Physics Quiz: Relate Wavelength Frequency Wave Speed

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Relate Wavelength Frequency Wave SpeedQuestion 1 of 20

A radio wave in air has wavelength $\lambda = 12\ \text{m}$ . Using $v = 3.0\times10^8\ \text{m/s}$ for electromagnetic waves in air, what is the frequency $f$ ? (Use $f=\frac{v}{\lambda}$ .)

2.5\times10^6\ \text{Hz}

2.5\times10^7\ \text{Hz}

3.6\times10^9\ \text{Hz}

4.0\times10^{-8}\ \text{Hz}

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Physics Quiz: Relate Wavelength Frequency Wave Speed

Practice Relate Wavelength Frequency Wave Speed in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Relate Wavelength Frequency Wave Speed, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics.

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Question 1

A radio wave in air has wavelength $\lambda = 12\ \text{m}$ . Using $v = 3.0\times10^8\ \text{m/s}$ for electromagnetic waves in air, what is the frequency $f$ ? (Use $f=\frac{v}{\lambda}$ .)

$2.5\times10^6\ \text{Hz}$
$2.5\times10^7\ \text{Hz}$ (correct answer)
$3.6\times10^9\ \text{Hz}$
$4.0\times10^{-8}\ \text{Hz}$

Explanation: 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 radio wave has wavelength λ = 12 m and travels at v = 3.0×10⁸ m/s, we calculate frequency using f = v/λ = (3.0×10⁸ m/s) / (12 m) = 2.5×10⁷ Hz. This frequency of 25 MHz falls in the shortwave radio band. Choice B is correct because it properly applies v = fλ with the correct values and units, calculating f = v/λ = 3.0×10⁸/12 = 2.5×10⁷ Hz. Choice A incorrectly calculates the numerical value, likely by dividing 3.0 by 12 to get 0.25 but then incorrectly handling the power of 10, resulting in 2.5×10⁶ Hz instead of 2.5×10⁷ 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 electromagnetic waves all travel at the same speed in vacuum or air (3.0×10⁸ m/s), so their frequency and wavelength are always inversely related—longer wavelengths mean lower frequencies.

Question 2

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 $f$ ?

240 Hz
490 Hz (correct answer)
343 Hz
0.0020 Hz

Explanation: 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.

Question 3

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 $\lambda = v/f$ .

0.50 m (correct answer)
2.0 m
4.99×10⁻³ m
1.01×10⁵ m

Explanation: 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.

Question 4

A wave traveling on a stretched rope has wavelength $\lambda = 2.5\ \text{m}$ and speed $v = 5.0\ \text{m/s}$ . What is the frequency $f$ of the wave? (Use $f=v/\lambda$ .)

$12.5\ \text{Hz}$
$2.0\ \text{Hz}$ (correct answer)
$0.50\ \text{Hz}$
$7.5\ \text{Hz}$

Explanation: 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 = 2.5 m and the wave speed is v = 5.0 m/s, we calculate frequency using f = v/λ = (5.0 m/s) / (2.5 m) = 2.0 Hz. This frequency of 2.0 Hz falls in the low range for waves on a rope, consistent with visible oscillations occurring twice 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 = 5.0 × 2.5 = 12.5 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.

Question 5

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 $f=\frac{v}{\lambda}$ .)

$268\ \text{Hz}$
$4.5\times10^2\ \text{m}$
$440\ \text{Hz}$ (correct answer)
$0.0023\ \text{Hz}$

Explanation: 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.

Question 6

Two sound waves travel through the same room-temperature air where the speed of sound is constant at $343\ \text{m/s}$ . Wave 1 has frequency $f_1 = 300\ \text{Hz}$ and Wave 2 has frequency $f_2 = 600\ \text{Hz}$ . How does the wavelength of Wave 2 compare to the wavelength of Wave 1?

$\lambda_2 = 2\lambda_1$
$\lambda_2 = \tfrac{1}{2}\lambda_1$ (correct answer)
$\lambda_2 = \lambda_1$
$\lambda_2 = 4\lambda_1$

Explanation: 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 = 343 m/s, the equation v = fλ shows that frequency and wavelength are inversely proportional: if frequency increases by a factor of 2 (from 300 Hz to 600 Hz), wavelength must decrease by the same factor to keep their product (wave speed) constant. Mathematically, fλ = constant, so f₁λ₁ = f₂λ₂, giving λ₂ = λ₁ × (f₁/f₂) = λ₁ × (1/2). 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 (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.

Question 7

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 $f=v/\lambda$ .)

$14.4\ \text{Hz}$
$0.90\ \text{Hz}$ (correct answer)
$1.1\ \text{Hz}$
$7.6\ \text{Hz}$

Explanation: 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.

Question 8

An FM radio station broadcasts at a frequency of $101.5\ \text{MHz}$ (that is, $1.015\times10^8\ \text{Hz}$ ). Assuming the radio wave travels at $v = 3.0\times10^8\ \text{m/s}$ in air, what is its wavelength $\lambda$ ? (Use $\lambda=v/f$ .)

$2.96\ \text{m}$ (correct answer)
$0.338\ \text{m}$
$296\ \text{m}$
$2.96\times10^{-8}\ \text{m}$

3.0×10^8

m/s) / (

1.015×10^8

Hz) ≈ 2.96 m. This wavelength of 2.96 m is consistent with FM radio wave characteristics, which typically have wavelengths in the meter range. Choice A is correct because it properly applies v = fλ with the correct values and units to calculate λ = v/f. 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 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 (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.

Question 9

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 $v=f\lambda$ .)

$0.40\ \text{m}$ (correct answer)
$291{,}550\ \text{m}$
$2.48\ \text{m}$
$0.0025\ \text{m}$

Explanation: 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.

Question 10

A water wave on a lake has a wavelength of $\lambda = 8.0\ \text{m}$ . A floating buoy measures a wave speed of $v = 4.0\ \text{m/s}$ . What is the wave frequency $f$ ? (Use $v=f\lambda$ .)

$32\ \text{Hz}$
$0.50\ \text{Hz}$ (correct answer)
$2.0\ \text{Hz}$
$0.125\ \text{Hz}$

Explanation: 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 = 8.0 m and the wave speed is v = 4.0 m/s, we calculate frequency using f = v/λ = (4.0 m/s) / (8.0 m) = 0.50 Hz. This frequency of 0.50 Hz falls in the low range for water waves. Choice B is correct because it properly applies v = fλ with the correct values and units. Choice A uses the equation backwards, calculating f = v×λ instead of f = v/λ, which multiplies when it should divide and produces an incorrect result of 32 Hz 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 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.

Question 11

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 $v=f\lambda$ .)

$0.40\ \text{m}$ (correct answer)
$291{,}550\ \text{m}$
$2.48\ \text{m}$
$0.0025\ \text{m}$

Explanation: 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.

Question 12

Two sound waves in air travel at the same speed $v = 343\ \text{m/s}$ . Wave 1 has frequency $f_1 = 440\ \text{Hz}$ and Wave 2 has frequency $f_2 = 880\ \text{Hz}$ . How does the wavelength of Wave 2 compare to the wavelength of Wave 1?

$\lambda_2 = 2\lambda_1$
$\lambda_2 = \tfrac{1}{2}\lambda_1$ (correct answer)
$\lambda_2 = \lambda_1$
$\lambda_2 = 4\lambda_1$

Explanation: 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 440 Hz to 880 Hz), wavelength must decrease by the same factor 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 C 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.

Question 13

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.)

$0.778\ \text{m}$ (correct answer)
$1.27\ \text{m}$
$150{,}920\ \text{m}$
$0.00128\ \text{m}$

Explanation: 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).

Question 14

Ocean surface waves are measured to have wavelength $\lambda = 12\ \text{m}$ . A buoy records a frequency of $f = 0.50\ \text{Hz}$ (0.50 waves per second). What is the wave speed $v$ ? (Answer in m/s.)

$6.0\ \text{m/s}$ (correct answer)
$24\ \text{m/s}$
$0.0417\ \text{m/s}$
$11.5\ \text{m/s}$

Explanation: 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 ocean wave has wavelength λ = 12 m and frequency f = 0.50 Hz, we calculate wave speed using v = fλ = (0.50 Hz) × (12 m) = 6.0 m/s. This wave speed of 6.0 m/s is reasonable for ocean surface waves, which typically travel at speeds of a few to tens of meters per second. Choice A is correct because it properly applies v = fλ with the correct values and units, multiplying frequency and wavelength to get speed. Choice C incorrectly calculates v = f/λ instead of v = fλ, which divides when it should multiply and produces an incorrect result of 0.0417 m/s, far too slow for ocean waves. 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 × m = m/s), and (4) verify the answer makes sense for that wave type (ocean waves travel at several m/s, not cm/s).

Question 15

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.)

$0.343\ \text{m}$
$1.50\ \text{m}$ (correct answer)
$4.37\ \text{m}$
$1500\ \text{m}$

Explanation: 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).

Question 16

A sound wave in air at room temperature travels at $v = 343\ \text{m/s}$ and has frequency $f = 686\ \text{Hz}$ . Using $v = f\lambda$ , what is the wavelength $\lambda$ ?

0.50 m (correct answer)
2.0 m
235,000 m
0.50 Hz

Explanation: 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 speed v = 343 m/s and frequency f = 686 Hz, we can find wavelength using λ = v/f = (343 m/s) / (686 Hz) = 0.50 m. This wavelength of 0.50 m (50 cm) is consistent with audible sound characteristics, as human-audible frequencies typically have wavelengths ranging from centimeters to meters. Choice A is correct because it properly applies λ = v/f with the correct values and units, dividing speed by frequency to get wavelength. Choice C (235,000 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 hundreds of kilometers long. 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—double the frequency means half the wavelength, which is why high-pitched sounds have short wavelengths and low-pitched sounds have long wavelengths.

Question 17

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 $\lambda$ ?

The wavelength decreases because the speed is higher in water.
The wavelength increases because the speed is higher in water while the frequency stays the same. (correct answer)
The wavelength stays the same because the frequency stays the same.
The wavelength increases because the frequency increases in water.

Explanation: 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.

Question 18

A light wave in vacuum travels at $c = 3.0\times10^8\ \text{m/s}$ . If its wavelength is $\lambda = 600\ \text{nm}$ , what is its frequency $f$ ? (Use $f=\frac{c}{\lambda}$ and $1\ \text{nm}=1\times10^{-9}\ \text{m}$ .)

$5.0\times10^{14}\ \text{Hz}$ (correct answer)
$5.0\times10^5\ \text{Hz}$
$1.8\times10^2\ \text{Hz}$
$2.0\times10^{15}\ \text{Hz}$

Explanation: 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 light wave has wavelength λ = 600 nm = 600×10⁻⁹ m = 6.0×10⁻⁷ m and travels at c = 3.0×10⁸ m/s, we calculate frequency using f = c/λ = (3.0×10⁸ m/s) / (6.0×10⁻⁷ m) = 5.0×10¹⁴ Hz. This frequency of 5.0×10¹⁴ Hz falls in the visible light range (orange-red light). Choice A is correct because it properly applies v = fλ with the correct values and units, converting nanometers to meters before calculating f = c/λ = 3.0×10⁸/(6.0×10⁻⁷) = 5.0×10¹⁴ Hz. Choice B incorrectly handles the unit conversion or powers of 10, resulting in 5.0×10⁵ Hz instead of 5.0×10¹⁴ Hz, which would be in the radio wave range rather than visible light. 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). Remember that visible light has extremely high frequencies (10¹⁴ to 10¹⁵ Hz) because of its very short wavelengths (400-700 nm) combined with the enormous speed of light.

Question 19

A sound wave with frequency $f = 1000\ \text{Hz}$ travels from air into water. Take $v_{\text{air}} = 343\ \text{m/s}$ and $v_{\text{water}} = 1500\ \text{m/s}$ . What happens to the wavelength when the sound enters water? (Use $v=f\lambda$ and assume the frequency stays the same.)

The wavelength decreases because the speed increases.
The wavelength increases because the speed increases. (correct answer)
The wavelength stays the same because frequency stays the same.
The wavelength stays the same because wave speed stays the same.

Explanation: 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 (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 in the faster medium. Choice B is correct because it accurately describes that wavelength increases when speed increases, maintaining the relationship v = fλ with constant frequency. Choice C incorrectly assumes wavelength stays constant when a wave crosses between media, when actually it's frequency that stays constant (determined by the source), while speed and wavelength both change. 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). 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 a wave enters a medium where it travels about 4.4 times faster and frequency stays the same, the wavelength must increase by the same factor to satisfy the equation.

Question 20

An FM radio wave travels through air at approximately $v = 3.0\times10^8\ \text{m/s}$ . A station broadcasts at $f = 99.5\ \text{MHz}$ . What is the wavelength $\lambda$ of this radio wave? (Use $v=f\lambda$ .)

$3.0\ \text{m}$ (correct answer)
$0.33\ \text{m}$
$3.0\times10^{-3}\ \text{m}$
$3.0\times10^6\ \text{m}$

Explanation: 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 FM radio wave has frequency f = 99.5 MHz = 99.5×10⁶ Hz and travels at v = 3.0×10⁸ m/s, we can find wavelength using λ = v/f = (3.0×10⁸ m/s) / (99.5×10⁶ Hz) = 3.0 m. This wavelength of 3.0 m is consistent with FM radio wave characteristics. Choice A is correct because it properly applies v = fλ with the correct values and units, converting MHz to Hz before calculating λ = v/f = 3.0×10⁸/(99.5×10⁶) ≈ 3.0 m. Choice C incorrectly calculates the numerical value, likely by making an error with the powers of 10, resulting in 3.0×10⁻³ m instead of 3.0 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). Remember that radio waves have wavelengths ranging from millimeters to kilometers, with FM radio typically in the meter range, which helps verify that 3.0 m is a reasonable answer.