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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

Verified Answer :

dexteright02

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:

\(f_o = 21 Hz\)

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

\(f_2 = 2 f_o\)

\(42 = 2 f_o\)

\(f_o = 21 Hz\)

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