Question

Question 12 A cable with a linear density of -0.103 kg/m is hung from telephone poles. The tension in the cable is 696 N. The distance between poles is 15.4 meters. The wind blows across the line, causing the cable resonate. A standing waves pattern is produced that has 4.5 wavelengths between the two poles. Assuming roonm temperature air, what is the frequency of the hum? is the frequency of the hum Enter answer here 18.267 Hz 8.267HzAnse Your

Answer 1

frequency is 24.03Hz

V = squareroot T/mu T = 696 N mu = 0.013 kg/m V = squareroot 696/0.013 = theta 2.20 m/s no of wave length produce = 4.5 in the distance b/w two poles (15, 4) So wave length = 15, 4/4, 5 = 154/45 = 3.42 m lambda = 3.42 f = v/lambda = theta 2.20/3.42 = 24.03 HZ (frequency of hum) lambda_S = 343.00/24.03 = 14.27 (wavelength of hum)