Periodic Waves - AP Physics 2

A wave machine generates a continuous transverse wave on a rope. The rope's tension and linear mass density remain fixed, so the wave speed is constant. The wave has period $T$ and wavelength $\lambda$ . The driving mechanism is adjusted so the period becomes $\tfrac{1}{2}T$ . Which statement correctly relates the new wavelength $\lambda'$ to $\lambda$ ?
Answer: $\lambda' = \tfrac{1}{2}\lambda$ because halving $T$ doubles $f$ at constant speed 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.
A ripple tank produces straight, continuous wavefronts on the surface of water at a fixed depth, so the wave speed is constant. The source oscillates with frequency $f$ and creates wavelength $\lambda$ . The source is adjusted so the frequency decreases to $\tfrac{1}{2}f$ while the water depth is unchanged. Which statement correctly relates the new wavelength $\lambda'$ to $\lambda$ ?
Answer: $\lambda' = 2\lambda$ because wave speed stays constant 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.
A continuous sinusoidal wave travels along a stretched string with fixed tension. The wave speed is $v$ , and the wave has frequency $f$ and wavelength $\lambda$ . The string is replaced with a different string under the same tension, but the wave speed is measured to be $2v$ while the driver keeps the same frequency $f$ . Which statement correctly relates $\lambda'$ to $\lambda$ ?
Answer: $\lambda' = 2\lambda$ because $\lambda=\tfrac{v}{f}$ and $f$ is unchanged 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.
A traveling wave moves along a rope with speed $v$ and period $T$ . The rope and tension are unchanged, but the source is adjusted so the period becomes $\tfrac{1}{2}T$ . Which statement correctly relates the new frequency to the original frequency?
Answer: The frequency becomes $2f$ because frequency is the inverse of period. 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.
A loudspeaker emits a steady pure tone in air, producing a continuous, repeating sound wave. The speed of sound in the room is approximately constant. The tone has frequency $f$ and period $T$ . The speaker is adjusted so the frequency becomes $3f$ without changing the medium. Which statement correctly relates the new period $T'$ to the original period $T$ ?
Answer: $T' = \tfrac{1}{3}T$ because $T=\tfrac{1}{f}$ 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.
A periodic wave travels on a string with constant speed $v$ . The source is adjusted so the frequency increases by a factor of 3. Which statement correctly relates the new period to the original period?
Answer: The period becomes $\tfrac{1}{3}T$ because period is the inverse of frequency. 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.
A sinusoidal wave travels along a taut string with constant tension, so the wave speed stays constant. The string is driven by a motor that produces a continuous, repeating oscillation. Initially the wave has frequency $f$ and wavelength $\lambda$ . The motor is adjusted so the frequency becomes $2f$ while the string tension and linear density remain unchanged. Which statement correctly relates the new wavelength $\lambda'$ to the original wavelength $\lambda$ ?
Answer: $\lambda' = \tfrac{1}{2}\lambda$ because the wave speed stays constant This question tests understanding of periodic waves. The fundamental wave equation states that wave speed equals frequency times wavelength: v = fλ. When wave speed remains constant (as stated for the string with constant tension), the product of frequency and wavelength must remain constant. If frequency doubles from f to 2f, then wavelength must be halved to maintain the same wave speed: v = f·λ = (2f)·(λ/2). Choice B incorrectly assumes wavelength increases with frequency, revealing the misconception that frequency and wavelength change in the same direction. When wave speed is fixed, frequency and wavelength are inversely proportional—doubling one halves the other.
A repeating wave pattern has wavelength $\lambda$ and period $T$ as it propagates. Which statement correctly relates wave speed to these quantities?
Answer: $v=\dfrac{\lambda}{T}$ because one wavelength passes in one period. 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.
Two waves travel in the same string with the same tension and linear density. Wave 1 has frequency $f_1$ and wave 2 has $2f_1$ . Which statement correctly compares their wavelengths?
Answer: $\lambda_2=\dfrac{1}{2}\lambda_1$ because the wave speed is the same. 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.
A continuous sinusoidal wave travels at constant speed $v$ in a uniform medium. If the frequency decreases by a factor of 3, which statement correctly describes the wavelength?
Answer: The wavelength increases by a factor of 3 because the speed stays constant. This question tests understanding of periodic waves. The wave equation v = fλ describes how wave speed, frequency, and wavelength relate for any periodic wave. With constant wave speed v in a uniform medium, we rearrange to λ = v/f, showing wavelength and frequency are inversely proportional. If frequency decreases by a factor of 3 (f becomes f/3), then wavelength must increase by a factor of 3 to maintain the same wave speed. Choice C incorrectly claims frequency doesn't affect wavelength, ignoring the fundamental wave equation. The strategy is: when wave speed is constant, frequency and wavelength change inversely—one increasing by any factor means the other decreases by that same factor.
A buoy bobs up and down as water waves pass. The time between successive crests at the buoy is $T$ . Which statement correctly relates $T$ and $f$ ?
Answer: $T=\dfrac{1}{f}$ because one cycle takes one period. This question tests understanding of periodic waves. Period T is the time for one complete wave cycle to pass a fixed point, while frequency f is the number of cycles passing per unit time. These quantities are reciprocals: T = 1/f and f = 1/T. If a wave crest passes every T seconds, then 1/T crests pass per second, which is the frequency. Choice A incorrectly equates period and frequency, missing that they have inverse units (time vs. 1/time). The fundamental relationship to remember is: period measures time per cycle, frequency measures cycles per time, making them reciprocals of each other.
A sound wave travels through air at a constant speed in a room where temperature is steady. The frequency is decreased from $f$ to $\tfrac{1}{2}f$ . Which statement correctly relates the new wavelength to the original wavelength?
Answer: The wavelength becomes $2\lambda$ because the wave speed stays constant. 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.
A laser emits a continuous light wave in a transparent material where the wave speed is constant at $v$ . The light has wavelength $\lambda$ and frequency $f$ . The light enters a second material where the speed becomes $\tfrac{1}{2}v$ , while the frequency remains the same at the boundary. Which statement correctly relates the new wavelength $\lambda'$ to $\lambda$ ?
Answer: $\lambda' = \tfrac{1}{2}\lambda$ because $v=f\lambda$ and $f$ is unchanged This question tests understanding of periodic waves. When light crosses a boundary between materials, frequency remains constant but wave speed changes, affecting wavelength through v = fλ. If speed halves from v to v/2 while frequency f stays constant, wavelength must also halve: v/2 = f·(λ/2), giving λ' = λ/2. Choice A incorrectly claims lower speed means longer wavelength, reversing the direct proportionality between speed and wavelength at constant frequency. When waves enter a new medium, frequency stays fixed while wavelength scales with the speed ratio.
A continuous periodic wave travels along a string with constant wave speed $v$ . At one setting, the wave has wavelength $0.80\ \text{m}$ and frequency $5.0\ \text{Hz}$ . The driver is adjusted so the frequency becomes $10\ \text{Hz}$ while the string remains unchanged. Which statement correctly describes the new wavelength?
Answer: It becomes $0.40\ \text{m}$ because doubling frequency halves wavelength at constant speed 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.
A speaker produces a steady tone in air where the sound speed is $v$ . If the frequency is increased, which statement correctly relates the wavelength change?
Answer: The wavelength decreases because the speed stays approximately constant. This question tests understanding of periodic waves. The fundamental relationship v = fλ describes how wave speed, frequency, and wavelength connect. For sound waves in air at a given temperature, the wave speed v is approximately constant (around 343 m/s at room temperature). Rearranging to λ = v/f shows that wavelength and frequency are inversely related when speed is constant. Choice B incorrectly claims amplitude determines wavelength, mixing up the wave's height with its length—amplitude affects loudness, not wavelength. The strategy is: for waves in a uniform medium, wave speed is set by the medium's properties, so changing the source frequency inversely changes wavelength.
A wave on a string has speed $v$ , frequency $f$ , and wavelength $\lambda$ . The amplitude is doubled while the string tension and driving frequency remain unchanged. Which statement correctly describes the wavelength?
Answer: The wavelength stays $\lambda$ because speed and frequency are unchanged. 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.
Water waves in a ripple tank travel at a constant speed set by the water depth. The period of the wave is changed from $T$ to $3T$ by adjusting the source. Which statement correctly relates the new wavelength to the original wavelength?
Answer: The wavelength becomes $3\lambda$ because the wave speed stays constant. 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.
A sinusoidal wave on a long spring is described by $y(x,t)=A\sin(kx-\omega t)$ and repeats continuously. The wavelength is $\lambda$ and the period is $T$ . Without changing the spring's properties, the source is adjusted so the angular frequency doubles: $\omega' = 2\omega$ . Which statement correctly relates the new period $T'$ to $T$ ?
Answer: $T' = \tfrac{1}{2}T$ because $\omega = \tfrac{2\pi}{T}$ 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.
A tuning fork produces a steady sound wave in air with frequency $440\ \text{Hz}$ , repeating continuously. Assume the speed of sound in the room is constant at $v=343\ \text{m/s}$ . If the same fork is replaced by one producing $880\ \text{Hz}$ in the same air, which statement correctly relates the new wavelength $\lambda'$ to the original wavelength $\lambda$ ?
Answer: $\lambda' = \tfrac{1}{2}\lambda$ because $v=f\lambda$ and $v$ is constant 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.
A wave machine produces identical pulses that repeat, creating a periodic wave traveling at constant speed $v$ . The wavelength is adjusted from $\lambda$ to $\tfrac{1}{2}\lambda$ by changing the driver. Which statement correctly relates the new frequency to the original frequency?
Answer: The frequency becomes $2f$ because wave speed stays constant. 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.

Start this quiz