Question

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 tension is 859 N, what is the fundamental frequency?

Answer 1

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.