Waves, Sound, and Physical Optics

1. One complete spatial cycle spans 0.80 m, and maximum displacement from equilibrium is 3.0 cm.
2. Frequency is cycles/time = 5/2.0 = 2.5 Hz; T = 1/f = 0.40 s; propagation speed is fλ = 2.0 m s⁻¹.
3. A single spatial profile can move either left or right. A later snapshot or local displacement–time information is needed to establish direction.
4. Superposition adds signed displacements: +3.0+(−2.0) = +1.0 cm upward.
5. Partial cancellation describes displacement at that location and instant. It does not mean the waves permanently destroy each other.
6. The pulses continue in the ideal linear model. Energy is redistributed between kinetic and elastic forms; displacement alone is insufficient to determine all local energy during overlap.

Answer (a) A = 3.0 cm; λ = 0.80 m; T = 0.40 s; f = 2.5 Hz; v = 2.0 m s⁻¹. (b) Resultant displacement is +1.0 cm; in the ideal linear medium both pulses continue through each other.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. For both ends fixed, L = nλ/2. Two loops identify 50 Hz as the second harmonic, not the fundamental.
2. The fundamental is 50/2 = 25 Hz. Wave speed is 2Lf₁ = 2(1.20)(25) = 60 m s⁻¹.
3. At 75 Hz, λ = 60/75 = 0.80 m; adjacent nodes are λ/2 = 0.40 m apart.
4. String wave speed follows v = √(T/μ), so multiplying tension by four doubles speed and all resonance frequencies.
5. At a node, destructive superposition keeps displacement zero; at an antinode the displacement amplitude is maximum.
6. The standing pattern arises from incident and reflected waves of equal frequency travelling in opposite directions. It is not a single wave carrying the pattern along the string.

Answer (a) Fundamental = 25 Hz; wave speed = 60 m s⁻¹; node spacing at 75 Hz = 0.40 m. (b) The three observed resonances become 100, 150 and 200 Hz.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. Consecutive closed–open resonances differ by λ/2, so λ = 2(0.50−0.16) = 0.68 m.
2. Wave speed is fλ = 500(0.68) = 340 m s⁻¹.
3. At the first resonance, L+e = λ/4 = 0.17 m, giving e = 0.17−0.16 = 0.010 m. The spacing method cancels the common end correction.
4. The water boundary restricts longitudinal air displacement, producing a displacement node. The open end supports an antinode slightly outside the mouth.
5. Allowed effective lengths are λ/4, 3λ/4, 5λ/4 and so on, so only odd harmonics of the fundamental occur for this ideal closed–open tube.
6. The next length is 0.84+0.34 = 1.18 m. Pressure variation has the opposite node/antinode pattern to air displacement.

Answer (a) λ = 0.68 m; sound speed = 340 m s⁻¹; e = 0.010 m. (b) Next measured length = 1.18 m; displacement node at the water and antinode near the open mouth.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. 500 lines mm⁻¹ is 5.0×10⁵ lines m⁻¹, so d = 1/(5.0×10⁵) = 2.0×10⁻⁶ m.
2. From d sinθ = nλ, λ/d is the graph gradient, 0.30; therefore λ = 6.0×10⁻⁷ m and f = c/λ = 5.0×10¹⁴ Hz.
3. The condition sinθ≤1 requires n≤d/λ = 3.33. Integer orders stop at n = 3; a fourth order would require sinθ = 1.2.
4. The grating equation predicts directions of principal maxima, while finite slit width determines their broad intensity envelope.
5. Measure both +n and −n peaks; half their angular separation reduces an offset in the estimated central direction.
6. Repeat and use higher visible orders where practical, while avoiding weak peaks and confusion between order number and wavelength. Never view a laser directly.

Answer (a) d = 2.0 μm; λ = 600 nm; f = 5.0×10¹⁴ Hz; maximum order = 3. (b) Diffraction spreads each slit contribution; interference between the many coherent slit contributions selects the narrow orders.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. For two-slit interference, w = λD/d. The slope w/D is 2.0 mm m⁻¹ = 0.0020.
2. λ = slope×d = 0.0020(0.00030) = 6.0×10⁻⁷ m.
3. Doubling d halves w at fixed λ and D. At 1.5 m, the original 3.0 mm spacing becomes 1.5 mm.
4. First-to-eleventh fringe centres contain ten intervals, not eleven. A 30 mm span is less sensitive to a fixed endpoint error than one 3 mm interval.
5. Keep wavelength and slit separation fixed while changing D; proportional spacing supports the wave model. Then change d while keeping D fixed and test the inverse relationship.
6. Record repeat endpoint readings, keep the screen perpendicular to the central ray and shield stray light. A ruler artefact would not track both controlled physical changes.

Answer (a) Wavelength = 600 nm; doubled-separation fringe spacing = 1.5 mm. (b) Measure a long span over many intervals and vary a physical input with a quantitative prediction.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

1. The 0.60 s interval covers the outward and return path, so distance is 340(0.60)/2 = 102 m.
2. For the moving source, fobserved = 680(340)/(340−20) = 722.5 Hz. Wave speed in the air remains 340 m s⁻¹.
3. Frequency determines pitch, while amplitude is associated with loudness. The weaker echo can retain the emitted frequency even though its amplitude is smaller.
4. To avoid envelope overlap, the returning pulse should begin after the outgoing pulse ends, requiring a delay exceeding its 0.05 s duration.
5. The round-trip distance is vΔt; the one-way limiting distance is 340(0.05)/2 = 8.5 m.
6. Real resolution also depends on detector response, noise and reflections from other surfaces. A shorter pulse improves time resolution but may be harder to detect.

Answer (a) Cliff distance = 102 m; approaching frequency = 722.5 Hz. (b) A delay greater than about 0.05 s is required; the corresponding distance is about 8.5 m.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.