Chapter 14: Waves Case Study Questions | physics Class 11 CBSE

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Chapter 14: Waves Case Study Questions | physics Class 11 CBSE | ByteNotes.in

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Rohan is conducting an experiment with a long stretched string fixed at both ends. He attaches a vibrating tuning fork of frequency 100 Hz to one end. The speed of transverse waves on the string is 400 m s⁻¹. He observes that at certain conditions, the string no longer shows a travelling pattern but instead forms a beautiful pattern with fixed points of zero displacement and points of maximum vibration. He notes that the length of the string is exactly 2.0 m. He also observes that the string can be made to vibrate in multiple such patterns by changing the frequency of the driving source.

Question 1: What is the type of wave pattern Rohan observes on the string, and what are the fixed points of zero displacement called?

Rohan observes a standing wave (stationary wave) pattern formed by the superposition of the incident wave travelling forward and the reflected wave travelling backward along the string.
The fixed points of zero displacement are called nodes; at nodes the amplitude of vibration is zero and the particles of the string never move from their mean position.
The points of maximum displacement in the pattern are called antinodes; the amplitude there equals twice the amplitude of each constituent travelling wave.

Question 2: Calculate the wavelength and the fundamental frequency of the standing wave on the 2.0 m string, given the wave speed is 400 m s⁻¹.

For the fundamental mode (first harmonic) of a string of length L = 2.0 m fixed at both ends, the condition is L = λ/2, giving λ = 2L = 2 × 2.0 = 4.0 m.
The fundamental frequency is ν₁ = v/λ = 400/4.0 = 100 Hz; alternatively, ν₁ = v/2L = 400/(2 × 2.0) = 100 Hz.
The tuning fork frequency of 100 Hz exactly matches the fundamental, which is why a clear standing wave pattern is observed — the string resonates at its first harmonic.

Question 3: List all harmonic frequencies up to 500 Hz for this string. At which of these harmonics would a tuning fork of 300 Hz cause resonance, and how many loops (antinodes) would be visible?

The normal mode frequencies for a string of length 2.0 m and wave speed 400 m s⁻¹ are νₙ = nv/2L = n × 100 Hz for n = 1, 2, 3, 4, 5: giving 100 Hz, 200 Hz, 300 Hz, 400 Hz, 500 Hz.
A tuning fork of 300 Hz matches ν₃ = 300 Hz (the third harmonic, n = 3); resonance will occur for this mode.
In the third harmonic, the string vibrates in 3 loops (3 antinodes), with nodes at x = 0, 2/3 m, 4/3 m, and 2.0 m; adjacent nodes are separated by λ/2 = 400/(2 × 300) m ≈ 0.67 m.

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Case Set 2

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A physics teacher demonstrates the Laplace correction to students using the following scenario: at standard temperature and pressure (0°C, 1.01 × 10⁵ Pa), the density of air is 1.29 kg m⁻³. Newton had originally calculated the speed of sound in air using the assumption that sound propagation is an isothermal process, obtaining a value of approximately 280 m s⁻¹. However, the experimentally measured value was 331 m s⁻¹. Laplace pointed out the error and proposed that the process is adiabatic, with γ = Cp/Cv = 7/5 for air. The teacher asks students to verify Laplace's formula and understand its physical basis.

Question 1: What was Newton's assumption in deriving his formula for speed of sound, and what formula did he obtain?

Newton assumed that the compression and rarefaction of air during sound propagation occur under isothermal conditions — the temperature of the gas remains constant throughout the process.
Under isothermal conditions for an ideal gas, the bulk modulus B equals the gas pressure P (since PV = constant gives B = −V dP/dV = P).
Substituting into v = √(B/ρ) gives Newton's formula: v = √(P/ρ); for air at STP, this yields v ≈ 280 m s⁻¹, about 15% less than the experimentally measured 331 m s⁻¹.

Question 2: Explain why Laplace argued that the process is adiabatic and not isothermal during sound propagation.

Laplace argued that the pressure variations (compressions and rarefactions) during sound propagation occur extremely rapidly — so fast that the compressed region has virtually no time to exchange heat with its surroundings.
Because there is no heat exchange with surroundings during such rapid changes, the process is adiabatic (no heat transfer), not isothermal (constant temperature).
For an adiabatic process in an ideal gas, the effective bulk modulus is B_adiabatic = γP (where γ = Cp/Cv), which is larger than the isothermal value P, leading to a higher — and correct — value of sound speed.

Question 3: Using the Laplace formula, verify that the corrected speed of sound in air at STP is approximately 331 m s⁻¹. Also predict whether the speed will increase or decrease if the temperature is raised to 27°C, and justify your answer.

Laplace formula: v = √(γP/ρ) = √(1.4 × 1.01 × 10⁵ / 1.29) = √(1.414 × 10⁵ / 1.29) = √(109612) ≈ 331 m s⁻¹, confirming agreement with the measured value.
The speed of sound in a gas can be written as v = √(γRT/M), where T is the absolute temperature and M is the molar mass; v is directly proportional to √T.
Raising temperature from 0°C (273 K) to 27°C (300 K): v_new = v_STP × √(300/273) = 331 × √1.099 ≈ 331 × 1.048 ≈ 347 m s⁻¹. The speed increases with temperature because faster-moving molecules transmit pressure disturbances more rapidly.

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Case Set 3

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A musician is tuning a sitar before a concert. She plays string A (frequency 440 Hz) and string B simultaneously and hears a periodic waxing and waning of loudness at a rate of 4 times per second. She suspects string B is slightly out of tune. To correct it, she tightens the peg of string B slightly, which increases its tension. After tightening, she plays both strings again and notices the waxing and waning now occurs only 2 times per second. She continues to tighten until the beat sound completely disappears. The concert hall has a temperature of 25°C.

Question 1: What phenomenon is the musician observing, and what is its physical cause?

The musician is observing beats — a phenomenon arising from the superposition (interference) of two sound waves of slightly different frequencies being heard simultaneously.
When two waves of close frequencies ν₁ and ν₂ superpose, the resultant amplitude varies periodically with a frequency equal to the difference |ν₁ − ν₂|; this causes the intensity to wax and wane rhythmically.
The beat frequency is νbeat = |νA − νB|; here νbeat = 4 Hz initially, meaning the two strings differ in frequency by 4 Hz.

Question 2: Find the original frequency of string B, explaining your reasoning about whether it was higher or lower than string A.

Tightening string B increases its tension, which increases its frequency (ν ∝ √T). After tightening, the beat frequency decreased from 4 Hz to 2 Hz, meaning B's frequency moved closer to A's frequency.
If νB had originally been higher than νA = 440 Hz, tightening (increasing νB further) would have increased the beat frequency, not decreased it. Since beats decreased, νB was below νA.
Therefore νA − νB = 4 Hz, giving νB = 440 − 4 = 436 Hz. After tightening, the gap narrowed to 2 Hz, consistent with νB rising toward 440 Hz.

Question 3: Derive the mathematical expression for beat frequency starting from the superposition of two cosine waves s₁ = a cos ω₁t and s₂ = a cos ω₂t, and explain the condition for beats to be clearly audible.

Superposing s₁ = a cos ω₁t and s₂ = a cos ω₂t: s = a(cos ω₁t + cos ω₂t). Using the identity cos A + cos B = 2 cos((A−B)/2) cos((A+B)/2): s = 2a cos(ωbt) cos(ωat), where ωb = (ω₁−ω₂)/2 and ωa = (ω₁+ω₂)/2.
The factor [2a cos(ωbt)] acts as a slowly varying amplitude oscillating at ωb. Since the amplitude reaches its maximum magnitude (+2a or −2a) twice per cycle of cos(ωbt), the intensity (∝ amplitude²) peaks twice per cycle. The beat frequency is therefore 2ωb = ω₁ − ω₂, giving νbeat = ν₁ − ν₂.
For beats to be clearly audible, two conditions must be met: (i) the two frequencies must be very close so that |ν₁ − ν₂| is small (roughly 1–10 Hz) — if the difference is too large, the ear cannot distinguish the individual waxing and waning; (ii) the amplitudes should be comparable, since a very weak second wave barely affects the loudness variation.

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Case Set 4

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An engineer is designing an organ pipe for a church. She considers two configurations: Pipe P is open at both ends with a length of 0.5 m, and Pipe Q is closed at one end with the same length of 0.5 m. The speed of sound in air at the operating temperature is 340 m s⁻¹. She needs to determine the fundamental frequencies and the harmonic series produced by each pipe in order to select the correct pipe for a particular musical note. She also notes that the resonance condition is critical for producing a sustained, loud tone.

Question 1: Calculate the fundamental frequency of Pipe P (open at both ends, L = 0.5 m, v = 340 m s⁻¹).

For a pipe open at both ends, both ends are antinodes and the fundamental mode has one complete half-wavelength fitting in the pipe length: L = λ/2, so λ = 2L = 2 × 0.5 = 1.0 m.
Fundamental frequency: ν₁ = v/λ = 340/1.0 = 340 Hz.
Alternatively, using the formula ν₁ = v/2L = 340/(2 × 0.5) = 340 Hz; all harmonics nν₁ (n = 1, 2, 3, ...) are present for an open pipe.

Question 2: Calculate the fundamental frequency of Pipe Q (closed at one end, L = 0.5 m) and state which harmonics are absent.

For a pipe closed at one end, the closed end is a node and the open end is an antinode; the fundamental mode fits a quarter wavelength: L = λ/4, so λ = 4L = 4 × 0.5 = 2.0 m.
Fundamental frequency: ν₁ = v/λ = 340/2.0 = 170 Hz (alternatively ν₁ = v/4L = 340/2.0 = 170 Hz).
Only odd harmonics are present: 170 Hz, 510 Hz, 850 Hz, ... The even harmonics (340 Hz, 680 Hz, ...) are entirely absent from Pipe Q.

Question 3: The engineer requires a pipe producing a fundamental of 340 Hz. Both pipes have the same length 0.5 m. Which pipe should she choose, and what would be the first three harmonics of the chosen pipe? Also explain the boundary conditions at each end of the chosen pipe.

Pipe P (open at both ends) produces a fundamental of 340 Hz, which matches the requirement; Pipe Q (closed at one end) produces a fundamental of 170 Hz. The engineer should choose Pipe P.
For Pipe P, the harmonic series is νₙ = n × 340 Hz: first harmonic (fundamental) = 340 Hz, second harmonic = 680 Hz, third harmonic = 1020 Hz. All harmonics (both odd and even multiples of 340 Hz) are present.
Boundary conditions for Pipe P: at each open end, the air pressure equals atmospheric pressure at all times, so each open end is a pressure node and simultaneously a displacement antinode. The fundamental mode has one antinode at each end and one node at the midpoint of the pipe; the pipe length equals exactly one half-wavelength (λ/2 = 0.5 m).

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Case Set 5

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A student stretches a rubber string between two walls 1.2 m apart and plucks it at its midpoint. She observes that the string vibrates beautifully with a large amplitude and that the midpoint is a point of maximum vibration. She measures the linear mass density of the string to be 0.01 kg m⁻¹ and the tension to be 40 N. She changes the plucking point and notices that different vibrational patterns emerge. The student wants to understand the mathematical basis for these patterns and their frequencies.

Question 1: Calculate the speed of transverse waves on this rubber string.

The speed of transverse waves on a string is given by v = √(T/μ), where T is tension and μ is linear mass density.
v = √(40 / 0.01) = √4000 = 63.2 m s⁻¹.
This speed depends only on the mechanical properties of the string (tension and mass per unit length) and is independent of the frequency or amplitude of the wave.

Question 2: When the student plucks at the midpoint, which harmonic mode is preferentially excited and why? What is its frequency?

Plucking at the midpoint creates maximum displacement at the centre of the string; this is exactly the antinode position for the fundamental mode (first harmonic, n = 1), which has one antinode at the midpoint and nodes only at the fixed ends.
The midpoint is a node for even harmonics (n = 2 has a node at L/2, n = 4 has nodes at L/4 and 3L/4, etc.); plucking at the midpoint suppresses these even modes since they require zero displacement there.
Frequency of the fundamental: ν₁ = v/2L = 63.2/(2 × 1.2) = 63.2/2.4 ≈ 26.3 Hz.

Question 3: Write the equation of the standing wave in the third harmonic mode for this string and identify the positions of all nodes and antinodes.

For the third harmonic (n = 3): λ₃ = 2L/3 = 2 × 1.2/3 = 0.8 m; angular wave number k₃ = 2π/λ₃ = 2π/0.8 = 2.5π rad m⁻¹; frequency ν₃ = 3v/2L = 3 × 26.3 ≈ 78.9 Hz; ω₃ = 2πν₃ = 2π × 78.9 rad s⁻¹. Standing wave equation: y(x,t) = 2a sin(2.5πx) cos(ω₃t).
Nodes (sin(k₃x) = 0, i.e., k₃x = nπ): x = nλ₃/2 = n × 0.4 m for n = 0, 1, 2, 3: positions are x = 0, 0.4 m, 0.8 m, 1.2 m (four nodes including both fixed ends).
Antinodes (|sin(k₃x)| = 1, i.e., k₃x = (n+½)π): x = (n+½)λ₃/2 = (n+½) × 0.4 m for n = 0, 1, 2: positions are x = 0.2 m, 0.6 m, 1.0 m (three antinodes, forming 3 loops in the third harmonic).

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Chapter 14: Waves | Case Study Questions

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Question 1: What is the type of wave pattern Rohan observes on the string, and what are the fixed points of zero displacement called?

Question 2: Calculate the wavelength and the fundamental frequency of the standing wave on the 2.0 m string, given the wave speed is 400 m s⁻¹.

Question 3: List all harmonic frequencies up to 500 Hz for this string. At which of these harmonics would a tuning fork of 300 Hz cause resonance, and how many loops (antinodes) would be visible?

Passage

Question 1: What was Newton's assumption in deriving his formula for speed of sound, and what formula did he obtain?

Question 2: Explain why Laplace argued that the process is adiabatic and not isothermal during sound propagation.

Question 3: Using the Laplace formula, verify that the corrected speed of sound in air at STP is approximately 331 m s⁻¹. Also predict whether the speed will increase or decrease if the temperature is raised to 27°C, and justify your answer.

Passage

Question 1: What phenomenon is the musician observing, and what is its physical cause?

Question 2: Find the original frequency of string B, explaining your reasoning about whether it was higher or lower than string A.

Question 3: Derive the mathematical expression for beat frequency starting from the superposition of two cosine waves s₁ = a cos ω₁t and s₂ = a cos ω₂t, and explain the condition for beats to be clearly audible.

Passage

Question 1: Calculate the fundamental frequency of Pipe P (open at both ends, L = 0.5 m, v = 340 m s⁻¹).

Question 2: Calculate the fundamental frequency of Pipe Q (closed at one end, L = 0.5 m) and state which harmonics are absent.

Question 3: The engineer requires a pipe producing a fundamental of 340 Hz. Both pipes have the same length 0.5 m. Which pipe should she choose, and what would be the first three harmonics of the chosen pipe? Also explain the boundary conditions at each end of the chosen pipe.

Passage

Question 1: Calculate the speed of transverse waves on this rubber string.

Question 2: When the student plucks at the midpoint, which harmonic mode is preferentially excited and why? What is its frequency?

Question 3: Write the equation of the standing wave in the third harmonic mode for this string and identify the positions of all nodes and antinodes.

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