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Lena [83]
3 years ago
11

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?
Physics
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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