To solve this problem we will use the definition of harmonics of a frequency for which we know that
[tex]f_n = \frac{nV}{2L}[/tex]
Here
n = Number of Harmonic
V = Velocity
L = Length
Our values are given as
[tex]f_1= 510Hz[/tex]
[tex]f_2= 680Hz[/tex]
[tex]v = 85m/s[/tex]
Now we have that the difference between two frequencies of different harmonics is
[tex]f_{n+1}-f_{n} = \frac{(n+1)V}{2L}-\frac{nv}{2L}[/tex]
[tex]f_{n+1}-f_{n} = \frac{V}{2L}[/tex]
[tex]L = \frac{V}{2(f_{n+1}-f_{n} )}[/tex]
[tex]L = \frac{85}{2(680-510)}[/tex]
[tex]L = 0.25m[/tex]
Now using this value in the expression of the fundamental frequency we have
[tex]f_n = \frac{nV}{2L}[/tex]
[tex]510 = \frac{n(85)}{2(0.25)}[/tex]
[tex]n = 3[/tex]
Order of two harmonies is 3 and 4.
