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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

21 Hz
42 Hz
84 Hz
126 Hz

Respuesta :

dexteright02
Data:
[tex]f_{2} = 42 Hz[/tex]
n (Wave node)
V (Wave belly) 
L (Wave length)
The number of bells is equal to the number of the harmonic emitted by the string.

[tex]f_{n} = \frac{nV}{2L} [/tex]

Wire 2 → 2º Harmonic → n = 2

[tex]f_{n} = \frac{nV}{2L} [/tex]
[tex]f_{2} = \frac{2V}{2L} [/tex]
[tex]2V = f_{2} *2L[/tex]
[tex]V = \frac{ f_{2}*2L }{2} [/tex]
[tex]V = \frac{42*2L}{2} [/tex]
[tex]V = \frac{84L}{2} [/tex]
[tex]V = 42L[/tex]

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


[tex]f_{n} = \frac{nV}{2L} [/tex]
[tex]f_{1} = \frac{1V}{2L} [/tex]
[tex]f_{1} = \frac{V}{2L} [/tex]

If, We have:
V = 42L
Soon:
[tex]f_{1} = \frac{V}{2L} [/tex]
[tex]f_{1} = \frac{42L}{2L} [/tex]
[tex]\boxed{f_{1} = 21 Hz}[/tex]

Answer:

The fundamental frequency of the string:
21 Hz

Answer:

[tex]f_o = 21 Hz[/tex]

Explanation:

As we know that frequency of wave in stretched string is given as

[tex]f= \frac{nv}{2L}[/tex]

here we know that

n = number of harmonics

So we have fundamental frequency given as

[tex]f_o = \frac{v}{2L}[/tex]

also for second harmonic we have

n = 2

so we have

[tex]f_2 = 2 f_o[/tex]

[tex]42 = 2 f_o[/tex]

[tex]f_o = 21 Hz[/tex]

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