As shown in the figure, an aluminium wire of length 60 cm is connected to a steel wire of length 80 cm , and the combination is stretched between two fixed supports, producing a tension of 40 N . The steel wire has a cross-sectional area of 1.0 mm ^2 , while the aluminium wire has a cross-sectional area of 3.0 mm ^2 . Determine the minimum frequency of a tuning fork that can set up standing waves in the system such t

Step-by-step solution

The linear mass density of the steel wire is given by: _s = _s A_s = (7.8 10^3 kg m ⁻³)(1.0 10⁻⁶ m ^2) = 7.8 10⁻³ kg m ⁻¹ The linear mass density of the aluminium wire is: _a = _a A_a = (2.6 10^3 kg m ⁻³)(3.0 10⁻⁶ m ^2) = 7.8 10⁻³ kg m ⁻¹ Since _s = _a = , the wave speed v is the same in both wires: v = T = 40 7.8 10⁻³ 71.6 m s ⁻¹ For the joint to act as a node, both wires must form standing waves with nodes at their ends. The frequency f must be the same for both wires: f = n_s v 2L_s = n_a v 2L_a where n_s and n_