Original practice sets for Waves on String are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. Speed of a transverse wave on a stretched string is:
Explanation: v = √(T/μ) where T = tension (N) and μ = linear mass density (kg/m). Higher tension or lower mass density → faster wave.
2. Linear mass density μ of a string is:
Explanation: μ = m/L = ρA where ρ is volume density and A is cross-sectional area. All three expressions give the same quantity.
3. The wave equation for a transverse wave moving in the +x direction is:
Explanation: y = A sin(ωt − kx) represents a wave moving in +x direction. y = A sin(ωt + kx) moves in −x direction.
4. A string of length 2 m, mass 10 g is stretched under tension 100 N. Wave speed is:
Explanation: μ = 0.01/2 = 0.005 kg/m. v = √(100/0.005) = √20000 = 141 m/s.
5. If tension in a string is quadrupled, wave speed:
Explanation: v ∝ √T. If T → 4T, v → 2v.
6. A wave pulse travelling from a lighter to a heavier string at a junction:
Explanation: At a denser boundary: part reflects with inversion (phase change π), part transmits without inversion. Speed decreases in denser (heavier) string.
7. Reflection at a free end (lighter to denser, then free end):
Explanation: At a free end, the string is unconstrained — the antinode forms at the end. Reflection occurs without phase inversion.
8. Average power transmitted by a sinusoidal wave on string is:
Explanation: P = ½μω²A²v. Power is proportional to square of amplitude and square of frequency.
9. In a guitar, thicker strings produce lower notes because:
Explanation: f₁ = v/(2L) = √(T/μ)/(2L). Higher μ → lower v → lower f₁. Same tension and length, thicker (heavier) string sounds lower.
10. Wave equation y = 0.01 sin(πx − 100πt) m. Amplitude, wavelength, and wave speed are:
Explanation: A = 0.01 m. k = π → λ = 2π/k = 2 m. ω = 100π → f = 50 Hz. v = ω/k = 100π/π = 100 m/s.
11. Power of wave on string is P. If amplitude doubles and frequency halves (with same tension):
Explanation: P ∝ ω²A² = (2πf)²A². If f → f/2: ω → ω/2. If A → 2A: A² → 4A². P ∝ (ω/2)²×(2A)² = (ω²/4)×4A² = ω²A². Power unchanged.
12. Two strings of same length, same tension: string A (μ_A) and string B (μ_B = 4μ_A). Ratio of fundamental frequencies f_A:f_B is:
Explanation: f ∝ v ∝ 1/√μ. f_A/f_B = √(μ_B/μ_A) = √4 = 2. So f_A:f_B = 2:1.
13. Transverse velocity of string element at x, t is:
Explanation: Transverse particle velocity = ∂y/∂t. For y = A sin(ωt−kx): ∂y/∂t = Aω cos(ωt−kx). Max transverse speed = Aω (not wave speed v).
14. A stretched rubber cord has mass 100 g, natural length 1 m. When stretched to 2 m, tension is 50 N. Wave speed at stretched length:
Explanation: μ = 0.1/2 = 0.05 kg/m (density doesn't change, mass stays constant). v = √(50/0.05) = √1000 ≈ 31.6... wait: v = √(T/μ) = √(50/0.05) = √1000 = 31.6? No: √1000 ≈ 31.6 m/s. Given options: √500 ≈ 22.4 and √1000 ≈ 31.6. If μ = 0.05: v = √(50/0.05) = √1000. Answer: √1000 m/s.
15. A wave with intensity I₁ reaches a string junction. If 40% is reflected, transmitted intensity is:
Explanation: Energy conservation: I_transmitted = I_incident − I_reflected = I₁ − 0.4I₁ = 0.6I₁ = 60%.
16. Slope of string at any position x, t is ∂y/∂x = −Ak cos(ωt−kx). This represents:
Explanation: ∂y/∂x is the slope at point x — the tangent angle. For small amplitudes, angle θ ≈ ∂y/∂x (in radians).
17. A wire under tension T vibrates at frequency f in fundamental. If tension increases to 4T and length doubles, new fundamental:
Explanation: f = v/(2L) = √(T/μ)/(2L). New: √(4T/μ)/(2×2L) = 2√(T/μ)/(4L) = √(T/μ)/(2L) × (1/√... wait: original f = v/(2L). New f' = √(4T/μ)/(2×2L) = 2v/(4L) = v/(2L) = f. Actually ratio = (√4T/(2×2L))/(√T/(2L)) = (2/2) = 1. f stays same? Let me recalculate: f = (1/(2L))√(T/μ). New: (1/(2×2L))√(4T/μ) = (1/4L)×2√(T/μ) = (1/2L)√(T/μ)/2... Actually (1/4L)×2 = 2/(4L) = 1/(2L). So new f = (1/2L)√(T/μ) = f. Same!
18. Relationship between transverse particle velocity (v_y) and wave speed (v) for a transverse wave:
Explanation: ∂y/∂t = −v × ∂y/∂x (from wave equation ∂²y/∂t² = v²∂²y/∂x²'s first integral). Transverse particle speed = −v × slope.
19. String is fixed at both ends. It vibrates in 5th harmonic. Number of nodes and antinodes:
Explanation: 5th harmonic: 5 loops. Nodes = 5+1 = 6 (including fixed ends). Antinodes = 5. So 6 nodes and 5 antinodes.
20. A transverse wave on string carries energy. At a node of a standing wave, instantaneous energy flow:
Explanation: At a node, displacement = 0 always and transverse velocity = 0. No energy flows past a node in a standing wave.
21. Standing waves are formed by superposition of:
Explanation: y = 2A sin(kx) cos(ωt). Standing wave formed when a travelling wave and its reflection superpose.
22. For a string fixed at both ends, fundamental frequency (1st harmonic) is:
Explanation: Both ends are nodes. λ/2 = L → λ = 2L. f₁ = v/λ = v/(2L).
23. For a string fixed at both ends, nth harmonic frequency is:
Explanation: f_n = nv/(2L) for n = 1, 2, 3,... All harmonics are allowed for a string fixed at both ends.
24. Standing wave equation y = 2A sin(kx) cos(ωt). Antinodes occur at kx equal to:
Explanation: Antinodes: sin(kx) = ±1 → kx = π/2, 3π/2, 5π/2 = (2n+1)π/2. Maximum amplitude = 2A at antinodes.
25. String fixed at both ends vibrates in 3rd harmonic. Number of loops (antinodes) is:
Explanation: nth harmonic has n loops (antinodes) and n+1 nodes. 3rd harmonic: 3 loops, 4 nodes.
26. A sonometer wire vibrates in fundamental. Law of lengths states frequency is:
Explanation: f₁ = v/(2L) ∝ 1/L. Shorter wire → higher frequency. Sonometer laws: f ∝ 1/L, f ∝ √T, f ∝ 1/√μ.
27. Sonometer: doubling the tension while keeping length same changes fundamental frequency by factor:
Explanation: f ∝ √T. T → 2T: f → f√2.
28. At a node in a standing wave on a string, which quantity is maximum?
Explanation: At nodes, displacement = 0 and velocity = 0. The slope ∂y/∂x is maximum at nodes (where the string is steepest), meaning maximum tension variation / restoring force.
29. Laws of vibration of a string: which one is correct?
Explanation: f = (1/2L)√(T/μ) ∝ 1/√μ. Heavier string (more mass per length) vibrates at lower frequency.
30. 5th overtone of a string fixed at both ends is the:
Explanation: Overtone numbering: 1st overtone = 2nd harmonic, 2nd overtone = 3rd harmonic, nth overtone = (n+1)th harmonic. 5th overtone = 6th harmonic.
31. A wire 1 m long, linear density 0.04 kg/m, under 100 N tension. Frequency of 3rd harmonic is:
Explanation: v = √(100/0.04) = √2500 = 50 m/s. f₁ = 50/(2×1) = 25 Hz. f₃ = 3×25 = 75 Hz.
32. y₁ = A sin(ωt−kx) and y₂ = A sin(ωt+kx). Resultant:
Explanation: Using superposition: A sin(ωt−kx) + A sin(ωt+kx) = 2A sin(ωt)cos(kx) = 2A cos(kx)sin(ωt).
33. String of 60 cm length vibrates at 180 Hz (4th harmonic). Wave speed is:
Explanation: f₄ = 4v/(2L) = 4v/(2×0.6) = 4v/1.2. 180 = 4v/1.2 → v = 180×1.2/4 = 54 m/s.
34. Two sonometer wires of lengths L₁ and L₂ are in unison. Tension in wire 1 is halved. To maintain unison, L₁ must become:
Explanation: f ∝ √T/L. Wire 1: f = √T₁/L₁ = √(T₁/2)/L₁'. For same f: L₁' = L₁ × √(T₁/2)/√T₁ = L₁/√2.
35. Average energy density in a standing wave compared to the original travelling wave (same amplitude):
Explanation: Standing wave = sum of two travelling waves. Total energy = twice that of one wave. Average energy density in standing wave = 2 × that of travelling wave.
36. String A (length L, μ) and String B (length 2L, 4μ) under same tension. Ratio of fundamental frequencies f_A:f_B is:
Explanation: f_A = v_A/(2L) = √(T/μ)/(2L). f_B = v_B/(4L) = √(T/4μ)/(4L) = √(T/μ)/(8L). f_A/f_B = (1/(2L))/(1/(8L)) = 4. Ratio 4:1.
37. A sonometer wire resonates with a tuning fork at 240 Hz. When 2 kg mass is added to the hanger (total becomes 5 kg), resonating length must change from L₁ to L₂ = L₁ × :
Explanation: f ∝ √T/L. f stays at 240 Hz. T₁ = 3g, T₂ = 5g. L₂/L₁ = √(T₂/T₁) = √(5/3).
38. In standing wave y = 2A sin(kx)cos(ωt), maximum energy is at antinodes and zero at nodes. Time-averaged KE and PE in one loop:
Explanation: Time-averaged KE = time-averaged PE for a standing wave over a complete cycle. Instantaneously they differ (KE max when PE = 0 at cos²ωt = 1, and vice versa).
39. String fixed at both ends vibrates in nth harmonic. Wavelength of the nth harmonic is:
Explanation: f_n = nv/(2L). v = fλ. λ_n = v/f_n = 2L/n.
40. String vibrates at f₁. Another string of same material and length under different tension vibrates at f₂. If f₂ > f₁ and 6 beats/second are heard, and increasing tension of string 1 by 1% increases beats to 7, then:
Explanation: f₂−f₁ = 6. Adding 1% tension: f₁ → f₁√1.01 ≈ f₁(1+0.005). Δf₁ = 0.005f₁. Beats become 7: f₂−f₁(new) = 7 → f₁ increased by 1 Hz → 0.005f₁ = 1 → f₁ = 200 Hz. So f₂ is 6 Hz higher and tension increase adds less than 6 Hz.
41. A string fixed at one end (free at other) has its fundamental with:
Explanation: Fixed end → node (zero displacement). Free end → antinode (max displacement). Fundamental: λ/4 = L.
42. Fundamental frequency of string fixed at one end, free at other (length L):
Explanation: Fundamental: λ/4 = L → λ = 4L. f₁ = v/(4L). Same as a closed organ pipe.
43. A string fixed at one end and free at other supports harmonics:
Explanation: f_n = nv/(4L) for n = 1, 3, 5,... Only odd harmonics allowed — analogous to closed organ pipe.
44. In Melde's transverse mode, when vibrator frequency doubles while tension is kept constant, resonating length of string:
Explanation: f = nv/(2L). For same n, if f doubles: L must halve to maintain resonance. Or more loops form at same length.
45. In Melde's longitudinal mode, string vibrates at half the frequency of vibrator. String is in resonance when vibrator gives n vibrations per second and string gives:
Explanation: In longitudinal mode, string frequency = half of vibrator frequency. One vibrator oscillation creates one complete loop.
46. If a string is struck transversely and a pulse is created, the pulse speed depends on:
Explanation: Wave speed v = √(T/μ) is a property of the medium, not of the disturbance. A hard strike creates a larger amplitude, not a faster pulse.
47. A 60 cm string is clamped at midpoint (making two 30 cm sections), both ends fixed. Possible resonant frequencies:
Explanation: Each 30 cm section vibrates independently as a string fixed at both ends: f = nv/(2×0.3) = nv/0.6. Allowed frequencies are multiples of v/0.6.
48. A heavy string hanging vertically: wave speed varies with position because:
Explanation: T(x) = μgx where x is measured from free end (bottom). v(x) = √(T/μ) = √(gx) increases upward. This creates non-uniform wave speed.
49. A string carries a travelling wave of power P. If wave reaches a fixed boundary and is totally reflected, power of reflected wave:
Explanation: For a perfectly rigid fixed boundary, all energy is reflected: reflected wave has same power P (and amplitude A). This is how standing waves store energy.
50. Sonometer wire resonates at 200 Hz in 1st harmonic. When length is reduced by 25%, resonant frequency in 1st harmonic:
Explanation: f₁ ∝ 1/L. L → 0.75L: f₁ → f/0.75 = 200/0.75 = 266.7 Hz.
51. 3rd harmonic of string fixed at one end, free at other of length 1 m (v = 200 m/s):
Explanation: f_n = nv/(4L) for odd n. f₃ = 3×200/4 = 150 Hz.
52. In Melde's transverse experiment, vibrator (200 Hz) makes string form 4 loops at tension T₁. At what tension T₂ will 5 loops form?
Explanation: f = nv/(2L) = n√(T/μ)/(2L). Same f and L: n₁√T₁ = n₂√T₂. 4√T₁ = 5√T₂ → T₂/T₁ = 16/25. T₂ = 16T₁/25.
53. String of length L has mass m attached at centre. Effective wave problem: the central mass causes:
Explanation: A mass on a string acts like an impedance discontinuity. It partially reflects and partially transmits waves, creating complex standing wave patterns with a partial node at the mass position.
54. A string shows harmonics f, 3f, 5f, 7f. This pattern indicates:
Explanation: Only odd harmonics f, 3f, 5f,... is characteristic of string fixed at one end (or closed organ pipe).
55. Dispersive waves on a deep water string: phase velocity v_p = √(g/k) and group velocity v_g = v_p/2. A wave packet (pulse) on deep water travels at:
Explanation: Energy and information travel at group velocity v_g = v_p/2 for deep water gravity waves. Individual wave crests move at v_p but the envelope (packet) moves at v_g.
56. Three sonometer wires of lengths 1:2:3 (same tension, same material) vibrate in harmonics such that all have same frequency. The harmonics are:
Explanation: f = nv/(2L) ∝ n/L. For same f: n₁/L₁ = n₂/L₂ = n₃/L₃. With L₁:L₂:L₃ = 1:2:3: n₁:n₂:n₃ = 1:2:3 → same ratios, so n:1, n:2, n:3 means 6:3:2 for simplest integers.
57. Transverse standing wave y = 2A sin(nπx/L)cos(ωt) on a string [0,L]. Number of nodes:
Explanation: Nodes at sin(nπx/L) = 0 → x = 0, L/n, 2L/n,...,L → n+1 nodes including both fixed ends.
58. Linear energy density (energy per unit length) of a travelling wave on string is:
Explanation: Total linear energy density = KE density + PE density = ½μω²A² + ½μω²A² = μω²A² (averaged over one wavelength).
59. A string of length 4 m resonates at 125 Hz in fundamental. At what position should it be touched (node) to sound 3rd harmonic?
Explanation: 3rd harmonic has 3 loops, 4 nodes at 0, 4/3, 8/3, 4 m. Touch at 4/3 m from either end (second node position, first interior node).
60. Speed of wave on a string doubles when tension is:
Explanation: v ∝ √T. To double v: T → 4T.
61. Fundamental mode of vibration of a string fixed at both ends has how many loops?
Explanation: Fundamental (1st harmonic) has exactly 1 loop (1 antinode) between the two nodes at fixed ends.
62. String of 64 cm gives 4 beats with a fork of 256 Hz. On shortening to 63.5 cm, beats decrease to 2. Frequency of string was:
Explanation: Shortening increases f. Beats decrease → string was higher than fork. f_string = 256+4 = 260 Hz. Check: at 63.5 cm, f → higher still, beats should increase... actually: if string was 260 Hz and fork 256 Hz, beats = 4. Shortening raises string → 262 Hz, beats = 262−256 = 6 — increases. So string must be below fork: 256−4 = 252 Hz. Shortening → 254 Hz, beats = 256−254 = 2. ✓ Answer: 252 Hz.
63. String fixed at both ends of length 3 m vibrates in 6th harmonic at 300 Hz. Wave speed is:
Explanation: f₆ = 6v/(2L) = 6v/6 = v. So v = 300 m/s.
64. String in resonance has maximum amplitude when driven at:
Explanation: Resonance occurs when driving frequency = natural frequency. Amplitude is maximum (limited by damping).
65. Law of mass density for sonometer: frequency is inversely proportional to:
Explanation: f = (1/2L)√(T/μ) ∝ 1/√μ. Frequency is inversely proportional to square root of linear density.
66. A wave y = 5×10⁻³ sin(100t − 2x) m. Amplitude, angular frequency, and wave number are:
Explanation: A = 5×10⁻³ m = 5 mm. ω = 100 rad/s. k = 2 rad/m. (f = ω/2π ≈ 15.9 Hz, not 100 Hz.)
67. Two travelling waves y₁ = A sin(ωt − kx + φ₁) and y₂ = A sin(ωt + kx + φ₂) superimpose. Resultant:
Explanation: Any two counter-propagating waves of same frequency form a standing wave. Different phases φ₁, φ₂ just shift the nodal positions along x.
68. Two strings of same tension. String 1 (μ₁ = μ) carries wave of amplitude A and frequency f. String 2 (μ₂ = 4μ) carries wave at same A and f. Ratio P₁:P₂:
Explanation: P = ½μω²A²v = ½μω²A²√(T/μ) = ½ω²A²√(μT). P ∝ √μ. P₁/P₂ = √(μ/4μ) = 1/2. So P₁:P₂ = 1:2.
69. Two strings A (L_A, μ_A) and B (L_B, μ_B) under same tension T. They are joined and the junction is free (antinode). If L_A = L_B and μ_A = μ_B/4, fundamental frequency of system:
Explanation: Joined strings have different wave speeds (v_A = 2v_B). The system's normal modes are complex and don't have simple closed-form frequencies without knowing boundary conditions.
70. A sinusoidal wave on string hits a denser string at junction. Reflected amplitude r and incident amplitude A are related by:
Explanation: Reflection coefficient: r/A = (v₂−v₁)/(v₂+v₁). If v₂
71. Sonometer wire of length L vibrates in p-th harmonic at f₁. Wire length changed to L/q. Frequency now:
Explanation: Original: f₁ = pv/(2L). New length L/q, fundamental of new length: f = v/(2×L/q) = qv/(2L). If comparing fundamental: f_new = qv/(2L) = q×(f₁/p)×... This depends on which harmonic is excited. If same harmonic p: f_new = pv/(2×L/q) = qpv/(2L) = qf₁.
72. Transverse wave on string: y = A sin(ωt − kx). If k = 0.5π rad/m and ω = 50π rad/s, wavelength and speed are:
Explanation: λ = 2π/k = 2π/(0.5π) = 4 m. v = ω/k = 50π/0.5π = 100 m/s.
73. A steel piano wire (μ = 0.005 kg/m, T = 500 N, L = 1 m) vibrates in 3rd harmonic. Frequency is:
Explanation: v = √(500/0.005) = √100000 = 100√10 m/s. f₃ = 3×100√10/(2×1) = 150√10 ≈ 474 Hz.
74. A string sounds simultaneously at 120 Hz, 180 Hz, 240 Hz. Fundamental frequency is:
Explanation: Harmonics 120, 180, 240 = 2×60, 3×60, 4×60. GCD = 60 Hz. Fundamental is 60 Hz (2nd, 3rd, and 4th harmonics are heard).
75. In Melde's transverse mode, a 1.5 m string forms 3 loops at a tension of 0.09 N and a frequency of 50 Hz. Its mass per unit length is:
Explanation: Using $f=n\sqrt{T/μ}/(2L)$ gives $50=3\sqrt{0.09/μ}/3=\sqrt{0.09/μ}$. Therefore $μ=0.09/50^2= 3.6×10^{-5}\text{ kg/m} $.
76. Two wires of ratio of radii 2:1, same material, length, tension vibrate at fundamental. Ratio of frequencies f₁:f₂ (thicker:thinner) is:
Explanation: μ = ρπr². μ ∝ r². f ∝ 1/√μ ∝ 1/r. f₁/f₂ = r₂/r₁ = 1/2. Thicker wire has half the frequency.
77. A wave in a lighter string (μ₁) reaches a junction with heavier string (μ₂ = 4μ₁). Reflection amplitude ratio:
Explanation: v₁ = √(T/μ₁), v₂ = √(T/4μ₁) = v₁/2. r/A = (v₂−v₁)/(v₂+v₁) = (v₁/2−v₁)/(v₁/2+v₁) = (−v₁/2)/(3v₁/2) = −1/3. Magnitude = 1/3.
78. String A (2 m, vibrates at 100 Hz in fundamental). String B of same material, same cross section, 1 m, tension 4 times. Fundamental of B:
Explanation: v ∝ √T. v_B = v_A√4 = 2v_A. L_B = 1 m = L_A/2. f_B = v_B/(2L_B) = 2v_A/(2×1) = v_A = v_A/(2×2)×4 = 4×f_A = 400 Hz.
79. A heavy uniform chain hangs vertically from a fixed point. Speed of transverse wave at a point distance x from free end is:
Explanation: Tension at height x from bottom = μgx (weight of chain below). v = √(T/μ) = √(μgx/μ) = √(gx).
80. Time for transverse pulse to travel from bottom to top of hanging chain (length L):
Explanation: v = √(gx). dt = dx/v = dx/√(gx). t = ∫₀ᴸ dx/√(gx) = (1/√g)×2√x |₀ᴸ = 2√(L/g).
81. Three waves of same amplitude A and frequency, with phases 0, 2π/3, 4π/3 superimpose. Resultant amplitude:
Explanation: Phasors at 0°, 120°, 240° — equally spaced on circle, they cancel completely. Resultant = 0.
82. For maximum power transmission from string 1 to string 2 at a junction:
Explanation: Acoustic impedance = μv. Maximum transmission occurs when Z₁ = Z₂ → no reflection. This is impedance matching.
83. String fixed at x=0 and x=L. For n=2 mode, nodes are at:
Explanation: 2nd harmonic (n=2): 3 nodes at x = 0, L/2, L. 2 antinodes at x = L/4, 3L/4.
84. Wave packet (pulse) travels on string. If string has some internal friction, the pulse:
Explanation: Internal friction (viscoelastic damping) converts wave energy to heat. Amplitude decreases exponentially. For most real strings, speed change is small but attenuation is observable.
85. A soliton (solitary wave) maintains its shape while travelling because:
Explanation: In nonlinear media, solitons are self-reinforcing wave packets where the nonlinearity prevents dispersion. Observed in optical fibres, water waves, and plasma physics.
86. Two identical strings are coupled by a spring. Normal modes of the coupled system:
Explanation: Coupling creates two normal modes: symmetric (both strings vibrate in phase — spring unstretched) and antisymmetric (out of phase — spring stretched, higher effective restoring force → higher frequency).
87. On a non-uniform string where μ = μ₀(1 + x/L), wave speed at x = L is (tension T throughout):
Explanation: At x = L: μ(L) = μ₀(1+1) = 2μ₀. v(L) = √(T/(2μ₀)).
88. A string is plucked at x = L/4 from one end. Which harmonics are absent in the sound produced?
Explanation: Plucking at x = L/4 excites all modes except those with a node at L/4. Modes with node at L/4: n = 4, 8, 12,... (where nπ×(L/4)/L = nπ/4 = mπ → n = 4m). Harmonics 4, 8, 12,... are suppressed.
89. For transverse waves on a string, in the long-wavelength limit (kL << 1 for a beaded string), the dispersion relation is:
Explanation: For a continuous string (or beaded string with λ >> bead spacing), the wave is non-dispersive: ω = vk = k√(T/μ). Dispersion (ω vs k non-linear) appears at short wavelengths comparable to the bead spacing.
90. Standing waves on a string are analogous to quantum mechanical particle-in-a-box because:
Explanation: In a quantum particle-in-a-box: ψ = 0 at walls → standing waves with λ_n = 2L/n → quantised k_n → quantised E_n = ℏ²k_n²/(2m). The mathematics is identical to string standing waves.
91. At resonance in a driven string with quality factor Q, amplitude relative to off-resonance amplitude is:
Explanation: At resonance, amplitude = Q × F₀/(mω₀²) = Q × static deflection. Higher Q (lower damping) → larger resonant amplitude.
92. A transverse string wave can be polarised because:
Explanation: String waves are transverse. Displacement can be in y-direction (vertical), z-direction (horizontal), or any linear combination. A groove or slot acts as a polariser, allowing only one direction of vibration to pass.
93. A string vibrates simultaneously in 1st and 3rd harmonics with amplitudes A₁ and A₃. The waveform is:
Explanation: Superposition of two modes: y = A₁ sin(πx/L)cos(ωt) + A₃ sin(3πx/L)cos(3ωt). Each mode has its own spatial pattern and temporal frequency.