1. The problem statement, all variables and given/known data An aluminium wire of length L1=60 cm and of cross sectional area 1.00×10-2cm2 is connected to a steel wire of the same cross sectional area. The compound wire; loaded with a block of mass m=10kg is arranged so it is hanging freely from a suspended pulley so that the distance L2 from the joint to the supporting pulley is 86.6 cm. Transverse waves are set up in the wire by using an external source of variable frequency. The density of Aluminium is 2.60 g cm-3, and that of steel is 7.80 g cm-3. (a) find the lowest frequency of excitation for which standing waves are observed such that the joint in the wire is a node. (b) What is the total number of nodes observed at this frequency, excluding the two at the ends of the wire? 2. Relevant equations 3. The attempt at a solution (a) v=fλ v=√(T/μ) T=9.8×10 = 98 μ= 2.6 v=√(98/2.6) = 1.941 1.914 = fλ λ= 0.6 × 1×10-2 = 0.006 f= 1.914/ 0.006 f = 323 Hz (b) length is 0.6 wavelength = 0.006 node every 2λ 2λ = 0.012 so 0.6 / 0.012 =50 take away the two nodes from each endpoint → 48 nodes observed Part (a) is right but part (b) should be 6? I really was not sure what to do here I think my reasoning must be off. Please help.