Dallas Jett Armstrong v (Student - section: 730) PHY183B, Fall 2023 - Physics for Scientists and Engineers I Messages Courses Help Logout Main Menu Contents | Grades |Chat Syllabus Course Contents > Homework Set 14 (due December 8) > 15.66 " Timer Notes Evaluate Feedback Print Calvin sloshes back and forth in his bathtub, producing a standing wave. What is the frequency of such a wave if the bathtub is 159 cm long and 93.3 cm wide and contains water that is 43.1 cm deep? .70 Hz Submit Answer Incorrect. Tries 1/99 Previous Tries Post Discussion Send FeedbackFor a standing wave in a rectangular bathtub, the wavelength (A) is related to the dimensions of the bathtub. The standing wave is formed by the interference of the incident and reflected waves, and it creates a pattern that repeats in both length and width. The wavelength can be expressed as: A = 2L where: ' L is the length of the bathtub. Given that the bathtub is 159 cm long, we can calculate the wavelength (A): A: 2 x 1590111 A: 318cm Now, let's consider the water depth (d) as this will affect the frequency. The relationship between the frequency (f), speed of the wave ('0), and wavelength (A) is given by the formula: 1) = f - A The speed of the wave (1)) can be expressed in terms of the square root of the acceleration due to gravity (9) and the water depth (d): v = wyd where: ' g is the acceleration due to gravity (approximately 980 (3111/82), ' d is the water depth. Now, we can equate the two expressions for v and solve for the frequency (f): m=w d. 1%? Substitute in the values: qr\" [980cn1f52)x43.1cm f : 318cm Calculating this gives: f m 0.982 Hz 80, the frequency of the standing wave in Calvin's bathtub is approximately 0.982 HZ