answer:
**(a)**
1. Loudness of a sound wave is directly proportional to the square of its amplitude (\( I \propto a^2 \)). Let the amplitudes of the two waves be \( a_1 \) and \( a_2 \), where \( \frac{a_1}{a_2} = \frac{3}{4} \). The ratio of their loudness is given by:
\( \frac{I_1}{I_2} = \frac{a_1^2}{a_2^2} = \frac{3^2}{4^2} = \frac{9}{16} \)
Hence, the ratio of their loudness is \( 9 : 16 \).
2. If the waves are assumed to have the same pitch, their frequencies will be identical. Therefore, the ratio of their frequencies is \( 1 : 1 \).
**(b)** The fundamental frequency of transverse vibrations in a stretched string is given by:
\( f = \frac{1}{2l} \sqrt{\frac{T}{m}} \)
where \( l \) is the string length, \( T \) is the tension, and \( m \) is the mass per unit length. To increase this frequency, one can:
1. Reduce the length of the vibrating string.
2. Increase the tension (\( T \)) applied to the string.
3. Decrease the radius or thickness of the string (which lowers \( m \)).
**(c)** Noise pollution represents the environmental disturbance generated by loud, harsh, and unwanted sounds from various activities. Standard sources of noise pollution include blaring loudspeakers, industrial machinery, heavy road traffic, and passing trains.
In simple words: If the height of two waves is in a 3:4 ratio, their volumes will be in a 9:16 ratio because volume depends on height squared. To make a guitar string vibrate faster (higher pitch), you can make it shorter, stretch it tighter, or use a thinner string. Loud, disturbing sounds like traffic or loudspeakers cause noise pollution.

Exam Tip: Always write down the general formula for a stretched string's frequency \( f = \frac{1}{2l} \sqrt{\frac{T}{m}} \) before explaining the factors, as this makes your reasoning clear and ensures you secure full marks.

Question 23. (a) :
1. Name the waves used for echo depth sounding.
2. Give one reason for their use for the above purpose.
3. Why are the waves mentioned by you not audible to us?
(b) :
1. What is an echo
2. State two conditions for an echo to take place.
(c) :
1. Name the phenomenon involved in tuning a radio set to a particular station.
2. Define the phenomenon named by you in part (i) above.
3. What do you understand by loudness of sound ?
4. In which units is the loudness of sound measured ?
Answer:
**(a)**
1. **Ultrasonic waves** (or ultrasound) are used for echo depth sounding.
2. These waves are chosen because they possess high energy, can be focused into a narrow beam, and can travel long distances underwater without significant deviation or absorption.
3. These waves are inaudible to human ears because their frequencies exceed \( 20,000\ \text{Hz} \), which is above the human audible limit of \( 20\ \text{Hz} \) to \( 20,000\ \text{Hz} \).
**(b)**
1. An **echo** is the secondary sound heard when a primary sound wave is reflected from a distant, rigid surface after the original sound has completely died out.
2. The essential requirements for hearing a distinct echo are:
- The minimum distance between the sound source and the reflecting surface must be at least \( 17\ \text{m} \) (in air).
- The reflecting surface must be large enough compared to the wavelength of the sound wave.
**(c)**
1. The phenomenon involved is **resonance**.
2. **Resonance** is a special category of forced vibrations that occurs when the frequency of an externally applied periodic force matches the natural frequency of a system, causing it to vibrate with a significantly increased amplitude.
3. **Loudness** is the subjective characteristic of sound that allows us to distinguish between a strong (loud) sound and a weak (faint) sound, even when they share the same pitch and quality.
4. Loudness is measured in **decibels (dB)**.
In simple words: Sonar uses ultrasound because these waves travel straight underwater without scattering. Echoes are bounced sounds that need a wall at least 17 meters away to be heard clearly. Adjusting a radio to a station works on resonance, and sound volume is measured in decibels.

Exam Tip: Remember to distinguish between 'loudness' (subjective sensation measured in decibels) and 'intensity' (objective power per unit area measured in \( \text{W/m}^2 \)), as examiners often test this distinction.