Question
Problem Solving (23 marks) 26. The distance between the second and sixth crests in a wave is 75 cm, and one crest travels a distance of 25 cm in 3.0 s. Find the frequency of the wave. (4 marks) 27. A sound wave with a frequency of 1.25times 10^4 Hz travels at 344m/s What is its wavelength? (3 marks) 28. A standing wave with five antinodes is generated in a string. If the waves travel at 17.5m/s with a frequency of 1.40times 10^2 Hz, how long is the string? (3 marks)
Solution
4.1
(191 Votes)
Xavier
Master · Tutor for 5 years
Answer
**26.*** **Concept:** The distance between two consecutive crests (or troughs) is the wavelength (λ). The speed (v) of a wave is related to its frequency (f) and wavelength by the equation v = fλ.* **Solution:** * The distance between the 2nd and 6th crest spans 4 wavelengths (6-2 = 4). * Therefore, 4λ = 75 cm, so λ = 75 cm / 4 = 18.75 cm = 0.1875 m. * The crest travels 25 cm (0.25 m) in 3.0 s, so the speed of the wave is v = 0.25 m / 3.0 s = 0.0833 m/s. * Now, using v = fλ, we can find the frequency: f = v/λ = (0.0833 m/s) / (0.1875 m) = 0.444 Hz.* **Answer:** The frequency of the wave is 0.444 Hz.**27.*** **Concept:** The speed (v) of a wave is related to its frequency (f) and wavelength (λ) by the equation v = fλ.* **Solution:** * We are given v = 344 m/s and f = 1.25 x 10⁴ Hz. * We need to find λ. Rearranging the equation, we get λ = v/f. * λ = (344 m/s) / (1.25 x 10⁴ Hz) = 0.02752 m = 2.752 cm.* **Answer:** The wavelength is 0.02752 m or 2.752 cm.**28.*** **Concept:** A standing wave with five antinodes represents the second overtone (or third harmonic). In this case, the length (L) of the string is related to the wavelength (λ) by the equation L = (3/2)λ. The speed (v) of the wave is related to its frequency (f) and wavelength by v = fλ.* **Solution:** * We are given v = 17.5 m/s and f = 1.40 x 10² Hz. * First, find the wavelength: λ = v/f = (17.5 m/s) / (1.40 x 10² Hz) = 0.125 m. * Now, find the length of the string: L = (3/2)λ = (3/2)(0.125 m) = 0.1875 m.* **Answer:** The length of the string is 0.1875 m.