Answer :
In this exercise we have to use frequency knowledge to calculate the diameter of the string, so:
A) [tex]d_2= 0.0749[/tex]
B) [tex]r_2= 0.0374545[/tex]
A) Calculate the diameter of string as follows:
[tex]d_2= 2r_2\\[/tex]
To continue this account it is necessary to know the value of R, then we will follow the calculations of letter B and then we will return to letter A.
B) Calculate the new frequency [tex]f'_2[/tex] as follows:
[tex]\frac{f_1}{f'_2} = \frac{r_2'}{r_1} \\f_2'= (\frac{r_1}{r'_2}) f_1\\=((\frac{d_1}{2})/(\frac{d'_2}{2}))f_1\\= (\frac{d_1}{d'_2}) f_1\\= (\frac{0.05}{0.06245})(82.4)= 65.97 Hz[/tex]
The frequency is directly proportional to the tension:
[tex]f \ \alpha \ \sqrt{T} \\\frac{f_2}{f'_2}= \sqrt{\frac{T_{before}}{T_{after}}} \\\frac{T_{before}}{T_{after}}= (\frac{f_2}{f'_2})^2[/tex]
Calculate the ratio as follows:
[tex]\frac{T_{before}}{T_{after}}= (\frac{55}{65.97})^2= 0.695[/tex]
The relation between the frequency, length, tension ans linear mass is:
[tex]f= \frac{1}{2L} \sqrt{\frac{T}{\mu} }[/tex]
The linear mass density is:
[tex]\mu= \rho r^2[/tex]
So:
[tex]f=\frac{1}{2L}=\sqrt{\frac{T}{\rho r^2} } \\f=\frac{1}{2Lr}=\sqrt{\frac{T}{\rho} } \\f\ \alpha \ \frac{1}{r} \\\frac{f_1}{f_2}= (\frac{1}{r_1} )/(\frac{1}{r_2})\\ \frac{f_1}{f_2} = \frac{r_2}{r_1}\\r_2= (\frac{f_1}{f_2})(r_1)\\r_2= (\frac{82.4}{55})(\frac{0.05}{2})= 0.0374545[/tex]
Returning to the letter A we find:
[tex]d_2= 2r_2\\= 2(0.0374545)= 0.0749[/tex]
See more about frequency at brainly.com/question/5102661