What is the fundamental frequency of the steel wire in FIGURE P17.48?

Frequency is a number that occurs in a repetitive factor that happens per unit of time.

The provided information is

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

Based on the given description the need is to calculate the tension in the string and its linear density.

Based on drawing information from the mechanic's chapter the bar pivoted to the left. in the sum of the moments, the pivot point is 0.

Also, knowing the weight of the bar, the uniform weight distribution concentrates on one point and stretches to the midway of the bar. by understanding the positive direction of the rotation is counterclockwise the equation could be,

$T\sin \left({45}^{\circ }\right)·L-mgL+m_{bar}g\frac{L}{2}\Rightarrow T\frac{g(m+0.5·m_{bar})}{\sin \left({45}^{\circ }\right)}$

Finding the value is easier than calculating and replacing the value to solve the equation parametrically. the equation then will be

$T=\frac{9.8·(8+0.5·4)}{\sin \left({45}^{\circ }\right)}=\frac{13.86N}{}$

The neglected weight of steel wire is 50 times less than the other weights of the part.

The total length of the string based on the linear density is $\mu =\frac{m_{wire}}{L_{string}}$

Thus the length of the wire will be

$L_{wire}=\frac{L}{\sin \left(45\circ \right)}=2.83m$

The final calculation of the frequency is

role="math" localid="1649071315341" $f=\frac{1}{2L}\sqrt{\frac{T{L}_{string}}{m_{string}}}=\frac{1}{2·}\sqrt{\frac{13.86·2.83}{0.075}}=5.73Hz$