Question

ReviewI ConstantsI Periodic Table A combination work of art/musical instrument is illustrated. (Figure 1) Six pieces of identical piano wire (cut to different lengths) are hung from the same support, and masses are hung from the free end of each wire. Each wire is 1, 2, or 3 units long, and each supports 1, 2, or 4 units of mass. The mass of each wire is negligible compared to the total mass hanging from it. When a strong breeze blows, the wires vibrate and create an eerie sound The correct ranking cannot be determined Previous Ans Correct Part C Rank each wire-mass system on the basis of its fundamental frequency Rank from largest to smallest. To rank items as equivalent, overlap them. View Available Hint(s) Reset Help largest smallest Figure 1 of 1 The correct ranking cannot be determined. SubmitPrevi Incorrect, Try Again; 9 attempts remaining The fundamental frequency depends on more than just the tension of the wire. I A B C D E F Provide Feedback

Answer 1

Velocity of wave in a stretched string is given by, V = squareroot T/mu: T = Tension in string mu = Mass per unit length of wire. In fundamental mode of vibration: length of wire, l = lambda/2: lambda = wavelength Fundamental frequency of vibration: f = v/lambda Therefore f_A = V_A/lambda_A = (squareroot 2 mg/mu)/2l_A = 1/4 squareroot 2mg/mu = x/2 squareroot 2 putting squareroot mg/mu = X f_B = V_B/lambda_B = (squareroot 2mg/mu/2l_B = 1/2 squareroot 2mg/mu = x/squareroot 2 f_c = V_c/lambda_c = (squareroot mg/mu)/2l_c = 1/4 (squareroot mg/mu) = x/4 f_D = V_D/lambda_D = (squareroot mg/mu)/2l_D = 1/6 (squareroot mg/mu) = x/6 f_E = V_E/lambda_E = (squareroot 4mg/mu)/2l_E = 1/4 (squareroot 4mg/mu) = x/2 f_F = V_F/lambda_F = (squareroot mg/mu)/2l_F = 1/2 (squareroot mg/mu) = x/2 Required Order: - Largest B Smallest E A C D F \r\n SB = squareroot l^2 + b^2 = squareroot 25 + 16 = 6.403m almostequalto 6.5m Wave velocity