Two wires are made of the same material but the second wire has twice the diameter and twice the length of the first wire. When the two wires are stretched, and the tension in the second wire is also twice the tension in the first wire. the funamental frequency of the first wire is 880 Hz. What is the fundamental frequency of the second wire?
Since the tension and size are different the two wire will have different wave speeds Since v= sqrt(T/mu) mu can be represented by density * area Since the second wire has twice the diameter it will have 4 times the mu and since Tension in wire 2 is twice that in wire 1 v2 = sqrt(2T/4mu) sqrt(.5)*v1 = 0.707*v1 Since f1 (wire 1) = v/2L1 then f2(wire 2) = v2/2L2 so replacing v2 and L2 yields 0.707v1/(2*2*L1) = 0.707/2(v1/2L) = 0.707/2 *f1 = 0.707/2*880Hz = 311Hz
