Maharashtra Board Class 12 Physics Chapter 6 Superposition of Waves Exercise Solutions

1. Choose the correct option.

(i) When an air column in a pipe closed at one end vibrates such that three nodes are formed in it, the frequency of its vibrations are.......times the fundamental frequency.
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (d) 5
In simple words: In a pipe closed at one end, the wave patterns skip even numbers. The first pattern has 1 node, the second has 2 nodes (3 times faster), and the third has 3 nodes (5 times faster).

📝 Teacher's Note: Draw the standing wave patterns for the fundamental, first overtone, and second overtone. Help students see that "three nodes" corresponds to the second overtone (the 5th harmonic).

🎯 Exam Tip: For pipes closed at one end, remember that only odd harmonics (\( 1, 3, 5, ... \)) are present. Each "node" added corresponds to the next odd harmonic.

(ii) If two open organ pipes of length 50 cm and 51 cm sounded together to produce 7 beats per second, the speed of sound is.
(a) 307 m/s
(b) 327 m/s
(c) 350 m/s
(d) 357 m/s
Answer: (d) 357 m/s
In simple words: When two sounds are slightly different in pitch, they create "beats." By knowing the lengths of the pipes and how many beats happen per second, we can calculate how fast sound is traveling.

📝 Teacher's Note: Use the formula for beat frequency \( n_1 - n_2 = 7 \), where \( n = v/2L \). This is a great practical application of the speed of sound formula.

🎯 Exam Tip: Always convert lengths from cm to meters (e.g., 50 cm = 0.5 m) before calculating the speed of sound to ensure the units are correct.

(iii) The tension in a piano wire is increased by 25%. Its frequency becomes ….. times the original frequency.
(a) 0.8
(b) 1.12
(c) 1.25
(d) 1.56
Answer: (b) 1.12
In simple words: Stretching a wire tighter makes it vibrate faster. Increasing the tightness (tension) by 25% makes the sound about 1.12 times higher in pitch.

📝 Teacher's Note: Explain that frequency is proportional to the square root of tension (\( n \propto \sqrt{T} \)). Since \( 1.25 \) is roughly \( 1.12^2 \), the frequency increases by that factor.

🎯 Exam Tip: For percentage changes in tension, always use the square root relationship. \( \sqrt{1.25} \approx 1.118 \), which rounds to 1.12.

(iv) Which of the following equations represents a wave travelling along the y-axis?
(a) \( x = A \sin(ky - \omega t) \)
(b) \( y = A \sin(kx - \omega t) \)
(c) \( y = A \sin(ky) \cos(\omega t) \)
(d) \( y = A \cos(ky) \sin(\omega t) \)
Answer: (a) \( x = A \sin(ky - \omega t) \)
In simple words: If a wave moves along the y-axis, the variable inside the sine function must be 'y', and the actual vibration happens in a different direction (like 'x').

📝 Teacher's Note: Distinguish between the direction of propagation (variable inside the function) and the direction of displacement (the subject of the equation).

🎯 Exam Tip: Look for the term \( (ky \pm \omega t) \) to identify a wave moving along the y-axis. Option (b) moves along the x-axis, and (c)/(d) are standing waves.

(v) A standing wave is produced on a string fixed at one end with the other end free. The length of the string
(a) must be an odd integral multiple of \( \lambda/4 \).
(b) must be an odd integral multiple of \( \lambda/2 \).
(c) must be an odd integral multiple of \( \lambda \).
(d) must be an even integral multiple of \( \lambda \).
Answer: (a) must be an odd integral multiple of \( \lambda/4 \).
In simple words: When one end is tied down and the other is loose, the wave must end in a "big swing" (antinode). This only happens at specific lengths like 1/4, 3/4, or 5/4 of a full wave.

📝 Teacher's Note: This is the "closed pipe" analogy for strings. The fixed end is a node and the free end is an antinode.

🎯 Exam Tip: Boundary conditions are key: Fixed-Fixed or Free-Free uses \( \lambda/2 \), while Fixed-Free uses \( \lambda/4 \).

2. Answer in brief.

(i) A wave is represented by an equation \( y = A \sin (Bx + Ct) \). Given that the constants A, B, and C are positive, can you tell in which direction the wave is moving?
Answer:
The wave is travelling along the negative x-direction.
In simple words: In wave equations, if the 'x' part and the 'time' part have the same sign (both positive here), the wave is moving to the left.

📝 Teacher's Note: Use the general form \( y = f(x \pm vt) \). Explain that \( + \) means negative direction and \( - \) means positive direction.

🎯 Exam Tip: Always check the sign between the position and time terms. Opposite signs = Positive direction; Same signs = Negative direction.

(ii) A string is fixed at the two ends and is vibrating in its fundamental mode. It is known that the two ends will be at rest. Apart from these, is there any position on the string which can be touched so as not to disturb the motion of the string? What will be the answer to this question if the string is vibrating in its first and second overtones?
Answer:
Nodes are the points where the vibrating string can be touched without disturbing its motion.
When the string vibrates in its fundamental mode, the string vibrates in one loop. There are no nodes formed between the fixed ends. Hence, there is no point on the string which can be touched without disturbing its motion.
When the string vibrates in its first overtone (second harmonic), there are two loops of the stationary wave on the string. Apart from the two nodes at the two ends, there is now a third node at its centre. Hence, the string can be touched at its centre without disturbing the stationary wave pattern.
When the string vibrates in its second overtone (third harmonic), there are three loops of the stationary wave on the string. So, apart from the two end nodes, there are two additional nodes in between, at distances one-third of the string length from each end. Thus, now the string can be touched at these two nodes.
In simple words: You can only touch a vibrating string at its "nodes" (the quiet spots that don't move). In the simplest mode, only the ends are quiet. In faster modes, quiet spots appear in the middle.

📝 Teacher's Note: Demonstrate this with a long spring or rope. Show how touching a node doesn't stop the vibration, but touching an antinode does.

🎯 Exam Tip: Define "nodes" clearly in your answer. For the nth harmonic, there are \( n-1 \) nodes between the fixed ends.

(iii) What are harmonics and overtones?
Answer:
A stationary wave is set up in a bounded medium in which the boundary could be a rigid support (i.e., a fixed end, as for instance a string stretched between two rigid supports) or a free end (as for instance an air column in a cylindrical tube with one or both ends open). The boundary conditions limit the possible stationary waves and only a discrete set of frequencies is allowed.
The lowest allowed frequency, \( n_1 \), is called the fundamental frequency of vibration. Integral multiples of the fundamental frequency are called the harmonics, the fundamental frequency being the fundamental or \( 2n_1 \), the third harmonic is \( 3n_1 \), and so on.
The higher allowed frequencies are called the overtones. Above the fundamental, the first allowed frequency is called the first overtone, the next higher frequency is the second overtone, and ‘so on. The relation between overtones and allowed harmonics depends on the system under consideration.
In simple words: The "fundamental" is the lowest note a string can play. "Harmonics" are exact multiples of that speed (2x, 3x, etc.), and "overtones" are simply any allowed notes higher than the fundamental.

📝 Teacher's Note: Clarify that while all harmonics are multiples of the fundamental, not all harmonics may be physically present in a specific system (like a closed pipe).

🎯 Exam Tip: Use the term "discrete set of frequencies" to show a deeper understanding of how boundaries restrict wave motion.

(iv) For a stationary wave set up in a string having both ends fixed, what is the ratio of the fundamental frequency to the second harmonic?
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (\( n_1 \)) is 1 : 2.
In simple words: The second harmonic vibrates exactly twice as fast as the first one, so the ratio is simply 1 to 2.

📝 Teacher's Note: Remind students that for a string fixed at both ends, all harmonics (\( n, 2n, 3n... \)) are present.

🎯 Exam Tip: Always state that the fundamental frequency is the "first harmonic" to justify the 1:2 ratio.

(v) The amplitude of a wave is represented by \( y = 0.2 \sin 4\pi [ \frac{t}{0.08} - \frac{x}{0.8} ] \) in SI units. Find (a) wavelength, (b) frequency and (c) amplitude of the wave. [(a) 0.4 m (b) 25 Hz (c) 0.2 m]
Answer:
\( y = 0.2 \sin 4\pi [ \frac{t}{0.08} - \frac{x}{0.8} ] \)
\( y = 0.2 \sin 2\pi [ \frac{2t}{0.08} - \frac{2x}{0.8} ] \)
\( y = 0.2 \sin 2\pi [ \frac{t}{0.04} - \frac{x}{0.4} ] \)
Let us compare above equation with the equation of a simple harmonic progressive wave:
\( y = A \sin 2\pi [ \frac{t}{T} - \frac{x}{\lambda} ] = 0.2 \sin 2\pi [ \frac{t}{0.04} - \frac{x}{0.4} ] \)
Comparing the quantities on both sides, we get,
\( A = 0.2 \) m, \( T = 0.04 \) s, \( \lambda = 0.4 \) m
\( \therefore \) (a) Wavelength \( (\lambda) = 0.4 \) m
(b) Frequency \( (n) = \frac{1}{T} = \frac{1}{0.04} = 25 \) Hz
(c) Amplitude \( (A) = 0.2 \) m
In simple words: By comparing the given math formula to a standard wave pattern, we can "pick out" the height of the wave (0.2m), the length of one wave (0.4m), and how many times it repeats per second (25 times).

📝 Teacher's Note: Teach students to factor out coefficients (like the \( 4\pi \) to \( 2\pi \) transition) to make the equation match the standard form perfectly.

🎯 Exam Tip: SI units are essential. Always include 'm' for distance and 'Hz' for frequency in your final answer.

Question 3. State the characteristics of progressive waves.
Answer:
Characteristics of a progressive wave :

1. Energy is transmitted from particle to particle without the physical transfer of matter.

2. The particles of the medium vibrate periodically about their equilibrium positions.

3. In the absence of dissipative forces, every particle vibrates with the same amplitude and frequency, but differs in phase from its adjacent particles. Every particle lags behind in its state of motion compared to the one before it.

4. A wave motion is doubly periodic, i.e., it is periodic in time and periodic in space.

5. The velocity of propagation through a medium depends upon the properties of the medium.

6. Progressive waves are of two types : transverse and longitudinal. In a transverse mechanical wave, the individual particles of the medium vibrate perpendicular to the direction of propagation of the wave. In a longitudinal mechanical wave, the individual particles of the medium vibrate along the line of propagation of the wave.

7. A transverse wave can propagate only through solids, but not through liquids and gases while a longitudinal wave can propagate through any material medium.

In simple words: A progressive wave is a wave that travels forward, carrying energy (like sound or light) while the particles themselves just wiggle in place.

📝 Teacher's Note: Use the analogy of a "stadium wave" where people stand up and sit down (vibrate) but the "wave" travels around the stadium (energy transfer).

🎯 Exam Tip: The "doubly periodic" characteristic (space and time) is a key phrase that examiners look for.

Question 4. State the characteristics of stationary waves.
Answer:
Characteristics of stationary waves :

1. Stationary waves are produced by the interference of two identical progressive waves travelling in opposite directions, under certain conditions.

2. The overall appearance of a standing wave is of alternate intensity maximum (displacement antinode) and minimum (displacement node).

3. The distance between adjacent nodes (or antinodes) is \( \lambda/2 \).

4. The distance between successive node and antinode is \( \lambda/4 \).

5. There is no progressive change of phase from particle to particle. All the particles in one loop, between two adjacent nodes, vibrate in the same phase, while the particles in adjacent loops are in opposite phase.

6. A stationary wave does not propagate in any direction and hence does not transport energy through the medium.

7. In a region where a stationary wave is formed, the particles of the medium (except at the nodes) perform SHM of the same period, but the amplitudes of the vibrations vary periodically in space from particle to particle.

[Note : Since the nodes are points where the particles are always at rest, energy cannot be transmitted across a node. The energy of the particles within a loop remains localized, but alternates twice between kinetic and potential energy during each complete vibration.]
In simple words: A stationary (standing) wave looks like it's trapped in place. Some spots never move (nodes), and energy just sloshes back and forth inside the loops instead of traveling away.

📝 Teacher's Note: Contrast this with Question 3. Emphasize that stationary waves do NOT transport energy, which is their most important difference from progressive waves.

🎯 Exam Tip: Mentioning the distance between nodes (\( \lambda/2 \)) and the lack of energy transport usually secures high marks.

Question 5. Derive an expression for equation of stationary wave on a stretched string.
Answer:
When two progressive waves having the same amplitude, wavelength and speed propagate in opposite directions through the same region of a medium, their superposition under certain conditions creates a stationary interference pattern called a stationary wave.
Consider two simple harmonic progressive waves, of the same amplitude A, wavelength \( \lambda \) and frequency \( n = \omega/2\pi \), travelling on a string stretched along the x-axis in opposite directions. They may be represented by
\( y_1 = A \sin (\omega t – kx) \) (along the + x-axis) … (1)
\( y_2 = A \sin (\omega t + kx) \) (along the – x-axis) …. (2)
where \( k = 2\pi/\lambda \) is the propagation constant.
By the superposition principle, the resultant displacement of the particle of the medium at the point at which the two waves arrive simultaneously is the algebraic sum
\( y = y_1 + y_2 = A [\sin (\omega t – kx) + \sin (\omega t + kx)] \)
Using the trigonometrical identity,
\( \sin C + \sin D = 2 \sin (\frac{C+D}{2}) \cos (\frac{C-D}{2}) \),
\( y = 2A \sin \omega t \cos (- kx) \)

\( \implies y = 2A \sin \omega t \cos kx \) [\( \because \cos(- kx) = \cos(kx) \)]

\( \implies y = 2A \cos kx \sin \omega t \) … (3)

\( \implies y = R \sin \omega t \), … (4)
where \( R = 2A \cos kx \). … (5)
Equation (4) is the equation of a stationary wave.
In simple words: We take two waves moving in opposite directions and add them together. Using some math tricks, we find a new formula that shows the wave stays in one spot, with the "bigness" of the wiggle depending on where you are on the string.

📝 Teacher's Note: Walk students through the trigonometry step-by-step. The separation of the 'x' part and 't' part in the final result is what mathematically defines a standing wave.

🎯 Exam Tip: Clearly define the Resultant Amplitude \( R = 2A \cos kx \) at the end of the derivation to show the spatial dependence of the amplitude.

Question 6. Find the amplitude of the resultant wave produced due to interference of two waves given as \( y_1 = A_1 \sin \omega t \) and \( y_2 = A_2 \sin (\omega t + \phi) \)
Answer:
The amplitude of the resultant wave produced due to the interference of the two waves is
\( A = \sqrt{A_1^2 + 2A_1A_2 \cos \phi + A_2^2} \).
In simple words: When two waves overlap, their combined height depends on how much they are "in sync" (\( \phi \)). This formula calculates the final height of the combined wave.

📝 Teacher's Note: Note the similarity of this formula to the resultant of two vectors. Interference is essentially vector addition of displacements.

🎯 Exam Tip: This formula is very common in both Wave Optics and Superposition of Waves. Memorize it as the "Vector Addition" equivalent for waves.

Question 7. State the laws of vibrating strings and explain how they can be verified using a sonometer.
Answer:
The fundamental is the first harmonic. Therefore, the ratio of the fundamental frequency (n) to the second harmonic (\( n_1 \)) is 1 : 2.
Here, \( L = 3 \frac{\lambda}{2} \)
\( \therefore \) Wavelength, \( \lambda = \frac{2L}{3} = \frac{2 \times 30}{3} = 20 \) cm.
In simple words: A string's pitch depends on its length, its tightness, and its weight. We use a tool called a sonometer to prove these rules by changing one thing at a time.

📝 Teacher's Note: Although the provided solution text in the source appears to be a numerical example, the "Laws of Vibrating Strings" usually refer to the Law of Length (\( n \propto 1/L \)), Law of Tension (\( n \propto \sqrt{T} \)), and Law of Linear Density (\( n \propto 1/\sqrt{m} \)).

🎯 Exam Tip: When asked for "Laws," always list the three proportional relationships between frequency, length, tension, and mass per unit length.

Question 8. Show that only odd harmonics are present in the vibrations of air column in a pipe closed at one end.
Answer:
Consider a narrow cylindrical pipe of length l closed at one end. When sound waves are sent down the air column, they are reflected at the closed end with a phase reversal and at the open end without phase reversal. Interference between the incident and reflected waves sets up stationary waves in the air column.
The stationary waves in the air column in this case are subject to two boundary conditions that there must be a node at the closed end and an antinode at the open end. Taking into account the end correction e at the open end, the resonating length of the air column is \( L = l + e \).
Let v be the speed of sound in air. In the simplest mode of vibration, there is a node at the closed end and an antinode at the open end. The distance between a node and a consecutive anti-node is \( \lambda/4 \).
\( \lambda = 4L \) and \( n = \frac{v}{\lambda} = \frac{v}{4L} = \frac{v}{4(l+e)} \) …… (1)
This is the fundamental mode or first harmonic.
In the first overtone, two nodes and two antinodes are formed. The corresponding wavelength \( \lambda_1 \) and frequency \( n_1 \) are
\( \lambda_1 = \frac{4L}{3} \) and \( n_1 = \frac{v}{\lambda_1} = \frac{3v}{4L} = 3n \) … (2)
Therefore, the frequency in the first overtone is three times the fundamental frequency, i.e., the third harmonic.
In the second overtone, three nodes and three antinodes are formed.
\( n_2 = 5n \), which is the fifth harmonic.
Therefore, in general, the frequency of the pth overtone is \( n_p = (2p + 1)n \).
In simple words: Because one end of the pipe is a wall (node) and the other is open (antinode), the wave can only fit in "jumps" of odd numbers (1, 3, 5...). It's like trying to step across a stream where you can only land on certain rocks.

📝 Teacher's Note: Emphasize the boundary conditions: the air cannot move at the closed end (Node) and moves most freely at the open end (Antinode).

🎯 Exam Tip: Always include the diagram and the general formula \( n_p = (2p + 1)n \) to prove that only odd multiples exist.

Question 9. Prove that all harmonics are present in the vibrations of the air column in a pipe open at both ends.
Answer:
Consider a cylindrical pipe of length l open at both the ends. When sound waves are sent down the air column in a cylindrical open pipe, they are reflected at the open ends without a change of phase. Interference between the incident and reflected waves under appropriate conditions sets up stationary waves in the air column.
The stationary waves in the air column in this case are subject to the two boundary conditions that there must be an antinode at each open end.
Taking into account the end correction e at each of the open ends, the resonating length of the air column is \( L = l + 2e \).
Let v be the speed of sound in air. In the simplest mode of vibration, the fundamental mode or first harmonic, there is a node midway between the two antinodes at the open ends. The distance between two consecutive antinodes is

\( \lambda/2 \), where \( \lambda \) is the wavelength of sound. The corresponding wavelength \( \lambda \) and the fundamental frequency \( n \) are
\( \lambda = 2L \) and \( n = \frac{v}{\lambda} = \frac{v}{2L} = \frac{v}{2(l+2e)} \) .... (1)

In the next higher mode, the first overtone, there are two nodes and three antinodes. The corresponding wavelength \( \lambda_1 \) and frequency \( n_1 \)
\( \lambda_1 = L \) and \( n_1 = \frac{v}{\lambda_1} = \frac{v}{L} = \frac{v}{(l+2e)} = 2n \) .... (2)
i.e., twice the fundamental. Therefore, the first overtone is the second harmonic.

In the second overtone, there are three nodes and four antinodes. The corresponding wavelength \( \lambda_2 \) and frequency \( n_2 \) are
\( \lambda_2 = \frac{2L}{3} \) and \( n_2 = \frac{v}{\lambda_2} = \frac{3v}{2L} = \frac{3v}{2(l+2e)} = 3n \) .... (3)
or thrice the fundamental. Therefore, the second overtone is the third harmonic.

Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, ...) is
\( n_p = (p + 1)n \) ... (4)
i.e., the pth overtone is the (p + 1)th harmonic.
Equations (1), (2) and (3) show that allowed frequencies in an air column in a pipe open at both ends are n, 2n, 3n, .... That is, all the harmonics are present as overtones.
In simple words: Since both ends are open, the air can swing wildly at both sides. This freedom allows every possible vibration speed (1, 2, 3, 4...) to fit perfectly inside the pipe.

📝 Teacher's Note: Point out that the "End Correction" is doubled (\( 2e \)) because there are two open ends, unlike the closed pipe which only has one.

🎯 Exam Tip: For open pipes, the distance between two antinodes in the fundamental mode is \( \lambda/2 \). This leads to \( n, 2n, 3n... \), which means all harmonics are present.

Question 10. A wave of frequency 500 Hz is travelling with a speed of 350 m/s.
(a) What is the phase difference between two displacements at a certain point at times 1.0 ms apart?
(b) what will be the smallest distance between two points which are 45º out of phase at an instant of time?
[Ans : \( \pi \), 8.75 cm ]
Answer:
Data : \( n = 500 \text{ Hz}, v = 350 \text{ m/s} \)
\( v = n \times \lambda \)

\( \therefore \lambda = \frac{350}{500} = 0.7 \text{ m} \)
(a) in \( t = 1.0 \text{ ms} = 0.001 \text{ s} \), the path difference is the distance covered \( v \times t = 350 \times 0.001 = 0.35 \text{ m} \)

\( \therefore \text{Phase difference} = \frac{2\pi}{\lambda} \times \text{Path difference} \)
\( = \frac{2\pi}{0.7} \times 0.35 = \pi \text{ rad} \)
(b) Phase difference = \( 45^\circ = \frac{\pi}{4} \text{ rad} \)

\( \therefore \text{Path difference} = \frac{\lambda}{2\pi} \times \text{Phase difference} \)
\( = \frac{0.7}{2\pi} \times \frac{\pi}{4} = 0.0875 \text{ m} \)
In simple words: Phase difference tells us how much two points in a wave cycle are "out of step" with each other. We use the wave's speed and frequency to convert time intervals or physical distances into this phase measurement.

📝 Teacher's Note: Remind students to always convert units like milliseconds to seconds before starting calculations. The relationship between path difference and phase difference is a fundamental wave property that applies to both light and sound.

🎯 Exam Tip: When dealing with phase, specify whether your answer is in degrees or radians. Most formula results in physics are in radians unless otherwise stated.

Question 11. A sound wave in a certain fluid medium is reflected at an obstacle to form a standing wave. The distance between two successive nodes is 3.75 cm. If the velocity of sound is 1500 m/s, find the frequency. [Ans : 20 kHz]
Answer:
Data : Distance between two successive nodes =
\( \frac{\lambda}{2} = 3.75 \times 10^{-2} \text{ m}, v = 1500 \text{ m/s} \)

\( \therefore \lambda = 7.5 \times 10^{-2} \text{ m} \)
\( v = n \times \lambda \)

\( \therefore n = \frac{1500}{7.5 \times 10^{-2}} = 20 \text{ kHz} \)
In simple words: In a standing wave, the distance between two quiet spots (nodes) is exactly half a wavelength. By finding the full wavelength and knowing the speed, we can calculate how many vibrations happen per second (frequency).

📝 Teacher's Note: Use a rope or a slinky in class to demonstrate that the distance between "still" points is half the length of one complete wave loop. This visual helps students remember the \( \lambda/2 \) factor.

🎯 Exam Tip: Be careful with powers of 10. Converting cm to m (using \( 10^{-2} \)) is a common step where simple calculation errors occur.

Question 12. Two sources of sound are separated by a distance 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180º out of phase. At what points between the two sources, will the sound intensity be maximum? (Take velocity of sound to be 330 m/s) [Ans: ± 0.25, ± 0.75, ± 1.25 and ± 1.75 m from the point at the center]
Answer:
\( \therefore \lambda = \frac{v}{n} = \frac{330}{330} = 1 \text{ m} \)
Directly at the centre of two sources of sound, path difference is zero. But since the waves are 180° out of phase, two maxima on either sides should be at a distance of \( \frac{\lambda}{4} \) from the point at the centre. Other maxima will be located each \( \frac{\lambda}{2} \) further along.
Thus, the sound intensity will be maximum at ± 0.25, ± 0.75, ± 1.25, ± 1.75 m from the point at the centre.
In simple words: Usually, the middle point between two speakers is the loudest if they are in phase. Because these are opposite (180° out of phase), the middle is quiet, and the loud spots (maxima) are shifted slightly to the left and right.

📝 Teacher's Note: Explain that "180° out of phase" means that when one source pushes air forward, the other pulls it back. This causes destructive interference at the exact geometric center.

🎯 Exam Tip: Always state the reference point for your distances (e.g., "from the center") to ensure your answer is clear to the examiner.

Question 13. Two sound waves travel at a speed of 330 m/s. If their frequencies are also identical and are equal to 540 Hz, what will be the phase difference between the waves at points 3.5 m from one source and 3 m from the other if the sources are in phase? [Ans : 1.636 \( \pi \)]
Answer:
Data : \( v = 330 \text{ m/s}, n_1 = n_2 = 540 \text{ Hz} \)
\( v = n \times \lambda \)

\( \therefore \lambda = \frac{330}{540} = 0.61 \text{ m} \)
Here, the path difference = \( 3.5 - 3 \text{ m} = 0.5 \text{ m} \)
Phase difference = \( \frac{2\pi}{\lambda} \times \text{Path difference} \)
\( = \frac{2\pi}{0.61} \times 0.5 = 1.64\pi \text{ rad} \)
In simple words: Because the paths from the two sources to the point have different lengths, the waves arrive at different times in their cycle. We calculate this "timing difference" as a phase angle.

📝 Teacher's Note: This is a classic interference problem. Highlight that if the sources were "out of phase" initially, you would need to add that initial phase to this calculated result.

🎯 Exam Tip: Keep your final answer in terms of \( \pi \) if the question suggests it, as it is often more precise than using decimal values like 3.14.

Question 14. Two wires of the same material and same cross-section are stretched on a sonometer. One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire. [Ans: 1.2 m]
Answer:
Data : \( m_1 = m_2 = m, L_1 = 60 \text{ cm} = 0.6 \text{ m}, T_1 = 1.5 \text{ kg} = 14.7 \text{ N}, T_2 = 6 \text{ kg} = 58.8 \text{ N} \)
\( n_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{m}} \) and \( n_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{m}} \)
But, \( n_1 = n_2 \)

\( \therefore \frac{1}{2L_1} \sqrt{\frac{T_1}{m}} = \frac{1}{2L_2} \sqrt{\frac{T_2}{m}} \)
\( L_2 = \sqrt{\frac{T_2}{T_1}} \times L_1 \)
\( = \sqrt{\frac{58.8}{14.7}} \times 0.6 = \sqrt{4} \times 0.6 = 1.2 \text{ m} \)
The vibrating length of the second wire is 1.2 m.
In simple words: If you increase the tension (tightness) of a wire, it naturally wants to vibrate faster. To keep the frequency the same while the wire is tighter, you must make the wire longer.

📝 Teacher's Note: This problem demonstrates the Law of Tension and Law of Length for vibrating strings. It's a great example of how different variables can be balanced to achieve the same result.

🎯 Exam Tip: Since both frequencies are equal, you can solve this using the ratio method: \( \frac{L_2}{L_1} = \sqrt{\frac{T_2}{T_1}} \). This simplifies the math and reduces the chance of errors.

Question 15. A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz and 1152 Hz. Calculate the fundamental frequency. [Ans: 128 Hz]
Answer:
The difference between the given frequencies of the overtones is 256 Hz. This implies that they are consecutive overtones. Let \( n_c \) be the fundamental frequency of the closed pipe and \( n_q, n_{q+1}, n_{q+2} \) = the frequencies of the q^th, (q + 1)^th and (q + 2)^th consecutive overtones, where q is an integer.
Data : \( n_q = 640 \text{ Hz}, n_{q+1} = 896 \text{ Hz}, n_{q+2} = 1152 \text{ Hz} \)
Since only odd harmonics are present as overtones, \( n_q = (2q +1) n_c \)
and \( n_{q+1} = [2(q + 1) + 1] n_c = (2q + 3) n_c \)

\( \therefore \frac{n_{q+1}}{n_q} = \frac{2q + 3}{2q + 1} = \frac{896}{640} = \frac{7}{5} \)

\( \therefore 14q + 7 = 10q + 15 \therefore 4q = 8 \therefore q = 2 \)
Therefore, the three given frequencies correspond to the second, third and fourth overtones, i.e., the fifth, seventh and ninth harmonics, respectively.

\( \therefore 5n_c = 640 \therefore n_c = 128 \text{ Hz} \)
In simple words: Pipes closed at one end only play "odd" notes (1, 3, 5, 7...). By looking at the gaps between the frequencies provided, we can work backward to find the very first, lowest note the pipe can make.

📝 Teacher's Note: Emphasize that the common difference between consecutive overtones in a closed pipe is always \( 2n_c \). This is a quick way for students to check their work.

🎯 Exam Tip: Always remember that closed pipes skip even harmonics. This is the most common trap in pipe resonance questions.

Question 16. A standing wave is produced in a tube open at both ends. The fundamental frequency is 300 Hz. What is the length of tube in the fundamental mode? (speed of the sound = 340 m s\( ^{-1} \)). [Ans: 0.5666 m]
Answer:
Data : For the tube open at both the ends, \( n = 300 \text{ Hz} \) and \( v = 340 \text{ m/s} \). Ignoring end correction, the fundamental frequency of the tube is
\( n = \frac{v}{2L} \)

\( \therefore L = \frac{v}{2n} = \frac{340}{2 \times 300} = 0.566 \text{ m} \)
The length of the tube open at both the ends is 0.5667 m.
In simple words: For a tube open at both ends, the wave looks like half a circle stretching from one end to the other in its simplest mode. We use the speed of sound and frequency to find how long that half-wavelength is.

📝 Teacher's Note: Explain that "ignoring end correction" means we assume the antinode is exactly at the tube's opening, though in reality, it's slightly outside.

🎯 Exam Tip: Read carefully if the question asks for the tube open at one end or both ends, as the formulas differ by a factor of 2.

Question 17. Find the fundamental, first overtone and second overtone frequencies of a pipe, open at both the ends, of length 25 cm if the speed of sound in air is 330 m/s. [Ans: 660 Hz, 1320 Hz, 1980 Hz]
Answer:
Data : Open pipe, \( L = 25 \text{ cm} = 0.25 \text{ m}, v = 330 \text{ m/s} \)
The fundamental frequency of an open pipe ignoring end correction,
\( n_0 = \frac{v}{\lambda} = \frac{v}{2L} \)

\( \therefore n_0 = \frac{330}{2 \times 0.25} = 660 \text{ Hz} \)
Since all harmonics are present as overtones, the first overtone is, \( n_1 = 2n_0 = 2 \times 660 = 1320 \text{ Hz} \)
The second overtone is \( n_2 = 3n_0 = 3 \times 660 = 1980 \text{ Hz} \)
In simple words: An open pipe produces a sequence of frequencies that are simple multiples of the first one (660, 660x2, 660x3, etc.). These are like the notes of a bugle or a flute.

📝 Teacher's Note: This is a great time to show how music works—how changing the length of an air column (like on a recorder) changes the fundamental frequency.

🎯 Exam Tip: If the question asks for frequencies of "harmonics," it's the same as the overtone sequence for open pipes, but the numbering starts at the fundamental.

Question 18. A pipe open at both the ends has a fundamental frequency of 600 Hz. The first overtone of a pipe closed at one end has the same frequency as the first overtone of the open pipe. How long are the two pipes? (Take velocity of sound to be 330 m/s) [Ans : 27.5 cm, 20.625 cm]
Answer:
Data : Open pipe, \( n_o = 600 \text{ Hz} \), \( n_{c,1} = n_{o,1} \) (first overtones)
For an open pipe, the fundamental frequency,
\( n_o = \frac{v}{2L_O} \)

\( \therefore \) The length of the open pipe is
\( L_O = \frac{v}{2n_o} = \frac{330}{2 \times 600} = 0.275 \text{ m} \)
For the open pipe, the frequency of the first overtone is
\( 2n_o = 2 \times 600 = 1200 \text{ Hz} \)
For the pipe closed at one end, the frequency of the first overtone is \( \frac{3v}{4L_C} \).
By the data, \( \frac{3v}{4L_C} = 1200 \)

\( \therefore L_C = \frac{3 \times 330}{4 \times 1200} = 0.206 \text{ m} \)
The length of the pipe open at both ends is 27.5 cm and the closed pipe is 20.6 cm.
In simple words: We are comparing two different types of pipes to see what lengths they need to be to play the same "high note". We first find the length of the open pipe from its basic note, then calculate what the other pipe needs.

📝 Teacher's Note: This comparison question tests if students can distinguish between overtone formulas for open and closed pipes simultaneously.

🎯 Exam Tip: Be very careful with the wording "first overtone". For an open pipe it's the 2nd harmonic (\( 2n \)), but for a closed pipe it's the 3rd harmonic (\( 3n \)).

Question 19. A string 1m long is fixed at one end. Transverse vibrations of frequency 15 Hz are imposed at the free end. Due to this, a stationary wave with four complete loops, is produced on the string. Find the speed of the progressive wave which produces the stationary wave.[Hint: Remember that the free end is an antinode.] [Ans: 6.67 m s\( ^{-1} \)]
Answer:
Data : \( L = 1 \text{ m}, n = 15 \text{ Hz} \).
The string is fixed only at one end. Hence, an antinode will be formed at the free end. Thus, with four and half loops on the string, the length of the string is
\( L = \frac{\lambda}{4} + 4(\frac{\lambda}{2}) = \frac{9}{4}\lambda \)

\( \therefore \lambda = \frac{4L}{9} = \frac{4}{9} \times 1 = \frac{4}{9} \text{ m} \)
\( v = n\lambda \)

\( \therefore \) Speed of the progressive wave
\( v = 15 \times \frac{4}{9} = \frac{60}{9} = 6.667 \text{ m/s} \)
In simple words: A string that is loose at one end creates a slightly different wave pattern than one tied down at both ends. By counting the loops and the extra half-loop at the free end, we find the total wavelength and then the wave speed.

📝 Teacher's Note: The "fixed-free" boundary condition is similar to a pipe closed at one end. Each "loop" is \( \lambda/2 \), and the bit at the free end is \( \lambda/4 \).

🎯 Exam Tip: Visualizing the wave helps—draw the four loops and the final antinode to ensure you don't miss the extra \( 1/4 \) wavelength.

Question 20. A violin string vibrates with fundamental frequency of 440Hz. What are the frequencies of first and second overtones? [Ans: 880 Hz, 1320 Hz]
Answer:
Data: \( n = 440 \text{ Hz} \)
The first overtone, \( n_1 = 2n = 2 \times 440 = 880 \text{ Hz} \)
The second overtone, \( n_2 = 3n = 3 \times 440 = 1320 \text{ Hz} \)
In simple words: On a violin string (fixed at both ends), the higher notes it can make naturally are just double and triple the frequency of the first note.

📝 Teacher's Note: For strings fixed at both ends, all harmonics are present. This makes string instruments sound "richer" than some wind instruments.

🎯 Exam Tip: Fundamental frequency is the 1st harmonic. The 1st overtone is the 2nd harmonic. Don't confuse the overtone number with the harmonic number.

Question 21. A set of 8 tuning forks is arranged in a series of increasing order of frequencies. Each fork gives 4 beats per second with the next one and the frequency of last for k is twice that of the first. Calculate the frequencies of the first and the last for k. [Ans: 28 Hz, 56 Hz]
Answer:
Data : \( n_8 = 2n_1 \), beat frequency = 4 Hz
The set of tuning fork is arranged in the increasing order of their frequencies.

\( \therefore n_2 = n_1 + 4 \)
\( n_3 = n_2 + 4 = n_1 + 2 \times 4 \)
\( n_4 = n_3 + 4 = n_1 + 3 \times 4 \)

\( \therefore n_8 = n_7 + 4 = n_1 + 7 \times 4 = n_1 + 28 \)
Since \( n_8 = 2n_1 \),
\( 2n_1 = n_1 + 28 \)

\( \therefore \) The frequency of the first fork, \( n_1 = 28 \text{ Hz} \)

\( \therefore \) The frequency of the last fork,
\( n_8 = n_1 + 28 = 28 + 28 = 56 \text{ Hz} \)
In simple words: Think of a set of steps where each step is 4 Hz higher than the last. If you take 7 steps from the first fork to get to the eighth, and the last is double the first, you can find exactly how high the first step started.

📝 Teacher's Note: This is an application of an Arithmetic Progression (AP). Students should note that between 8 forks, there are 7 gaps (beats).

🎯 Exam Tip: A common mistake is to multiply the beat frequency by the total number of forks (8) instead of the number of intervals (7). Always count the intervals.

Question 22. A sonometer wire is stretched by tension of 40 N. It vibrates in unison with a tuning fork of frequency 384 Hz. How many numbers of beats get produced in two seconds if the tension in the wire is decreased by 1.24 N? [Ans: 12 beats]
Answer:
Data : \( T_1 = 40 \text{ N}, n_1 = 384 \text{ Hz}, T_2 = 40 - 1.24 = 38.76 \text{ N} \)
\( n_1 = \frac{1}{2l} \sqrt{\frac{T_1}{m}} \) and \( n_2 = \frac{1}{2l} \sqrt{\frac{T_2}{m}} \)

\( \therefore \frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} \)

\( \therefore \frac{n_2}{384} = \sqrt{\frac{38.76}{40}} = \sqrt{0.969} = 0.9844 \)

\( \therefore n_2 = 384 \times 0.9844 = 378.0 \text{ Hz} \)

\( \therefore n_1 - n_2 = 384 - 378 = 6 \text{ Hz} \)

\( \therefore \) The number of beats produced in two seconds \( = 2 \times 6 = 12 \)
In simple words: When the tension is lowered, the wire vibrates slightly slower. The difference between the original frequency and the new one creates "beats," and we just multiply that per-second difference by two to get the total beats.

📝 Teacher's Note: Beats per second is simply the absolute difference between two frequencies. Remind students that frequency is beats *per second*, so for 2 seconds, they must multiply by 2.

🎯 Exam Tip: Be careful to calculate the *decrease* in tension correctly before plugging it into the ratio formula.

Question 23. A sonometer wire of length 0.5 m is stretched by a weight of 5 kg. The fundamental frequency of vibration is 100 Hz. Calculate linear density of wire. [Ans: 4.9 \( \times 10^{-3} \text{ kg/m} \)]
Answer:
Data : \( L = 0.5 \text{ m}, T = 5 \text{ kg} = 5 \times 9.8 = 49 \text{ N}, n = 100 \text{ Hz} \)
\( n = \frac{1}{2L} \sqrt{\frac{T}{m}} \)

\( \therefore \) Linear density, \( m = \frac{T}{4L^2n^2} \)
\( = \frac{49}{4(0.5)^2(100)^2} \)
\( = 4.9 \times 10^{-3} \text{ kg/m} \)
In simple words: Linear density is basically how "heavy" the wire is per meter. We can calculate this by knowing the length, the tension pulling it, and the sound it makes when plucked.

📝 Teacher's Note: Linear density (mass per unit length) is often confused with volume density. Use a heavy string and a light string to show how mass per unit length changes the pitch.

🎯 Exam Tip: Don't forget to convert the load (kg) to tension (Newtons) by multiplying by \( g = 9.8 \text{ m/s}^2 \).

Question 24. The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency of 160 Hz, where should the person press the string? [Ans : 56 cm from one end]
Answer:
Data : \( L_1 = 80 \text{ cm}, n_1 = 112 \text{ Hz}, n_2 = 160 \text{ Hz} \)
According to the law of length, \( n_1L_1 = n_2L_2 \).

\( \therefore \) The vibrating length to produce the fundamental frequency of 160 Hz,
\( L_2 = \frac{n_1L_1}{n_2} = \frac{112(80)}{160} = 56 \text{ cm} \)
In simple words: To get a higher note on a guitar, you press down on the string to make the vibrating part shorter. This calculation shows exactly how short the string needs to be to reach the target note.

📝 Teacher's Note: This is a direct application of the inverse relationship between length and frequency. If frequency goes up, length must go down.

🎯 Exam Tip: State clearly that the person should press the string such that the *vibrating length* becomes the calculated value.

Can You Tell? (Textbook Page No. 132)

Question 1. What is the minimum distance between any two particles of a medium which always have the same speed when a sinusoidal wave travels through the medium ?
Answer:
When a sinusoidal wave travels through a medium the minimum distance between any two particles of the medium which always have the same speed is \( \frac{\lambda}{2} \).
Such particles are opposite in phase, i.e., their instantaneous velocities are opposite in direction.
[Note : The minimum distance between any two particles which have the same velocity is \( \lambda \)]
In simple words: Even if two particles are moving in opposite directions, they can have the same speed. This happens at points that are half a wavelength apart. To have the exact same velocity (speed AND direction), they must be a full wavelength apart.

📝 Teacher's Note: This distinction between speed and velocity is crucial. Speed is a scalar (doesn't care about direction), while velocity is a vector.

🎯 Exam Tip: Pay close attention to the terms "speed" and "velocity" in wave physics questions; they lead to different answers (\( \lambda/2 \) vs \( \lambda \)).

Do You Know? (Textbook Page No. 140)

Question 1. What happens if a simple pendulum is pulled aside and released ?
Answer:
If a simple pendulum is pulled aside and released, it oscillates freely about its equilibrium position at its natural frequency which is inversely proportional to the square root of its length and directly proportional to the square root of the acceleration of gravity at the place. These oscillations, called as free oscillations, are periodic and tautochronous if the displacement of its bob is small and the dissipative forces can be ignored.
In simple words: The pendulum swings back and forth on its own at a specific rhythm that depends mostly on its length. This is called its "natural" way of moving.

📝 Teacher's Note: "Tautochronous" means the time taken for one swing is the same regardless of the width of the swing (for small angles).

🎯 Exam Tip: Mention "natural frequency" and "free oscillations" as these are the technical terms examiners look for.

Question 2. What happens when a guitar string is plucked ?
Answer:
When a guitar string is plucked, two wave pulses of the same amplitude, frequency and phase move out from that point towards the fixed ends of the string where they get reflected. For certain ratios of wavelength to length of the string, these reflected pulses moving towards each other will meet in phase to form standing waves on the string. The vibrations of the string cause the air molecules to oscillate, forming sound waves that radiate away from the string. The frequency of the sound waves is equal to the frequency of the vibrating string. In general, the wavelengths of the sound waves and the waves on the string are different because their speeds in the two mediums are not the same.
In simple words: Plucking the string sends ripples to both ends. When they bounce back and meet correctly, they form a "standing wave" that makes the string vibrate steadily. This vibration then pushes the air around it to create the sound we hear.

📝 Teacher's Note: Emphasize that while frequency stays the same between the string and air, the speed (and therefore wavelength) changes significantly when moving from solid string to gas air.

🎯 Exam Tip: Remember: Frequency is determined by the source, but speed is determined by the medium.

Question 3. Have you noticed vibrations in a drill machine or in a washing machine ? How do they differ from vibrations in the above two cases ?
Answer:
Vibrations in the body of a drill machine or that of a washing machine are forced vibrations induced by the vibrations of the motors of these machines. On the other hand, the oscillations of a simple pendulum or a guitar string are free oscillations, produced when they are disturbed from their equilibrium position and released.
In simple words: A drill vibrates because a motor is constantly forcing it to. A guitar string vibrates because you plucked it once and let it go. One is "forced," the other is "free."

📝 Teacher's Note: Forced vibrations occur at the frequency of the driving force, whereas free vibrations occur at the object's own natural frequency.

🎯 Exam Tip: Use the comparison between "natural frequency" and "driving frequency" to explain forced vibrations clearly.

Question 4. A vibrating tuning fork of certain frequency is held in contact with a tabletop and its vibrations are noticed and then another vibrating tuning fork of different frequency is held on the tabletop. Are the vibrations produced in the tabletop the same for both the tuning forks ? Why ?
Answer:
No. Because the tuning forks have different frequencies, the forced vibrations in the tabletop differ both in frequency and amplitude. The tuning fork whose frequency is closer to a natural frequency of the tabletop induces forced vibrations of a larger amplitude.
In simple words: The table will vibrate at whatever frequency the fork tells it to. However, if a fork's frequency matches the table's own natural frequency, the table will vibrate much more strongly (resonance).

📝 Teacher's Note: This is a lead-in to the concept of resonance. The tabletop acts as a sounding board, amplifying the sound of the tuning fork through forced vibrations.

🎯 Exam Tip: Mention "amplitude" when talking about why different forks cause different vibration levels on the table.