The frequencies of the n-harmonics of a string is given by
[tex]f_n = \frac{n}{2L} \sqrt{ \frac{T}{\mu} } [/tex]
where n is the number of the harmonic, L the length of the wire, T the tension and [tex]\mu = \frac{m}{L} [/tex] is the linear density, with m being the mass of the string.
Let's calculate the linear density first, using the mass [tex]m=5 g=5 \cdot 10^{-3}kg[/tex]:
[tex]\mu = \frac{m}{L} = \frac{5\cdot 10^-3 kg}{0.5 m}=0.01 kg/m [/tex]
The problem says that the computer is able to analyze frequencies up to 10000 Hz. This means that we have to find the highest number of harmonic that generates a frequency smaller than this value. So, using [tex]f=10000 Hz[/tex], the tension of the string [tex]T=1000 N[/tex] and the mass m and the length of the string L, we can re-arrange the previous formula to find which n corresponds to this frequency:
[tex]n = 2f_n L \sqrt{ \frac{\mu}{T} }=2 (10000 Hz)(0.5 m) \sqrt{ \frac{0.01 kg/m}{1000 N} }=31.6 [/tex]
And since n can only be integer, the highest harmonic that can be analyzed by the computer is n=31.
