Question

A steel wire in a piano has a length of 0.5000 m and a mass of 4.200 10-3 kg. To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C (fC = 261.6 Hz on the chromatic musical scale)?

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Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each fixed at both ends and stretched so that the tension in the longer wire is five times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 42 Hz, what is the frequency of the second harmonic in the longer wire?

Answer 1

Being fixed at both ends the modes of vibration will have a node at each end. The longest wavelength/lowest frequency that will fulfil this condition is when the length is one-half a wavelength λ=2L.
If the required frequency is 261.6Hz, then by the wave equation v=fλ
v=f*2L=261.6*1.000= 261.6 m/s

Now the speed of a wive on a wire is
v=√(T/μ) where T=tension and μ=mass per unit length ("linear density") μ=0.0042/0.5=0.0084kg/m
Rearranging
v²=T/μ
T=v²μ
T=261.6²*0.0084=574.85 N