Chapter 14: Waves Long Questions | physics Class 11 CBSE

ByteNotes

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

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Compare transverse and longitudinal waves on the basis of particle motion, medium requirements, examples, and wave speed. Explain why ocean waves are a special case.

In transverse waves, particles oscillate perpendicular to wave propagation; in longitudinal waves, particles oscillate parallel (along) the direction of propagation, forming compressions and rarefactions.
Transverse waves require media that can sustain shearing stress; only solids possess this property, so transverse waves cannot propagate through fluids (liquids or gases).
Longitudinal waves require only bulk modulus (compressibility), which all elastic media possess, so they propagate in solids, liquids, and gases alike.
Examples of transverse waves: waves on a stretched string, seismic S-waves. Examples of longitudinal waves: sound in air, ultrasonic waves produced by a quartz crystal.
Ocean (gravity) waves are a combination of both transverse and longitudinal motion; water particles trace elliptical paths — both up-down (transverse) and back-forth (longitudinal) — with amplitude decreasing with depth.
In the same medium, transverse and longitudinal waves generally travel with different speeds, reflecting the different elastic moduli (shear versus bulk) involved in each type of wave.

ByteNotes

Long Answer

Medium difficulty • Structured explanation

Medium

Question 2

Long Form

Derive the general speed formula v = λν for a progressive wave starting from the phase velocity concept, and explain why wave speed is determined by the medium and not by the source frequency.

For a progressive wave y(x,t) = a sin(kx − ωt + φ), any fixed phase point satisfies kx − ωt = constant; differentiating gives k(dx/dt) − ω = 0, so the phase velocity v = dx/dt = ω/k.
Substituting ω = 2π/T and k = 2π/λ gives v = (2π/T)/(2π/λ) = λ/T = λν, the general wave speed relation valid for all progressive waves.
This speed v is the speed at which any fixed-phase feature (crest, trough, node) of the wave moves through the medium.
The speed of a mechanical wave is determined solely by medium properties: for a string, v = √(T/μ); for sound in a fluid, v = √(B/ρ). These depend on tension, elasticity, and density — all medium characteristics.
The source determines only the frequency ν of the wave. Once v is fixed by the medium, the wavelength λ = v/ν adjusts accordingly; a high-frequency source produces short-wavelength waves in the same medium.
This is why, for example, the speed of sound in air does not change when a louder (higher amplitude) or higher-pitch (higher frequency) source is used — the medium alone governs the speed.

ByteNotes

Long Answer

Medium difficulty • Structured explanation

Medium

Question 3

Long Form

Using dimensional analysis, derive the formula for the speed of a transverse wave on a stretched string. Why does dimensional analysis leave a constant undetermined?

The speed of a transverse wave on a string depends on two medium properties: tension T (dimension [MLT⁻²]) and linear mass density μ (dimension [ML⁻¹]).
We need to combine T and μ to get dimensions of speed [LT⁻¹]; examining T/μ gives [MLT⁻²]/[ML⁻¹] = [L²T⁻²], whose square root is [LT⁻¹] — the correct dimension of speed.
Therefore, by dimensional analysis v = C√(T/μ), where C is a dimensionless constant that cannot be determined by dimensional analysis alone.
Dimensional analysis only checks that the formula has consistent dimensions; it cannot determine pure numbers, so C remains undetermined until the problem is solved using Newton's second law of motion.
The exact derivation (beyond scope of this chapter) shows that C = 1, giving v = √(T/μ) as the exact formula for transverse wave speed on a string.
This result is significant: it shows speed increases with tension (stronger restoring force) and decreases with linear mass density (greater inertia resisting acceleration), consistent with physical intuition.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 4

Long Form

Explain Newton's formula for speed of sound in a gas, identify the error in his assumption, and describe how Laplace corrected it. Compare the values obtained.

Newton assumed that the compression and rarefaction of air during sound propagation occur under isothermal conditions (temperature remains constant), giving bulk modulus B = P (the gas pressure).
Using v = √(B/ρ) with B = P gives Newton's formula: v = √(P/ρ). For air at STP (P = 1.01 × 10⁵ Pa, ρ = 1.29 kg m⁻³), this yields v ≈ 280 m/s.
The experimental value of the speed of sound in air at STP is 331 m/s; Newton's formula underestimates this by about 15%, indicating an error in the assumption.
Laplace corrected Newton by pointing out that sound propagation is adiabatic, not isothermal: pressure changes occur so rapidly that the compressed (or rarefied) region has no time to exchange heat with surroundings.
For an adiabatic process, PV^γ = constant, giving the adiabatic bulk modulus B_ad = γP, where γ = Cp/Cv. For air, γ = 7/5 = 1.4.
The corrected Laplace formula v = √(γP/ρ) gives v ≈ √(1.4) × 280 ≈ 331.3 m/s for air at STP, in excellent agreement with the measured value.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 5

Long Form

Derive the equation of a standing wave on a string by superposing two waves travelling in opposite directions. Identify nodes and antinodes and find their positions.

Consider two identical waves of amplitude a travelling in opposite directions: y1 = a sin(kx − ωt) in the +x direction and y2 = a sin(kx + ωt) in the −x direction.
By the principle of superposition, the resultant is y = y1 + y2 = a[sin(kx − ωt) + sin(kx + ωt)]. Using the identity sin A + sin B = 2 sin((A+B)/2)cos((A−B)/2):
y(x, t) = 2a sin(kx) cos(ωt). The terms kx and ωt appear separately, confirming the wave pattern does not propagate; the amplitude at any point x is 2a sin(kx).
Nodes (amplitude = 0): 2a sin(kx) = 0 requires sin(kx) = 0, i.e., kx = nπ, giving x = nλ/2 for n = 0, 1, 2, 3, ... Adjacent nodes are separated by λ/2.
Antinodes (amplitude = maximum = 2a): |sin(kx)| = 1 requires kx = (n + 1/2)π, giving x = (n + 1/2)λ/2 for n = 0, 1, 2, ... Adjacent antinodes are also separated by λ/2.
Unlike a progressive wave where energy flows, a standing wave stores energy — particles between two nodes all oscillate in phase but with position-dependent amplitudes, and no net energy is transported along the string.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 6

Long Form

Determine the normal modes of vibration of a stretched string of length L fixed at both ends. Explain the concepts of fundamental mode and harmonics.

For a string of length L fixed at both ends, both ends (x = 0 and x = L) must be nodes of the standing wave pattern. The node condition is sin(kx) = 0, giving kx = nπ.
At x = L: kL = nπ, so (2π/λ)L = nπ, giving L = nλ/2 or λ = 2L/n for n = 1, 2, 3, ...
The corresponding frequencies are νn = v/λ = nv/2L for n = 1, 2, 3, ..., where v = √(T/μ) is the wave speed in the string. These are the normal modes (natural frequencies).
The fundamental mode (n = 1, first harmonic) has frequency ν1 = v/2L; the string vibrates as a single loop with one antinode in the middle and nodes only at the fixed ends.
The second harmonic (n = 2) has frequency ν2 = 2ν1 = v/L; the string vibrates in two loops with a node at the midpoint. All harmonics (n = 1, 2, 3, ...) are present for this system.
In practice, a vibrating string (as in sitar or violin) produces a superposition of multiple modes; the prominence of each mode (timbre of the instrument) depends on where and how the string is plucked or bowed.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 7

Long Form

Determine the normal modes of an air column in a pipe closed at one end and open at the other. Show that only odd harmonics are present, and compare with a pipe open at both ends.

For a pipe closed at one end (x = 0) and open at the other (x = L), the boundary conditions are: node at x = 0 (closed end, no displacement) and antinode at x = L (open end, maximum displacement).
Node at x = 0 is automatically satisfied by the standing wave y = 2a sin(kx) cos(ωt). Antinode at x = L requires |sin(kL)| = 1, so kL = (n + 1/2)π for n = 0, 1, 2, ...
This gives L = (2n + 1)λ/4, or λ = 4L/(2n + 1). The allowed frequencies are νn = v/λ = (2n + 1)v/4L = v/4L, 3v/4L, 5v/4L, ...
Only odd multiples of the fundamental frequency v/4L are allowed; even harmonics are absent. The harmonics present are the 1st, 3rd, 5th, 7th, ... (odd harmonics only).
For a pipe open at both ends, both ends are antinodes; the boundary condition gives L = nλ/2, so νn = nv/2L for n = 1, 2, 3, ... All harmonics (both odd and even) are present, with fundamental frequency v/2L.
The fundamental of the open pipe (v/2L) is twice that of the closed pipe of the same length (v/4L), which is why the open pipe produces a richer, brighter tone with more harmonics.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 8

Long Form

Derive the mathematical expression for beats and explain how musicians use this phenomenon to tune instruments. Mention the condition for beats to be audible.

Consider two harmonic sound waves of nearly equal frequencies ω1 and ω2, both of amplitude a, at the same location x = 0: s1 = a cos ω1t and s2 = a cos ω2t.
By superposition: 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 is the slowly varying beat angular frequency and ωa = (ω1 + ω2)/2 is the average angular frequency.
The term [2a cos(ωbt)] acts as the time-varying amplitude of the resultant wave oscillating at ωa. Since the amplitude is maximum (±2a) twice per cycle of cos(ωbt), the intensity waxes and wanes at frequency 2ωb = ω1 − ω2.
Therefore the beat frequency is νbeat = ν1 − ν2 (the difference of the two frequencies), and the resultant sound oscillates at the average frequency (ν1 + ν2)/2.
Musicians use beats to tune instruments: they play two strings or pipes simultaneously and listen for beats; they adjust tension or length until the beat frequency reduces to zero, indicating both sources have identical frequencies.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 9

Long Form

Analyse the role of the principle of superposition in producing both constructive and destructive interference. Derive the condition for each and discuss the energy implications.

Consider two harmonic waves of equal amplitude a and angular frequency ω, travelling in the same direction but differing in phase by φ: y1 = a sin(kx − ωt) and y2 = a sin(kx − ωt + φ).
By superposition: y = y1 + y2 = a[sin(kx − ωt) + sin(kx − ωt + φ)]. Using sin A + sin B = 2 sin((A+B)/2) cos((A−B)/2): y = 2a cos(φ/2) sin(kx − ωt + φ/2).
The resultant is a harmonic wave with amplitude A = 2a cos(φ/2). This shows the amplitude depends entirely on the phase difference φ between the two component waves.
Constructive interference occurs when φ = 0, 2π, 4π, ...; cos(φ/2) = 1, so A = 2a (maximum). The waves reinforce each other, producing the largest possible displacement.
Destructive interference occurs when φ = π, 3π, 5π, ...; cos(φ/2) = 0, so A = 0. The waves completely cancel, with zero displacement everywhere at all times.
Energy is conserved: in constructive interference, intensity (∝ A²) increases to 4 times the individual intensity; in destructive regions, it is zero. When averaged over all phase differences, the mean intensity equals the sum of individual intensities, consistent with energy conservation.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 10

Long Form

Explain how the speed of sound in a gas depends on temperature, pressure, and humidity using the formula v = √(γP/ρ). Predict how each factor affects the speed.

The Laplace formula for speed of sound in a gas is v = √(γP/ρ), where γ = Cp/Cv is the ratio of specific heats, P is pressure, and ρ is the mass density of the gas.
Effect of pressure at constant temperature: For an ideal gas, P/ρ = RT/M (where R is gas constant, T is absolute temperature, M is molar mass) so P/ρ is constant when T is constant. Therefore v = √(γRT/M) is independent of pressure — doubling pressure also doubles density at constant temperature, leaving v unchanged.
Effect of temperature: Since v = √(γRT/M), the speed of sound is directly proportional to √T (absolute temperature). As temperature rises, molecular speeds increase, so pressure disturbances propagate faster.
Effect of humidity: Water vapour (H2O, molar mass 18 g/mol) is lighter than dry air (molar mass ≈ 29 g/mol). When humidity increases, the average molar mass M of the air-water vapour mixture decreases.
Since v = √(γRT/M), a decrease in M increases v; therefore, sound travels faster in humid air than in dry air at the same temperature and pressure.
Combining effects: The speed of sound increases with temperature and humidity, and is independent of pressure — important for acoustic engineering and atmospheric science applications.

ByteNotes

Premium Access

Unlock all Long Questions

You are viewing 10 free long questions from this chapter. Upgrade to Premium to unlock the remaining 10 questions.

Upgrade to Premium10 questions locked

Chapter sections

Chapter 14: Waves | Long Questions

Compare transverse and longitudinal waves on the basis of particle motion, medium requirements, examples, and wave speed. Explain why ocean waves are a special case.

Derive the general speed formula v = λν for a progressive wave starting from the phase velocity concept, and explain why wave speed is determined by the medium and not by the source frequency.

Using dimensional analysis, derive the formula for the speed of a transverse wave on a stretched string. Why does dimensional analysis leave a constant undetermined?

Explain Newton's formula for speed of sound in a gas, identify the error in his assumption, and describe how Laplace corrected it. Compare the values obtained.

Derive the equation of a standing wave on a string by superposing two waves travelling in opposite directions. Identify nodes and antinodes and find their positions.

Determine the normal modes of vibration of a stretched string of length L fixed at both ends. Explain the concepts of fundamental mode and harmonics.

Determine the normal modes of an air column in a pipe closed at one end and open at the other. Show that only odd harmonics are present, and compare with a pipe open at both ends.

Derive the mathematical expression for beats and explain how musicians use this phenomenon to tune instruments. Mention the condition for beats to be audible.

Analyse the role of the principle of superposition in producing both constructive and destructive interference. Derive the condition for each and discuss the energy implications.

Explain how the speed of sound in a gas depends on temperature, pressure, and humidity using the formula v = √(γP/ρ). Predict how each factor affects the speed.

Unlock all Long Questions