Wave motion, sound, standing waves, resonance and beats.
CHAPTER 14
Waves
Follow a disturbance through a medium, derive wave speed, then connect interference, resonance and beats.
14.1
What waves carry
A wave carries a disturbance, energy and information through space or a medium.
A wave is a disturbance that propagates and transports energy without transporting matter as a whole. Medium particles oscillate about equilibrium; they do not travel with the wave.
Mechanical waves need an elastic material medium. Electromagnetic waves need no medium and travel in vacuum at c=2.99792458×10⁸ m s⁻¹. Matter waves are associated with particles. This chapter treats mechanical waves.
Coupled oscillators pass a disturbance to neighbours. In sound, a vibrating source creates compressions and rarefactions that move through the medium.
Wave types: transverse (particle displacement ⟂ propagation) and longitudinal (particle displacement ∥ propagation). Solids support both; fluids support longitudinal bulk waves. Surface water waves combine vertical and horizontal particle motion.
Short-wavelength capillary waves are restored mainly by surface tension; longer gravity waves by gravity.
The medium oscillates locally; the disturbance and its energy move along the string.Particle displacement is parallel to propagation; alternating dense and sparse regions travel.
NCERT Example 14.1
Classifying wave motion
Question. For each, state whether the wave motion is transverse, longitudinal or a combination: (a) a kink in a longitudinal spring when one end is displaced sideways; (b) waves in a liquid cylinder when its piston moves back and forth; (c) waves produced by a motorboat; (d) ultrasonic waves in air from a vibrating quartz crystal.
Step 1. (a) The sideways kink and along-spring compression give transverse and longitudinal components
Step 3. (c) Water-surface motion combines transverse and longitudinal components
Step 4. (d) Sound in air is longitudinal.
14.2
Progressive wave and its equation
A progressive wave shifts the same disturbance along the medium while each particle oscillates.
A sinusoidal wave travelling in +x is y(x,t)=a sin(kx−ωt+φ); in −x use kx+ωt (or x+vt). Here a is amplitude, φ initial phase, k wave number and ω angular frequency.
At fixed t, the shape varies sinusoidally with x; at fixed x, each medium particle executes SHM with angular frequency ω. A crest/trough is a maximum/minimum transverse displacement. Longitudinal displacement can be written s(x,t)=a sin(kx−ωt+φ).
Wavelength λ is the shortest distance between points in the same phase: k=2π/λ. Period T=2π/ω; frequency ν=1/T=ω/(2π). Phase at (x,t) is kx−ωt+φ.
Derive speed: follow a fixed phase kx−ωt+φ=C. Differentiating gives k dx/dt−ω=0; hence v=ω/k=λν=λ/T. The medium sets wave speed; source frequency then fixes λ.
A general pulse has y=f(x−vt) for +x, or f(x+vt) for −x. The profile translates unchanged at speed v.
NCERT Example 14.2
Read a harmonic wave
Question. For y=0.005 sin(80x−3t) (SI units), find amplitude, wavelength, period, frequency, and y at x=0.30 m,t=20 s.
Step 1. Compare with a sin(kx−ωt): a=.005 m, k=80 m⁻¹, ω=3 s⁻¹
Step 2. λ=2π/k=.0785 m; T=2π/ω=2.09 s; ν=ω/2π=.477 Hz
Step 3. At the stated point, phase=24−60=−36 rad, so y=.005 sin(−36)≈0.00496 m.
14.3
Wave speed in a string and in sound
Wave speed follows from the medium’s restoring elasticity and inertia.
For a stretched string, linear density μ=m/L and tension F give v=√(F/μ). Dimensional analysis: [F/μ]=L²T⁻², so v=C√(F/μ); the transverse-force derivation gives C=1.
For a longitudinal wave in a medium, v=√(B/ρ), where B is bulk modulus and ρ density. In a thin solid rod use Young’s modulus Y: v=√(Y/ρ).
Bulk modulus B=−ΔP/(ΔV/V). In an isothermal ideal gas B=P (Newton’s estimate v=√(P/ρ), about 280 m s⁻¹ at STP, too low). Sound compression/rarefaction is rapid and approximately adiabatic: B=γP, so Laplace’s result v=√(γP/ρ); γ=7/5 for air gives about 331.3 m s⁻¹ at STP.
For an ideal gas, P/ρ=RT/M, hence v=√(γRT/M): independent of pressure at fixed temperature, increases with absolute temperature. Humid air is less dense at equal P,T and sound travels faster. Greater elastic modulus generally makes sound faster; density opposes it.
NCERT Example 14.3
Wave speed on a steel wire
Question. A steel wire 0.72 m long has mass 5.0×10⁻³ kg and tension 60 N. Find transverse wave speed.
Step 1. μ=m/L=0.005/0.72=6.94×10⁻³ kg m⁻¹
Step 2. v=√(F/μ)=√(60/0.00694)≈93.0 m s⁻¹.
NCERT Example 14.4
Newton and Laplace sound speeds
Question. Estimate sound speed in air at STP using Newton’s isothermal model, then Laplace’s correction.
Step 1. Newton: B=P, v=√(P/ρ)≈280 m s⁻¹
Step 2. Sound changes are adiabatic, so B=γP
Step 3. For air γ=1.4, v=√γ times Newton’s result ≈331.3 m s⁻¹.
14.4
Superposition and interference
When waves overlap in a linear medium, their displacements add algebraically.
Principle of superposition: y=Σfᵢ(x±vᵢt). It applies while disturbances overlap; each continues through the other. It underlies interference.
For equal-amplitude waves a with same k and ω travelling in +x, y₁=a sin(kx−ωt), y₂=a sin(kx−ωt+δ). Using sin A+sin B gives y=2a cos(δ/2) sin(kx−ωt+δ/2).
The resultant keeps the same frequency and wavelength; amplitude depends on phase difference. δ=0 gives constructive interference, amplitude 2a. δ=π gives destructive interference, zero resultant displacement.
14.5
Reflection, standing waves and normal modes
Boundary reflections can combine with incident waves to form fixed vibration patterns.
At a rigid end, the reflected pulse has phase change π so displacement at the boundary remains zero. At an ideal free/open end there is no phase change. At an interface, part reflects and part transmits; oblique reflection/refraction follow their usual laws. Echo is reflection.
Equal waves travelling in opposite directions y₁=a sin(kx−ωt), y₂=a sin(kx+ωt) give y=2a sin kx cosωt. Space and time factors separate: this is a standing wave, with no net propagation of the pattern.
Amplitude at position x is 2a|sin kx|. Nodes: x=nλ/2; antinodes: x=(n+½)λ/2. Successive nodes (or antinodes) are λ/2 apart; adjacent loops oscillate oppositely. Every non-node element has the same frequency; ideal standing-wave elements are in phase within one loop and π out of phase in adjacent loops.
String fixed at both ends has nodes at x=0,L. L=nλ/2, so λₙ=2L/n and νₙ=nv/(2L), n=1,2,… . These natural modes are harmonics; the lowest is the fundamental.
Open pipe: displacement antinodes at both ends, all harmonics νₙ=nv/(2L). Closed-open pipe: displacement node at closed end, antinode at open end, λ=4L/(2n+1), ν=(2n+1)v/(4L); only odd harmonics. Pressure nodes and antinodes are reversed relative to displacement.
Forced vibration near a natural frequency gives resonance. A clamped circular membrane has two-dimensional normal modes; the circumference is a node. Sitar/violin timbre depends on the mix of modes.
Standing-wave nodes remain fixed; adjacent antinodes oscillate in opposite phase.
14.6
Sound in pipes and resonance
Boundary conditions select the allowed wavelengths of an air column.
At a closed pipe end air displacement is zero (node) and pressure variation is maximum (antinode). At an open end displacement is maximum (antinode) and pressure variation is minimum (node).
Geometry gives the modes: open-open ends are displacement antinodes, so L=nλ/2 and νₙ=nv/(2L). For closed-open, node-to-antinode spans an odd number of quarter wavelengths: L=(2n+1)λ/4 and νₙ=(2n+1)v/(4L). Hence a closed pipe has no even harmonics. End corrections can shift real resonances; neglect them only when the question says so.
A resonant mode is driven strongly when source frequency matches a natural frequency. Consecutive resonance lengths in a closed pipe differ by λ/2, which lets speed be determined without the end correction.
NCERT Example 14.5
Open and closed pipe modes
Question. A 30 cm pipe is open at both ends. Does a 1.1 kHz source resonate? Would it resonate if one end were closed? Take v=330 m s⁻¹.
Step 1. Open pipe fundamental ν₁=v/(2L)=330/.60=550 Hz; 1100 Hz=2ν₁, so second harmonic resonates
Step 2. Closed pipe fundamental is v/(4L)=275 Hz and allowed frequencies are 275,825,1375… Hz; 1100 Hz is absent, so no resonance.
14.7
Beats
Two close frequencies alternately reinforce and cancel, producing a periodic loudness change.
At a fixed point, use cos A+cos B=2cos((A−B)/2)cos((A+B)/2): s=a cosω₁t+a cosω₂t=2a cos(ωᵦt)cos(ωₐt), where ωₐ=(ω₁+ω₂)/2 and ωᵦ=(ω₁−ω₂)/2.
The rapid factor cos(ωₐt) is the audible oscillation; the slowly varying envelope sets loudness. Beat frequency is νᵦ=|ν₁−ν₂| (beats per second).
Musicians tune instruments by reducing the beat frequency. Increasing string tension increases frequency; decreasing tension lowers it.
NCERT Example 14.6
Sitar strings and beats
Question. Two sitar strings A and B playing “Dha” produce 5 Hz beats. The tension of B is slightly increased and the beat frequency decreases to 3 Hz. If νA=427 Hz, find the original frequency of B.
Step 1. Increasing B tension raises νB
Step 2. If B were above A, that increase would widen the beat gap
Step 3. Since the gap shrinks, B must initially be below A
Step 4. Thus νB=427−5.
MODEL
Watch a wave carry a pattern
The dots oscillate in place while the sinusoidal disturbance moves along the string. Change amplitude and frequency to see how the wave profile responds.
Particle motion is perpendicular to the direction of wave travel.
35 QUESTIONS · ALL SECTIONS
Quiz
Choose an answer to see why it is right.
0 / 35 AnsweredCorrect: 0 (0%)
1. A wave transports through a medium:
The particles oscillate locally while disturbance and energy propagate.
2. Which wave can propagate in vacuum?
Electromagnetic waves do not require a material medium.
3. In a longitudinal wave, particle displacement is:
Longitudinal displacement is parallel.
4. Capillary waves are restored mainly by:
Surface tension dominates for short wavelengths.
5. A wave travelling in +x has phase:
Constant kx−ωt moves toward increasing x.
6. Wave number equals:
k=2π/λ.
7. A phase point satisfies kx−ωt=C. Its speed is:
Differentiate phase: dx/dt=ω/k.
8. If medium speed is fixed and source frequency doubles, wavelength:
v=λν is fixed by the medium.
9. Transverse speed on a string is:
Tension supplies restoring force and μ is inertia per length.
10. Sound speed in an elastic medium is:
v=√(elastic modulus/density).
11. Laplace corrected Newton’s gas result by using:
Rapid sound propagation is approximately adiabatic.
12. At fixed temperature ideal-gas sound speed is independent of pressure because:
P/ρ=RT/M.
13. The superposition principle says displacements:
The net displacement is the sum of individual displacements.
14. Two equal in-phase waves of amplitude a produce amplitude:
Constructive interference doubles amplitude.
15. For equal waves, phase difference π gives:
Their displacements cancel everywhere.
16. Reflection at a rigid string end changes phase by:
The reflected disturbance cancels displacement at the fixed boundary.
17. A standing wave forms from waves of equal frequency moving:
Their superposition fixes nodes and antinodes.
18. Adjacent nodes in a standing wave are separated by:
Nodes occur at x=nλ/2.
19. A string fixed at both ends has fundamental wavelength:
For n=1, L=λ/2.
20. A clamped circular membrane has boundary nodes because:
The clamped boundary cannot move.
21. At the closed end of an air column, displacement is a:
Air cannot move through the closed boundary.
22. A closed-open pipe supports:
Allowed modes are odd multiples of fundamental.
23. For an open-open pipe fundamental frequency is:
The fundamental has λ=2L.
24. Successive closed-pipe resonances differ in length by:
Adjacent allowed resonances differ by half wavelength.
25. Beat frequency is:
Loudness maxima recur at the difference frequency.
26. Two sounds of 256 Hz and 260 Hz produce:
Beat rate is the absolute frequency difference.
27. Reducing string tension generally makes its frequency:
String speed and modal frequency increase with √tension.
28. A surface water wave has particle motion:
Surface particles have vertical and horizontal motion.
29. At fixed position, a harmonic wave particle oscillates at:
The time dependence is sinusoidal with ω.
30. At a sound displacement node, pressure variation is:
Pressure change follows the spatial gradient of displacement.
31. A sum of same-frequency waves travelling together remains:
The resultant has common k,ω and phase.
32. A solid rod’s longitudinal speed uses:
For a thin rod v=√(Y/ρ).
33. Two close tones sound different in loudness because their envelope:
The difference frequency sets the amplitude envelope.
34. A wave profile f(x+vt) travels toward:
Constant x+vt moves toward decreasing x.
35. A standing wave transports energy steadily along the string:
Equal counter-propagating waves produce no net transport.
NCERT · EXERCISES 14.1–14.19
Exercises
Try each textbook question before opening its worked solution.
NCERT 14.1
A string of mass 2.50 kg is under a tension of 200 N. How long would a transverse wave take to travel the length of the string, 20.0 m?
Step 1. A travelling profile must depend on x and t through x−vt or x+vt only. (a) and (c) do; in (b), x₀ is constant scale, so the argument still depends only on x+vt.
Step 2. All three are travelling forms (where defined); (a),(c) travel +x/−x respectively? f(x−vt) moves +x; f(x+vt) moves −x. (b) also moves −x.
NCERT 14.6
A bat emits ultrasound of frequency 1000 kHz into water. Find wavelengths in (a) reflected sound in air and (b) transmitted sound in water; speeds 340 and 1486 m s⁻¹.
For y=3 sin(36t+0.018x+π/4), x,y in cm and t in s, identify wave type, speed/direction, amplitude, frequency, phase at origin and least crest separation.
Step 2. All have amplitude 3 cm and frequency 36/(2π) Hz. Their phase constants are π/4, π/4+.036, π/4+.072 rad; draw equal sinusoids shifted progressively in phase.
NCERT 14.10
For y=2 cos 2π(10t−0.0080x+0.35), x,y in cm and t in s, find phase difference for separations (a) 4 m, (b) .5 m, (c) λ/2, (d) 3λ/4.
A string fixed at both ends has y=.06 sin(2πx/3)cos(120πt), length 1.5 m, mass .030 kg. Is it travelling or standing? Interpret as two waves; find λ, frequency, speed and tension.
Step 1. It factors into space and time, so it is standing. Use sin kx cosωt=½[sin(kx−ωt)+sin(kx+ωt)].
Step 2. Each component amplitude .03 m; k=2π/3, λ=3 m; f=60 Hz; v=fλ=180 m s⁻¹. μ=.030/1.5=.020 kg m⁻¹; F=μv²=648 N.
NCERT 14.12
(i) For the wave on a string described in Exercise 15.11 [as numbered in the provided reprint; apparently 14.11], do all points oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain. (ii) Find amplitude 0.375 m from one end.
Step 1. For y=.06 sin(2πx/3)cos(120πt), every moving point shares ω=120π. Spatial factor sets sign and amplitude; nodes do not oscillate.
Step 2. (a) Same frequency 60 Hz except nodes at rest. (b) Same phase within each loop, adjacent loops differ by π. (c) Amplitude varies with x. At x=.375 m: .06|sin(π/4)|=.0424 m.
NCERT 14.13
Classify each displacement: (a) y=2cos(3x)sin(10t); (b) y=2√(x−vt); (c) y=3sin(5x−.5t)+4cos(5x−.5t); (d) y=cos x sin t+cos2x sin2t.
Step 1. (a) is a product of space/time factors, so standing. (b) is f(x−vt), so travelling. (c) combines same phase into one sinusoid: 5 sin(5x−.5t+phase), travelling. (d) is a sum of two standing modes with different frequencies.
Step 2. (a) stationary; (b),(c) travelling +x; (d) none of the three as a single wave or single-frequency stationary pattern.
NCERT 14.14
A wire fixed at both ends vibrates in its fundamental at 45 Hz. Its mass is 3.5×10⁻² kg and μ=4.0×10⁻² kg m⁻¹. Find wave speed and tension.
Step 1. Length L=m/μ=.035/.040=.875 m. Fundamental λ=2L=1.75 m.
Step 2. v=fλ=78.75 m s⁻¹; F=μv²≈2.48×10² N.
NCERT 14.15
A one-metre tube open at one end and piston-closed at the other resonates at lengths 25.5 cm and 79.3 cm with a 340 Hz source. Estimate sound speed; neglect edge effects.
Step 1. The clamp is a displacement node; each free end is an antinode. Center-to-end distance is 0.50 m=λ/4, so λ=2.00 m.
Step 2. v=fλ=2530×2=5.06×10³ m s⁻¹.
NCERT 14.17
A 20 cm pipe is closed at one end. Which harmonic resonates at 430 Hz? Would the same source resonate if both ends were open? Take sound speed 340 m s⁻¹.
Step 1. Closed-pipe fundamental=340/(4×.20)=425 Hz; 430 Hz is approximately the fundamental (first odd harmonic). Open-pipe fundamental=340/(2×.20)=850 Hz.
Step 2. Closed pipe: fundamental resonance (within given approximation). Open pipe: no, 430 Hz is not an integer harmonic of 850 Hz.
NCERT 14.18
Two sitar strings A and B playing “Ga” produce 6 Hz beats. Slightly reducing A tension reduces beats to 3 Hz. If A originally has frequency 324 Hz, find B.
Step 1. Reducing A tension lowers νA. Since the beat gap shrinks, A was initially above B.
Step 2. νB=324−6=318 Hz.
NCERT 14.19
Explain: (a) sound displacement node is pressure antinode and vice versa; (b) bats use echoes to locate obstacles; (c) violin and sitar can share pitch but sound different; (d) solids support transverse and longitudinal waves but gases only longitudinal; (e) a pulse distorts in a dispersive medium.
Step 1. (a) Pressure variation is proportional to the negative spatial gradient of displacement: where displacement is zero by a node, its gradient is largest. (b) Echo delay and direction, strength and spectrum give distance, direction, size and surface clues. (c) The harmonic mixture and amplitudes (timbre) differ although fundamental frequency (pitch) is equal. (d) Solids have shear and bulk elasticity; fluids at rest have no shear restoring modulus. (e) In a dispersive medium speed depends on frequency, so pulse components separate in phase.
Step 2. These follow from boundary/constitutive properties, superposition and frequency-dependent phase speed.
POINT-WISE SUMMARY
Revision
Definitions, equations, assumptions and key results by concept.