Question

10a. A string of length 1 m and linear density of 0.035 kg/m is stretched between 2 posts with a tension of 667 N. What is the frequency of the first 5 modes? Sketch these 5 modes. b. Now assume that you are plucking the string at a distance of 20 cm from one end. What will be the first four lowest harmonic frequencies? (Hint: You don't have to recalculate use the results from part a)

Answer 1

For a string fixed at bath end, the different frequency mode is given by f_n = n/2 L squareroot T/mu = n f_0 where, T = Tension in the string mu = linear mass density of the string L = string length f_0 = fundamental frequency = 1/2 L squareroot T/mu and n = 1, 2, 3... since, T = 667 N, mu = 0.035 kg/m L = 1 m Therefore, the fundamental frequency is f_0 = 2/L squareroot T/mu \r\n f_0 = 1/2 times 1 squareroot 667/0.035 f_0 = 1/2 squareroot 19057.1429 f_0 = 1/2 times 138.0476 f_0 = 69.023 H_z f_0 = 69 H_z Therefore, the frequency of the first 5 modes are f_1 = 1 times 69 = 69 H_z f_1 = 69 H_z f_2 = 2 times 69 = 138 H_z f_2 = 138 H_z f_3 = 3 times 69 = 207 H_z f_3 = 207 H_z \r\n f_4 = 4 times 69 = 276 H_z