Waves solutions

Superposition adds instantaneous signed displacements: y = y₁ + y₂.

Why: At one place and time, the string’s signed displacements add, rather than its wave amplitudes alone.

The phase angles are 9(4) − 4(4) = 20 rad and 4(4) − 4 = 12 rad.

Why: Insert the same x and t into each wave’s own phase before evaluating either sine.

Evaluate in radians: y = 4U sin(20) + 6U sin(12) = 3.651781U − 3.219438U = 0.432343U.

Why: Evaluate both sines in radians and retain their opposite signs when adding the two displacements.

The minus time sign describes travel in +x; the plus time sign describes travel in −x. They still superpose at each point.

Why: The phase’s time sign identifies travel direction, but both waves still contribute to the local displacement.

Their phases at the stated event are 1 − 6 = −5 rad and 1 + 6 = 7 rad.

Why: Substitute the specified position and time separately into the counter-traveling waves.

y = 6U sin(−5) + 8U sin(7) = 5.753546U + 5.255893U = 11.009438U.

Why: Add the two instantaneous sine values with their given amplitudes; both contributions happen to be positive here.

For a receding source and stationary listener, f_observed = f_source v/(v + v_source).

Why: A receding source stretches the spacing between sound fronts, so the observed frequency uses v + v_source in its denominator.

The ratio is f_observed/f_source = 340/(340 + 40) = 17/19 = 0.894737.

Why: Divide by the emitted frequency to find the dimensionless heard-to-emitted ratio.

The fractional decrease is 1 − 17/19 = 2/19 = 0.105263.

Why: Subtract the heard fraction from one to report the decrease, rather than the fraction still heard.

Light in vacuum has no material medium, so the sound Doppler formula cannot be used.

Why: Vacuum light propagation has no air-rest-frame sound speed, so this problem needs the relativistic light relation.

For recession, λ_observed/λ_emitted = √[(1 + β)/(1 − β)], where β = u/c.

Why: Recession stretches wavelength, corresponding to a ratio greater than one in the given longitudinal Doppler formula.

Let r = 700/500 = 1.4. Squaring and solving gives β = (r² − 1)/(r² + 1) = 0.96/2.96 = 0.324324.

Why: Square the wavelength ratio and isolate β algebraically to determine speed as a fraction of c.

u = βc = (0.324324)(299792458 m/s) = 9.72300 × 10⁷ m/s.

Why: Multiply the dimensionless β by vacuum light speed to report the recession speed in m/s.

Apply Snell’s law at the first interface: sin θ_water = (1.00/1.33)sin25°, giving θ_water = 18.5274° to the normal.

Why: Use the angle to the normal in Snell’s law to get the intermediate water-ray direction.

For parallel interfaces, n_air sin25° = n_water sinθ_water = n_glass sinθ_glass.

Why: Parallel boundaries share one normal direction, so n sinθ carries through both refractions unchanged.

θ_glass = arcsin(sin25°/1.52) = 16.1432° to the normal.

Why: Solve the air-to-glass invariant for the glass angle to that common normal.

The labelled source θ₂ is drawn from the horizontal interface, so θ₂ = 90° − 16.1432° = 73.8568°.

Why: The diagram labels an angle to the surface, so take the complement of the glass normal-angle.

Snell’s law uses angles to the normal, so incidence is 90° − 30° = 60°.

Why: Convert the pictured surface-angle to a normal-angle before applying Snell’s law.

At the first interface, sinθ_transmitted = (1/0.8)sin60° = 1.08253, which exceeds 1. Equivalently, θ_critical = arcsin0.8 = 53.1301° < 60°.

Why: The required transmitted sine is greater than one, equivalently putting incidence above the first boundary’s critical angle.

There is total internal reflection in the stated ray model, so the transmitted multi-layer path drawn in the source does not exist and its exit angle θ is undefined.

Why: The first interface already blocks the drawn propagating path in this ray model, so there is no later exit angle to calculate.

For monochromatic light, E_total = N hc/λ. A total energy alone leaves both N and λ unknown.

Why: Total light energy includes the number of photons, so without N it cannot isolate energy per photon or wavelength.

Only under a single-photon interpretation does λ = hc/E = (6.62607015 × 10⁻³⁴)(299792458)/20 = 9.93223 × 10⁻²⁷ m.

Why: If the entire 20 J belongs to one photon, the photon energy relation alone gives the formal wavelength.

Convert λ = 5.00 × 10⁻⁷ m and electron mass = 9.11 × 10⁻³¹ kg.

Why: Convert nanometers to meters and grams to kilograms so the photon and electron calculations use compatible SI units.

Each photon has energy E = hc/λ = 3.97289 × 10⁻¹⁹ J.

Why: The incident wavelength fixes the energy available from one photon through hc/λ.

The maximum kinetic-energy expression would be K_max = E − φ = −2.42711 × 10⁻¹⁹ J. A negative result means no single-photon emission, not a negative physical kinetic energy.

Why: The work function exceeds that photon energy, so a negative energy remainder rejects emission in the single-photon model.

Thus p = √(2mK_max) cannot be evaluated for an emitted electron. Increasing the number of incident photons does not make a subthreshold photon eject an electron in this model.

Why: Without an emitted electron there is no momentum to calculate; more separate subthreshold photons do not fix that deficit in this model.

The cutoff wavelength is λ_cutoff = hc/φ = 3.10382 × 10⁻⁷ m = 310.382 nm. The incident wavelength would need to be at or below that cutoff.

Why: Setting photon energy equal to the work function identifies the longest wavelength that can reach the emission threshold.

Convert water density: 1 g/cm³ = 1000 kg/m³. Use v = √(B/ρ).

Why: The sound-speed formula needs density in kg/m³; bulk modulus provides stiffness and density supplies inertia.

Air: v = √[(1.42 × 10⁵)/1.275] = 333.725 m/s, hence d = vt = 1668.63 m.

Why: Calculate air’s propagation speed from its supplied properties, then multiply by the common five-second travel time.

Water: v = √[(2.2 × 10⁹)/1000] = 1483.24 m/s, hence d = 7416.20 m.

Why: Repeat for water’s different stiffness and density; its larger speed gives a longer five-second travel distance.

Levels cannot be added directly. For N equal incoherent sources, I_total = NI_one and β_total = β_one + 10log₁₀N.

Why: Independent sources add intensities, while decibels are logarithms; equal sources increase the level by 10 log₁₀N.

140 − 65 = 10log₁₀N, so N = 10^7.5 = 31622776.6.

Why: Use the 75 dB difference to solve the logarithmic equation for the number of equal intensity contributions.

At least 31622777 idealized identical sources would be required if an integer count is wanted.

Why: Round upward because a fractional source cannot provide the minimum integer count needed to match or exceed the jet.

Convert level to intensity: I = I₀10^(β/10) = 10⁻¹²10^12.5 = 3.16228 W/m².

Why: Invert the logarithmic decibel definition to obtain physical intensity in watts per square meter.

Convert area: 55 mm² = 55 × 10⁻⁶ m² = 5.5 × 10⁻⁵ m².

Why: Square-length units require the 10⁻⁶ conversion from mm² to m².

Power crossing one eardrum area: P = IA = (3.16228)(5.5 × 10⁻⁵) = 1.73925 × 10⁻⁴ W.

Why: Multiply intensity by one eardrum’s area to find the incident power crossing that area.

Emitted source power would also require distance and radiation geometry. This calculation gives incident power, not absorbed power or total power emitted.

Why: A detector samples only part of a sound field, so distance and spreading geometry are needed to infer total emitted power.

k = 2π/λ = 8.37758 rad/m; ω = 2πf = 75.3982 rad/s.

Why: One wavelength and one period correspond to 2π phase radians, setting the spatial and temporal phase rates.

Travel toward +x uses kx − ωt. Thus y = (0.00400 m)sin[(8.37758 rad/m)x − (75.3982 rad/s)t].

Why: A constant phase of kx − ωt moves toward +x; put the amplitude in meters to keep the whole equation in SI units.

T = 1/f = 0.0833333 s and v = fλ = 9.00 m/s.

Why: The period is the reciprocal of frequency, and one wavelength per period gives propagation speed.

The phases are 1.25π and 1.75π.

Why: Insert the event’s x and t into the sine phase and cosine phase separately.

y₁ = (3.00 mm)(−√2/2) = −2.12132 mm; y₂ = (4.00 mm)(√2/2) = +2.82843 mm.

Why: Evaluate the quadrant signs carefully: the sine contribution is negative and the cosine contribution positive.

Add the signed displacements: y_total = +0.707107 mm.

Why: Superposition adds those signed instantaneous displacements, leaving the smaller positive difference.

For equal component amplitudes, A_result = 2A|cos(Δφ/2)|.

Why: Equal-frequency waves with a phase offset combine as phase vectors; their resultant is smaller than the sum of amplitudes unless they line up.

A_result = 2(7.00 mm)cos(π/6) = 12.1244 mm.

Why: Insert half the phase difference into the equal-amplitude result to calculate the combined amplitude.

For fixed ends, λ_n = 2L/n. With n = 3, λ₃ = 0.800 m.

Why: Fixed endpoints must be nodes, so three half-wavelengths fit into the length for the third harmonic.

f₃ = v/λ₃ = 90.0 Hz.

Why: Divide propagation speed by the permitted wavelength to find the third-harmonic frequency.

Nodes are separated by λ₃/2 = 0.400 m, including both fixed ends.

Why: Adjacent nodes are half a wavelength apart, and both fixed endpoints belong in the node list.

μ = 0.00600 kg/m.

Why: Convert linear density to kg/m so the tension-over-density speed formula uses SI units.

v = √(96.0/0.00600) = 126.491 m/s.

Why: Tension provides the restoring effect and linear density the inertia, setting the pulse propagation speed.

Travel time t = 3.50/126.491 = 0.0276699 s.

Why: Divide the path length by that wave speed to obtain the pulse’s travel time.

Beat frequency is the absolute difference: f_beat = |442 − 436| = 6 Hz.

Why: The relative phase cycles at the difference of the two frequencies, setting the beat frequency.

Successive maxima are separated by T_beat = 1/f_beat = 1/6 s.

Why: One beat cycle separates successive loudness peaks, so its duration is the reciprocal of that frequency difference.

v = √(B/ρ) = √[(1.65 × 10⁹)/950] = 1317.89 m/s.

Why: Use the adiabatic stiffness and density because the supplied model describes rapid compressions during sound propagation.

λ = v/f = 1317.89/2500 = 0.527157 m.

Why: One wave cycle travels one wavelength, so divide speed by the 2500 Hz frequency.

ω = 2πf = 4712.39 rad/s.

Why: The pressure-displacement relation uses angular frequency, so convert cycles per second to radians per second.

s_max = p_max/(ρvω) = 0.180/[(1.18)(346)(4712.39)] = 9.35563 × 10⁻⁸ m.

Why: Isolate displacement amplitude in the supplied acoustic relation, keeping pressure amplitude distinct from intensity.

The power spreads over area 4πr², so I = 0.120/[4π(7.50)²] = 1.69765 × 10⁻⁴ W/m².

Why: Uniform all-direction emission spreads the fixed acoustic power over a sphere of radius 7.50 m.

β = 10log₁₀(I/I₀) = 82.2985 dB.

Why: Convert that local intensity to a decibel level using the stated common reference intensity.

For inverse-square intensity, β₂ − β₁ = −20log₁₀(r₂/r₁).

Why: Inverse-square spreading becomes a factor of −20 when expressed as a logarithmic level change versus distance.

Thus r₂/r₁ = 10^[(88.0 − 70.0)/20] = 7.94328.

Why: Invert the logarithm of the 18 dB drop to find how many times farther away the detector must be.

r₂ = (2.50)(7.94328) = 19.8582 m.

Why: Multiply the starting distance by that ratio to locate the 70 dB point.

Convert each level to intensity relative to I₀: I₁/I₀ = 10^6.2 and I₂/I₀ = 10^6.7.

Why: Decibel values cannot be added directly, so turn each into a linear intensity relative to the same reference.

Add intensities, then convert back: β = 10log₁₀(10^6.2 + 10^6.7) = 68.1933 dB.

Why: Add the independent intensity contributions, then take the logarithm once to obtain the combined level.

Approach compresses the wave spacing: f_app = fv/(v − u) = (780)(343)/(343 − 28.0) = 849.333 Hz.

Why: Approach squeezes successive sound fronts closer together, raising frequency with the v − u source denominator.

Recession stretches it: f_rec = fv/(v + u) = (780)(343)/(343 + 28.0) = 721.132 Hz.

Why: Recession spaces the sound fronts farther apart, lowering frequency with the v + u source denominator.

For a receding observer and stationary source, f_obs = f(v − u_obs)/v.

Why: A listener moving away meets fewer fronts per second; the source stays still, so wave spacing itself is unchanged.

f_obs = (520)(344 − 14.0)/344 = 498.837 Hz.

Why: Use the listener’s reduced front-encounter speed v − u_obs to calculate the heard frequency.

The moving reflector first detects a shifted frequency and then acts as a moving source for the reflection.

Why: The reflector’s motion shifts frequency once when receiving and again when sending the reflected wave back.

f_return/f_emit = (1540 + 1.80)/(1540 − 1.80).

Why: The given two-stage Doppler ratio contains the approaching reflector’s speed in both numerator and denominator.

Δf = 4.00 × 10⁶[(1541.80/1538.20) − 1] = 9361.59 Hz.

Why: Multiply the ratio’s excess above one by the emitted frequency to isolate the upward frequency shift.

Snell’s law gives sinθ₂ = sin38.0°/1.46, so θ₂ = 24.9411°.

Why: Snell’s law uses normal-angles and the index ratio to find how far the ray bends toward the normal.

The frequency stays fixed across the stationary boundary while speed changes to c/n.

Why: A stationary boundary keeps oscillation frequency continuous, while the higher index slows propagation.

λ_silica = λ_vacuum/n = 632/1.46 = 432.877 nm.

Why: Because frequency stays fixed, the reduced light speed shortens wavelength by the same index factor.

The glass index exceeds the water index, so a critical angle exists.

Why: Total internal reflection is possible only when the ray attempts to leave the higher-index medium.

θ_c = arcsin(n_water/n_glass) = arcsin(1.33/1.62) = 55.1838°.

Why: Set the transmitted angle to 90° in Snell’s law to obtain the critical incident angle.

58.0° > θ_c, so there is total internal reflection in the ray model.

Why: The given incidence exceeds that limit, so no propagating transmitted ray exists in the stated ray model.

For the pulse, v_group = c/n_group.

Why: A pulse envelope travels at group speed, so use the group index rather than a phase-index value.

t = L/v_group = n_group L/c = (1.48)(2400)/(299792458) = 1.18482 × 10⁻⁵ s.

Why: Convert the path length to meters and divide it by group speed to find one-way transit time.

Each photon has E_photon = hc/λ = 4.90480 × 10⁻¹⁹ J.

Why: The monochromatic wavelength fixes the energy carried by each photon.

Photon rate = optical power/E_photon = 0.00450/(4.90480 × 10⁻¹⁹) = 9.17468 × 10¹⁵ s⁻¹.

Why: Optical power is joules emitted per second, so dividing it by joules per photon gives photons per second.

Photon energy E = hc/λ = 8.27686 × 10⁻¹⁹ J, which exceeds the work function.

Why: Compare the incident photon energy with the work function to check that single-photon emission is allowed.

K_max = E − φ = 4.07686 × 10⁻¹⁹ J.

Why: The work function uses part of the photon energy; the largest electron kinetic energy is the remaining amount.

Use K = p²/(2m): p_max = √(2mK_max) = 8.61831 × 10⁻²⁵ kg·m/s.

Why: Solve the nonrelativistic kinetic-energy relation for momentum using the electron mass.

This kinetic energy is much smaller than the electron rest energy, consistent with the nonrelativistic approximation.

Why: Compare kinetic energy with rest energy to check that the nonrelativistic momentum formula is appropriate.

Let r = λ_obs/λ_emit = 1.080. Relativistic recession gives r² = (1 + β)/(1 − β).

Why: The wavelength ratio is greater than one for recession, so use the relativistic longitudinal redshift relation.

β = (r² − 1)/(r² + 1) = (1.080² − 1)/(1.080² + 1) = 0.0768095.

Why: Solve that squared ratio equation for β, the dimensionless recession speed.

u = βc = 2.30269 × 10⁷ m/s.

Why: Multiply β by vacuum light speed to express the same recession in meters per second.