Relate Wavelength Frequency Wave Speed - Physics

A sound wave in air has wavelength $\lambda = 0.50\ \text{m}$ and travels at speed $v = 343\ \text{m/s}$ . What is the frequency $f$ of this sound? (Answer in Hz.)
Answer: $686\ \text{Hz}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the sound wave has wavelength λ = 0.50 m and travels at v = 343 m/s, we calculate frequency using f = v/λ = (343 m/s) / (0.50 m) = 686 Hz. This frequency of 686 Hz falls in the audible range for humans (20 Hz to 20,000 Hz), producing a relatively high-pitched sound. Choice B is correct because it properly applies f = v/λ with the correct values and units, dividing speed by wavelength to get frequency. Choice A incorrectly calculates f = v/2λ instead of f = v/λ, dividing by twice the wavelength and producing half the correct frequency (171.5 Hz instead of 686 Hz). When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (m/s ÷ m = 1/s = Hz), and (4) verify the answer makes sense for that wave type (sound frequencies in the audible range, with 686 Hz being a moderately high pitch).
A light (electromagnetic) wave in vacuum has wavelength $\lambda = 500\ \text{nm}$ . Using $c = 3.0\times 10^8\ \text{m/s}$ , what is its frequency $f$ ? (Use $v=f\lambda$ and note $500\ \text{nm} = 5.00\times 10^{-7}\ \text{m}$ .)
Answer: $6.0\times 10^{14}\ \text{Hz}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the wavelength is λ = 500 nm = 5.00× $10^{-7}$ m and the wave speed is v = 3.0× $10^8$ m/s, we calculate frequency using f = v/λ = ( $3.0×10^8$ m/s) / ( $5.00×10^{-7}$ m) = 6.0× $10^{14}$ Hz. This frequency of 6.0× $10^{14}$ Hz falls in the visible light range for electromagnetic waves. Choice B is correct because it properly applies v = fλ with the correct values and units. Choice C uses the wrong conversion for nm to m or misplaces the exponent, calculating an overly large frequency like 1.5× $10^{15}$ Hz instead of 6.0× $10^{14}$ Hz. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (convert MHz to Hz, nm to m, etc.), and (4) verify the answer makes sense for that wave type (sound wavelengths in cm-to-m range, visible light in nanometer range). The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.
A sound wave from a tuning fork has frequency $f = 440\ \text{Hz}$ and travels through air at $v = 343\ \text{m/s}$ . Using $v=f\lambda$ , what is the wavelength $\lambda$ of the sound in air? (Answer in meters.)
Answer: $0.778\ \text{m}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the sound wave has frequency f = 440 Hz and travels at v = 343 m/s, we calculate wavelength using λ = v/f = (343 m/s) / (440 Hz) = 0.780 m ≈ 0.778 m. This wavelength of about 78 cm is consistent with audible sound characteristics, as the 440 Hz A note is a standard tuning pitch. Choice A is correct because it properly applies λ = v/f with the correct values and units, yielding a wavelength in the expected range for audible sound. Choice C uses the equation backwards, calculating λ = v × f instead of λ = v/f, which multiplies when it should divide and produces an incorrect result of 150,920 m instead of 0.78 m. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (Hz = 1/s, so m/s ÷ Hz = m), and (4) verify the answer makes sense for that wave type (sound wavelengths in cm-to-m range, visible light in nanometer range).
A 1000 Hz sound wave is produced by a speaker. It travels in air at $v_{\text{air}} = 343\ \text{m/s}$ and then enters water where the speed of sound is $v_{\text{water}} = 1500\ \text{m/s}$ . What is the wavelength of the sound wave in water? (Answer in meters.)
Answer: $1.50\ \text{m}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. When a wave crosses from air to water, the frequency remains constant at 1000 Hz (determined by the source), but the wave speed changes from v₁ = 343 m/s to v₂ = 1500 m/s. Since v = fλ and f is constant, the wavelength must change proportionally: λ₂ = v₂/f = (1500 m/s) / (1000 Hz) = 1.50 m, so wavelength increases in the faster medium. Choice B is correct because it properly calculates the wavelength using λ = v/f with the water's wave speed, recognizing that frequency stays constant across media while wavelength changes with speed. Choice A incorrectly uses the air speed (343 m/s) instead of the water speed (1500 m/s) to calculate wavelength, failing to recognize that wavelength changes when the wave enters a new medium. Remember that when waves cross from one medium to another, frequency always stays constant (the source determines how many waves are created per second), but wave speed changes (depends on medium properties), so wavelength must adjust accordingly: if speed increases, wavelength increases proportionally (λ₂ = λ₁ × v₂/v₁).
A wave travels along a rope with speed $v = 12\ \text{m/s}$ . If the person shaking the rope produces a frequency of $f = 3.0\ \text{Hz}$ , what is the wavelength $\lambda$ of the wave on the rope?
Answer: 4.0 m This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the wave on the rope has speed v = 12 m/s and frequency f = 3.0 Hz, we can find wavelength using λ = v/f = (12 m/s) / (3.0 Hz) = 4.0 m. This wavelength of 4.0 m means that consecutive wave crests on the rope are separated by 4 meters, which is reasonable for a rope wave with relatively low frequency. Choice B is correct because it properly applies λ = v/f with the correct values and units, dividing speed by frequency to get wavelength. Choice A (36 m) uses the equation backwards, calculating v×f instead of v/f, which multiplies when it should divide and produces an incorrect result that would be unreasonably long for a typical rope. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (m/s ÷ Hz = m), and (4) verify the answer makes sense for that wave type. The key insight is that at constant wave speed, frequency and wavelength are inversely related—if you shake the rope faster (higher frequency), the wavelength becomes shorter, and if you shake it slower (lower frequency), the wavelength becomes longer.
A sonar device sends a sound wave of frequency $f = 1.0\ \text{kHz}$ into both air and water. The speed of sound is about $343\ \text{m/s}$ in air and $1500\ \text{m/s}$ in water. When the sound enters water from air, which statement best describes what happens to its wavelength $\lambda$ ?
Answer: The wavelength increases because the speed is higher in water while the frequency stays the same. This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. When a wave crosses from air to water, the frequency remains constant at f = 1.0 kHz (determined by the source), but the wave speed changes from v₁ = 343 m/s to v₂ = 1500 m/s. Since v = fλ and f is constant, the wavelength must change proportionally: λ₂/λ₁ = v₂/v₁ = 1500/343 ≈ 4.4, so wavelength increases by a factor of about 4.4 in the faster medium (water). Choice B is correct because it accurately describes that wavelength increases when the wave enters the faster medium (water) while frequency stays constant. Choice A incorrectly claims wavelength decreases in the faster medium, when actually the relationship λ = v/f shows that at constant frequency, wavelength increases with speed. Remember that when waves cross from one medium to another, frequency always stays constant (the source determines how many waves are created per second), but wave speed changes (depends on medium properties), so wavelength must adjust accordingly: if speed increases, wavelength increases proportionally (λ₂ = λ₁ × v₂/v₁). This principle explains why sonar systems must account for different wavelengths in air versus water even though the frequency remains the same.
A weather siren produces a sound wave with frequency $f = 850\ \text{Hz}$ . Assuming the speed of sound in air is $v = 343\ \text{m/s}$ , what is the wavelength $\lambda$ of the sound in air? (Use $v=f\lambda$ .)
Answer: $0.40\ \text{m}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that a sound wave has frequency f = 850 Hz and travels at v = 343 m/s, we can find wavelength using λ = v/f = (343 m/s) / (850 Hz) ≈ 0.4035 m, which rounds to 0.40 m. This wavelength of 0.40 m is consistent with audible sound characteristics, as human hearing ranges involve wavelengths from about 0.017 m to 17 m in air. Choice A is correct because it properly applies v = fλ with the correct values and units to calculate λ = v/f. Choice B uses the equation backwards, calculating λ = v×f instead of λ = v/f, which multiplies when it should divide and produces an incorrect result with wrong units (m²/s instead of m). When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent, and (4) verify the answer makes sense for that wave type (sound wavelengths in cm-to-m range, visible light in nanometer range). The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.
A sound wave from a tuning fork has wavelength $\lambda = 0.78\ \text{m}$ in air. If the speed of sound is $v = 343\ \text{m/s}$ , what is the frequency $f$ ? (Use $f=\frac{v}{\lambda}$ .)
Answer: $440\ \text{Hz}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the sound wave has wavelength λ = 0.78 m and the wave speed is v = 343 m/s, we calculate frequency using f = v/λ = (343 m/s) / (0.78 m) = 440 Hz. This frequency of 440 Hz falls in the audible range and is actually the standard tuning pitch A4. Choice C is correct because it properly applies v = fλ with the correct values and units, rearranging to f = v/λ = 343/0.78 = 440 Hz. Choice D incorrectly inverts the equation, calculating f = λ/v instead of f = v/λ, which produces an incorrect result of 0.0023 Hz by dividing the smaller number by the larger one. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (convert MHz to Hz, nm to m, etc.), and (4) verify the answer makes sense for that wave type (sound wavelengths in cm-to-m range, visible light in nanometer range). The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.
A sound wave in air has wavelength $\lambda = 0.70\ \text{m}$ and travels at $v = 343\ \text{m/s}$ . Using $v = f\lambda$ , what is the frequency $f$ ?
Answer: 490 Hz This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the sound wave has wavelength λ = 0.70 m and travels at v = 343 m/s, we calculate frequency using f = v/λ = (343 m/s) / (0.70 m) = 490 Hz. This frequency of 490 Hz falls in the audible range for humans (20 Hz to 20,000 Hz) and corresponds roughly to the musical note B4. Choice B is correct because it properly applies f = v/λ with the correct values and units, dividing speed by wavelength to get frequency. Choice A (240 Hz) appears to use an incorrect calculation, possibly using λ = 1.43 m instead of 0.70 m, while choice D (0.0020 Hz) incorrectly calculates λ/v instead of v/λ, which inverts the correct relationship. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (m/s ÷ m = Hz), and (4) verify the answer makes sense for that wave type. The key insight is that at constant wave speed, frequency and wavelength are inversely related—a wavelength of 0.70 m (70 cm) corresponds to a mid-range audible frequency, neither very high nor very low.
Two electromagnetic waves travel in a vacuum at $3.0\times 10^8\ \text{m/s}$ . Wave 1 has frequency $f_1 = 5.0\times 10^{14}\ \text{Hz}$ and Wave 2 has frequency $f_2 = 1.0\times 10^{14}\ \text{Hz}$ . Which wave has the longer wavelength?
Answer: Wave 2 has the longer wavelength. This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. For EM spectrum comparison: Wave 1 has frequency f₁ = 5.0 × $10^{14}$ Hz and Wave 2 has frequency f₂ = 1.0 × $10^{14}$ Hz. Since all electromagnetic waves travel at c = 3.0 × $10^8$ m/s in vacuum, using λ = c/f gives λ₁ = 6.0 × $10^{-7}$ m and λ₂ = 3.0 × $10^{-6}$ m, showing that the lower frequency wave has the longer wavelength. Choice B is correct because it correctly identifies that frequency stays constant but lower frequency means longer wavelength. Choice A reverses the wavelength order in the electromagnetic spectrum, claiming Wave 1 has longer wavelength than Wave 2, when actually higher frequency means shorter wavelength. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (convert MHz to Hz, nm to m, etc.), and (4) verify the answer makes sense for that wave type (visible light in nanometer range).
A guitar string produces a sound in air. If the pitch (frequency) increases while the speed of sound in air remains constant at $343\ \text{m/s}$ , what happens to the wavelength $\lambda$ ? (Use $v = f\lambda$ .)
Answer: The wavelength decreases. This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. For inverse relationship: For waves traveling at constant speed v, the equation v = fλ shows that frequency and wavelength are inversely proportional: if frequency increases, wavelength must decrease to keep their product (wave speed) constant. Mathematically, fλ = constant, so f₁λ₁ = f₂λ₂. Choice B is correct because it accurately describes the inverse relationship between f and λ. Choice A incorrectly claims frequency and wavelength are directly proportional (both increase together), when actually they're inversely proportional: as frequency increases, wavelength must decrease to maintain constant wave speed. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent, and (4) verify the answer makes sense for that wave type. Common mistake: assuming all three quantities can change independently—they cannot. The equation v = fλ means if you know how two quantities change, the third is determined: for example, if frequency doubles and speed stays constant, the wavelength must halve.
Two sound waves travel through the same room air (so their speed is the same, $v \approx 343\ \text{m/s}$ ). Wave 1 has frequency $f_1 = 250\ \text{Hz}$ and Wave 2 has frequency $f_2 = 1000\ \text{Hz}$ . How does the wavelength of Wave 1 compare to the wavelength of Wave 2?
Answer: $\lambda_1$ is 4 times $\lambda_2$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. For waves traveling at constant speed v, the equation v = fλ shows that frequency and wavelength are inversely proportional: if frequency increases by a factor of 4 (from 250 Hz to 1000 Hz), wavelength must decrease by the same factor to keep their product (wave speed) constant. Mathematically, since v₁ = v₂, we have f₁λ₁ = f₂λ₂, so λ₁/λ₂ = f₂/f₁ = 1000/250 = 4, meaning λ₁ is 4 times λ₂. Choice A is correct because it accurately describes the inverse relationship between f and λ, recognizing that the lower frequency wave (250 Hz) has 4 times the wavelength of the higher frequency wave (1000 Hz). Choice D incorrectly claims wavelength and frequency are independent in the same medium, when actually they're inversely proportional: as frequency increases, wavelength must decrease to maintain constant wave speed. The key insight is that at constant wave speed, frequency and wavelength are inversely related—quadruple the frequency means one-quarter the wavelength, which is why low-pitched sounds have long wavelengths and high-pitched sounds have short wavelengths.
A water wave travels at a constant speed. If the frequency increases from $0.40\ \text{Hz}$ to $0.80\ \text{Hz}$ while the wave speed stays the same, what happens to the wavelength $\lambda$ ?
Answer: The wavelength is cut in half This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. For waves traveling at constant speed v, the equation v = fλ shows that frequency and wavelength are inversely proportional: if frequency increases by a factor of 2 (from 0.40 Hz to 0.80 Hz), wavelength must decrease by the same factor to keep their product (wave speed) constant. Mathematically, fλ = constant, so f₁λ₁ = f₂λ₂, which means λ₂ = λ₁ × (f₁/f₂) = λ₁ × (0.40/0.80) = λ₁/2. Choice B is correct because it accurately describes the inverse relationship between f and λ, recognizing that doubling the frequency halves the wavelength when wave speed remains constant. Choice A incorrectly claims frequency and wavelength are directly proportional (both increase together), when actually they're inversely proportional: as frequency increases, wavelength must decrease to maintain constant wave speed. The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which explains why higher frequency waves always have shorter wavelengths when traveling at the same speed.
A seismic P-wave travels through rock with speed $v = 6.0\ \text{km/s}$ and has wavelength $\lambda = 2.0\ \text{km}$ . Using $v = f\lambda$ , what is the frequency $f$ (in Hz)?
Answer: 3.0 Hz This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the seismic P-wave has speed v = 6.0 km/s and wavelength λ = 2.0 km, we calculate frequency using f = v/λ = (6.0 km/s) / (2.0 km) = 3.0 s⁻¹ = 3.0 Hz. This frequency of 3.0 Hz is consistent with seismic waves, which typically have very low frequencies (often below human hearing range) due to their long wavelengths. Choice A is correct because it properly applies f = v/λ with the correct values and units, noting that km/s ÷ km = s⁻¹ = Hz. Choice B (12 Hz) uses the equation backwards, calculating v×λ instead of v/λ, which multiplies when it should divide and produces an incorrect result. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (km cancels out, leaving 1/s = Hz), and (4) verify the answer makes sense for that wave type. Remember that seismic waves have much lower frequencies than sound waves because they have much longer wavelengths, even though they travel faster than sound in air.
A laser emits red light with wavelength $\lambda = 650\ \text{nm}$ . Assuming the light travels at $v = 3.0\times10^8\ \text{m/s}$ , what is the frequency $f$ of this light? (Answer in Hz.)
Answer: $4.6\times10^{14}\ \text{Hz}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the red light has wavelength λ = 650 nm = 650 × 10⁻⁹ m and travels at v = 3.0 × 10⁸ m/s, we calculate frequency using f = v/λ = (3.0 × 10⁸ m/s) / (650 × 10⁻⁹ m) = 4.6 × 10¹⁴ Hz. This frequency of 4.6 × 10¹⁴ Hz falls in the visible light range of the electromagnetic spectrum, consistent with red light. Choice A is correct because it properly applies f = v/λ with the correct values and units, converting nanometers to meters before calculation. Choice B incorrectly calculates f = λ/v instead of f = v/λ, which inverts the equation and produces an incorrect result with wrong units (m²/s instead of Hz). When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (convert nm to m: 1 nm = 10⁻⁹ m), and (4) verify the answer makes sense for that wave type (visible light frequencies are in the 10¹⁴ Hz range).
A sound wave in air at room temperature travels at $v = 343\ \text{m/s}$ and has frequency $f = 686\ \text{Hz}$ . What is its wavelength $\lambda$ ? Use $\lambda = v/f$ .
Answer: 0.50 m This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. For calculating wavelength: Given that sound waves have frequency f = 686 Hz and travel at v = 343 m/s, we can find wavelength using λ = v/f = (343 m/s) / (686 Hz) ≈ 0.50 m. This wavelength of 0.50 m is consistent with audible sound characteristics. Choice A is correct because it properly applies v = fλ with the correct values and units. Choice D uses the wrong wave speed for this wave type (perhaps light speed) instead of the correct speed (343 m/s), leading to an incorrect wavelength calculation. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent, and (4) verify the answer makes sense for that wave type (sound wavelengths in cm-to-m range). The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.
A water wave has speed $v = 10\ \text{m/s}$ . If the wave speed stays constant but the frequency increases from $0.50\ \text{Hz}$ to $1.0\ \text{Hz}$ , what happens to the wavelength?
Answer: The wavelength is cut in half. This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. For waves traveling at constant speed v, the equation v = fλ shows that frequency and wavelength are inversely proportional: if frequency increases by a factor of 2 (from 0.50 Hz to 1.0 Hz), wavelength must decrease by the same factor to keep their product (wave speed) constant. Mathematically, fλ = constant, so f₁λ₁ = f₂λ₂. Choice C is correct because it accurately describes the inverse relationship between f and λ. Choice A incorrectly claims frequency and wavelength are directly proportional (both increase together), when actually they're inversely proportional: as frequency increases, wavelength must decrease to maintain constant wave speed. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (convert MHz to Hz, nm to m, etc.), and (4) verify the answer makes sense for that wave type (sound wavelengths in cm-to-m range, visible light in nanometer range). The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.
A surface water wave has speed $v = 3.6\ \text{m/s}$ and wavelength $\lambda = 4.0\ \text{m}$ . What is the frequency $f$ of the wave? (Use $f=v/\lambda$ .)
Answer: $0.90\ \text{Hz}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that the wavelength is λ = 4.0 m and the wave speed is v = 3.6 m/s, we calculate frequency using f = v/λ = (3.6 m/s) / (4.0 m) = 0.90 Hz. This frequency of 0.90 Hz falls in the low range for surface water waves, consistent with waves passing about once per second. Choice B is correct because it properly applies v = fλ with the correct values and units to find f = v/λ. Choice A incorrectly calculates f = v×λ = 3.6 × 4.0 = 14.4 Hz, which multiplies when it should divide and produces an incorrect result with wrong units (m²/s instead of Hz). When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent, and (4) verify the answer makes sense for that wave type (sound wavelengths in cm-to-m range, visible light in nanometer range). The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.
A weather siren produces a sound wave in air with frequency $f = 850\ \text{Hz}$ . Assuming the speed of sound in air is $v = 343\ \text{m/s}$ , what is the wavelength $\lambda$ of the sound? (Use $v=f\lambda$ .)
Answer: $0.40\ \text{m}$ This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. Given that sound wave has frequency f = 850 Hz and travels at v = 343 m/s, we can find wavelength using λ = v/f = (343 m/s) / (850 Hz) = 0.40 m. This wavelength of 0.40 m is consistent with audible sound characteristics. Choice A is correct because it properly applies v = fλ with the correct values and units. Choice C uses the equation backwards, calculating λ = f/v instead of λ = v/f, which inverts the division and produces an incorrect result of 2.48 m. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent (convert MHz to Hz, nm to m, etc.), and (4) verify the answer makes sense for that wave type (sound wavelengths in cm-to-m range, visible light in nanometer range). The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.
Two sound waves travel through the same air at room temperature (so the wave speed is the same for both). Wave 1 has frequency $f_1 = 250\ \text{Hz}$ and Wave 2 has frequency $f_2 = 500\ \text{Hz}$ . How does the wavelength of Wave 1 compare to the wavelength of Wave 2?
Answer: Wave 1 has twice the wavelength of Wave 2. This question tests understanding of the relationship between wavelength, frequency, and wave speed, described by the equation v = fλ. The wave equation v = fλ states that wave speed (v) equals the product of frequency (f, measured in Hz or cycles per second) and wavelength (λ, the distance between successive wave crests), and this relationship can be rearranged to solve for any of the three quantities: f = v/λ or λ = v/f. For waves traveling at constant speed v, the equation v = fλ shows that frequency and wavelength are inversely proportional: if frequency increases by a factor of 2 (from 250 Hz to 500 Hz), wavelength must decrease by the same factor to keep their product (wave speed) constant. Mathematically, f₁λ₁ = f₂λ₂, so λ₁/λ₂ = f₂/f₁ = 500/250 = 2, meaning Wave 1 has twice the wavelength of Wave 2. Choice C is correct because it accurately describes the inverse relationship between f and λ: since Wave 1 has half the frequency of Wave 2, it must have twice the wavelength. Choice A incorrectly reverses the relationship, claiming the lower frequency wave has shorter wavelength, when actually lower frequency means longer wavelength. When solving v = fλ problems: (1) identify which two quantities are given, (2) rearrange the equation to solve for the unknown (λ = v/f, f = v/λ, or v = fλ), (3) check that units are consistent, and (4) verify the answer makes sense for that wave type. The key insight is that at constant wave speed, frequency and wavelength are inversely related—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.

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