Periodic Waves
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AP Physics 2 › Periodic Waves
A repeating wave pattern has wavelength $\lambda$ and period $T$ as it propagates. Which statement correctly relates wave speed to these quantities?
$v=\lambda T$ because distance and time multiply.
$v=\dfrac{1}{\lambda T}$ because speed is inverse of wave size.
$v=\dfrac{\lambda}{T}$ because one wavelength passes in one period.
$v=\dfrac{T}{\lambda}$ because speed is time per distance.
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.
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?
$\lambda_2=\dfrac{1}{2}\lambda_1$ because the amplitude is larger.
$\lambda_2=\dfrac{1}{2}\lambda_1$ because the wave speed is the same.
$\lambda_2=2\lambda_1$ because higher frequency means longer wavelength.
$\lambda_2=\lambda_1$ because the string sets wavelength directly.
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.
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?
The wavelength decreases by a factor of 3 because the speed decreases.
The wavelength increases by a factor of 3 because the speed stays constant.
The wavelength increases by a factor of 3 because amplitude increases.
The wavelength is unchanged because frequency does not affect wavelength.
Explanation
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 $f$?
$T=\dfrac{1}{f}$ because one cycle takes one period.
$T=\dfrac{f}{2}$ because a crest is half a cycle.
$T=\dfrac{1}{2f}$ because two crests make one cycle.
$T=f$ because both measure how often crests pass.
Explanation
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 sinusoidal wave travels along a taut string with constant tension, so the wave speed is constant. A student increases the driving frequency from $f$ to $2f$ while keeping the same string. Which statement correctly relates the new wavelength to the original wavelength?
The wavelength becomes $\tfrac{1}{4}\lambda$ because both speed and frequency double.
The wavelength becomes $\tfrac{1}{2}\lambda$ because speed stays constant.
The wavelength stays $\lambda$ because amplitude determines wavelength.
The wavelength becomes $2\lambda$ because speed increases with frequency.
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.
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 $\lambda$?
$\lambda' = \tfrac{1}{2}\lambda$ because halving $T$ doubles $f$ at constant speed
$\lambda' = 2\lambda$ because halving $T$ halves the frequency
$\lambda' = 4\lambda$ because the wave speed increases as $T$ decreases
$\lambda' = \lambda$ because the rope properties determine wavelength directly
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.
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?
It becomes $0.20\ \text{m}$ because doubling frequency quarters wavelength
It becomes $1.6\ \text{m}$ because frequency and wavelength increase together
It becomes $0.40\ \text{m}$ because doubling frequency halves wavelength at constant speed
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.
Ocean waves move past a pier at constant speed $v$. If the period doubles, which statement correctly relates the new wavelength to the old?
The new wavelength is twice the old because frequency halves.
The new wavelength is twice the old because the wave speed doubles.
The new wavelength is unchanged because amplitude is unchanged.
The new wavelength is half the old because frequency doubles.
Explanation
This question tests understanding of periodic waves. The wave equation v = fλ connects wave speed, frequency, and wavelength. Since period T = 1/f, when period doubles, frequency halves (f becomes f/2). With constant wave speed v (as stated), we can rearrange to λ = v/f, showing wavelength is inversely proportional to frequency. If frequency halves, wavelength must double to maintain the same wave speed. Choice C incorrectly assumes amplitude determines wavelength, confusing wave height with wave length. Remember: when wave speed stays constant, period and wavelength change in the same direction—both double or both halve together.
A wave on a long rope is driven at frequency $f$ while the rope’s tension and linear density stay constant. Which statement correctly relates wavelength to frequency?
The wavelength increases as $f$ increases because the wave speed increases.
The wavelength is unchanged as $f$ increases because amplitude sets wavelength.
The wavelength decreases as $f$ increases because the period increases.
The wavelength decreases as $f$ increases because the wave speed stays constant.
Explanation
This question tests understanding of periodic waves. The fundamental wave equation is v = fλ, where v is wave speed, f is frequency, and λ is wavelength. For waves on a rope with constant tension and linear density, the wave speed v remains constant. When we rearrange the equation to λ = v/f, we see that wavelength and frequency are inversely proportional when speed is constant. Choice C incorrectly assumes amplitude determines wavelength, which is a common misconception—amplitude affects the wave's energy but not its wavelength. The key strategy is: when wave speed is fixed by the medium's properties, increasing frequency always decreases wavelength proportionally.
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?
The wavelength becomes $\tfrac{1}{3}\lambda$ because frequency increases.
The wavelength becomes $3\lambda$ because the wave speed stays constant.
The wavelength stays $\lambda$ because period does not affect wavelength.
The wavelength becomes $9\lambda$ because both period and speed triple.
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.