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3 years ago

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A string is stretched to a length of 339 cm and both ends are fixed. If the density of the string is 0.0073 g/cm, and its tensio

n is 859 N, what is the fundamental frequency?

1 answer:

sashaice [31]3 years ago

7 0

Answer:

160 Hz.

Explanation:

For nth harmonic, the fundamental frequency,

$f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu} }$

Here T is the tension in string, \mu is the mass/unit of length of the string and L is the string length.

Given n = 1 frequency of the 1st harmonic (the Fundamental), T = 859 N,

L= 339 cm =3.39 m and $\mu=0.0073 g/cm =0.00073 kg /m$ .

Substituting these values, we get

$f_1=\frac{1}{2\times3.39m} \sqrt{\frac{859N}{0.00073\ kg/m } }$

$f_1= 0.147 \times1084.76=159.99 Hz$

$f_1=160 Hz$

Thus, the fundamental frequency is 160 Hz.

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