A sound wave in air has wavelength and travels at . Using , what is the frequency ?
- 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.