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 electromagnetic waves traveling at constant speed c = 3.0 × 10⁸ m/s in vacuum, the equation c = fλ shows that frequency and wavelength are inversely proportional: if frequency increases, wavelength must decrease to keep their product (wave speed) constant. Mathematically, since λ = c/f, as f increases, λ decreases proportionally. Choice C is correct because it accurately describes the inverse relationship and correctly identifies that wavelength decreases when frequency increases, using the proper equation λ = c/f. 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 electromagnetic waves in vacuum, speed is always c, so frequency and wavelength must vary inversely.

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.

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.

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 ≈ 4.37, so wavelength increases in the faster medium. Choice C is correct because it accurately describes the inverse relationship between f and λ. Choice D incorrectly assumes frequency increases in water, 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 twice as fast and frequency stays the same, the wavelength must double to satisfy the equation.

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.

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 = $3.0×10^8$ m/s, 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 an increase in f leads to a decrease in λ. Choice C is correct because it accurately describes the inverse relationship between f and λ. Choice D 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.

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 ≈ 4.37, so wavelength increases in the faster medium. Choice C is correct because it accurately describes that wavelength increases with speed when frequency is constant across media. Choice D incorrectly assumes frequency changes in water, when actually it's frequency that stays constant (determined by the source), while speed and wavelength both change. 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). 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 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.

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 = $532×10^{-9}$ m and the wave speed is c = $3.0×10^8$ m/s, we calculate frequency using f = c/λ = $(3.0×10^8$ m/s) / $(532×10^{-9}$ m) ≈ $5.64×10^{14}$ Hz. This frequency of $5.64×10^{14}$ Hz falls in the visible light range, consistent with green light. Choice A is correct because it properly applies v = fλ with the correct values and units, including converting nm to m. Choice B uses the equation backwards, calculating f = λ/c instead of c/λ, which divides incorrectly and produces a very small frequency. 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.

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.