Exercises
1: String of mass 2.50 kg, tension 200 N, length 20.0 m. Time for disturbance to travel from one end to other?
Answer
Given: m = 2.50 kg, T = 200 N, L = 20.0 m
Linear mass density: μ = m/L = 2.50/20.0 = 0.125 kg/m
Speed of transverse wave: v = √(T/μ) = √(200/0.125) = √1600 = 40 m/s
Time = L/v = 20.0/40 = 0.5 s
2: Stone dropped from tower 300 m. When is splash heard at top? Speed of sound = 340 m/s, g = 9.8 m/s².
Answer
Time for stone to fall: h = (1/2)gt₁²
⇒ t₁ = √(2h/g) = √(2 × 300/9.8) = √61.22 = 7.82 s
Time for sound to travel back: t₂ = h/v = 300/340 = 0.882 s
Total time = t₁ + t₂ = 7.82 + 0.882 = 8.70 s
3: Steel wire 12.0 m, mass 2.10 kg. Tension for transverse wave speed = 343 m/s (speed of sound in air at 20 °C)?
Answer
μ = m/L = 2.10/12.0 = 0.175 kg/m
v = √(T/μ)
⇒ T = μv² = 0.175 × (343)² = 0.175 × 117,649 = 20,588 N ≈ 2.06 × 10⁴ N
4: Use v = √(γP/ρ) to explain why speed of sound in air is (a) independent of pressure, (b) increases with temperature, (c) increases with humidity.
Answer
(a) Independent of pressure:
From ideal gas law: P = ρRT/M
⇒ ρ = PM/(RT). So P/ρ = RT/M = constant at fixed T.
Since v = √(γP/ρ) = √(γRT/M), speed depends only on temperature, not pressure. If P increases, ρ increases proportionally, keeping P/ρ unchanged.
(b) Increases with temperature:
v = √(γRT/M)
⇒ v ∝ √T. As temperature increases, v increases.
At 0 °C: v ≈ 331 m/s; at 27 °C: v ≈ 347 m/s.
(c) Increases with humidity:
Moist air (with water vapor H₂O, M = 18) is less dense than dry air (average M ≈ 29). Since v = √(γP/ρ) and moisture decreases average molar mass
⇒ decreases density
⇒ increases v.
5: Is every function f(x ± vt) a travelling wave? Examine: (a) (x − vt)², (b) log[(x + vt)/x₀], (c) 1/(x + vt).
Answer
For a function to represent a travelling wave, it must be finite, continuous, and single-valued for all x and t, and the waveshape must remain constant as it propagates.
(a) y = (x − vt)²:
At t = 0: y = x² (parabola extending to infinity). As t increases, the parabola shifts right. This does not represent a physically valid travelling wave — it grows unboundedly and doesn't maintain a finite, localized shape.
(b) y = log[(x + vt)/x₀]:
This is defined only for x + vt > 0, and it grows without bound as x + vt
⇒ ∞. Not a valid travelling wave.
(c) y = 1/(x + vt):
Defined except at x + vt = 0, where it diverges. Has a discontinuity. Not a valid travelling wave.
Conclusion: The converse is not true — not every function of (x ± vt) represents a physically valid travelling wave. Valid waves must be finite, continuous, and single-valued.
6: Bat emits 1000 kHz in air, meets water surface. Wavelengths of (a) reflected sound, (b) transmitted sound? vair = 340 m/s, vwater = 1486 m/s.
Answer
Frequency remains same across boundary (boundary condition), f = 1000 kHz = 10⁶ Hz
(a) Reflected sound (in air): λreflected = vair/f = 340/(10⁶) = 3.4 × 10⁻⁴ m = 0.34 mm
(b) Transmitted sound (in water): λtransmitted = vwater/f = 1486/(10⁶) = 1.486 × 10⁻³ m = 1.49 mm
7: Ultrasonic scanner frequency = 4.2 MHz. Speed of sound in tissue = 1.7 km/s. Wavelength in tissue?
Answer
λ = v/f = (1.7 × 10³ m/s) / (4.2 × 10⁶ Hz) = 1.7 × 10³ / (4.2 × 10⁶) = 4.05 × 10⁻⁴ m ≈ 0.40 mm
8: y(x, t) = 3.0 sin(36t + 0.018x + π/4), x and y in cm, t in s. Is it travelling? Speed and direction? Amplitude, frequency? Initial phase? Least distance between crests?
Answer
(a) Travelling wave? Yes — the argument is (ωt + kx + φ), which represents a wave travelling in the −x direction (positive sign before x means propagation in −x direction).
Speed: v = ω/k = 36/0.018 = 2000 cm/s = 20 m/s, direction: −x direction (right to left)
(b) Amplitude and frequency:
Amplitude A = 3.0 cm
Angular frequency ω = 36 rad/s
⇒ f = ω/(2π) = 36/(2π) = 5.73 Hz
(c) Initial phase at origin (x = 0, t = 0): Phase = π/4 = 45°
(d) Wavelength (distance between crests):
k = 0.018 cm⁻¹
⇒ λ = 2π/k = 2π/0.018 = 2 × 3.14159/0.018 = 349 cm ≈ 349 cm
9: For the wave in Exercise 14.8, plot y vs t at x = 0, 2, 4 cm. How do oscillations differ?
Answer
y(0, t) = 3.0 sin(36t + π/4)
y(2, t) = 3.0 sin(36t + 0.036 + π/4)
y(4, t) = 3.0 sin(36t + 0.072 + π/4)
All three plots are identical sinusoidal curves with:
- Same amplitude: 3.0 cm
- Same frequency: 5.73 Hz
- Different phases: each successive point is shifted in phase by 0.036 rad
In a travelling wave, all points oscillate with the same amplitude and frequency, but different phases.
10: y(x, t) = 2.0 cos 2π(10t − 0.0080x + 0.35). Phase difference between points separated by: (a) 4 m, (b) 0.5 m, (c) λ/2, (d) 3λ/4.
Answer
Phase = 2π(10t − 0.0080x + 0.35)
Phase difference between two points: Δφ = 2π × 0.0080 × Δx (with x in cm)
Wait — x is in cm. So k = 2π × 0.0080 cm⁻¹ (rad/cm).
λ = 1/0.0080 = 125 cm = 1.25 m
Phase difference Δφ = (2π/λ) × Δx
(a) Δx = 4 m = 400 cm: Δφ = (2π/125) × 400 = 2π × 3.2 = 6.4π rad
(b) Δx = 0.5 m = 50 cm: Δφ = (2π/125) × 50 = 2π × 0.4 = 0.8π rad
(c) Δx = λ/2 = 62.5 cm: Δφ = (2π/125) × 62.5 = π rad
(d) Δx = 3λ/4 = 93.75 cm: Δφ = (2π/125) × 93.75 = 3π/2 rad
11: y(x,t) = 0.06 sin(2πx/3) cos(120πt). String length 1.5 m, mass 3.0 × 10⁻² kg.
Answer
(a) Type of wave: The function is a product of a function of x only and a function of t only
⇒ Stationary (standing) wave.
(b) Interpretation as superposition:
Using: sin A cos B = (1/2)[sin(A+B) + sin(A−B)]
y = 0.03[sin(2πx/3 + 120πt) + sin(2πx/3 − 120πt)]
Two waves of equal amplitude 0.03 m, both with:
k = 2π/3 rad/m → λ = 3 m
ω = 120π rad/s → f = 60 Hz
Speed = ω/k = 120π/(2π/3) = 120π × 3/(2π) = 180 m/s
(c) Tension in string:
μ = m/L = 3.0 × 10⁻² / 1.5 = 0.02 kg/m
v = √(T/μ)
⇒ T = μv² = 0.02 × (180)² = 0.02 × 32,400 = 648 N
12: For the wave in 14.11: (i) Do all points oscillate with same (a) frequency, (b) phase, (c) amplitude? (ii) Amplitude at x = 0.375 m?
Answer
(i)
(a) Frequency: Yes — all points oscillate at the same frequency (60 Hz). This is a property of standing waves.
(b) Phase: No — points in the same segment (between two nodes) oscillate in phase; points in adjacent segments oscillate in antiphase (180° phase difference). Phase is not uniform along the string.
(c) Amplitude: No — amplitude varies with position: A(x) = 0.06|sin(2πx/3)| m. It is zero at nodes and maximum at antinodes.
(ii) Amplitude at x = 0.375 m:
A(0.375) = 0.06 × |sin(2π × 0.375/3)| = 0.06 × |sin(0.25π)| = 0.06 × sin(45°) = 0.06 × 0.707 = 0.042 m = 4.24 cm
13: Classify each function as (i) travelling wave, (ii) stationary wave, (iii) neither.
Answer
(a) y = 2 cos(3x) sin(10t):
Product of function of x alone and function of t alone
⇒ (ii) Stationary wave
(b) y = 2√(x − vt):
Function of (x − vt) — but y must be real for x ≥ vt only; undefined for x < vt. (iii) Neither (not a valid physical wave — not finite for all x, t).
(c) y = 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t):
Both terms are functions of (5x − 0.5t) = 5(x − 0.1t): y = 5 sin(5x − 0.5t + φ)
⇒ (i) Travelling wave in +x direction with amplitude 5 (since √(3² + 4²) = 5).
(d) y = cos x sin t + cos 2x sin 2t:
Sum of two standing waves with different frequencies
⇒ (ii) Stationary wave (superposition of standing modes; not a pure travelling wave).
14: Wire between rigid supports vibrates at fundamental frequency 45 Hz. Mass = 3.5 × 10⁻² kg, linear density = 4.0 × 10⁻² kg/m. (a) Speed of transverse wave, (b) tension.
Answer
Length of wire: L = m/μ = (3.5 × 10⁻²)/(4.0 × 10⁻²) = 0.875 m
(a) Speed: For fundamental mode, L = λ/2
⇒ λ = 2L = 1.75 m
v = fλ = 45 × 1.75 = 78.75 m/s ≈ 79 m/s
(b) Tension: v = √(T/μ)
⇒ T = μv² = 4.0 × 10⁻² × (78.75)² = 0.04 × 6201.6 = 248 N
15: Tube 1 m open at one end, piston at other. Resonates at 340 Hz when tube length is 25.5 cm or 79.3 cm. Estimate speed of sound.
Answer
For a tube open at one end, closed at other — resonant lengths: L = (2n−1)λ/4
Difference between successive resonant lengths = λ/2
λ/2 = 79.3 − 25.5 = 53.8 cm
⇒ λ = 107.6 cm = 1.076 m
v = fλ = 340 × 1.076 = 365.8 m/s ≈ 366 m/s
16: Steel rod 100 cm clamped at middle. Fundamental frequency of longitudinal vibrations = 2.53 kHz. Speed of sound in steel?
Answer
When clamped at middle, the rod has a node at centre and antinodes at both free ends.
For fundamental: L = λ/2
⇒ λ = 2L = 2 × 1.00 = 2.00 m
v = fλ = 2530 × 2.00 = 5060 m/s ≈ 5.06 km/s
17: Pipe 20 cm closed at one end. Which harmonic is excited by 430 Hz? Will same source resonate if both ends open? vsound = 340 m/s.
Answer
Closed pipe: Resonant frequencies = (2n−1) × v/(4L) = (2n−1) × 340/(4 × 0.20) = (2n−1) × 425 Hz
n = 1: 425 Hz ✗; checking 430 ≈ 425 — not exactly resonant.
Actually: fn = (2n−1) × v/4L = (2n−1) × 425 Hz
For 430 Hz to be a harmonic: 430 = (2n−1) × 425
⇒ 2n−1 ≈ 1.01
⇒ n ≈ 1 (1st harmonic, very close).
More precisely — the source at 430 Hz will excite near resonance at the first harmonic (fundamental) of the closed pipe (425 Hz ≈ 430 Hz is close but not exact).
The NCERT answer: resonance occurs for n = 1 (1st mode, fundamental). The pipe would be in resonance if L were adjusted.
Open pipe: Resonant frequencies = n × v/(2L) = n × 340/(2 × 0.20) = n × 850 Hz
430 Hz is not an integer multiple of 850 Hz
⇒ No, the source will NOT resonate with the open pipe.
18: Sitar strings A and B playing note 'Ga'. Beat frequency = 6 Hz. Tension in A reduced → beats reduce to 3 Hz. Original frequency of A = 324 Hz. Find frequency of B.
Answer
Beat frequency = |fA − fB| = 6 Hz
⇒ fB = 324 ± 6 = 330 Hz or 318 Hz
When tension in A is reduced, fA decreases (f ∝ √T).
If fB = 330 Hz: reducing fA brings fA closer to fB
⇒ beats decrease
⇒ consistent ✓
If fB = 318 Hz: reducing fA from 324 would move fA further from 318
⇒ beats increase
⇒ inconsistent ✗
∴ fB = 330 Hz
19: Explain why (a) displacement node = pressure antinode in sound; (b) bats echolocate without eyes; (c) violin and sitar can be distinguished at same frequency; (d) solids support transverse waves but gases don't; (e) pulse shape distorts in dispersive medium.
Answer
(a) Displacement node = pressure antinode: At a displacement node, molecules cannot move (displacement is zero), so they are maximally compressed or rarefied by neighboring molecules — creating maximum pressure variation. Conversely, at a displacement antinode where molecules move most, pressure changes are minimal (molecules spread out evenly).
(b) Bat echolocation: Bats emit ultrasonic pulses and listen to the echoes. The time delay gives distance, the direction of the echo gives direction, and the change in frequency (Doppler effect) gives the relative velocity. Changes in echo intensity and frequency give information about size, shape, and surface texture of objects.
(c) Violin vs. sitar — same fundamental frequency but different timbre: The two instruments produce the same fundamental frequency (pitch) but different overtone spectra (harmonics). The violin's body shape and string/wood characteristics amplify certain harmonics differently than the sitar. Our ears detect these different harmonic compositions as different timbres or quality of sound.
(d) Transverse waves need shear rigidity: Transverse waves require the medium to have rigidity (resistance to shear stress). Solids have both bulk and shear moduli, supporting both longitudinal and transverse waves. Gases and liquids have no shear rigidity (they flow and cannot sustain shear forces) — so only longitudinal (compressional) waves propagate.
(e) Pulse distortion in dispersive medium: In a dispersive medium, wave speed depends on frequency (v = v(f)). A pulse is a superposition of many frequency components. If different frequencies travel at different speeds, they arrive at different times, spreading out the pulse and distorting its shape over distance. In a non-dispersive medium (v same for all f), all components travel together and the pulse shape is preserved.
Important Extra Questions
Extra Q1: What is the Doppler effect? Derive the formula for frequency heard when source moves towards a stationary observer.
Answer:
The Doppler effect is the change in observed frequency of a wave when there is relative motion between the source and the observer.
When source moves toward stationary observer at speed vs
The source emits f₀ waves in time T = 1/f₀. Distance between consecutive wave crests ahead of source:
λ' = λ − vsT = (v − vs)/f₀
Observed frequency: f = v/λ' = f₀ × v/(v − vs)
Since v − vs < v, f > f₀ (frequency increases as source approaches).
General formula: f = f₀ × (v + vO)/(v − vs)
where vO = observer speed (+ if towards source), vs = source speed (+ if towards observer), v = speed of sound.
Extra Q2: Explain standing waves. What are nodes and antinodes? Derive the condition for nodes and antinodes.
Answer:
When two identical waves travel in opposite directions and superpose, the result is a standing (stationary) wave:
y₁ = A sin(kx − ωt), y₂ = A sin(kx + ωt)
y = y₁ + y₂ = 2A sin(kx) cos(ωt)
Amplitude at position x: A(x) = 2A|sin(kx)|
Nodes (zero amplitude): sin(kx) = 0 → kx = nπ → x = nλ/2 (n = 0, 1, 2, ...)
Antinodes (maximum amplitude = 2A): sin(kx) = ±1 → kx = (2n+1)π/2 → x = (2n+1)λ/4
Between adjacent nodes: all points oscillate in phase. Nodes and antinodes alternate at spacing λ/4.
Extra Q3: A tuning fork of frequency 512 Hz is held above a resonance tube. First resonance at 16 cm, second at 49 cm. Find: (a) speed of sound, (b) end correction.
Answer:
For tube open at top, closed at bottom:
1st resonance: L₁ + e = λ/4 ... (1)
2nd resonance: L₂ + e = 3λ/4 ... (2)
(2) − (1): L₂ − L₁ = λ/2 = 49 − 16 = 33 cm → λ = 66 cm = 0.66 m
(a) Speed of sound: v = fλ = 512 × 0.66 = 337.9 m/s ≈ 338 m/s
(b) End correction: From (1): e = λ/4 − L₁ = 16.5 − 16 = 0.5 cm
Extra Q4: What are the conditions for a string fixed at both ends to produce standing waves? Derive the frequencies of harmonics.
Answer:
For a string fixed at both ends, displacement must be zero at both ends (boundary conditions: nodes at both ends).
Condition: L = nλ/2, where n = 1, 2, 3, ...
→ λn = 2L/n
Frequency: fn = v/λn = nv/(2L) = n × f₁
where f₁ = v/(2L) = (1/2L)√(T/μ) is the fundamental frequency.
The harmonics are: f₁, 2f₁, 3f₁, ... (all integer multiples of fundamental)
A fixed-fixed string produces all harmonics (both even and odd), unlike a closed pipe which produces only odd harmonics.
Extra Q5: Two strings A (f = 440 Hz) and B (f = 441 Hz) sound together. (a) Beat frequency? (b) How long until beats occur 10 times?
Answer:
(a) Beat frequency: fbeat = |fA − fB| = |440 − 441| = 1 Hz
One beat per second — the sound rises and falls in loudness once every second.
(b) Time for 10 beats: Time = number of beats / beat frequency = 10/1 = 10 seconds
Physical explanation: Beats arise from constructive and destructive interference. The two waves periodically come in phase (constructive → loud) and out of phase (destructive → soft). The beat period = 1/fbeat = 1 s; beat frequency = fA − fB.