RBSE Solutions Class 11 Physics Chapter 9 Wave Motion

Question 1. What is transferred in wave motion?
Answer: In wave motion, energy moves from one place to another. This happens because the particles of the medium vibrate and pass the energy along. Think of a ripple in water; the water itself doesn't travel far, but the wave's energy does.
In simple words: In a wave, energy moves forward because small parts of the material shake and pass it on.

🎯 Exam Tip: Remember that in wave motion, it's energy that transfers, not matter. Particles of the medium only oscillate around their mean positions.

Question 2. If the length of a stretched string is doubled and the tension is increased to four times then what would be the relation between new frequency and the old frequency?
Answer: The frequency of a stretched string is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is length, \( T \) is tension, and \( \mu \) is mass per unit length. If the length \( L \) is doubled to \( 2L \), and tension \( T \) is increased to \( 4T \), the new frequency \( f' \) will be:
\( f' = \frac{1}{2(2L)} \sqrt{\frac{4T}{\mu}} \)
\( f' = \frac{1}{4L} \sqrt{4} \sqrt{\frac{T}{\mu}} \)
\( f' = \frac{1}{4L} \times 2 \sqrt{\frac{T}{\mu}} \)
\( f' = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \)
Therefore, the new frequency \( f' \) is equal to the old frequency \( f \). This means the frequency remains the same despite the changes.
In simple words: If you double the string's length and make the tension four times stronger, the new sound frequency will be exactly the same as the old one.

🎯 Exam Tip: For problems involving ratios of physical quantities, always write down the initial and final states clearly and use the relevant formula to find the relationship, simplifying step-by-step.

Question 3. Write the relation between angular frequency, angular wave number and wave velocity.
Answer: The angular frequency \( \omega \) is related to the frequency \( n \) (or \( f \)) by the equation \( \omega = 2\pi n \). This describes how fast a wave oscillates in time.
\( \implies \) So, \( n = \frac{\omega}{2\pi} \).
The angular wave number \( k \) is related to the wavelength \( \lambda \) by \( k = \frac{2\pi}{\lambda} \). This tells us how many wave cycles fit into a given length.
\( \implies \) So, \( \lambda = \frac{2\pi}{k} \).
Wave velocity \( v \) is the product of frequency and wavelength: \( v = n\lambda \).
Now, substituting the expressions for \( n \) and \( \lambda \) into the wave velocity equation:
\( v = \left(\frac{\omega}{2\pi}\right) \times \left(\frac{2\pi}{k}\right) \)
\( v = \frac{\omega}{k} \)
This fundamental relationship shows that wave velocity is the ratio of angular frequency to angular wave number, linking how quickly the wave oscillates in time and space.
In simple words: Wave velocity is found by dividing how fast a wave wiggles (angular frequency) by how packed it is (angular wave number).

🎯 Exam Tip: Remember the basic definitions for angular frequency (\( \omega \)), angular wave number (\( k \)), and wave velocity (\( v \)), and practice deriving their relationships to ensure you understand the underlying concepts.

Question 4. What does the number of vibrations executed by the particle of a medium in one second is called?
Answer: The number of vibrations a particle of a medium completes in one second is called its frequency. Frequency is a measure of how often a repetitive event occurs per unit of time.
In simple words: The number of times something wiggles in one second is called its frequency.

🎯 Exam Tip: Distinguish between frequency (vibrations per second) and time period (seconds per vibration), as they are inverse to each other.

Question 5. What is the time to complete one the vibration called?
Answer: The time it takes for a particle to complete one full vibration or oscillation is called the time period. This is the inverse of frequency.
In simple words: The time taken for one full back-and-forth movement is called the time period.

🎯 Exam Tip: Understanding the relationship between time period and frequency (\( T = 1/f \)) is crucial for solving wave motion problems.

Question 6. Write the equation of wave velocity.
Answer: The equation for wave velocity \( v \) is given by the product of its frequency \( n \) (or \( f \)) and its wavelength \( \lambda \). This fundamental equation connects how fast a wave travels with how often it repeats and its spatial extent.
\( v = n\lambda \)
Here, \( n \) represents the frequency (number of waves passing a point per second) and \( \lambda \) represents the wavelength (distance between two consecutive crests or troughs).
In simple words: Wave speed is found by multiplying how many waves pass per second by the length of one wave.

🎯 Exam Tip: This equation is fundamental to all wave phenomena; make sure you know its components and what they represent.

Question 7. What is the velocity of sound in air at standard temperature and pressure?
Answer: At standard temperature and pressure (STP), the velocity of sound in air is approximately 332 meters per second. This value can vary slightly with changes in temperature and humidity. Understanding this benchmark is important for many acoustic calculations.
In simple words: Sound travels through the air at about 332 meters every second when the weather is normal.

🎯 Exam Tip: Remember this standard value for calculations, but also know that sound velocity is affected by factors like temperature (increases with temperature) and the medium itself.

Question 9. From which medium a wave is reflected so that the phase changes for a reflected wave?
Answer: When a wave is reflected from a denser medium, its phase changes by \( \pi \) radians (or 180 degrees). This means the reflected wave inverts or flips upside down compared to the incident wave. This phenomenon is critical in understanding standing waves and interference patterns.
In simple words: A wave changes its phase by 180 degrees when it bounces off something much thicker or denser than itself.

🎯 Exam Tip: A phase change of \( \pi \) (or 180°) occurs upon reflection from a denser medium (fixed end), while no phase change occurs from a rarer medium (free end).

Question 10. What will be the frequency of the beats if two tuning forks of frequency 400 and 402 are vibrated together?
Answer: When two sound waves with slightly different frequencies are sounded together, they produce beats. The frequency of these beats is equal to the absolute difference between the frequencies of the two waves. This creates a pulsating sound.
Given frequencies are \( N_1 = 400 \) Hz and \( N_2 = 402 \) Hz.
Beat frequency \( n = |N_1 - N_2| \)
\( n = |400 - 402| \)
\( n = |-2| \)
\( n = 2 \) beats per second.
In simple words: If you have two sounds that are almost the same pitch, like 400 and 402, you will hear two "beats" of sound every second.

🎯 Exam Tip: The beat frequency is always the absolute difference between the two frequencies. Ensure you take the absolute value so the frequency is always positive.

Question 11. What will be the ratio of fundamental frequencies of an open and closed organ pipes of same length?
Answer: For an open organ pipe, which is open at both ends, the fundamental frequency \( n \) is given by \( n = \frac{v}{2L} \), where \( v \) is the speed of sound and \( L \) is the length of the pipe. This pipe allows for all harmonics.
For a closed organ pipe, which is closed at one end and open at the other, the fundamental frequency \( n' \) is given by \( n' = \frac{v}{4L} \). This type of pipe only produces odd harmonics.
To find the ratio of their fundamental frequencies, we divide the frequency of the open pipe by that of the closed pipe:
\( \frac{n}{n'} = \frac{\frac{v}{2L}}{\frac{v}{4L}} \)
\( \frac{n}{n'} = \frac{v}{2L} \times \frac{4L}{v} \)
\( \frac{n}{n'} = \frac{4}{2} \)
\( \frac{n}{n'} = \frac{2}{1} \)
Thus, the ratio \( n:n' = 2:1 \). This means an open pipe of the same length produces a fundamental frequency twice that of a closed pipe.
In simple words: An open pipe makes a sound twice as high (double the frequency) as a closed pipe of the same length.

🎯 Exam Tip: Remember the fundamental frequency formulas for both open and closed organ pipes, and be careful with the lengths (open pipe has an antinode at both ends, closed pipe has a node at the closed end and an antinode at the open end).

Question 12. In which closed or open organ pipe only odd harmonics are generated?
Answer: Only in a closed organ pipe are odd harmonics generated. This is due to the boundary conditions, with a node at the closed end and an antinode at the open end, which restricts the possible wavelengths.
In simple words: Only a pipe that is closed at one end makes sounds that are only odd multiples of the basic note.

🎯 Exam Tip: An open organ pipe can produce all harmonics (fundamental, 2nd, 3rd, etc.), while a closed organ pipe can only produce odd harmonics (fundamental, 3rd, 5th, etc.).

Question 14. Is there transfer of energy by the standing waves?
Answer: No, there is no net transfer of energy by standing waves. While energy is continuously exchanged between kinetic and potential forms within the wave, it does not propagate from one point to another. Standing waves store energy in a specific region.
In simple words: Standing waves do not carry energy from one place to another; they just hold it in one spot.

🎯 Exam Tip: A key characteristic of standing waves is the absence of net energy transfer, unlike progressive waves which transport energy.

Question 15. Which waves are produced in a resonance air column?
Answer: Longitudinal standing waves are produced in a resonance air column. These waves are formed when sound waves traveling through the air column reflect off its ends and interfere with each other, creating fixed patterns of vibration. The air particles oscillate parallel to the wave direction.
In simple words: When air vibrates inside a special tube (resonance air column), it creates sound waves that stay in place, called longitudinal standing waves.

🎯 Exam Tip: Recall that sound waves are longitudinal, and in a confined space like an air column, they can form standing wave patterns at specific frequencies.

Question 16. How much is the distance between an antinode and its consecutive node?
Answer: The distance between an antinode and its consecutive node in a standing wave is exactly one-quarter of a wavelength, or \( \frac{\lambda}{4} \). An antinode is a point of maximum displacement, while a node is a point of zero displacement. This relationship is fundamental to understanding standing wave patterns.
In simple words: The space from a spot that wiggles the most to the next spot that does not wiggle at all is a quarter of one full wave.

🎯 Exam Tip: Visualize a standing wave to easily remember that the distance between adjacent nodes (or antinodes) is \( \frac{\lambda}{2} \), and between a node and an adjacent antinode is \( \frac{\lambda}{4} \).

Question 17. What is the effect of temperature on velocity of sound?
Answer: The velocity of sound in a gas increases as the temperature increases. This is because at higher temperatures, gas molecules move faster and collide more frequently, leading to quicker transmission of sound energy. The relationship is often expressed as \( u \propto \sqrt{T} \), meaning velocity is proportional to the square root of the absolute temperature. For small changes in temperature, the velocity \( u_t \) at temperature \( t \) (in Celsius) can be approximated by the formula:
\( u_t = u_0 + 0.61t \)
where \( u_0 \) is the velocity of sound at \( 0^\circ \text{C} \).
In simple words: Sound travels faster when the air is hotter. For every degree Celsius the temperature goes up, sound speed increases a little bit.

🎯 Exam Tip: Remember the direct relationship between sound velocity and the square root of absolute temperature. The given linear approximation is useful for practical calculations within a small temperature range.

Question 18. From loud or shrill sound whose pitch is more?
Answer: Shrill sounds have a higher pitch. Pitch is determined by the frequency of the sound wave; higher frequency waves produce higher pitched sounds, while lower frequency waves produce lower pitched sounds. Loudness, on the other hand, relates to the amplitude of the sound wave.
In simple words: A shrill sound has a higher pitch, meaning its sound waves vibrate more quickly.

🎯 Exam Tip: Differentiate between pitch (related to frequency) and loudness (related to amplitude) for sound characteristics.

RBSE Class 11 Physics Chapter 9 Short Answer Type Questions

Question 1. What are elastic waves?
Answer: Elastic waves are disturbances that travel through a medium where the particles, once displaced, experience a restoring force that pushes them back to their original position. This restoring force is directly related to the displacement. If a material has elasticity and its particles are set into vibration, an elastic wave will move through it. For example, sound travels through gas as an elastic wave because gas molecules can be compressed and expanded, allowing the disturbance to propagate.
In simple words: Elastic waves are waves that travel in materials that can spring back to their original shape after being pushed or pulled. Sound waves are a common type of elastic wave.

🎯 Exam Tip: The key property for elastic wave propagation is the presence of an elastic restoring force that attempts to bring displaced particles back to equilibrium.

Question 2. For effective propagation of wave what should be the properties of a medium?
Answer: For waves to spread well through a medium, it needs to have certain qualities:
1. The medium must have **elasticity**. This means if a particle is moved, a force should pull it back to its starting place, and it should then push on nearby particles to continue the wave.
2. The medium should have **inertia**. This allows it to carry the energy of the wave. Without inertia, the energy would just be lost instead of being passed on.
3. The medium should have **very low resistance**. All materials resist motion to some degree, which makes the particle's vibrations weaker over time. If this resistance is too high, the wave's energy will quickly fade, and it won't travel far. So, for effective propagation, resistance should be minimal.
In simple words: A good material for waves to travel through must be stretchy (elastic), have mass (inertia), and not slow the waves down too much (low resistance).

🎯 Exam Tip: Remember these three essential properties: elasticity for restoring force, inertia for energy storage/transfer, and low resistance for efficient propagation over distance.

Question 3. Define wave propagation constant.
Answer: The wave propagation constant, often denoted by \( k \) (also called the angular wave number), describes how the phase of a wave changes over a unit distance. It tells us how many radians of phase shift occur per meter (or other unit of length) as the wave travels. It's a crucial parameter for characterizing the spatial variation of a wave. The propagation constant is related to wavelength \( \lambda \) by \( k = \frac{2\pi}{\lambda} \).
In simple words: The wave propagation constant tells you how much the wave's pattern changes when you move a certain distance along its path.

🎯 Exam Tip: The wave propagation constant (\( k \)) is essentially a spatial frequency, similar to how angular frequency (\( \omega \)) is a temporal frequency. Both are expressed in radians per unit (distance for \( k \), time for \( \omega \)).

Question 4. Differentiate between transverse and longitudinal waves.
Answer:

🎯 Exam Tip: Remembering key examples like light (transverse) and sound (longitudinal) helps visualize and distinguish their particle motion relative to wave propagation.

Question 5. Write about the reflection of waves.
Answer: Wave reflection happens when a wave hits a boundary or an obstacle and bounces back into the same medium. The way a wave reflects depends on the type of boundary it encounters.

For **Transverse Waves**: Imagine a stretched string with one end tied to a rigid wall (a fixed end). When a wave pulse travels towards this fixed end, it exerts a force on the wall. The wall, in turn, exerts an equal and opposite reaction force on the string. This reaction force generates a new pulse that travels back along the string in the opposite direction. Crucially, the reflected pulse is inverted (flipped upside down) compared to the original incoming pulse. So, a crest reflects as a trough. This is like Figure 9.9, which shows a reflected disturbance from a rigid base.

For **Longitudinal Waves**: Consider longitudinal waves, like sound waves, traveling through a medium. When these waves encounter a boundary between two different media, they can also reflect. For instance, sound waves hitting a wall will reflect. The reflection depends on the impedance mismatch at the boundary. If the wave reflects from a denser medium, it experiences a phase change, similar to a transverse wave reflecting from a fixed end. If it reflects from a rarer medium, there is no phase change. Figure 9.13 shows a conceptual setup with light and heavy balls to illustrate reflection of longitudinal waves.
In simple words: Reflection is when a wave hits a wall or another boundary and bounces back. If the wall is stiff (like a fixed end for a string or a denser medium), the wave flips over when it comes back.

🎯 Exam Tip: The key point in reflection is the phase change: a wave reflecting from a denser medium undergoes a 180° phase change, while reflection from a rarer medium causes no phase change.

Question 6. Differentiate between progressive and standing waves with their definitions.
Answer:

🎯 Exam Tip: Focus on the energy transfer and the variation in amplitude to clearly differentiate between progressive and standing waves in your explanation.

Question 7. What do you understand by superposition of waves? Explain.
Answer: Superposition of waves happens when two or more waves travel through the same medium at the same time. When this occurs, the total displacement of any particle in the medium at any given instant is the vector sum of the displacements caused by each individual wave. This principle is fundamental to understanding phenomena like interference, diffraction, and beats.

For example, if two disturbances (waves) are moving towards each other in a stretched string:
1. If the waves are in the same phase (meaning their crests and troughs align), their effects add up, and the displacement of particles where they meet will be larger than the individual displacements. After they pass each other, each wave continues as if the other hadn't been there.
2. If the waves are in opposite phase (meaning a crest meets a trough), their effects subtract from each other. The displacement of particles where they meet will be smaller, or even zero if the amplitudes are equal. Again, after passing, they continue unchanged.

Therefore, when multiple waves act on a particle at the same time, they combine their effects on that particle's displacement. The principle states that "when more than one wave is imposed on a particle at the same time, it is called superposition of waves."
In simple words: Superposition is when two or more waves meet and combine their effects at one point. The total movement at that point is simply the sum of the movements from each wave.

🎯 Exam Tip: The principle of superposition is crucial for understanding interference patterns, where waves combine to either reinforce (constructive interference) or cancel (destructive interference) each other.

Question 8. Analyse the Laplace's correction for the wave velocity expression in gas.
Answer: Newton initially proposed that sound waves travel isothermally (at constant temperature) in a gas. However, Laplace corrected this by stating that sound propagation is an adiabatic process (no heat exchange).

**Laplace Correction:**
According to Laplace, when sound waves move through the air, they create areas of compression and rarefaction. In **compressions**, air particles get closer, which makes the temperature rise. In **rarefactions**, air particles spread out, causing the temperature to drop. This process of compression and rarefaction happens so quickly that heat does not have enough time to move in or out of the regions to equalize the temperature. Therefore, the temperature does not remain constant during sound propagation.

Since temperature is not constant, Newton's isothermal assumption was incorrect. Laplace applied the principles of adiabatic changes to derive the velocity of sound. For an adiabatic process, the relationship between pressure \( P \) and volume \( V \) is given by \( PV^\gamma = \text{constant} \), where \( \gamma \) is the adiabatic index (ratio of specific heats \( C_p/C_v \)).
From this, the bulk modulus for an adiabatic process is \( E_{adiabatic} = \gamma P \).
The velocity of sound \( v \) in a gas is given by \( v = \sqrt{\frac{E}{\rho}} \), where \( E \) is the bulk modulus and \( \rho \) is the density. Substituting \( E_{adiabatic} \):
\( v = \sqrt{\frac{\gamma P}{\rho}} \)
This is Laplace's corrected formula for the velocity of sound in a gas. This equation better matches experimental observations compared to Newton's original isothermal formula, which gave \( v = \sqrt{\frac{P}{\rho}} \). The factor \( \gamma \) accounts for the temperature variations in compressions and rarefactions, as \( \gamma \) is always greater than 1.
Here, \( \gamma = \frac{C_p}{C_v} \), where \( C_p \) is the specific heat of gas at constant pressure and \( C_v \) is the specific heat of gas at constant volume.
In simple words: Laplace corrected Newton's idea, saying sound waves make air hot and cold very fast, so no heat escapes. This means sound speed needs a special "gamma" factor in the formula to be correct.

🎯 Exam Tip: Understand that sound propagation is an adiabatic process due to rapid compressions and rarefactions, and remember Laplace's corrected formula \( v = \sqrt{\frac{\gamma P}{\rho}} \).

Question 9. Write the rules for the transverse vibrations in a stretched string?
Answer: Transverse vibrations in a stretched string are a classic example of transverse standing waves. These waves are created when a string, fixed at both ends, is plucked or vibrated. The rules governing these vibrations are:
1. **Formation of Standing Waves:** When a stretched string is disturbed, progressive transverse waves travel along it. Upon reaching the fixed ends, these waves reflect. The incident and reflected waves then superimpose to form standing waves. These standing waves persist as long as their energy isn't lost to friction or other damping forces.
2. **Nodes at Fixed Ends:** Since both ends of the string are tied, they cannot move. These fixed points always act as nodes (points of zero displacement) in the standing wave pattern.
3. **Vibration in Loops:** When the string is vibrated, it does so in one or more segments, called loops. For instance, if the string is plucked in the middle, it vibrates as a single loop, with the fixed ends as nodes and the midpoint as an antinode (point of maximum displacement). This is the simplest way to produce a standing wave, where the string vibrates in one loop.
4. **Velocity of Transverse Waves:** The velocity \( v \) of transverse waves in a stretched string is determined by the tension \( T \) in the string and its mass per unit length \( m \). The formula is \( v = \sqrt{\frac{T}{m}} \). This velocity is independent of the wave's amplitude or wavelength, assuming the string is ideal (perfectly elastic and uniform density) and its length doesn't change during vibration.
In simple words: When a string is stretched and vibrates, it makes standing waves with no movement at the ends. The wave speed depends on how tight the string is and how heavy it is per length.

🎯 Exam Tip: Remember that for a stretched string fixed at both ends, the ends are always nodes. The number of loops formed determines the harmonic or overtone.

Question 11. Define damped vibrations and maintained vibrations.
Answer: Vibrations are the back-and-forth movements of an object. These can be classified based on how their energy changes over time.

**Damped Vibrations:**
When an object is set into vibration (like plucking a guitar string) and then left alone, it initially vibrates at its natural (fundamental) frequency. However, due to resistive forces like air friction or internal friction within the medium, some of the object's energy is gradually converted into heat. As a result, the amplitude (the maximum displacement from the equilibrium position) of the vibrations slowly decreases over time until the object eventually stops moving. These types of vibrations, whose amplitude diminishes over time, are called damped vibrations. They represent a natural energy loss in any real oscillating system.

**Maintained Vibrations:**
Maintained vibrations occur when an external periodic force continuously supplies energy to a vibrating object, precisely compensating for the energy lost due to damping. If the energy supplied matches the energy lost, the object continues to vibrate with a constant (unchanging) amplitude. This type of vibration, where the amplitude is kept constant by an external energy source, is called a maintained vibration. A classic example is a clock pendulum kept swinging by a spring or weight mechanism.

**Forced Vibrations & Resonance:**
A vibrating object can also force another object to vibrate. These are called forced vibrations. If the frequency of the forcing object is equal to the natural frequency of the second object, a special condition called **resonance** occurs. At resonance, the amplitude of the forced vibrations increases significantly because every small push from the forcing frequency helps the system vibrate more strongly. This is why resonance can lead to very large oscillations, like a bridge collapsing due to strong wind at its natural frequency.
In simple words: Damped vibrations slowly lose energy and stop, like a swinging pendulum slowing down. Maintained vibrations keep a steady swing because energy is constantly added to replace what is lost.

🎯 Exam Tip: Differentiating between free (natural), damped, maintained, and forced vibrations, along with understanding resonance, is key. Remember that damping causes amplitude to decrease, while maintained vibrations sustain constant amplitude.

Question 12. If the wavelength of a wave is 2500 A then what will be the wave number?
Answer: The wave number \( k \) (also called the angular wave number) is defined as \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength. This tells us how many radians of phase change occur per unit length.
Given wavelength \( \lambda = 2500 \text{ Å} \).
First, convert the wavelength from Angstroms (Å) to meters (m):
\( 1 \text{ Å} = 10^{-10} \text{ m} \)
\( \lambda = 2500 \times 10^{-10} \text{ m} = 2.5 \times 10^3 \times 10^{-10} \text{ m} = 2.5 \times 10^{-7} \text{ m} \)
Now, calculate the wave number:
\( k = \frac{2\pi}{\lambda} \)
\( k = \frac{2 \times 3.14159}{2.5 \times 10^{-7} \text{ m}} \)
\( k = \frac{6.28318}{2.5 \times 10^{-7}} \text{ m}^{-1} \)
\( k \approx 2.513 \times 10^7 \text{ m}^{-1} \). This value indicates a large number of radians of phase change per meter, consistent with a very small wavelength.
In simple words: To find the wave number, you divide \( 2\pi \) by the wavelength of the wave, after changing the wavelength into meters.

🎯 Exam Tip: Always ensure consistent units in your calculations, especially when converting from Angstroms to meters for wavelength before calculating wave number.

Question 13. Define antinodes and nodes.
Answer: In a standing wave pattern, nodes and antinodes are specific points with distinct characteristics:
1. **Antinodes:** These are points in the medium where the particles always vibrate with the maximum possible amplitude. They are the points of greatest displacement from the equilibrium position. At antinodes, the velocity of particles (\( \frac{dy}{dt} \)) is maximum, and there is no change in pressure or density (pressure/density change is zero). For example, \( A_1, A_2, A_3 \ldots \) are antinodes. The condition \( \frac{dy}{dx} = 0 \) often describes maximum displacement at antinodes.
2. **Nodes:** These are points in the medium where the particles remain permanently at rest, meaning their amplitude of vibration is always zero. Nodes are located between consecutive antinodes and are spaced at regular intervals. At nodes, the velocity of particles (\( \frac{dy}{dt} \)) is zero, and the pressure or density change is maximum. For example, \( N_1, N_2, N_3 \ldots \) are nodes.
In simple words: Nodes are still points in a standing wave, while antinodes are the points where the wave wiggles the most.

🎯 Exam Tip: Remember that nodes have zero displacement and maximum pressure/density change, while antinodes have maximum displacement and zero pressure/density change.

Question 14. On what factors does the intensity of a wave depends upon?
Answer: The intensity of a wave, denoted by \( I \), is a measure of the power carried by the wave per unit area. It represents how much energy the wave transfers per second through a specific area. The intensity of a wave depends on several factors:
The general formula for wave intensity is \( I = 2\pi^2 n^2 a^2 \rho v \), where:
\( n \) = frequency of the wave
\( a \) = amplitude of the wave
\( \rho \) = density of the medium
\( v \) = velocity of the wave

Based on this formula, the intensity of a wave depends on:
(i) **Frequency (\( n \)):** Intensity is directly proportional to the square of the frequency (\( I \propto n^2 \)). This means a higher frequency wave carries significantly more energy.
(ii) **Amplitude (\( a \)):** Intensity is directly proportional to the square of the amplitude (\( I \propto a^2 \)). A larger amplitude means a more powerful wave.
(iii) **Density of a medium (\( \rho \)):** Intensity is directly proportional to the density of the medium (\( I \propto \rho \)). Denser mediums can transmit more energy.
(iv) **Velocity of a wave (\( v \)):** Intensity is directly proportional to the velocity of the wave (\( I \propto v \)). Faster waves also carry more energy.
In simple words: How strong a wave is (its intensity) depends on how fast it wiggles (frequency), how big it wiggles (amplitude), how heavy the material it's in is (density), and how fast it travels (velocity).

🎯 Exam Tip: The most significant dependencies are on the square of frequency and the square of amplitude, so small changes in these quantities lead to large changes in intensity.

Question 15. How can the frequency of a tuning fork be calculated by the methods of beats?
Answer: The method of beats is a very useful technique in acoustics, particularly for tuning musical instruments and determining unknown frequencies. Beats are produced when two sound waves with slightly different frequencies interfere with each other, resulting in a periodic variation in loudness.

To find the frequency of an unknown tuning fork using beats, you would use a tuning fork with a known frequency, say \( n_{known} \). You then sound both the known and the unknown tuning forks together and listen for beats. The beat frequency, \( n_{beat} \), is the number of times the loudness pulsates per second. This beat frequency is equal to the absolute difference between the two tuning forks' frequencies:
\( n_{beat} = |n_{unknown} - n_{known}| \)

This implies two possibilities for the unknown frequency:
\( n_{unknown} = n_{known} + n_{beat} \)
or
\( n_{unknown} = n_{known} - n_{beat} \)

To resolve this ambiguity, you can slightly alter the frequency of one of the tuning forks (e.g., by adding a small mass to the unknown fork, which lowers its frequency, or by filing its prongs, which raises its frequency) and observe how the beat frequency changes. If the beat frequency decreases, you know which of the two possibilities is correct. When beats are absent, it means the musical instrument is "sametone," or in unison, meaning their frequencies are identical.
In simple words: To find an unknown sound frequency, you play it with a known sound. You count the "beats" you hear, which tells you the difference in their frequencies. Then you adjust one sound a bit to figure out if the unknown frequency is higher or lower than the known one.

🎯 Exam Tip: The crucial step in using beats to find an unknown frequency is resolving the ambiguity ( \( n_{known} \pm n_{beat} \)) by slightly loading or filing one of the forks and observing the change in beat frequency.

Question 16. What are the limitations of Doppler's Effect?
Answer: Doppler's Effect explains how the perceived frequency of a wave changes when there is relative motion between the source and the observer. However, this effect has certain limitations:
1. **Relative Velocity less than Wave Velocity:** For the Doppler's effect to be accurately observed and applied, the relative velocities of the source, the observer, and the medium must be significantly less than the velocity of the wave itself (e.g., the speed of sound for sound waves).
2. **Shock Waves and Distortion:** If the velocity of the source or observer becomes comparable to or exceeds the wave velocity, the Doppler's effect model breaks down. In such cases, the wave velocity graph gets severely distorted, leading to phenomena like shock waves (e.g., a sonic boom from a jet plane exceeding the speed of sound). When a jet plane moves faster than sound, the Doppler's effect, as normally understood, is not observed because the underlying wave patterns are fundamentally changed.
In essence, the standard formulas for Doppler's effect are valid only for non-relativistic speeds, meaning speeds much lower than the wave speed.
In simple words: The Doppler's Effect only works well if the things moving (like a car making sound) are going slower than the sound itself. If they go too fast, like a supersonic jet, the effect doesn't apply in the same way because shock waves are created.

🎯 Exam Tip: Remember that the Doppler's effect formula is for speeds much less than the wave speed. When speeds are comparable or higher, other phenomena like shock waves occur.

Question 17. On what factors, does the Doppler's Effect in sound waves depend upon?
Answer: The Doppler's Effect in sound waves describes the change in observed frequency due to relative motion between the source, the observer, and the medium. The perceived frequency depends on several factors:
1. **Velocities of the source and the observer:** The speed and direction of both the sound source and the listener relative to the medium are critical. If either moves, the frequency changes.
2. **Relative velocities of the source and the observer:** It is the relative speed and direction of the source and observer that cause the frequency shift. Whether they are moving towards or away from each other dictates if the frequency is heard as higher or lower.
3. **Whether the source or/and the observer is in motion:** The effect occurs if either the source, the observer, or both are moving. Even if the relative speed is the same, the actual observed frequency can differ depending on who is moving relative to the stationary medium.
4. **Velocity of the medium:** If the medium itself (e.g., air with wind) is moving, its velocity must also be considered. This adds another layer of complexity to the relative velocities, as the source and observer velocities are then measured relative to the medium. This makes the observed frequency unique.
In simple words: The Doppler's Effect depends on how fast the sound maker is moving, how fast the listener is moving, and even how fast the air itself is moving, and in what directions.

🎯 Exam Tip: Always consider the velocities of the source, observer, and medium relative to each other, assigning positive or negative signs based on the direction of motion (e.g., towards the observer is often positive for source, away is positive for observer).

Question 18. With what velocity does source move towards the observer such that the apparent frequency become double?
Answer: When a sound source moves towards a stationary observer, the apparent (observed) frequency \( n' \) increases. The formula for the apparent frequency when the source moves towards a stationary observer is:
\( n' = n \left(\frac{v}{v - v_s}\right) \)
where \( n \) is the original frequency, \( v \) is the speed of sound in the medium, and \( v_s \) is the velocity of the source.
We want the apparent frequency to be double the original frequency, so \( n' = 2n \).
Substitute this into the formula:
\( 2n = n \left(\frac{v}{v - v_s}\right) \)
Divide both sides by \( n \):
\( 2 = \frac{v}{v - v_s} \)
Now, solve for \( v_s \):
\( 2(v - v_s) = v \)
\( 2v - 2v_s = v \)
\( 2v - v = 2v_s \)
\( v = 2v_s \)
\( v_s = \frac{v}{2} \)
This means the source must move towards the observer at half the speed of sound for the apparent frequency to be doubled. The source is effectively 'catching up' to its own sound waves enough to compress them to double the rate.
In simple words: For the sound to seem twice as high pitched, the source must move towards you at half the speed of sound.

🎯 Exam Tip: Be careful with the signs in Doppler's effect formulas. When the source moves towards the observer, the denominator \( (v - v_s) \) makes the apparent frequency higher.

RBSE Class 11 Physics Chapter 9 Long Answer Type Questions

Question 1. What do you understand by wave motion? Clearly explain. Establish the expression and the one dimensional differential equation for a progressive wave.
Answer: Wave motion is a process that transfers energy from one place to another without permanently moving the particles of the medium. It means the medium's particles vibrate but do not travel along with the wave itself. This is important because it allows energy to travel through a material without the material itself moving. Many types of waves exist, like electromagnetic waves (e.g., light, gamma rays) and mechanical waves (e.g., sound waves). Mechanical waves need a physical medium (like air or water) to travel through. When a wave moves through a medium, the particles in that medium vibrate in a simple back-and-forth motion, known as simple harmonic motion. Let's consider a progressive wave moving in the positive x-direction. The displacement equation for a particle at point P, which is at a distance \(x\) from the origin O, is: \[ y = a \sin (\omega t - \Phi) \] Here, \(a\) is the amplitude of the wave, \(\omega\) is the angular frequency, \(t\) is time, and \(\Phi\) is the phase difference. We know that the angular frequency \(\omega = \frac{2\pi}{T}\) and wave velocity \(v = \frac{\lambda}{T}\), where \(T\) is the time period and \(\lambda\) is the wavelength. The phase difference \(\Phi\) between particles at O and P is \(\frac{2\pi}{\lambda} x\). So, the displacement equation can be written as: \[ y = a \sin \left[ \frac{2\pi}{T} t - \frac{2\pi}{\lambda} x \right] \] We can substitute \(T = \frac{\lambda}{v}\) into the equation: \[ y = a \sin \frac{2\pi}{\lambda} \left( \frac{\lambda}{T} t - x \right) \]
\( \implies \) \[ y = a \sin \frac{2\pi}{\lambda} (vt-x) \] This is the equation for a progressive wave moving in the positive x-direction. For a wave moving in the negative x-direction, the equation would be \(y = a \sin (\omega t + kx)\). Now, let's find the velocity and acceleration of a particle in the medium. Differentiating the displacement equation with respect to time: \[ u = \frac{dy}{dt} = \frac{d}{dt} [a \sin (\omega t - kx)] \]
\( \implies \) \[ u = \omega a \cos (\omega t - kx) \] This shows the velocity of a particle depends on both time and its position, while the wave velocity itself is constant. The maximum particle velocity is \(\omega a\). Differentiating the particle velocity with respect to time gives acceleration: \[ \text{Acceleration} = \frac{du}{dt} = \frac{d^2y}{dt^2} = -\omega^2 a \sin (\omega t - kx) \]
\( \implies \) \[ \frac{d^2y}{dt^2} = -\omega^2 y \] This equation confirms that the particle's acceleration is directly proportional to its displacement but in the opposite direction, which is characteristic of simple harmonic motion. To establish the one-dimensional differential equation for a wave, we differentiate the displacement equation twice with respect to position \(x\): \[ \frac{d^2y}{dx^2} = -k^2 a \sin (\omega t - kx) = -k^2 y \] We know \(k = \frac{2\pi}{\lambda}\) and \(\omega = \frac{2\pi}{T}\). Then, \(\frac{k^2}{\omega^2} = \frac{(2\pi/\lambda)^2}{(2\pi/T)^2} = \frac{T^2}{\lambda^2} = \frac{1}{v^2}\). So, from \(\frac{d^2y}{dx^2} = -k^2 y\) and \(\frac{d^2y}{dt^2} = -\omega^2 y\), we can write: \[ \frac{d^2y}{dx^2} = \frac{k^2}{\omega^2} \frac{d^2y}{dt^2} \]
\( \implies \) \[ \frac{d^2y}{dx^2} = \frac{1}{v^2} \frac{d^2y}{dt^2} \]
\( \implies \) \[ \frac{d^2y}{dt^2} = v^2 \frac{d^2y}{dx^2} \] This is the one-dimensional differential equation for a progressive wave. It links the change in displacement over time to the change in displacement over distance, through the wave speed.In simple words: Wave motion is how energy moves without carrying the material itself. A wave equation helps us understand how the wave's shape changes over time and space. We can find formulas for how fast tiny parts of the material move and speed up, and also a main equation that connects how the wave moves in space and time.

🎯 Exam Tip: Remember to clearly define wave motion, state the general displacement equation, and show the steps for deriving both particle kinematics (velocity and acceleration) and the wave differential equation. Ensure proper use of MathJax for all equations and symbols.

Question 2. Derive the formula for velocity of wave in liquid.
Answer: Let's derive the formula for the velocity of transverse waves in a stretched string first, as this concept is fundamental to understanding wave propagation. For a stretched string, the velocity \(v\) of a transverse wave is determined by the tension (\(T\)) in the string and its mass per unit length (\(m\)). The formula is given by: \[ v = \sqrt{\frac{T}{m}} \] This equation shows that the wave velocity increases with greater tension and decreases if the string is heavier (more mass per unit length). An ideal string is perfectly elastic and has uniform density, and its length does not change during vibration. Now, let's consider the propagation of longitudinal waves in a liquid. Longitudinal waves, like sound waves, involve compressions (where particles are closer) and rarefactions (where particles are farther apart). Imagine a piston pushing into a tube filled with liquid. When the piston pushes, it compresses the liquid, increasing its pressure and density. This compressed layer then pushes the next layer, creating a wave of compression moving through the liquid. When the piston pulls back, it creates a rarefaction, where pressure and density decrease, and this rarefaction also travels through the liquid. To derive the wave velocity in a liquid, we consider a small element of the liquid moving through these compressions and rarefactions. Let \(\Delta P\) be the pressure difference across the compression/rarefaction, and \(\Delta v\) be the change in velocity of the liquid element. The acceleration \(a\) of a liquid element is given by \(a = -\frac{\Delta v}{\Delta t}\). From Newton's second law, the force \(F\) acting on a cross-sectional area \(A\) is \(F = A\Delta P\). The force on a liquid element (mass \(\rho u A \Delta t\)) causes its acceleration: \[ A\Delta P = (\rho u A \Delta t) \times \left(-\frac{\Delta v}{\Delta t}\right) \]
\( \implies \) \[ \Delta P = -\rho u \Delta v \] This relationship connects the pressure change to the velocity change and density. For a longitudinal wave, the wave velocity \(v\) is often used in place of \(u\). We can rewrite this in terms of volume elasticity. The bulk modulus or volume elasticity \(E\) of a liquid is defined as the ratio of stress (\(\Delta P\)) to volumetric strain (\(\frac{\Delta V}{V}\)): \[ E = \frac{\Delta P}{\left(\frac{\Delta V}{V}\right)} \] Where \(\Delta V\) is the change in volume and \(V\) is the original volume. From the wave equation, we can show a relationship where \(\rho v^2 = E\).
\( \implies \) \[ v^2 = \frac{E}{\rho} \]
\( \implies \) \[ v = \sqrt{\frac{E}{\rho}} \] This formula gives the velocity of a longitudinal wave in a liquid. It shows that wave velocity depends on the liquid's elasticity (how much it resists compression) and its density. A liquid that is harder to compress (higher \(E\)) or less dense (lower \(\rho\)) will transmit waves faster.In simple words: The speed of a wave in a liquid depends on how easily the liquid can be squeezed (its elasticity) and how heavy it is for its size (its density). If a liquid is harder to squeeze or lighter, sound waves will travel faster through it. This is similar to how a tight, light string makes faster waves.

🎯 Exam Tip: When deriving wave velocity, clearly distinguish between transverse waves in a string and longitudinal waves in a liquid. For liquids, focus on concepts of elasticity (bulk modulus) and density, explaining how compressions and rarefactions propagate.

Question 5. Derive the relationship between the amplitude of a wave and its intensity.
Answer: The intensity of a wave tells us how much energy it carries per unit area per unit time. When a wave travels, it transfers energy. The relationship between a wave's intensity and its amplitude is very important. The intensity (\(I\)) of a wave is directly proportional to the square of its amplitude (\(a\)) and also to the square of its frequency (\(n\)). It also depends on the density (\(\rho\)) of the medium and the wave's velocity (\(v\)). The general formula for the intensity of a wave is: \[ I = 2\pi^2 n^2 a^2 \rho v \] From this formula, we can clearly see the proportionality: (i) Intensity is proportional to the square of the frequency: \(I \propto n^2\) (ii) Intensity is proportional to the square of the amplitude: \(I \propto a^2\) (iii) Intensity is proportional to the density of the medium: \(I \propto \rho\) (iv) Intensity is proportional to the velocity of the wave: \(I \propto v\) This means if you double the amplitude of a wave, its intensity will become four times greater. This is why louder sounds (which have higher amplitude) carry much more energy.In simple words: The power or strength of a wave (its intensity) is directly linked to how big its "swing" is (its amplitude). If a wave swings twice as high, it carries four times more energy. It also gets stronger if it vibrates faster, or travels through a denser material.

🎯 Exam Tip: Remember that wave intensity is proportional to the square of the amplitude and the square of the frequency. This is a key relationship in wave physics and applies to various types of waves, including sound and light.

Question 6. What are standing waves? Derive the resultant wave equation for standing waves.
Answer: Standing waves, also known as stationary waves, are formed when two progressive waves with the same amplitude and frequency travel in opposite directions along the same line within a confined medium and superimpose (combine). The resulting wave appears to stand still, meaning it does not seem to move forward or backward. Unlike progressive waves that carry energy, standing waves primarily store energy within the medium. There are two main types of standing waves: 1. **Transverse standing waves:** These are produced in stretched strings, such as those in musical instruments like a sitar, violin, piano, or guitar. In these waves, the particles of the medium vibrate perpendicular to the wave's direction of travel. 2. **Longitudinal standing waves:** These are produced in air columns, for example, in wind instruments like a flute or whistle, or in percussive instruments like a tabla. Here, the particles vibrate parallel to the wave's direction of travel. Standing waves are crucial for the production of musical sounds in many instruments. They have specific points called nodes (where particle displacement is always zero) and antinodes (where particle displacement is maximum).In simple words: Standing waves happen when two waves moving in opposite directions meet and combine. They look like they are fixed in place, with some points always still (nodes) and other points swinging the most (antinodes). This type of wave doesn't carry energy forward but is important for making music in instruments.

🎯 Exam Tip: Clearly define standing waves as the superposition of two identical waves traveling in opposite directions. Be sure to mention the characteristics like nodes and antinodes. While the derivation of the resultant wave equation is often expected, if not explicitly provided in the source, focus on the definition and properties.

Question 7. What is resonance? By drawing the diagram of resonance tube calculate the formula for velocity of sound in air.
Answer: **Resonance** is a phenomenon where a vibrating object, called a driver, causes another object to vibrate with maximum amplitude. This happens when the frequency of the driving object matches the natural frequency of the second object. When this match occurs, even a small input of energy from the driver can cause large vibrations in the driven object, leading to a significant increase in amplitude. For example, if you push a swing at its natural speed, it goes higher and higher. That’s resonance. Let's discuss related concepts first: * **Free Vibrations:** When an object is displaced from its resting position and then left alone, it vibrates at its natural (fundamental) frequency due to restoring forces. This is called free vibration. In an ideal situation, the amplitude would be constant. * **Damped Vibrations:** In reality, free vibrations lose energy due to friction or air resistance, converting it into heat. As a result, the amplitude of the vibrations slowly decreases over time until the object stops. These are called damped vibrations. * **Maintained Vibrations:** If we continuously supply energy to a vibrating object to compensate for the energy lost due to damping, its amplitude can be kept constant. These are called maintained vibrations. * **Forced Vibrations:** When one vibrating object causes another to vibrate, it's called forced vibration. If the frequency of the first object matches the second object's natural frequency, then resonance occurs. **Resonance Tube:** A resonance tube is an apparatus used to find the velocity of sound in air by using the principle of resonance. It typically consists of a long glass tube (like 1 meter long, 5 cm diameter) and a water reservoir. The water level in the tube can be adjusted, changing the length of the air column above the water. This air column acts like a closed organ pipe, which means it has its own natural frequencies of vibration. When a vibrating tuning fork (the driver) is held near the open end of the tube, it forces the air column inside to vibrate. By slowly changing the water level, we can adjust the length of the air column. When the air column's natural frequency matches the tuning fork's frequency, resonance occurs. At this point, the air column vibrates with maximum amplitude, and a loud, clear (shrill) sound is heard. The length of the air column at which this happens is called the resonant length. By finding the first and second resonant lengths (\(l_1\) and \(l_2\)), and knowing the frequency of the tuning fork (\(n\)), we can calculate the velocity of sound (\(v\)) in air. The formula for the velocity of sound in air using a resonance tube (considering the end correction \(e\)) is: \[ v = 2n (l_2 - l_1) \] The end correction accounts for the fact that the antinode does not form exactly at the open end of the tube but slightly above it.In simple words: Resonance is when one vibrating object makes another object vibrate very strongly, because their natural speeds of vibration match. A resonance tube helps us measure the speed of sound by making an air column inside it vibrate loudly when it matches the sound from a tuning fork.

🎯 Exam Tip: Define resonance clearly, distinguishing it from damped and forced vibrations. Explain how a resonance tube works to find resonant lengths. Ensure to present the formula for the velocity of sound derived from resonance tube experiments, even if the source description focuses more on the setup.

Question 8. What are beats? By mathematical analysis prove that the number of beats per second is equal to the difference of frequencies of the sources.
Answer: **Beats** are formed when two sound waves with slightly different frequencies travel in the same direction and combine (superimpose). Because their frequencies are not exactly the same, their phase relationship changes over time. Sometimes the waves are in phase, making a louder sound (constructive interference), and sometimes they are out of phase, making a softer sound or no sound at all (destructive interference). This causes the intensity of the sound to rise and fall periodically. Each rise and fall in intensity is called a beat. This phenomenon is important in music and for tuning instruments. Let's understand this with an example: Imagine two waves, one vibrating 5 times in one second and another vibrating 4 times in one second. When these two waves combine: * At some moments, their crests and troughs align, making the sound louder. * At other moments, a crest from one wave aligns with a trough from the other, making the sound softer or silent. Over a period of one second, these alignment changes create a pattern of loudness and softness. If one wave makes 5 vibrations and the other makes 4 vibrations in a second, their difference is 1 vibration. This means they will align and misalign to produce one full beat (one cycle of loud-soft-loud) in that one second. The number of beats heard per second is always equal to the absolute difference between the frequencies of the two sources. For example, if two tuning forks have frequencies \(n_1\) and \(n_2\), the beat frequency (\(n_{\text{beat}}\)) is given by: \[ n_{\text{beat}} = |n_1 - n_2| \] This principle helps in tuning musical instruments: when two instruments are played together, and beats are heard, the instruments are slightly out of tune. As they are brought closer to being in tune, the beat frequency decreases, and when no beats are heard, they are perfectly in tune (their frequencies match).In simple words: Beats are like a "wobble" or "pulse" in sound when two notes that are almost but not quite the same are played together. The sound gets loud, then soft, then loud again. The number of these "wobbles" you hear in one second is exactly the difference between the two sound frequencies.

🎯 Exam Tip: Clearly define beats and explain how they arise from the superposition of waves with slightly different frequencies. State the formula \(n_{\text{beat}} = |n_1 - n_2|\) and explain its significance, particularly for applications like tuning instruments.

Question 9. Describe Doppler's effect for sound waves and calculate the formula for experienced frequency when : (a) Source is moving towards the stationary observer. (b) Observer is moving towards the stationary source.
Answer: **Doppler's Effect** is a change in the observed frequency (or pitch) of a wave when the source of the wave or the observer (or both) are moving relative to the medium. This effect is not just for sound waves but also for electromagnetic waves like light. For example, a car horn sounds higher pitched as it approaches you and lower pitched as it moves away. For the Doppler effect to be noticeable, the speeds of the source and observer must be less than the speed of the wave in the medium. If their speeds are equal to or greater than the wave speed, the wave pattern can become distorted, creating shock waves (like a sonic boom). One important application of Doppler's effect for sound waves is in calculating the velocity of submarines using **SONAR** (Sound Navigation and Ranging). Sound waves are sent from a station (e.g., on a seashore) towards a target (like a submarine). These waves reflect off the submarine and return to the station. By analyzing the change in the wavelength (or frequency) of the reflected waves, the submarine's velocity can be determined. Let's consider the change in wavelength when the submarine (acting as a source of reflected waves) is moving. If the submarine is moving with velocity \(u_s\) and the speed of sound in water is \(v\), the observed wavelength (\(\lambda'\)) will change from the original wavelength (\(\lambda\)). The general formula for the change in wavelength (\(\Delta\lambda\)) is: \[ \lambda' = \frac{v \mp u_s}{n} = \frac{v \mp u_s}{v/\lambda} = \left( 1 \mp \frac{u_s}{v} \right) \lambda \]
\( \implies \) \[ \lambda' - \lambda = \mp \frac{u_s}{v} \lambda \] So, the change in wavelength is: \[ \Delta\lambda = \lambda' - \lambda = \pm \frac{u_s}{v} \lambda \] Here, the negative sign (\(-\frac{u_s}{v}\)) is used when the source (submarine) is moving towards the observer (SONAR station), causing the observed wavelength to decrease. The positive sign (\(+\frac{u_s}{v}\)) is used when the source is moving away, causing the observed wavelength to increase. Using this formula, the velocity of the submarine (\(u_s\)) can be calculated, and its position relative to the SONAR station can also be determined.In simple words: Doppler's effect is when a sound changes pitch (or frequency) if the thing making the sound or the person hearing it is moving. For example, a car horn sounds different when it comes towards you compared to when it moves away. This effect helps calculate how fast things like submarines are moving by looking at how their reflected sound waves change.

🎯 Exam Tip: Define Doppler's effect clearly, mentioning relative motion between source and observer. While the question asks for apparent frequency formulas for specific cases (a) and (b), the provided answer focuses on wavelength change for submarines. If asked for general frequency formulas, remember to use the standard formulas for approaching/receding sources and observers.

Question 3. The equation of a wave propagating on a string is as follows: \( y = 10 \sin \pi(0.01x - 2.00t) \) where y and x are in cm and t is in second. Calculate the amplitude, frequency and velocity of the wave. Calculate the phase difference between the two particles at a distance 40.0 cm at any instant.
Answer: The given wave equation is \( y = 10 \sin \pi(0.01x - 2.00t) \). This can be written as \( y = 10 \sin(0.01\pi x - 2\pi t) \). Comparing this to the standard wave equation \( y = A \sin(kx - \omega t) \):
Amplitude, \( A = 10 \) cm.
Wave number, \( k = 0.01\pi \) rad/cm.
Angular frequency, \( \omega = 2\pi \) rad/s.

We can find the frequency \( n \) using \( \omega = 2\pi n \):
\( 2\pi = 2\pi n \implies n = 1 \) Hz.

We can find the wavelength \( \lambda \) using \( k = \frac{2\pi}{\lambda} \):
\( 0.01\pi = \frac{2\pi}{\lambda} \implies \lambda = \frac{2}{0.01} = 200 \) cm.

The wave velocity \( v \) is calculated as \( v = n\lambda \):
\( v = (1 \text{ Hz})(200 \text{ cm}) = 200 \) cm/s.

Now, we calculate the phase difference \( \Delta \phi \) between two particles at a distance \( \Delta x = 40.0 \) cm. The formula for phase difference is \( \Delta \phi = k \Delta x \):
\( \Delta \phi = (0.01\pi \text{ rad/cm})(40.0 \text{ cm}) = 0.4\pi \) rad.
This is equal to \( \frac{2\pi}{5} \) radians, which is \( 72^\circ \). Understanding wave characteristics is essential for analyzing wave behavior in various mediums.
In simple words: We found out how big the wave is (10 cm), how often it vibrates (1 Hz), and how fast it travels (200 cm/s). We also calculated how "out of step" two points on the wave are at a certain distance, which is 72 degrees.

🎯 Exam Tip: Remember to clearly state the units for each calculated quantity and to convert all given values to a consistent system (like SI units) before starting calculations.

Question 4. The fundamental frequency of 1.0 meter long stretched steel wire is 250 Hz. Density of Steel is 8000 kg/m³.
(i) Calculate the speed of a transverse wave in the wire.
(ii) Calculate the longitudinal stress of the wire.
(iii) If the tension of the wire is increased by 2% then calculate the percentage change in the frequency.
Answer: Given values are:
Length of wire \( l = 1.0 \) m
Fundamental frequency \( n = 250 \) Hz
Density of steel \( \rho = 8000 \) kg/m\(^3\)

(i) To calculate the speed of a transverse wave in the wire:
For the fundamental frequency of a stretched wire, the length of the wire is half the wavelength: \( l = \frac{\lambda}{2} \).
So, wavelength \( \lambda = 2l = 2 \times 1.0 \text{ m} = 2 \) m.
The speed of the transverse wave \( v \) is given by \( v = n\lambda \):
\( v = (250 \text{ Hz})(2 \text{ m}) = 500 \) m/s.

(ii) To calculate the longitudinal stress of the wire:
The speed of a transverse wave in a string is also given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is tension and \( \mu \) is mass per unit length.
We know \( \mu = \text{Area} \times \rho \). So, \( v = \sqrt{\frac{T}{\text{Area} \times \rho}} \).
Squaring both sides gives \( v^2 = \frac{T}{\text{Area} \times \rho} \).
Longitudinal stress is defined as \( \text{Stress} = \frac{T}{\text{Area}} \).
Therefore, \( \text{Stress} = v^2 \rho \).
\( \text{Stress} = (500 \text{ m/s})^2 \times (8000 \text{ kg/m}^3) \)
\( \text{Stress} = 250000 \times 8000 = 2 \times 10^9 \) N/m\(^2\).

(iii) To calculate the percentage change in frequency if tension increases by 2%:
The frequency \( n \) of a stretched wire is proportional to the square root of its tension \( T \) (assuming length and mass per unit length are constant), i.e., \( n \propto \sqrt{T} \).
If the initial tension is \( T_1 \) and the new tension is \( T_2 \), where \( T_2 = T_1 + 0.02T_1 = 1.02 T_1 \).
The ratio of new frequency \( n_2 \) to initial frequency \( n_1 \) is: \( \frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{1.02 T_1}{T_1}} = \sqrt{1.02} \).
Using the binomial approximation \( (1+x)^p \approx 1+px \) for small \( x \):
\( \sqrt{1.02} = (1+0.02)^{1/2} \approx 1 + \frac{1}{2}(0.02) = 1 + 0.01 = 1.01 \).
So, \( \frac{n_2}{n_1} = 1.01 \).
The percentage change in frequency \( = \left(\frac{n_2 - n_1}{n_1}\right) \times 100\% = \left(\frac{n_2}{n_1} - 1\right) \times 100\% \).
\( = (1.01 - 1) \times 100\% = 0.01 \times 100\% = 1\% \). This small change shows how frequency is quite sensitive to tension in a string.
In simple words: First, we found the wave speed in the wire to be 500 m/s. Then, we calculated the pulling force per unit area, called stress, which was \( 2 \times 10^9 \) N/m\(^2\). Finally, if the pulling force on the wire increases by 2%, the wire's vibration frequency will go up by 1%.

🎯 Exam Tip: Remember that for fundamental mode, wavelength is twice the length of the string. Also, tension directly affects wave speed, and thus frequency, so understanding these relationships is key.

Question 5. Calculate the wavelength of a longitudinal wave vibrating with 400 vibrations/second in a metal of density 5.5 x 10³ kg/m³. The Young's modulus of metal is Y = 8.8 × 10¹⁰N/m².
Answer: Given values are:
Frequency of vibration \( n = 400 \) Hz
Density of metal \( \rho = 5.5 \times 10^3 \) kg/m\(^3\)
Young's modulus \( Y = 8.8 \times 10^{10} \) N/m\(^2\)

First, we calculate the velocity \( v \) of a longitudinal wave in a solid using the formula:
\( v = \sqrt{\frac{Y}{\rho}} \)
\( v = \sqrt{\frac{8.8 \times 10^{10} \text{ N/m}^2}{5.5 \times 10^3 \text{ kg/m}^3}} \)
\( v = \sqrt{\frac{8.8}{5.5} \times 10^{10-3}} = \sqrt{1.6 \times 10^7} = \sqrt{16 \times 10^6} \)
\( v = 4 \times 10^3 \) m/s \( = 4000 \) m/s.

Next, we calculate the wavelength \( \lambda \) using the wave speed formula \( v = n\lambda \):
\( \lambda = \frac{v}{n} \)
\( \lambda = \frac{4000 \text{ m/s}}{400 \text{ Hz}} \)
\( \lambda = 10 \) m. This wavelength represents the spatial extent of one complete cycle of the wave in the metal.
In simple words: We first found how fast the sound travels through the metal, which was 4000 meters per second. Then, using that speed and how often the wave vibrates (400 times a second), we calculated that each wave cycle is 10 meters long.

🎯 Exam Tip: Remember the specific formulas for wave velocity in different mediums (solids, liquids, gases) and how Young's modulus relates to elasticity in solids.

Question 6. Calculate the wavelength, wave number and frequency of a wave whose propagation constant is 2.8 × 10⁴ per metre and its velocity is 400 m/s.
Answer: Given values are:
Propagation constant (wave number) \( k = 2.8 \times 10^4 \) m\(^{-1}\)
Velocity of wave \( v = 400 \) m/s

(i) To calculate the wavelength \( \lambda \):
The propagation constant \( k \) is related to wavelength \( \lambda \) by \( k = \frac{2\pi}{\lambda} \):
\( \lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{2.8 \times 10^4 \text{ m}^{-1}} \)
\( \lambda = \frac{6.28}{2.8 \times 10^4} = 2.2428 \times 10^{-4} \approx 2.24 \times 10^{-4} \) m.

(ii) To calculate the frequency \( n \):
The frequency \( n \) is related to wave velocity \( v \) and wavelength \( \lambda \) by \( v = n\lambda \):
\( n = \frac{v}{\lambda} = \frac{400 \text{ m/s}}{2.24 \times 10^{-4} \text{ m}} \)
\( n = \frac{400}{2.24} \times 10^4 \approx 178.57 \times 10^4 = 1.7857 \times 10^6 \approx 1.78 \times 10^6 \) Hz.

(iii) To calculate the wave number \( \bar{v} \):
The wave number \( \bar{v} \) is also defined as the reciprocal of the wavelength: \( \bar{v} = \frac{1}{\lambda} \). (Note: sometimes \( k \) is also referred to as wave number, but here a distinct symbol is used for clarity.)
\( \bar{v} = \frac{1}{2.24 \times 10^{-4} \text{ m}} \approx 0.4464 \times 10^4 = 4.46 \times 10^3 \) m\(^{-1}\). Wave number tells us how many wave cycles fit into a unit of length.
In simple words: We found the length of one wave cycle, which is about 0.000224 meters. The wave vibrates about 1.78 million times every second. Finally, we calculated that about 4460 wave cycles fit into every meter.

🎯 Exam Tip: Distinguish between propagation constant (\( k = \frac{2\pi}{\lambda} \)) and wave number (\( \bar{v} = \frac{1}{\lambda} \)), as both terms are sometimes used interchangeably for \( k \) in different contexts, but are distinct. Pay attention to units.

Question 7. Calculate the frequency, wave number and propagation constant of a light wave of wavelength 5000 Å.
Answer: Given values are:
Wavelength \( \lambda = 5000 \) Å = \( 5000 \times 10^{-10} \) m = \( 5 \times 10^{-7} \) m.
For a light wave, the velocity is the speed of light \( c = 3 \times 10^8 \) m/s.

(i) To calculate the frequency \( n \):
Using the formula \( c = n\lambda \):
\( n = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{5 \times 10^{-7} \text{ m}} = \frac{3}{5} \times 10^{15} = 0.6 \times 10^{15} = 6 \times 10^{14} \) Hz.

(ii) To calculate the wave number \( \bar{v} \):
The wave number \( \bar{v} \) is the reciprocal of the wavelength:
\( \bar{v} = \frac{1}{\lambda} = \frac{1}{5 \times 10^{-7} \text{ m}} = 0.2 \times 10^7 = 2 \times 10^6 \) m\(^{-1}\).

(iii) To calculate the propagation constant \( k \):
The propagation constant \( k \) is given by \( k = \frac{2\pi}{\lambda} \):
\( k = \frac{2\pi}{5 \times 10^{-7} \text{ m}} = (2 \times 3.14) \times 0.2 \times 10^7 = 1.256 \times 10^7 \) m\(^{-1}\). The propagation constant describes how a wave's phase changes over distance.
In simple words: For a light wave 5000 Angstroms long, we found it vibrates \( 6 \times 10^{14} \) times each second. It has \( 2 \times 10^6 \) wave cycles per meter, and its propagation constant is about \( 1.256 \times 10^7 \) per meter, telling us how its phase changes through space.

🎯 Exam Tip: Always remember the speed of light (\( c \)) for calculations involving light waves unless specified otherwise. Distinguish between frequency (temporal) and wave number/propagation constant (spatial) properties of a wave.

Question 8. A simple harmonic wave is moving towards the positive ac-direction with 100 m/s velocity. Amplitude of wave is 22 cm and the frequency is 100 vibrations/second (Hz). Calculate the displacement, velocity and acceleration of a particle which is at a distance x = 2 m from the origin at t = 5 seconds.
Answer: Given values are:
Wave velocity \( u = 100 \) m/s
Amplitude \( a = 22 \) cm = \( 0.22 \) m
Frequency \( n = 100 \) Hz
Particle position \( x = 2 \) m
Time \( t = 5 \) s

First, calculate the wavelength \( \lambda \):
\( \lambda = \frac{u}{n} = \frac{100 \text{ m/s}}{100 \text{ Hz}} = 1 \) m.

Calculate the angular frequency \( \omega \):
\( \omega = 2\pi n = 2\pi (100) = 200\pi \) rad/s.

Calculate the wave number \( k \):
\( k = \frac{2\pi}{\lambda} = \frac{2\pi}{1} = 2\pi \) rad/m.

The general equation for a progressive wave moving in the positive x-direction is \( y = a \sin(\omega t - kx) \).
Substituting the values: \( y = 0.22 \sin(200\pi t - 2\pi x) \).

(i) To calculate the displacement \( y \) at \( x = 2 \) m and \( t = 5 \) s:
\( y = 0.22 \sin(200\pi (5) - 2\pi (2)) \)
\( y = 0.22 \sin(1000\pi - 4\pi) = 0.22 \sin(996\pi) \).
Since \( \sin(N\pi) = 0 \) for any integer N, \( \sin(996\pi) = 0 \).
Therefore, displacement \( y = 0.22 \times 0 = 0 \) m.

(ii) To calculate the velocity of the particle \( v_p \) at \( x = 2 \) m and \( t = 5 \) s:
The particle velocity \( v_p = \frac{dy}{dt} = a\omega \cos(\omega t - kx) \).
\( v_p = (0.22)(200\pi) \cos(200\pi t - 2\pi x) = 44\pi \cos(200\pi t - 2\pi x) \).
At \( x = 2 \) m, \( t = 5 \) s:
\( v_p = 44\pi \cos(1000\pi - 4\pi) = 44\pi \cos(996\pi) \).
Since \( \cos(N\pi) = 1 \) for an even integer N, \( \cos(996\pi) = 1 \).
Therefore, \( v_p = 44\pi \times 1 = 44\pi \) m/s. (Approximately \( 138.16 \) m/s).

(iii) To calculate the acceleration of the particle \( A_p \) at \( x = 2 \) m and \( t = 5 \) s:
The particle acceleration \( A_p = \frac{dv_p}{dt} = -a\omega^2 \sin(\omega t - kx) \).
\( A_p = -(0.22)(200\pi)^2 \sin(200\pi t - 2\pi x) \).
At \( x = 2 \) m, \( t = 5 \) s:
\( A_p = -(0.22)(200\pi)^2 \sin(996\pi) \).
Since \( \sin(996\pi) = 0 \).
Therefore, \( A_p = -(0.22)(200\pi)^2 \times 0 = 0 \) m/s\(^2\). At an integer multiple of the period, the particle returns to its equilibrium position with maximum velocity.
In simple words: For a particle on this wave at the given time and place, its position is exactly at the middle (zero displacement). It is moving very fast at that moment, about 138 meters per second. However, its acceleration is zero at this point.

🎯 Exam Tip: When dealing with wave equations, remember that displacement, particle velocity, and particle acceleration are functions of both position and time. Pay close attention to the sine and cosine values at multiples of \( \pi \).

Question 9. Calculate the frequency of the fundamental tone of a 0.50 m long wire stretched by a weight of 10 kg where weight of 1 m long wire is 2.45 g. (g= 980 cm/s²)
Answer: Given values are:
Length of wire \( l = 0.50 \) m
Mass of load \( M = 10 \) kg
Mass per unit length \( \mu = 2.45 \) g/m \( = 2.45 \times 10^{-3} \) kg/m
Acceleration due to gravity \( g = 980 \) cm/s\(^2\) \( = 9.8 \) m/s\(^2\)

First, calculate the tension \( T \) in the wire:
\( T = Mg = (10 \text{ kg})(9.8 \text{ m/s}^2) = 98 \) N.

The frequency \( n \) of the fundamental tone of a stretched wire is given by the formula:
\( n = \frac{1}{2l} \sqrt{\frac{T}{\mu}} \)
\( n = \frac{1}{2 \times 0.50 \text{ m}} \sqrt{\frac{98 \text{ N}}{2.45 \times 10^{-3} \text{ kg/m}}} \)
\( n = \frac{1}{1} \sqrt{\frac{98 \times 10^3}{2.45}} \)
\( n = \sqrt{40 \times 10^3} = \sqrt{40000} \)
\( n = 200 \) Hz. This fundamental frequency is the lowest possible vibration mode for the wire.
In simple words: We calculated the pulling force (tension) on the wire. Then, using the wire's length, its weight per meter, and the pulling force, we found that the wire vibrates at a basic frequency of 200 times per second.

🎯 Exam Tip: Ensure consistent units (SI is preferred) for all quantities. Remember that 'weight of 1 m long wire' directly gives the linear mass density \( \mu \).

Question 10. Calculate the frequency of the fundamental tone of a 100 cm long wire with 1.8 mm diameter (density 8.4) and stretched by 20 kg weight.
Answer: Given values are:
Length of wire \( l = 100 \) cm = \( 1.00 \) m
Diameter \( D = 1.8 \) mm \( \implies \) Radius \( r = 0.9 \) mm = \( 0.9 \times 10^{-3} \) m \( = 9 \times 10^{-4} \) m
Density \( \rho = 8.4 \times 10^3 \) kg/m\(^3\) (assuming 8.4 g/cm\(^3\) as this is a common density for metals like iron/steel, converted to SI)
Mass of load \( M = 20 \) kg
Assuming \( g = 9.8 \) m/s\(^2\)

First, calculate the tension \( T \) in the wire:
\( T = Mg = (20 \text{ kg})(9.8 \text{ m/s}^2) = 196 \) N.

Next, calculate the mass per unit length \( \mu \) of the wire:
\( \mu = \text{Area} \times \text{Density} = (\pi r^2) \times \rho \)
\( \mu = \pi (9 \times 10^{-4} \text{ m})^2 \times (8.4 \times 10^3 \text{ kg/m}^3) \)
\( \mu = 3.14159 \times (81 \times 10^{-8}) \times 8.4 \times 10^3 \)
\( \mu = 2136.456 \times 10^{-5} \approx 0.021385 \) kg/m.

Finally, calculate the frequency \( n \) of the fundamental tone:
\( n = \frac{1}{2l} \sqrt{\frac{T}{\mu}} \)
\( n = \frac{1}{2 \times 1.00 \text{ m}} \sqrt{\frac{196 \text{ N}}{0.021385 \text{ kg/m}}} \)
\( n = \frac{1}{2} \sqrt{9165.20} = \frac{1}{2} \times 95.735 \approx 47.8675 \) Hz.
Rounding to the nearest whole number, \( n = 48 \) Hz. This is the lowest natural frequency at which the wire can vibrate, producing a clear musical tone.
In simple words: We first found the pulling force on the wire. Then we calculated how heavy the wire is per unit length. Using these values, along with the wire's total length, we found that its basic vibration speed is about 48 times per second.

🎯 Exam Tip: For problems involving wire frequency, ensure correct calculation of linear mass density \( \mu \), which depends on both the wire's cross-sectional area and material density.

Question 11. A 25 cm stretched wire and a tuning fork has same frequency. If the length of wire is changed to 25.5 cm
Answer: Let the frequency of the tuning fork be \( n \).
Initially, the length of the stretched wire is \( l_1 = 25 \) cm.
Since the wire and tuning fork have the same frequency, the initial frequency of the wire is \( n_1 = n \).
When the length of the wire is changed to \( l_2 = 25.5 \) cm, its frequency changes to \( n_2 \).
For a stretched wire, frequency \( n \) is inversely proportional to its length \( l \), i.e., \( n \propto \frac{1}{l} \).
So, \( \frac{n_1}{n_2} = \frac{l_2}{l_1} \).
We are given that when the length is changed, the tuning fork produces 3 beats/s (implied from OCR solution, using \( n-3 \) as the new wire frequency). This means \( |n - n_2| = 3 \).
Since \( l_2 > l_1 \), it implies \( n_2 < n_1 \). Therefore, the new wire frequency \( n_2 \) is \( n-3 \).
Substituting the values into the ratio equation:
\( \frac{n}{n-3} = \frac{25.5 \text{ cm}}{25 \text{ cm}} \)
\( \frac{n}{n-3} = \frac{51}{50} \)
Now, cross-multiply to solve for \( n \):
\( 50n = 51(n-3) \)
\( 50n = 51n - 153 \)
\( 51n - 50n = 153 \)
\( n = 153 \) Hz. This tuning fork frequency sets the base vibration for the wire, changing its sound as the wire length changes.
In simple words: The tuning fork vibrates 153 times per second. When the wire's length was slightly increased, its vibration rate dropped slightly, causing it to make 3 beats per second with the tuning fork.

🎯 Exam Tip: When dealing with beats, remember that the observed beat frequency is the absolute difference between the two interacting frequencies. For a string, frequency is inversely proportional to length.

Question 12. A sonometer wire gives sound of frequency 150 Hz. If wire's tension is changed in the ratio 9 : 16 and length in the ratio 1 : 2. Calculate the frequency of the new wire.
Answer: Given values are:
Initial frequency \( n_1 = 150 \) Hz.
Ratio of tensions \( \frac{T_2}{T_1} = \frac{16}{9} \).
Ratio of lengths \( \frac{l_2}{l_1} = \frac{2}{1} \).
We assume the mass per unit length \( \mu \) of the wire remains constant.

The formula for the frequency \( n \) of a sonometer wire is:
\( n = \frac{1}{2l} \sqrt{\frac{T}{\mu}} \)
To find the new frequency \( n_2 \), we can set up a ratio with the initial frequency \( n_1 \):
\( \frac{n_2}{n_1} = \frac{\frac{1}{2l_2} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2l_1} \sqrt{\frac{T_1}{\mu}}} \)
\( \frac{n_2}{n_1} = \left(\frac{l_1}{l_2}\right) \sqrt{\frac{T_2}{T_1}} \)
Substitute the given ratios:
\( \frac{n_2}{n_1} = \left(\frac{1}{2}\right) \sqrt{\frac{16}{9}} \)
\( \frac{n_2}{n_1} = \frac{1}{2} \times \frac{4}{3} = \frac{4}{6} = \frac{2}{3} \).
Now, calculate the new frequency \( n_2 \):
\( n_2 = \frac{2}{3} n_1 = \frac{2}{3} \times 150 \text{ Hz} \)
\( n_2 = 2 \times 50 = 100 \) Hz. The frequency changes because both the length and tension affect how the wire vibrates.
In simple words: The wire's frequency was 150 Hz. When its pulling force was increased and its length was doubled, the new frequency became 100 Hz.

🎯 Exam Tip: Remember that frequency is inversely proportional to length and directly proportional to the square root of tension for a stretched string. Be careful with ratios and square roots in calculations.

Question 13. Equation of transverse vibrations of a string tied at both the ends is \( y = (x, t) = 0.3\sin \left[ \frac{2\pi}{3} x \right] \cos (120 \pi t) \). Where y and x are in m and t in s. Length of the wire is 1.5 m and mass 0.002 kg, then,
(i) Calculate the maximum displacement at x = 0.5 m.
(ii) Mark the places of nodes on the wire.
(iii) Calculate the wave velocity.
(iv) Calculate the velocity of the particle at x = 0.75 m and t = 0.25 s.
Answer: The given wave equation is \( y = 0.3\sin \left( \frac{2\pi}{3} x \right) \cos (120 \pi t) \).
This is a standing wave equation of the form \( y = A_0 \sin(kx) \cos(\omega t) \), where:
Overall Amplitude \( A_0 = 0.3 \) m
Wave number \( k = \frac{2\pi}{3} \) rad/m
Angular frequency \( \omega = 120\pi \) rad/s
Length of the wire \( L = 1.5 \) m

(i) To calculate the maximum displacement (amplitude) at \( x = 0.5 \) m:
The amplitude of a particle at position \( x \) in a standing wave is \( A(x) = A_0 \sin(kx) \).
\( A(0.5) = 0.3 \sin \left( \frac{2\pi}{3} \times 0.5 \right) \)
\( A(0.5) = 0.3 \sin \left( \frac{\pi}{3} \right) \)
Since \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \):
\( A(0.5) = 0.3 \times \frac{\sqrt{3}}{2} \approx 0.3 \times \frac{1.732}{2} = 0.3 \times 0.866 = 0.2598 \) m.
Thus, the maximum displacement at \( x = 0.5 \) m is approximately \( 0.26 \) m.

(ii) To mark the places of nodes on the wire:
Nodes are points where the amplitude is always zero, meaning \( \sin(kx) = 0 \).
\( \sin \left( \frac{2\pi}{3} x \right) = 0 \).
This implies \( \frac{2\pi}{3} x = n\pi \), where \( n \) is an integer (0, 1, 2, ...).
\( x = \frac{3n}{2} \).
Considering the length of the wire \( L = 1.5 \) m:
For \( n = 0, x = \frac{3 \times 0}{2} = 0 \) m.
For \( n = 1, x = \frac{3 \times 1}{2} = 1.5 \) m.
For \( n = 2, x = \frac{3 \times 2}{2} = 3 \) m (This is outside the wire length).
So, the nodes on the wire are located at \( x = 0 \) m and \( x = 1.5 \) m. These are the fixed points where the string does not move.

(iii) To calculate the wave velocity \( v \):
From the wave number \( k = \frac{2\pi}{3} \), we find the wavelength \( \lambda \):
\( \frac{2\pi}{\lambda} = \frac{2\pi}{3} \implies \lambda = 3 \) m.
From the angular frequency \( \omega = 120\pi \), we find the frequency \( n \):
\( 2\pi n = 120\pi \implies n = \frac{120\pi}{2\pi} = 60 \) Hz.
The wave velocity \( v \) is given by \( v = n\lambda \):
\( v = (60 \text{ Hz})(3 \text{ m}) = 180 \) m/s. This is the speed at which the constituent progressive waves travel to form the standing wave.

(iv) To calculate the velocity of the particle \( v_p \) at \( x = 0.75 \) m and \( t = 0.25 \) s:
The velocity of a particle in a standing wave is \( v_p = \frac{\partial y}{\partial t} \).
\( v_p = \frac{\partial}{\partial t} [A_0 \sin(kx) \cos(\omega t)] = -A_0 \omega \sin(kx) \sin(\omega t) \).
\( v_p = -0.3 \times (120\pi) \sin\left(\frac{2\pi}{3} x\right) \sin(120\pi t) \).
\( v_p = -36\pi \sin\left(\frac{2\pi}{3} x\right) \sin(120\pi t) \).
Substitute \( x = 0.75 \) m and \( t = 0.25 \) s:
\( \frac{2\pi}{3} x = \frac{2\pi}{3} \times 0.75 = \frac{2\pi}{3} \times \frac{3}{4} = \frac{\pi}{2} \).
\( 120\pi t = 120\pi \times 0.25 = 120\pi \times \frac{1}{4} = 30\pi \).
So, \( v_p = -36\pi \sin\left(\frac{\pi}{2}\right) \sin(30\pi) \).
Since \( \sin(\frac{\pi}{2}) = 1 \) and \( \sin(30\pi) = 0 \) (as 30 is an even integer).
\( v_p = -36\pi \times 1 \times 0 = 0 \) m/s.
In simple words: At 0.5 meters from the start, the biggest a particle can move is about 0.26 meters. The string is tied at 0 and 1.5 meters, so these are the fixed points. The wave itself travels at 180 meters per second. At 0.75 meters and after 0.25 seconds, a particle on the string is momentarily still.

🎯 Exam Tip: For standing waves, remember that particle velocity and acceleration are functions of both position and time. Nodes are fixed points, while antinodes have maximum displacement and particle velocity at specific times.

Question 14. 41 tuning forks are so arranged that every tuning fork generates 5 beat/s to its nearest tuning fork. The frequency of the last tuning fork is double the frequency of first tuning fork. Calculate the frequency of first and last tuning fork.
Answer: Let the frequency of the first tuning fork be \( f_1 \).
There are 41 tuning forks in total, meaning there are \( 41 - 1 = 40 \) intervals between the forks.
Each tuning fork generates 5 beats/s with its nearest neighbor. This indicates that the frequencies of the tuning forks form an arithmetic progression.
So, the difference between consecutive frequencies is 5 Hz.
The frequency of the last tuning fork, \( f_{41} \), can be expressed as:
\( f_{41} = f_1 + (41-1) \times 5 = f_1 + 40 \times 5 = f_1 + 200 \).

We are given that the frequency of the last tuning fork is double the frequency of the first tuning fork:
\( f_{41} = 2 f_1 \).

Now, we can set up an equation by equating the two expressions for \( f_{41} \):
\( f_1 + 200 = 2 f_1 \)
Subtract \( f_1 \) from both sides:
\( 200 = 2 f_1 - f_1 \)
\( f_1 = 200 \) Hz.

Now, calculate the frequency of the last tuning fork \( f_{41} \):
\( f_{41} = 2 f_1 = 2 \times 200 = 400 \) Hz.
This shows a clear progression in frequency across the series of tuning forks.
In simple words: We have 41 tuning forks, and each one is 5 Hz different from the next. The last fork vibrates twice as fast as the first one. We found that the first fork vibrates 200 times per second, and the last fork vibrates 400 times per second.

🎯 Exam Tip: When beat frequencies are constant between successive elements in a series, it implies an arithmetic progression of frequencies. Clearly define variables and set up equations based on the given relationships.

Question 15. Calculate the velocity of sound in that gas in which two waves of wavelengths 1 m and 1.01 m produce 10 beats in 3 s.
Answer: Given values are:
Wavelength of the first wave \( \lambda_1 = 1 \) m.
Wavelength of the second wave \( \lambda_2 = 1.01 \) m.
Number of beats produced = 10 beats in 3 seconds.

First, calculate the beat frequency \( n_b \):
\( n_b = \frac{\text{Number of beats}}{\text{Time taken}} = \frac{10}{3} \) Hz.

Let \( v \) be the velocity of sound in the gas. The frequencies of the two waves are \( n_1 = \frac{v}{\lambda_1} \) and \( n_2 = \frac{v}{\lambda_2} \).
Since \( \lambda_2 > \lambda_1 \), it implies that \( n_2 < n_1 \).
The beat frequency is the absolute difference between the two frequencies:
\( n_b = n_1 - n_2 \)
\( n_b = \frac{v}{\lambda_1} - \frac{v}{\lambda_2} = v \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \).

Substitute the given values into the equation:
\( \frac{10}{3} = v \left( \frac{1}{1 \text{ m}} - \frac{1}{1.01 \text{ m}} \right) \)
\( \frac{10}{3} = v \left( \frac{1.01 - 1}{1.01 \times 1} \right) \)
\( \frac{10}{3} = v \left( \frac{0.01}{1.01} \right) \).

Now, solve for \( v \):
\( v = \frac{10}{3} \times \frac{1.01}{0.01} = \frac{10}{3} \times 101 = \frac{1010}{3} \) m/s.
\( v \approx 336.67 \) m/s. This calculation demonstrates how beat frequencies can be used to determine unknown wave properties.
In simple words: We know two sound waves have slightly different lengths and create 10 beats in 3 seconds. By using this information, we found that the speed of sound in that gas is about 336.67 meters per second.

🎯 Exam Tip: For beat frequency problems, always remember that \( n_b = |n_1 - n_2| \). If one wavelength is clearly larger, its frequency will be smaller, so you can drop the absolute value and subtract the smaller frequency from the larger one directly.

Question 16. Calculate the frequency of the tuning fork which when vibrated with the tuning fork of frequency 256 Hz produces 6 beat/s and when vibrated with tuning fork having frequency 253 Hz produces 3 beats/s.
Answer: Let the unknown frequency of the tuning fork be \( f \).

We analyze the two given cases:
Case 1: When vibrated with a tuning fork of frequency \( f_A = 256 \) Hz, it produces 6 beats/s.
According to the concept of beats, the absolute difference between the two frequencies is the beat frequency:
\( |f - f_A| = 6 \)
\( |f - 256| = 6 \)
This means \( f - 256 = 6 \) or \( f - 256 = -6 \).
So, \( f = 256 + 6 = 262 \) Hz or \( f = 256 - 6 = 250 \) Hz.

Case 2: When vibrated with a tuning fork of frequency \( f_B = 253 \) Hz, it produces 3 beats/s.
Again, applying the beat frequency concept:
\( |f - f_B| = 3 \)
\( |f - 253| = 3 \)
This means \( f - 253 = 3 \) or \( f - 253 = -3 \).
So, \( f = 253 + 3 = 256 \) Hz or \( f = 253 - 3 = 250 \) Hz.

By comparing the possible frequencies from both cases, the common frequency is 250 Hz.
Therefore, the frequency of the unknown tuning fork is 250 Hz. This method uses the observed beat patterns to precisely determine an unknown frequency.
In simple words: When the tuning fork was tested with a 256 Hz fork, it made 6 beats per second, meaning its frequency could be 250 Hz or 262 Hz. When tested with a 253 Hz fork, it made 3 beats per second, meaning its frequency could be 250 Hz or 256 Hz. The only frequency common to both possibilities is 250 Hz.

🎯 Exam Tip: For problems involving beat frequencies, always list all possible frequencies from each condition. The true frequency will be the one common to all conditions.

Question 17. A tuning fork produces 4 beats/second from the stretched string of a sonometer of length 0.49 m and 0.50 m. Calculate the frequency of the tuning fork.
Answer: Let \( f \) be the frequency of the tuning fork.
Let \( n_1 \) be the frequency of the sonometer wire when its length is \( l_1 = 0.49 \) m.
Let \( n_2 \) be the frequency of the sonometer wire when its length is \( l_2 = 0.50 \) m.

For a sonometer wire, the frequency \( n \) is inversely proportional to its length \( l \) (assuming tension and linear mass density are constant): \( n \propto \frac{1}{l} \).
So, \( \frac{n_1}{n_2} = \frac{l_2}{l_1} = \frac{0.50}{0.49} \).

The tuning fork produces 4 beats/second with the stretched string, regardless of whether its length is 0.49 m or 0.50 m. This means the difference between the tuning fork's frequency and the string's frequency is always 4 Hz.
As the length of the string increases from 0.49 m to 0.50 m, its frequency decreases (since \( n \propto \frac{1}{l} \)).
So, we can set up the conditions as:
\( n_1 = f + 4 \) (when length is shorter, frequency is higher, so it's above the tuning fork's frequency)
\( n_2 = f - 4 \) (when length is longer, frequency is lower, so it's below the tuning fork's frequency)

Substitute these into the ratio equation:
\( \frac{f+4}{f-4} = \frac{0.50}{0.49} \).
Now, cross-multiply:
\( 0.49(f+4) = 0.50(f-4) \)
\( 0.49f + 0.49 \times 4 = 0.50f - 0.50 \times 4 \)
\( 0.49f + 1.96 = 0.50f - 2.00 \)
Rearrange the terms to solve for \( f \):
\( 1.96 + 2.00 = 0.50f - 0.49f \)
\( 3.96 = 0.01f \)
\( f = \frac{3.96}{0.01} = 396 \) Hz. This illustrates how small changes in length can reveal an unknown frequency through beat phenomena.
In simple words: A tuning fork makes 4 beats every second with a sonometer wire. When the wire is 0.49 meters long, its frequency is 4 Hz higher than the fork. When the wire is 0.50 meters long, its frequency is 4 Hz lower. From this, we figured out that the tuning fork vibrates 396 times per second.

🎯 Exam Tip: For sonometer problems involving beats and length changes, remember that increasing length decreases frequency. This helps determine whether the string's frequency is \( f+n_b \) or \( f-n_b \) relative to the tuning fork.

Question 18. Two engines cross each other in opposite direction. One engine produce 540 Hz sound. People sitting in the second engine will hear sound of which frequency before and after passing from one another? The velocity of both the engines is 40 m/s. Velocity of sound is 340 m/s.
Answer: Given values are:
Source frequency \( n = 540 \) Hz (sound produced by the first engine).
Velocity of the source \( v_s = 40 \) m/s (speed of the first engine).
Velocity of the observer \( v_o = 40 \) m/s (speed of the second engine, where people are listening).
Velocity of sound \( v = 340 \) m/s.

We will use the Doppler's effect formula for apparent frequency \( n' = n \left( \frac{v \pm v_o}{v \mp v_s} \right) \).
Here, the signs depend on whether the source/observer are approaching or receding.

(i) Frequency heard BEFORE passing (when engines are approaching each other):
In this scenario, the source is approaching the observer, and the observer is approaching the source. So, we use \( +v_o \) in the numerator and \( -v_s \) in the denominator.
\( n'_{\text{approaching}} = n \left( \frac{v + v_o}{v - v_s} \right) \)
\( n'_{\text{approaching}} = 540 \text{ Hz} \left( \frac{340 \text{ m/s} + 40 \text{ m/s}}{340 \text{ m/s} - 40 \text{ m/s}} \right) \)
\( n'_{\text{approaching}} = 540 \left( \frac{380}{300} \right) \)
\( n'_{\text{approaching}} = 540 \times \frac{38}{30} = 18 \times 38 = 684 \) Hz.

(ii) Frequency heard AFTER passing (when engines are receding from each other):
In this scenario, the source is receding from the observer, and the observer is receding from the source. So, we use \( -v_o \) in the numerator and \( +v_s \) in the denominator.
\( n'_{\text{receding}} = n \left( \frac{v - v_o}{v + v_s} \right) \)
\( n'_{\text{receding}} = 540 \text{ Hz} \left( \frac{340 \text{ m/s} - 40 \text{ m/s}}{340 \text{ m/s} + 40 \text{ m/s}} \right) \)
\( n'_{\text{receding}} = 540 \left( \frac{300}{380} \right) \)
\( n'_{\text{receding}} = 540 \times \frac{30}{38} = \frac{8100}{19} \approx 426.32 \) Hz.
The Doppler effect causes a noticeable change in pitch as the engines pass each other.
In simple words: Before the engines pass, the people in the second engine will hear the sound at a higher pitch, around 684 Hz, because the engines are moving closer. After they pass and move away, the sound will drop to a lower pitch, about 426.32 Hz.

🎯 Exam Tip: Carefully apply the signs in the Doppler effect formula: for relative approach, both numerator and denominator terms decrease the effective distance (increasing observed frequency). For relative recession, both terms increase effective distance (decreasing observed frequency).

Question 19. How can we calculate the velocity of submarine by Doppler's effect?
Answer: The Doppler's Effect is mostly used to find the speed of submarines. Sound waves are sent from the seashore into the sea. These waves bounce back from submarines and are picked up by a SONAR station on the shore. By measuring how much the wavelength of the reflected wave changes, we can figure out the submarine's speed. When sound waves travel from the SONAR station (which is not moving) towards the submarine, their wavelength does not change. But when the waves reflect off the submarine, the submarine itself acts like a moving sound source. If the submarine is coming closer to the SONAR station, the wavelength of the reflected sound appears to change. The apparent wavelength \( \lambda' \) of the reflected waves can be calculated using the formula \( \lambda' = \frac{v - u_s}{v} \lambda \), where \( v \) is the speed of sound, \( u_s \) is the speed of the submarine, and \( \lambda \) is the original wavelength. So, the change in wavelength \( \Delta \lambda = \lambda' - \lambda \) is given by \( \Delta \lambda = -\frac{u_s}{v}\lambda \). From this, we can find the submarine's velocity. This principle is vital for underwater navigation and detection, allowing ships to map the ocean floor and track objects.
In simple words: The Doppler's Effect helps us find a submarine's speed. We send sound waves that hit the submarine and reflect back. By looking at how the sound waves' length changes when they return, we can calculate how fast the submarine is moving.

🎯 Exam Tip: Remember that the Doppler effect for a stationary source and moving reflector changes the wavelength because the reflector acts as a new moving source.

RBSE Class 11 Physics Chapter 9 Long Answer Type Questions

Question 1. What do you understand by wave motion? Clearly explain. Establish the expression and the one dimensional differential equation for a progressive wave.
Answer: Wave motion is a process where energy moves from one point to another without the medium's particles moving permanently from their positions; the particles only vibrate around their fixed spots. This is how energy is transferred in many situations. Waves can be of different types:
1. **Electromagnetic Waves:** These waves do not need a medium to travel. Examples include light, X-rays, and gamma rays.
2. **Matter Waves:** These are waves linked to moving particles like electrons, protons, and neutrons. They are used in tools like electron microscopes.
3. **Mechanical Waves:** These waves need a physical medium (like air or water) to travel. Sound waves are a good example.
When a wave travels through a medium, the particles in that medium move back and forth in a simple harmonic motion.
Let's consider a wave moving along the x-axis. The displacement (y) of a particle at a distance x from the origin at time t can be described by the equation:
\( y = a \sin (wt - \Phi) \)
Here, \( a \) is the amplitude (maximum displacement), \( \omega \) is the angular frequency, and \( \lambda \) is the wavelength.
The phase difference \( \Phi \) between particles at origin O and at distance x is given by \( \Phi = \frac{2\pi}{\lambda} x \).
So, the equation for a progressive wave moving in the positive x direction becomes:
\( y = a \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \)
We know that \( \omega = \frac{2\pi}{T} \) (where T is the time period) and \( v = \frac{\lambda}{T} \) (where v is wave velocity). Using these relations, the wave equation can also be written as:
\( y = a \sin \left( \frac{2\pi}{T} t - \frac{2\pi}{\lambda} x \right) \)
\( y = a \sin \left( 2\pi \left( \frac{t}{T} - \frac{x}{\lambda} \right) \right) \)
\( y = a \sin \left( \frac{2\pi}{\lambda} (vt - x) \right) \) ...(9.10)
This equation shows how the displacement changes with both time and position for a progressive wave moving in the positive x direction.
To find the velocity of a particle in the wave, we differentiate the displacement equation with respect to time:
\( u = \frac{dy}{dt} = \frac{d}{dt} \left( a \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \right) \)
\( u = a \omega \cos \left( \omega t - \frac{2\pi}{\lambda} x \right) \) ...(9.11)
The maximum velocity of a particle is \( a \omega \).
To find the acceleration of a particle, we differentiate the particle velocity with respect to time:
\( \text{acceleration} = \frac{d^2y}{dt^2} = \frac{d}{dt} \left( a \omega \cos \left( \omega t - \frac{2\pi}{\lambda} x \right) \right) \)
\( \text{acceleration} = -a \omega^2 \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \)
This simplifies to \( \frac{d^2y}{dt^2} = -\omega^2 y \) ...(9.12), which tells us that the acceleration of a particle is directly proportional to its displacement but in the opposite direction, a key feature of simple harmonic motion.
Now, to find the one-dimensional differential equation for the wave, we use both time and space derivatives. After differentiating the displacement equation twice with respect to time and twice with respect to position, and combining them, we get:
We know that \( \frac{d^2y}{dt^2} = -\omega^2 y \).
Also, differentiating \( y = a \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \) twice with respect to x:
\( \frac{\partial y}{\partial x} = a \cos \left( \omega t - \frac{2\pi}{\lambda} x \right) \left( -\frac{2\pi}{\lambda} \right) \)
\( \frac{\partial^2 y}{\partial x^2} = -a \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \left( -\frac{2\pi}{\lambda} \right)^2 \)
\( \frac{\partial^2 y}{\partial x^2} = -a \left( \frac{2\pi}{\lambda} \right)^2 \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \)
Since \( \omega = \frac{2\pi v}{\lambda} \), we can write \( \left( \frac{2\pi}{\lambda} \right)^2 = \frac{\omega^2}{v^2} \).
Substituting this:
\( \frac{\partial^2 y}{\partial x^2} = -a \frac{\omega^2}{v^2} \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \)
\( \frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \left( -a \omega^2 \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \right) \)
Finally, by substituting \( \frac{d^2y}{dt^2} = -a \omega^2 \sin \left( \omega t - \frac{2\pi}{\lambda} x \right) \), we get:
\( \frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2} \) or \( \frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2} \) ...(9.14)
This is the fundamental one-dimensional differential equation for a progressive wave. It links how fast the wave changes over time with how its shape changes over distance. Understanding wave motion is fundamental to many fields, from acoustics and optics to quantum mechanics, as it describes how disturbances propagate through space and matter.
In simple words: Wave motion is when energy moves without the actual particles of the medium moving from their places; they just wiggle. This includes light waves, waves in matter, and sound waves. We can describe a wave with a special equation that shows where each particle is at any time. This equation helps us find how fast particles move and accelerate. From this, we can get a main equation that explains how wave changes over time and space.

🎯 Exam Tip: When deriving the differential equation, clearly state assumptions (like one-dimensional propagation) and correctly use partial derivatives for position and time. Remember that the wave velocity `v` is constant for a given medium, while particle velocity `u` varies.