Waves · quiz worked solutions

Tried the quiz first? Good. Here’s how to work through each answer. Follow the link beside a question to jump back to its original spot. Questions marked Extra exploration go beyond this AP Physics 1 course; Book controls questions explain the interface.

Light · question 1

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Question: If a radio station broadcasts on station 101.3FM, that equates to 101.3 MHz . What is the wavelength of this station’s broadcast?

Answer: 2.96 m.

1. Convert frequency 101.3 MHz to 101.3×10⁶ Hz. Why: Hertz counts cycles per second.

2. Use λ = c/f = (2.99792458×10⁸)/(101.3×10⁶) ≈ 2.9595 m. Why: Radio is electromagnetic radiation traveling at c in the vacuum approximation.

Light · question 2

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Question: If the sun were to emit a solar flare, how long would the light take to reach the earth? The distance from the earth to the sun is approximately 1.496 × 10 11 m .

Answer: About 8 min 19 s.

1. Use t = distance/speed = (1.496×10¹¹)/(2.99792458×10⁸) ≈ 499.01 s. Why: The question asks the propagation delay of light, not the arrival of flare matter.

2. Convert 499 s to 8 minutes and 19 seconds. Why: There are 60 seconds per minute.

Light · question 3

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Question: The nearest star to earth, Proxima Centauri, is 4.243 light years away. That means it takes its light 4.243 years to reach earth. Determine the distance from earth to that star in km .

Answer: 4.014×10¹³ km.

1. Use one light-year as c times one Julian year, 365.25 days. Why: A light-year is a distance, despite its name.

2. Find d = (4.243)(299792.458 km/s)(365.25·86400 s) ≈ 4.014×10¹³ km. Why: The year-to-second conversion supplies the distance unit.

Light · question 4

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Question: Sound travels at 340.29 m/s If you see lightning strike near you, and you hear the accompanying thunder after 5 seconds, how far away is the strike?

Answer: 1.70 km.

1. Neglect the much shorter light travel time relative to sound. Why: The observed 5 s delay is then approximately the sound travel time.

2. Find d = (340.29)(5) = 1701.45 m = 1.701 km. Why: This is an approximate lightning-distance estimate under the stated uniform sound speed.

Light · question 5

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Question: For an object receding along the line of sight, emitted visible spectral features shift toward which end of the visible spectrum?

Answer: Toward red.

1. Recession stretches the received wavelength of a spectral feature. Why: This is a Doppler shift of the spectrum.

2. Longer visible wavelengths lie toward the red end. Why: The object’s emitted spectrum is shifted rather than replaced by one pure color.

Mechanical Waves · question 1

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Question: Sound waves in air are mechanical transverse waves.

Answer: False.

1. Air sound consists of pressure compressions and rarefactions. Why: Its particles oscillate along the propagation direction.

2. That makes ordinary sound in air longitudinal, not transverse. Why: Mechanical means it requires a medium; it does not imply transverse motion.

Mechanical Waves · question 2

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Question: Find the resultant displacement at x = 1 m and t = 4 s for the two waves below. Take each coefficient as a displacement in mm and all phase arguments in radians. Wave 1: y = 5 sin ( 0.5 t - 2 x ) Wave 2: y = 2 sin ( 0.4 t - x )

Answer: 1.13 mm.

1. At x = 1 and t = 4, calculate y₁ = 5 sin(0.5·4 − 2·1) = 0 mm. Why: The phase must be evaluated before taking the sine.

2. Calculate y₂ = 2 sin(0.4·4 − 1) = 2 sin0.6 ≈ 1.1293 mm and add. Why: Superposition adds signed displacements in the stipulated linear-wave model.

Mechanical Waves · question 3

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Question: If a wave can be given by the following equation, what is the equation of the wave that would completely cancel this wave? (2 answers) Wave: y = 3 sin ( 5 t - 31 x )

Answer: Options 1 and 3.

1. Write the cancelling wave as −3 sin(5t − 31x). Why: Equal magnitude and opposite displacement cancel at every x and t.

2. Use sin(φ + π) = −sinφ and sin(φ + 2π) = sinφ. Why: Both 3 sin(φ + π) and −3 sin(φ + 2π) match the required opposite wave.

Mechanical Waves · question 4

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Question: For the two waves below, can one interference label describe their sum at all positions and times? Wave 1: y = 2 sin ( 6 t - 2 x + π ) Wave 2: y = sin ( 3 t - x + π / 2 )

Answer: It depends on position and time.

1. Compare the two frequencies and wave numbers: they are different. Why: The relative phase cannot stay fixed everywhere.

2. Their signed displacements can reinforce or oppose at different x and t. Why: One constant interference label therefore cannot describe the whole sum.

Mechanical Waves · question 5

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Question: A wave with a period of 0.75 s is moving along a string. Determine the wave equation if the length between each crest is 2 m . Assume the graph starts at the origin and A m a x = 4.5 m .

Answer: y = 4.5 sin(3.14x − 8.38t).

1. Find k = 2π/λ = π rad/m and ω = 2π/T = 2π/0.75 ≈ 8.378 rad/s. Why: Crest separation gives wavelength, not wave number.

2. Use amplitude 4.5 m and the stated zero phase to write y = 4.5 sin(πx − 8.378t). Why: The displayed form represents propagation in +x; phase arguments are dimensionless.

Sound Waves · question 1

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Question: When sound travels through air, the air particles:

Answer: Along the direction of propagation.

1. Describe sound in air as alternating compressions and rarefactions. Why: These disturbances require back-and-forth air motion along the wave direction.

2. Choose the longitudinal-motion answer. Why: The particles oscillate locally rather than traveling the entire distance with the sound pulse.

Sound Waves · question 2

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Question: A source of frequency of 500 Hz emits waves of wavelength 0.4 m , how long does the waves take to travel 600 m ?

Answer: 3 s.

1. Find wave speed v = fλ = (500)(0.4) = 200 m/s. Why: Frequency times wavelength gives propagation speed.

2. Find t = d/v = 600/200 = 3 s. Why: The particle oscillation speed is not the propagation speed used here.

Sound Waves · question 3

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Question: For a sound wave of fixed frequency in a fixed medium, which quantity is most directly related to its intensity?

Answer: Amplitude.

1. Hold frequency and medium properties fixed as the prompt requires. Why: This removes changes in the other intensity factors.

2. Sound intensity is proportional to squared pressure or displacement amplitude. Why: Doubling the relevant amplitude would quadruple intensity under these conditions.

Sound Waves · question 4

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Question: Runners stand along a straight line with the starting gun at one end. In this hypothetical medium, sound speed is 480 m/s. What line length gives a 1 s difference between arrival at the nearest end and the farthest end?

Answer: 480 m.

1. Use Δt = Δd/v for the same propagation speed. Why: The nearest end is at the gun and the farther end has the extra travel distance.

2. Find line length Δd = (480)(1) = 480 m. Why: The stipulated medium has a hypothetical speed; do not substitute ordinary air’s speed.

Sound Waves · question 5

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Question: A sinusoidal sound wave of wavelength 26.0 cm in air has intensity level 45.0 dB. Find its peak pressure variation about ambient pressure. Use I₀ = 1.00 × 10⁻¹² W/m², sound speed 344 m/s, and air density 1.20 kg/m³.

Answer: 5.11×10⁻³ Pa peak pressure variation.

1. Convert intensity level: I = I₀10^(45/10) = 10⁻¹²·10^4.5 ≈ 3.1623×10⁻⁸ W/m². Why: Decibels are logarithmic, not a direct intensity unit.

2. Use peak pressure p_max = √(2ρvI) = √[2(1.20)(344)(3.1623×10⁻⁸)] ≈ 5.110×10⁻³ Pa. Why: The factor of 2 distinguishes peak from rms pressure; wavelength is unnecessary with the supplied v and ρ.

FFR - Wave-Particles · question 1

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Question: A photon behaves like a particle and like a wave.

Answer: True.

1. Light shows wave behavior in interference and diffraction. Why: A wave model explains those experiments.

2. Photons also transfer energy and momentum in discrete interactions. Why: Particle-like and wave-like descriptions each capture measured behavior, rather than requiring a tiny classical ball with a wavy path.

FFR - Wave-Particles · question 2

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Question: What chiefly enables a focused high-power laser to cut a material?

Answer: Absorbed optical energy concentrated into a small area.

1. A focused beam raises energy delivery per unit area. Why: High power and a small spot can heat, melt or vaporize material.

2. The material must absorb optical energy for this cutting mechanism. Why: The explanation does not require photon rest mass or faster-than-c light.

FFR - Wave-Particles · question 3

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Question: A police siren is behind a robber’s car, and both move forward through still air. Identify the source and detector. Use f′ = f(v − u_D)/(v − u_S), with both u_D and u_S positive in the forward direction. Match −u_D to the detector and −u_S to the source.

Answer: Target 1: source; 2: detector; 3: −u_D; 4: −u_S.

1. The police siren emits sound and the robber receives it. Why: These roles identify the source and detector regardless of who is chasing whom.

2. In the supplied forward-direction formula, the detector term is in v − u_D and the source term in v − u_S. Why: A detector moving away meets fewer fronts, while a forward-moving source reduces forward front spacing in the subsonic regime.

FFR - Wave-Particles · question 4

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Question: A detector moves directly away from a stationary alarm at 25 m/s while the stipulated sound speed is only 5 m/s. What problem arises when applying the usual subsonic Doppler model to ask for a lower observed frequency?

Answer: The usual subsonic model cannot provide the requested answer.

1. Insert the stated speeds into (v − u_D)/v = (5 − 25)/5 = −4. Why: The detector outruns forward sound fronts in this geometry.

2. Do not interpret −4f as an ordinary lower positive heard frequency. Why: A negative signed result flags use outside the assumed subsonic setup; the usual formula is not a valid answer model here.

FFR - Wave-Particles · question 5

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Question: The police siren source has speed 12.5 m/s in a medium with stipulated sound speed 8 m/s. Can the usual subsonic source–detector Doppler formula be used to infer a unique robber speed from a doubled detected frequency?

Answer: No valid unique robber speed follows from that usual model.

1. Compare source speed 12.5 m/s with stipulated sound speed 8 m/s. Why: The source is supersonic, outside the usual moving-source formula’s assumptions.

2. Do not solve the subsonic expression as though its negative denominator described ordinary front arrival. Why: Shock-wave geometry and a different model are needed; the supplied data do not justify a unique conventional answer.

Model references

These original worked steps use the stated question models. Physics references: Wave interference, Sound intensity, Doppler effect. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.