AP Physics 2 Quiz: Properties Of Wave Pulses And Waves

What this quiz covers

This quiz focuses on Properties Of Wave Pulses And Waves, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It doubles because the wave speed doubles when frequency doubles.
2. It stays the same because wavelength depends only on amplitude.
3. It is halved because the wave speed stays constant in the same medium. (correct answer)
4. It increases because higher frequency means higher energy, so crests spread out.

Explanation: This question tests understanding of properties of wave pulses and waves. For waves on a string, the wave speed v is determined by the medium properties (tension and linear density), which remain constant here. The fundamental wave relationship is v = fλ, where f is frequency and λ is wavelength. When frequency doubles and wave speed stays constant, the wavelength must be halved to maintain this relationship. Choice A incorrectly assumes wave speed depends on frequency, reflecting the misconception that the source controls wave speed rather than the medium. When analyzing wave behavior, always remember that wave speed is set by the medium, not the source.

Question 2

1. It becomes $0.40\,\text{m}$ because wavelength is proportional to wave speed at fixed frequency. (correct answer)
2. It becomes $0.90\,\text{m}$ because wavelength is inversely proportional to wave speed.
3. It stays $0.60\,\text{m}$ because wavelength depends only on amplitude.
4. It becomes $0.30\,\text{m}$ because the frequency halves when the speed decreases.

Explanation: This question tests understanding of properties of wave pulses and waves. The wave relationship v = fλ governs how wavelength changes when wave speed changes. Initially, the frequency is f = v/λ = 9.0/0.60 = 15 Hz. When tension is adjusted to reduce wave speed to 6.0 m/s while the driver frequency remains constant at 15 Hz, the new wavelength becomes λ = v/f = 6.0/15 = 0.40 m. Choice C incorrectly assumes wavelength depends only on amplitude, confusing wave propagation properties with wave energy. Always use v = fλ to relate changes in wave speed, frequency, and wavelength.

Question 3

1. The frequency doubles while the wavelength stays the same.
2. The wavelength doubles while the frequency stays the same. (correct answer)
3. Both frequency and wavelength stay the same because the driver is unchanged.
4. The amplitude halves because faster waves carry less energy.

Explanation: This question tests understanding of properties of wave pulses and waves. The wave relationship v = fλ connects wave speed, frequency, and wavelength. When the string is replaced with one where waves travel twice as fast, the wave speed doubles from 2.0 m/s to 4.0 m/s. Since the same driver continues oscillating, the frequency remains 5.0 Hz (the source determines frequency). To maintain v = fλ with doubled speed and constant frequency, the wavelength must double from 0.40 m to 0.80 m. Choice A incorrectly assumes frequency changes, reflecting the misconception that frequency depends on the medium rather than the source. Always remember that frequency is set by the source, while wavelength adjusts based on the medium's wave speed.

Question 4

1. The period doubles.
2. The frequency doubles.
3. The amplitude doubles.
4. The wavelength doubles. (correct answer)

Explanation: This question tests understanding of properties of wave pulses and waves. The wave equation v = fλ relates wave speed, frequency, and wavelength. When tension increases and wave speed doubles from 4.0 m/s to 8.0 m/s while the source maintains constant frequency at 5.0 Hz, the wavelength must change to satisfy the wave equation: λ_new = v_new/f = 8.0/5.0 = 1.6 m, which is double the original 0.80 m. Choice B incorrectly suggests frequency doubles, reflecting the misconception that changing the medium properties affects the source frequency. Remember that the source controls frequency, while wavelength adjusts to maintain v = fλ when wave speed changes.

Question 5

1. Its wavelength increases while its frequency stays the same. (correct answer)
2. Its frequency decreases because the medium changed.
3. Its speed stays the same while its amplitude decreases.
4. Its frequency increases while its wavelength stays the same.

Explanation: This question tests understanding of properties of wave pulses and waves. When a wave crosses a boundary between media, its frequency remains constant because it's determined by the source, but wavelength changes to accommodate the new wave speed. Using v = fλ, the initial wavelength is λ₁ = 343/440 = 0.78 m, and in warmer air λ₂ = 360/440 = 0.82 m, showing wavelength increases. Choice D incorrectly claims frequency decreases when entering a new medium, reflecting the misconception that frequency depends on medium properties rather than being fixed by the source. Remember that frequency is invariant across boundaries while wavelength adjusts to the new wave speed.

Question 6

1. Both frequency and wavelength increase.
2. Wave speed stays the same because amplitude is unchanged.
3. Wavelength decreases while frequency stays the same. (correct answer)
4. Frequency decreases and wavelength stays the same.

Explanation: This question tests understanding of properties of wave pulses and waves. When a wave crosses a boundary between two media, the frequency remains constant because it's determined by the source, which hasn't changed. Using v = fλ, we can find the wavelengths: in Rope 1, λ₁ = v₁/f = 3.0/6 = 0.50 m; in Rope 2, λ₂ = v₂/f = 2.0/6 = 0.33 m. Since the speed decreased while frequency stayed constant, the wavelength must decrease proportionally. Choice A incorrectly suggests frequency changes at boundaries, which violates the principle that the source controls frequency. When waves cross boundaries, remember that frequency stays constant while wavelength adjusts to maintain v = fλ.

Question 7

1. It becomes $0.60\ \text{m}$ because wavelength depends only on amplitude.
2. It becomes $1.2\ \text{m}$ because wavelength is proportional to frequency.
3. It becomes $0.30\ \text{m}$ because wavelength is inversely proportional to frequency. (correct answer)
4. It becomes $0.15\ \text{m}$ because wave speed halves when frequency doubles.

Explanation: This question tests understanding of properties of wave pulses and waves. Using v = fλ, the initial frequency is f₁ = v/λ₁ = 15/0.60 = 25 Hz. When frequency doubles to f₂ = 50 Hz while the string properties (and thus wave speed) remain unchanged at 15 m/s, the new wavelength becomes λ₂ = v/f₂ = 15/50 = 0.30 m. Choice D incorrectly claims wave speed halves when frequency doubles, reflecting the misconception that changing source frequency affects the medium's wave speed. Remember that wave speed depends only on medium properties; when frequency changes, wavelength adjusts inversely to maintain constant speed.

Question 8

1. It doubles because wave speed on a string scales as $\sqrt{T}$. (correct answer)
2. It quadruples because wave speed is directly proportional to tension.
3. It is unchanged because tension affects only amplitude.
4. It halves because higher tension reduces oscillation of the rope.

Explanation: This question tests understanding of properties of wave pulses and waves. Wave speed on a string is given by v = √(T/μ), where T is tension and μ is linear mass density. When tension increases by a factor of 4, the new speed becomes v_new = √(4T/μ) = 2√(T/μ) = 2v, so the speed doubles. Choice B incorrectly assumes direct proportionality between speed and tension, reflecting the misconception that wave speed scales linearly with tension rather than with the square root of tension. Remember that wave speed on a string varies as the square root of tension.

Question 9

1. Only the energy transported increases. (correct answer)
2. Only the wave speed increases.
3. Only the frequency decreases.
4. Only the wavelength increases.

Explanation: This question tests understanding of properties of wave pulses and waves. Wave speed on a string depends only on the medium properties (tension and linear mass density), not on amplitude, frequency, or wavelength. Since the string and tension remain unchanged, the wave speed stays at 12 m/s regardless of amplitude changes. The frequency f = v/λ = 12/3 = 4 Hz also remains constant because both speed and wavelength are unchanged. However, wave energy is proportional to the square of amplitude (E ∝ A²), so doubling the amplitude quadruples the energy transported by the wave. Choice B incorrectly assumes amplitude affects wave speed, conflating wave properties with energy transport. Remember: amplitude affects energy but not speed, frequency, or wavelength.

Question 10

1. It stays the same because speed changes to match frequency.
2. It becomes zero because increasing frequency stops oscillations.
3. It becomes one-third as large because $\lambda=v/f$. (correct answer)
4. It triples because wavelength is proportional to frequency.

Explanation: This question tests understanding of properties of wave pulses and waves. The wave equation v = fλ can be rearranged to λ = v/f, showing that wavelength is inversely proportional to frequency when wave speed is constant. Since the medium is unchanged, the wave speed v remains constant. When frequency is tripled (f → 3f), the wavelength becomes λ = v/(3f) = (1/3)(v/f), or one-third of its original value. Choice A incorrectly assumes direct proportionality between wavelength and frequency, which would violate the wave equation. When frequency changes but the medium stays the same, wavelength must change inversely to maintain constant wave speed.

Question 11

1. It is $12\ \text{m/s}$ because wave speed depends only on the source frequency.
2. It is $24\ \text{m/s}$ because wave speed increases with mass density.
3. It is $\dfrac{12}{\sqrt{2}}\ \text{m/s}$ because wave speed scales as $1/\sqrt{\mu}$. (correct answer)
4. It is $6\ \text{m/s}$ because wave speed decreases by a factor of 2.

Explanation: This question tests understanding of properties of wave pulses and waves. The speed of a transverse wave on a string is given by v = √(T/μ), where T is tension and μ is linear mass density. When the linear mass density doubles while tension remains constant, the wave speed becomes v_new = √(T/2μ) = √(1/2) × √(T/μ) = v/√2 = 12/√2 m/s. Choice C incorrectly assumes wave speed depends on source frequency, reflecting the misconception that the source controls wave speed rather than the medium. Remember that wave speed in a medium is determined solely by the medium's properties, not by the source characteristics.

Question 12

1. It becomes 2 times larger because energy is proportional to amplitude.
2. It is unchanged because energy depends only on wave speed.
3. It becomes 4 times larger because energy is proportional to $A^2$. (correct answer)
4. It becomes 1/2 as large because larger amplitude reduces frequency.

Explanation: This question tests understanding of properties of wave pulses and waves. The energy transported by a wave is proportional to the square of its amplitude (E ∝ A²) and also proportional to the square of frequency (E ∝ f²). Since frequency remains constant and amplitude doubles from A to 2A, the energy becomes proportional to (2A)² = 4A², making it 4 times larger. Choice A incorrectly assumes linear proportionality with amplitude, reflecting the misconception that wave energy scales directly with amplitude rather than with amplitude squared. Remember that wave energy depends on both amplitude squared and frequency squared.

Question 13

1. It decreases because shorter wavelength reduces speed.
2. It is unchanged because speed depends on the medium properties. (correct answer)
3. It becomes zero because energy transfer requires increasing amplitude.
4. It increases because higher frequency waves travel faster.

Explanation: This question tests understanding of properties of wave pulses and waves. Wave speed in a medium like a slinky depends only on the medium's properties (tension and mass density), not on wave characteristics like frequency, wavelength, or amplitude. When frequency increases while the slinky's physical properties remain constant, the wave speed stays the same, though wavelength will decrease according to v = fλ. Choice A incorrectly suggests higher frequency waves travel faster, reflecting the misconception that wave characteristics affect propagation speed. Remember that wave speed is determined by the medium, not by the source or wave properties.

Question 14

1. The frequency increases and the wavelength decreases.
2. The frequency decreases and the wavelength increases.
3. The frequency is unchanged and the wavelength increases. (correct answer)
4. The wavelength is unchanged and the frequency increases.

Explanation: This question tests understanding of properties of wave pulses and waves. Sound waves travel faster in warmer air because the molecules move faster, increasing the speed of sound. When sound waves enter a new medium, the frequency remains constant at 440 Hz because it's determined by the source—the source continues vibrating at the same rate. Using v = fλ, if wave speed v increases (in warmer air) and frequency f stays constant, then wavelength λ must increase proportionally. This is analogous to water waves entering deeper water. Choice A incorrectly claims frequency increases and wavelength decreases, misunderstanding that frequency is locked to the source. Remember: when waves change media, frequency stays constant while speed and wavelength change in the same direction.

Question 15

1. The frequency increases to match the higher speed.
2. The wavelength increases to $3.0\ \text{m}$. (correct answer)
3. The amplitude must decrease to conserve energy.
4. The wave speed stays the same because the medium is a string.

Explanation: This question tests understanding of properties of wave pulses and waves. Initially, the wave has f = 10 Hz and v = 20 m/s, giving wavelength λ = v/f = 20/10 = 2.0 m. When tension increases, wave speed on the string increases to 30 m/s because v = √(T/μ). The driver continues at 10 Hz, so frequency remains unchanged—frequency is always determined by the source. Using v = fλ with the new speed: λ = 30/10 = 3.0 m. The wavelength increases from 2.0 m to 3.0 m to accommodate the higher speed at the same frequency. Choice A incorrectly assumes frequency changes with tension, confusing medium properties with source properties. Remember: changing the medium affects speed and wavelength, but frequency is locked to the source.

Question 16

1. It doubles because $v_{\max}$ is proportional to amplitude for fixed $\omega$. (correct answer)
2. It halves because doubling amplitude reduces frequency.
3. It stays the same because wave speed along the string is unchanged.
4. It becomes zero because points only move when a pulse passes, not for a continuous wave.

Explanation: This question tests understanding of properties of wave pulses and waves. For a sinusoidal wave, the maximum transverse speed of a point on the string is vmax = Aω, where A is amplitude and ω is angular frequency. This represents how fast the string moves perpendicular to the wave direction, not the wave speed along the string. When amplitude doubles while ω remains constant, the maximum transverse speed also doubles. Choice C confuses transverse particle motion with longitudinal wave propagation speed, a common misconception. Remember to distinguish between wave speed (horizontal propagation) and particle speed (vertical oscillation).

Question 17

1. It is smaller because frequency depends on wave speed in the medium.
2. It is larger because energy must be conserved across the boundary.
3. It is unchanged because the source sets the oscillation rate at the boundary. (correct answer)
4. It becomes zero because the wave cannot enter a slower medium.

Explanation: This question tests understanding of properties of wave pulses and waves. Frequency is determined by the source and remains constant as waves cross boundaries between different media. When a continuous wave enters string 2 where the wave speed is smaller, the wavelength decreases (from v = fλ) but the frequency stays the same. The boundary cannot create or destroy wave cycles; it must transmit them at the same rate they arrive. Choice A incorrectly assumes frequency depends on wave speed, confusing medium-dependent properties (speed, wavelength) with source-dependent properties (frequency). Remember that frequency is always conserved across boundaries between media.

Question 18

1. It increases because the frequency stays the same.
2. It is unchanged because the source frequency is unchanged.
3. It decreases because the wave speed decreases. (correct answer)
4. It decreases because the frequency decreases in shallow water.

Explanation: This question tests understanding of properties of wave pulses and waves. Water waves travel slower in shallow water than in deep water due to interaction with the bottom. When waves cross from one medium to another, frequency remains constant because it's determined by the source—the source continues producing waves at the same rate. Using the wave equation v = fλ, if wave speed v decreases and frequency f stays constant, then wavelength λ must decrease proportionally. This is why ocean waves appear to bunch up as they approach shore. Choice D incorrectly claims frequency changes in shallow water, misunderstanding that frequency is set by the source, not the medium. Remember: when waves change media, frequency stays constant while speed and wavelength change together.

Question 19

1. The frequency is unchanged and the wavelength decreases. (correct answer)
2. The frequency increases and the wavelength increases.
3. The frequency decreases and the wave speed increases.
4. The wave speed increases because the amplitude is unchanged.

Explanation: This question tests understanding of properties of wave pulses and waves. When a wave travels on a string, the wave speed depends on the medium properties: v = √(T/μ), where T is tension and μ is linear mass density. A heavier rope has larger μ, so the wave speed decreases while tension T remains constant. Since the source continues oscillating at 5.0 Hz, the frequency remains unchanged—frequency is always determined by the source, not the medium. Using v = fλ, if speed decreases and frequency stays constant, wavelength must decrease proportionally. Choice B incorrectly assumes both frequency and wavelength increase, confusing the role of the source versus the medium. Remember: frequency is set by the source, while wave speed is set by the medium properties.

Question 20

1. $\lambda=0.25\,\text{m}$ because wavelength equals $f/v$.
2. $\lambda=4.0\,\text{m}$ because wavelength equals $v/f$. (correct answer)
3. $\lambda=36\,\text{m}$ because wavelength equals $vf$.
4. $\lambda=9.0\,\text{m}$ because wavelength equals $v-f$.

Explanation: This question tests understanding of properties of wave pulses and waves. The fundamental wave relationship is v = fλ, which can be rearranged to find wavelength: λ = v/f. This relationship holds for all types of waves, whether mechanical or electromagnetic. Given v = 12 m/s and f = 3.0 Hz, the wavelength is λ = 12/3.0 = 4.0 m. This makes physical sense: if 3 complete waves pass a point each second, and the wave travels 12 meters in that second, each wave must be 4 meters long. Choice A incorrectly multiplies v and f, confusing the wave equation with other physics formulas and yielding units of m²/s rather than meters. Remember: wavelength equals wave speed divided by frequency (λ = v/f), giving the spatial extent of one complete oscillation.