Question

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

Answer 1

Data:

`f_(2) = 42 Hz`
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.


`f_(n) = (nV)/(2L)`

Wire 2 → 2º Harmonic → n = 2


`f_(n) = (nV)/(2L)`

`f_(2) = (2V)/(2L)`

`2V = f_(2) *2L`

`V = ( f_(2)*2L )/(2)`

`V = (42*2L)/(2)`

`V = (84L)/(2)`

`V = 42L`

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



`f_(n) = (nV)/(2L)`

`f_(1) = (1V)/(2L)`

`f_(1) = (V)/(2L)`

If, We have:
V = 42L
Soon:

`f_(1) = (V)/(2L)`

`f_(1) = (42L)/(2L)`

`\boxed{f_(1) = 21 Hz}`

Answer:

The fundamental frequency of the string:
21 Hz

Answer 2

`f_o = 21 Hz`

Step-by-step explanation:

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

`f= (nv)/(2L)`

here we know that

n = number of harmonics

So we have fundamental frequency given as

`f_o = (v)/(2L)`

also for second harmonic we have

n = 2

so we have

`f_2 = 2 f_o`

`42 = 2 f_o`

`f_o = 21 Hz`