Respuesta :
Given data:
* The length of the string is L = 1 m.
* The initial value of tension in the string is,
[tex]T_1=1\times10^2\text{ N}[/tex]* The final value of tension of the string is,
[tex]T_2=1.5\times10^2\text{ N}[/tex]* The new frequency of the string is,
[tex]f_2=400\text{ Hz}[/tex]Solution:
The frequency in terms of length and tension is,
[tex]f_1=\frac{\sqrt[]{T_1}}{L}\ldots\ldots.(1)[/tex]The frequency in terms of length and tension in the final state is,
[tex]f_2=\frac{\sqrt[]{T_2}}{L}\ldots\ldots\ldots..(2)[/tex]Dividing equations (1) and (2),
[tex]\begin{gathered} \frac{f_1}{f_2}=\frac{\frac{\sqrt[]{T_1}}{L}}{\frac{\sqrt[]{T_2}}{L}}_{} \\ \frac{f_1}{f_2}=\frac{\sqrt[]{T_1}}{\sqrt[]{T_2}} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} \frac{f_1}{400}=\frac{\sqrt[]{1\times10^2}}{\sqrt[]{1.5\times10^2}} \\ \frac{f_1}{400}=\frac{1}{\sqrt[]{1.5}} \\ f_1=400\times\frac{1}{\sqrt[]{1.5}} \\ f_1=326.6\text{ Hz} \end{gathered}[/tex]Thus, the original frequency of the string is 326.6 Hz.
