Question

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

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.

$f_{n} = \frac{nV}{2L}$

Wire 2 → 2º Harmonic → n = 2

$f_{n} = \frac{nV}{2L}$
$f_{2} = \frac{2V}{2L}$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
$V = 42L$

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


$f_{n} = \frac{nV}{2L}$
$f_{1} = \frac{1V}{2L}$
$f_{1} = \frac{V}{2L}$

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

Answer:

The fundamental frequency of the string:
21 Hz

Answer 2

Answer:

$f_o = 21 Hz$

Explanation:

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

$f= \frac{nv}{2L}$

here we know that

n = number of harmonics

So we have fundamental frequency given as

$f_o = \frac{v}{2L}$

also for second harmonic we have

n = 2

so we have

$f_2 = 2 f_o$

$42 = 2 f_o$

$f_o = 21 Hz$