Periodic Waves
Help Questions
AP Physics 2 › Periodic Waves
A periodic wave has frequency $5.0\ \text{Hz}$ and travels at $2.0\ \text{m/s}$ in a uniform medium. Which statement correctly gives its wavelength?
$\lambda=2.5\ \text{m}$ because $\lambda=\dfrac{f}{v}$.
$\lambda=0.40\ \text{m}$ because $\lambda=\dfrac{v}{f}$.
$\lambda=0.10\ \text{m}$ because $\lambda=\dfrac{1}{vf}$.
$\lambda=10\ \text{m}$ because $\lambda=vf$.
Explanation
This question tests understanding of periodic waves. The wave equation v = fλ connects wave speed, frequency, and wavelength for any periodic wave. Given v = 2.0 m/s and f = 5.0 Hz, we solve for wavelength by rearranging to λ = v/f. Substituting values: λ = (2.0 m/s)/(5.0 Hz) = 0.40 m. Choice A incorrectly multiplies v and f, reversing the relationship—this would give units of m²/s instead of meters. The reliable strategy is: always check units when using v = fλ; wavelength (m) equals speed (m/s) divided by frequency (1/s).
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?
The frequency stays $f$ because amplitude determines frequency.
The frequency becomes $4f$ because both speed and frequency must increase together.
The frequency becomes $\tfrac{1}{2}f$ because shorter wavelength means lower frequency.
The frequency becomes $2f$ because wave speed stays constant.
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.
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?
The wavelength stays $\lambda$ because amplitude sets the wavelength.
The wavelength becomes $2\lambda$ because the wave speed stays constant.
The wavelength becomes $4\lambda$ because the speed doubles when frequency halves.
The wavelength becomes $\tfrac{1}{2}\lambda$ because wavelength follows frequency directly.
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.
A traveling water wave has wavelength $\lambda$ and period $T$ in a tank where the wave speed is constant. The source is adjusted so the wavelength becomes $4\lambda$. Which statement correctly relates the new period to the original period?
The period becomes $4T$ because the wave speed stays constant.
The period stays $T$ because wavelength does not affect period.
The period becomes $16T$ because both wavelength and speed increase by 4.
The period becomes $\tfrac{1}{4}T$ because longer wavelength means higher frequency.
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.
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?
The frequency becomes $2f$ because frequency is the inverse of period.
The frequency becomes $4f$ because halving period doubles speed and frequency.
The frequency stays $f$ because wave speed is unchanged.
The frequency becomes $\tfrac{1}{2}f$ because frequency is proportional to period.
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.
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 stays $\lambda$ because amplitude determines wavelength.
The wavelength becomes $\tfrac{1}{2}\lambda$ because speed stays constant.
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.
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 $9\lambda$ because both period and speed triple.
The wavelength becomes $\tfrac{1}{3}\lambda$ because frequency increases.
The wavelength stays $\lambda$ because period does not affect wavelength.
The wavelength becomes $3\lambda$ because the wave speed stays constant.
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.
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?
The wavelength halves because larger amplitude increases frequency.
The wavelength stays $\lambda$ because speed and frequency are unchanged.
The wavelength increases because wave speed increases with amplitude.
The wavelength doubles because amplitude determines wavelength.
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.
A periodic wave on a string has frequency $f$ and wavelength $\lambda$. The string is replaced with one that makes the wave speed double, while the driver keeps the same frequency. Which statement correctly relates the new wavelength to the original wavelength?
The wavelength stays $\lambda$ because frequency sets wavelength regardless of speed.
The wavelength becomes $2\lambda$ because the frequency stays the same.
The wavelength becomes $\tfrac{1}{2}\lambda$ because higher speed means shorter wavelength.
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.
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?
The period becomes $\tfrac{1}{9}T$ because speed and frequency both triple.
The period becomes $3T$ because period increases with frequency.
The period becomes $\tfrac{1}{3}T$ because period is the inverse of frequency.
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.