Consider a string stretched between two supports. The speed of the propagation of transverse waves is 85 m/s, and two successive standing-wave frequencies are 510 Hz and 680 Hz. a) What is the order of these two harmonics (first, second, etc.)? b) What is the length of the string?

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cjmejiab cjmejiab · Answer 1 · 2020-01-22T07:16:36+01:00

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.