AP Physics 2 Quiz: Periodic Waves

What this quiz covers

This quiz focuses on Periodic 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. $v=\dfrac{T}{\lambda}$ because speed is time per distance.
2. $v=\lambda T$ because distance and time multiply.
3. $v=\dfrac{\lambda}{T}$ because one wavelength passes in one period. (correct answer)
4. $v=\dfrac{1}{\lambda T}$ because speed is inverse of wave size.

Explanation: This question tests understanding of periodic waves. Wave speed v represents how fast a wave pattern propagates through space. In one period T, exactly one wavelength λ passes any fixed point, giving the relationship v = λ/T. This can also be written as v = fλ since frequency f = 1/T. Choice A incorrectly inverts the relationship, giving units of s/m instead of m/s for speed. The fundamental principle is: wave speed equals the distance traveled (one wavelength) divided by the time taken (one period), making v = λ/T the correct relationship.

Question 2

1. $\lambda_2=2\lambda_1$ because higher frequency means longer wavelength.
2. $\lambda_2=\lambda_1$ because the string sets wavelength directly.
3. $\lambda_2=\dfrac{1}{2}\lambda_1$ because the wave speed is the same. (correct answer)
4. $\lambda_2=\dfrac{1}{2}\lambda_1$ because the amplitude is larger.

Explanation: This question tests understanding of periodic waves. For waves in the same string with identical tension and linear density, the wave speed v is the same for both waves. Using v = fλ and rearranging to λ = v/f, we see wavelength is inversely proportional to frequency when speed is constant. If wave 2 has frequency 2f₁ (double that of wave 1), then λ₂ = v/(2f₁) = (1/2)(v/f₁) = λ₁/2. Choice A incorrectly claims higher frequency means longer wavelength, reversing the inverse relationship. Remember: in the same medium, doubling frequency always halves wavelength because their product must equal the constant wave speed.

Question 3

1. The wavelength becomes $\tfrac{1}{2}\lambda$ because speed stays constant. (correct answer)
2. The wavelength stays $\lambda$ because amplitude determines wavelength.
3. The wavelength becomes $2\lambda$ because speed increases with frequency.
4. The wavelength becomes $\tfrac{1}{4}\lambda$ because both speed and frequency double.

Explanation: This question tests understanding of periodic waves. The fundamental wave equation states that wave speed equals frequency times wavelength (v = fλ), which means these three quantities are always related. When wave speed is constant (as stated for this string), the product of frequency and wavelength must remain constant. Since frequency doubles from f to 2f, the wavelength must be halved to λ/2 to maintain the same wave speed. Choice B incorrectly assumes that wave speed increases with frequency, which violates the given condition that tension (and thus wave speed) is constant. When solving wave problems, always identify which quantity remains constant—here it's speed, so doubling frequency requires halving wavelength.

Question 4

1. $\lambda' = \tfrac{1}{2}\lambda$ because halving $T$ doubles $f$ at constant speed (correct answer)
2. $\lambda' = 2\lambda$ because halving $T$ halves the frequency
3. $\lambda' = \lambda$ because the rope properties determine wavelength directly
4. $\lambda' = 4\lambda$ because the wave speed increases as $T$ decreases

Explanation: This question tests understanding of periodic waves. Period T and frequency f are related by f = 1/T, so halving the period doubles the frequency. With constant wave speed (fixed rope properties), the wave equation v = fλ requires wavelength to change inversely with frequency. When period becomes T/2, frequency doubles, so wavelength must halve: v = f·λ = (2f)·(λ/2), giving λ' = λ/2. Choice B reverses the relationship, incorrectly claiming that halving period halves frequency—a misconception about the inverse relationship T = 1/f. To solve period-wavelength problems, first convert period to frequency, then apply v = fλ.

Question 5

1. It becomes $0.40\ \text{m}$ because doubling frequency halves wavelength at constant speed (correct answer)
2. It becomes $0.20\ \text{m}$ because doubling frequency quarters wavelength
3. It becomes $1.6\ \text{m}$ because frequency and wavelength increase together
4. It becomes $0.80\ \text{m}$ because wave speed fixes wavelength

Explanation: This question tests understanding of periodic waves. Using v = fλ, we can find the constant wave speed: v = 5.0 Hz × 0.80 m = 4.0 m/s. When frequency doubles to 10 Hz while wave speed remains constant at 4.0 m/s, the new wavelength becomes λ' = v/f' = 4.0/10 = 0.40 m. This is half the original wavelength, confirming the inverse relationship. Choice A shows the misconception that frequency and wavelength increase together, ignoring that their product must equal the constant wave speed. To solve numerical wave problems, first calculate wave speed, then use it to find the new wavelength.

Question 6

1. The wavelength stays $\lambda$ because period does not affect wavelength.
2. The wavelength becomes $9\lambda$ because both period and speed triple.
3. The wavelength becomes $\tfrac{1}{3}\lambda$ because frequency increases.
4. The wavelength becomes $3\lambda$ because the wave speed stays constant. (correct answer)

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ connects wave speed, frequency, and wavelength, where frequency equals 1/period (f = 1/T). In this ripple tank, wave speed is constant because it depends only on water depth, not on the source. When period increases from T to 3T, frequency decreases to f/3, since frequency and period are inversely related. To maintain constant wave speed with one-third the frequency, wavelength must triple to 3λ. Choice A incorrectly assumes wavelength decreases when frequency decreases, missing that constant speed requires them to change oppositely. Remember: when wave speed is fixed, frequency and wavelength change inversely.

Question 7

1. The frequency becomes $4f$ because both speed and frequency must increase together.
2. The frequency becomes $2f$ because wave speed stays constant. (correct answer)
3. The frequency becomes $\tfrac{1}{2}f$ because shorter wavelength means lower frequency.
4. The frequency stays $f$ because amplitude determines frequency.

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ requires that when wave speed is constant, frequency and wavelength must change inversely. If wavelength is halved from λ to λ/2 while wave speed v remains constant, then frequency must double to maintain the equation: v = f·λ becomes v = (2f)·(λ/2). This gives a new frequency of 2f. Choice A incorrectly assumes shorter wavelength means lower frequency, but at constant speed, they change oppositely. When wave speed is fixed by the medium, adjusting the driver to create shorter wavelengths necessarily increases the frequency.

Question 8

1. The period becomes $\tfrac{1}{9}T$ because speed and frequency both triple.
2. The period becomes $3T$ because period increases with frequency.
3. The period becomes $\tfrac{1}{3}T$ because period is the inverse of frequency. (correct answer)
4. The period stays $T$ because wave speed is unchanged.

Explanation: This question tests understanding of periodic waves. Period and frequency are reciprocals related by T = 1/f, so when frequency increases by a factor of 3 (from f to 3f), period must decrease by the same factor. The new period becomes T/3, since T(new) = 1/(3f) = (1/3)(1/f) = T/3. Wave speed remaining constant is irrelevant to the period-frequency relationship. Choice A incorrectly claims period increases with frequency, when they always change inversely. To handle period-frequency conversions, remember they are reciprocals: tripling one means dividing the other by three.

Question 9

1. The wavelength becomes $\tfrac{3}{2}\lambda$ because the frequency stays the same. (correct answer)
2. The wavelength becomes $\tfrac{2}{3}\lambda$ because higher speed means higher frequency.
3. The wavelength becomes $\tfrac{9}{4}\lambda$ because both speed and period increase.
4. The wavelength stays $\lambda$ because the medium does not affect wavelength.

Explanation: This question tests understanding of periodic waves. Using v = fλ, when frequency remains constant but wave speed increases, wavelength must increase proportionally. Originally v = fλ, and with the new gas, (3v/2) = f(new wavelength). Solving gives new wavelength = (3/2)λ, since wavelength scales directly with speed when frequency is fixed. This makes physical sense: faster waves at the same frequency must have crests farther apart. Choice A incorrectly assumes wavelength decreases with higher speed, confusing the relationship when frequency is held constant. Remember: at fixed frequency, wavelength and wave speed change together proportionally.

Question 10

1. $T' = 3T$ because period increases with frequency
2. $T' = \tfrac{1}{3}T$ because $T=\tfrac{1}{f}$ (correct answer)
3. $T' = T$ because wave speed stays constant
4. $T' = \tfrac{1}{9}T$ because wavelength decreases by a factor of 3

Explanation: This question tests understanding of periodic waves. Period T is the time for one complete oscillation, and it relates to frequency by T = 1/f. This inverse relationship means that when frequency increases, period decreases proportionally. If frequency triples from f to 3f, then the new period becomes T' = 1/(3f) = (1/3)·(1/f) = T/3. Choice A shows the misconception that period and frequency change in the same direction, when they are actually inversely related. To find period changes, use the relationship T = 1/f and recognize that tripling frequency means dividing period by three.

Question 11

1. $\lambda' = 4\lambda$ because speed increases as frequency decreases
2. $\lambda' = 2\lambda$ because wave speed stays constant (correct answer)
3. $\lambda' = \lambda$ because amplitude sets the spacing of crests
4. $\lambda' = \tfrac{1}{2}\lambda$ because wavelength is proportional to frequency

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ connects wave speed, frequency, and wavelength for any periodic wave. When wave speed stays constant (fixed water depth), frequency and wavelength must change inversely to maintain their product. If frequency decreases to half (f/2), wavelength must double to keep v constant: v = f·λ = (f/2)·(2λ). Choice A incorrectly claims wavelength is proportional to frequency, a common misconception that ignores the constraint of constant wave speed. When wave speed is fixed, halving frequency doubles wavelength—they change in opposite directions.

Question 12

1. $\lambda' = \tfrac{1}{4}\lambda$ because the speed doubles with frequency
2. $\lambda' = 2\lambda$ because higher frequency means longer spacing
3. $\lambda' = \tfrac{1}{2}\lambda$ because $v=f\lambda$ and $v$ is constant (correct answer)
4. $\lambda' = \lambda$ because wavelength depends only on loudness

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ governs all periodic waves, including sound waves in air. With constant sound speed (343 m/s), doubling frequency from 440 Hz to 880 Hz requires halving the wavelength to maintain v = fλ. The new wavelength becomes λ' = v/f' = 343/(880) = (343/440)/2 = λ/2. Choice C incorrectly suggests wavelength depends only on loudness (amplitude), confusing wave properties—amplitude affects energy but not wavelength. When analyzing frequency changes at constant wave speed, use v = fλ to find that frequency and wavelength change inversely.

Question 13

1. $T' = 2T$ because angular frequency is inversely related to wavelength
2. $T' = \tfrac{1}{4}T$ because doubling $\omega$ quadruples $f$
3. $T' = T$ because amplitude $A$ is unchanged
4. $T' = \tfrac{1}{2}T$ because $\omega = \tfrac{2\pi}{T}$ (correct answer)

Explanation: This question tests understanding of periodic waves. Angular frequency ω relates to period through ω = 2π/T, which can be rearranged to T = 2π/ω. This inverse relationship means doubling angular frequency halves the period. When ω becomes 2ω, the new period is T' = 2π/(2ω) = (1/2)·(2π/ω) = T/2. Choice D incorrectly suggests that doubling ω quadruples frequency, confusing the linear relationship between ω and f with a quadratic one. To analyze period changes with angular frequency, use T = 2π/ω and recognize the inverse proportionality.

Question 14

1. $\lambda' = 4\lambda$ because doubling $f$ doubles $c$
2. $\lambda' = 2\lambda$ because wavelength increases with frequency
3. $\lambda' = \lambda$ because speed changes with amplitude
4. $\lambda' = \tfrac{1}{2}\lambda$ because $c=f\lambda$ remains constant (correct answer)

Explanation: This question tests understanding of periodic waves. For electromagnetic waves in vacuum, the wave equation c = fλ applies, where c is the constant speed of light. Since wave speed cannot change in vacuum, frequency and wavelength must vary inversely to maintain their product. When frequency doubles from f to 2f, wavelength must halve: c = f·λ = (2f)·(λ/2), giving λ' = λ/2. Choice B reflects the misconception that wavelength increases with frequency, ignoring that their product must equal the fixed speed c. For waves with constant speed, increasing frequency always decreases wavelength proportionally.

Question 15

1. $\lambda' = \lambda$ because frequency determines wavelength regardless of speed
2. $\lambda' = 2\lambda$ because $\lambda=\tfrac{v}{f}$ and $f$ is unchanged (correct answer)
3. $\lambda' = \tfrac{1}{2}\lambda$ because wavelength decreases when speed increases
4. $\lambda' = 4\lambda$ because doubling speed doubles frequency and wavelength

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ shows that wavelength depends on both wave speed and frequency. When frequency stays constant but wave speed doubles, wavelength must also double to maintain the relationship: 2v = f·(2λ). This gives λ' = (2v)/f = 2(v/f) = 2λ. Choice A incorrectly assumes wavelength decreases when speed increases, missing that frequency is held constant—only when frequency changes at constant speed does wavelength change inversely. When wave speed changes but frequency remains fixed, wavelength changes proportionally with speed.

Question 16

1. The wavelength stays $\lambda$ because amplitude sets the wavelength.
2. The wavelength becomes $\tfrac{1}{2}\lambda$ because wavelength follows frequency directly.
3. The wavelength becomes $2\lambda$ because the wave speed stays constant. (correct answer)
4. The wavelength becomes $4\lambda$ because the speed doubles when frequency halves.

Explanation: This question tests understanding of periodic waves. Sound waves in air travel at a constant speed determined by temperature, following the relationship v = fλ. When frequency is halved from f to f/2 while wave speed remains constant, the wavelength must double to maintain the equation's balance. This gives a new wavelength of 2λ, making the product (f/2)(2λ) equal the original fλ. Choice A incorrectly assumes wavelength and frequency change in the same direction, but they actually change inversely when speed is constant. The key strategy is recognizing that in a uniform medium, changing the source frequency changes wavelength proportionally but inversely.

Question 17

1. The period becomes $\tfrac{1}{4}T$ because longer wavelength means higher frequency.
2. The period stays $T$ because wavelength does not affect period.
3. The period becomes $4T$ because the wave speed stays constant. (correct answer)
4. The period becomes $16T$ because both wavelength and speed increase by 4.

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ can be rewritten using T = 1/f to give v = λ/T, showing that at constant wave speed, wavelength and period are directly proportional. When wavelength increases from λ to 4λ while wave speed remains constant, the period must also increase by a factor of 4, giving 4T. This makes sense physically: longer waves take more time to pass a fixed point. Choice A incorrectly assumes longer wavelength means higher frequency (and thus shorter period), missing that at constant speed, wavelength and period change together. Strategy: when speed is constant, wavelength and period change proportionally.

Question 18

1. The frequency becomes $\tfrac{1}{2}f$ because frequency is proportional to period.
2. The frequency becomes $2f$ because frequency is the inverse of period. (correct answer)
3. The frequency stays $f$ because wave speed is unchanged.
4. The frequency becomes $4f$ because halving period doubles speed and frequency.

Explanation: This question tests understanding of periodic waves. Frequency and period are inversely related by the equation f = 1/T, meaning when one doubles, the other halves. When the period changes from T to T/2, the frequency must change from f to 2f, since (1/(T/2)) = 2/T = 2f. The wave speed remains constant because the rope and tension are unchanged—only the driving source is adjusted. Choice A incorrectly claims frequency is proportional to period, when they are actually inversely proportional. To solve period-frequency problems, always remember they are reciprocals: halving period doubles frequency.

Question 19

1. The wavelength stays $\lambda$ because speed and frequency are unchanged. (correct answer)
2. The wavelength halves because larger amplitude increases frequency.
3. The wavelength doubles because amplitude determines wavelength.
4. The wavelength increases because wave speed increases with amplitude.

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ shows that wavelength depends only on wave speed and frequency, not on amplitude. Since string tension (which determines wave speed) and driving frequency both remain unchanged, the wavelength must stay λ regardless of amplitude changes. Amplitude affects only the wave's height or energy, not its spatial or temporal characteristics. Choice B incorrectly claims amplitude determines wavelength, a common misconception that confuses wave height with wave length. Remember: amplitude is independent of the v = fλ relationship—changing amplitude never changes wavelength, frequency, or speed.

Question 20

1. The wavelength becomes $2\lambda$ because the frequency stays the same. (correct answer)
2. The wavelength becomes $\tfrac{1}{2}\lambda$ because higher speed means shorter wavelength.
3. The wavelength stays $\lambda$ because frequency sets wavelength regardless of speed.
4. The wavelength becomes $4\lambda$ because both speed and frequency double.

Explanation: This question tests understanding of periodic waves. The wave equation v = fλ shows that wavelength depends on both wave speed and frequency. When the string is replaced with one that doubles the wave speed while the driver maintains the same frequency f, we can analyze using the equation: originally v = fλ, and now 2v = f(new wavelength). Solving for the new wavelength gives 2λ, since doubling speed while keeping frequency constant requires doubling wavelength. Choice A incorrectly assumes higher speed means shorter wavelength, but this is only true if frequency increases. When wave speed changes but frequency stays constant, wavelength changes proportionally with speed.