Question

A string oflength L-2.1 m and mass m 0.065 kg is fixed between two stationary points, and when the string is lucked a transverse wave of frequency fe 112 Hz is generated. andomized Variables = 2.1 m - 0.065 kg 112 Hz

Answer 1

Answer:

$\mu = 0.031 kg/m$

$T = 6859.6 N$

Explanation:

As we know that

mass of the string is

$m = 0.065 kg$

length of the string is given as

$L = 2.1 m$

now we can find the linear mass density of the string as

$\mu = \frac{m}{L}$

$\mu = \frac{0.065}{2.1}$

$\mu = 0.031 kg/m$

Now we know the frequency of the string wave as

$f = 112 Hz$

now we have

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

$112 = \frac{1}{2(2.1)}\sqrt{\frac{T}{0.031}}$

$T = 6859.6 N$