Given data:
* The length of the first string is,
[tex]\begin{gathered} d_1=50\text{ cm} \\ d_1=0.5\text{ m} \end{gathered}[/tex]* The diameter of first string is d_1 = 0.5 mm.
* The length of the second string is L_2 = 1 m.
* The diameter of second string is d_2 = 0.25 mm.
* The frequency of first string is f_1 = 320 Hz.
Solution:
The frequency of the first string in terms of length and diameter is,
[tex]f_1\propto\frac{1}{L_1d_1}\ldots\ldots(1)[/tex]The frequency of the second string in terms of length and diameter is,
[tex]f_2\propto\frac{1}{L_2d_2}\ldots\ldots....(2)[/tex]By dividing (2) equation by (1) equation,
[tex]\begin{gathered} \frac{f_2}{f_1}=\frac{\frac{1}{L_2d_2}}{\frac{1}{L_1d_1}} \\ \frac{f_2}{f_1}=\frac{L_1d_1}{L_2d_2} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} \frac{f_2}{320}=\frac{0.5\times0.5}{1\times0.25} \\ \frac{f_2}{320}=\frac{0.25}{0.25} \\ \frac{f_2}{320}=1 \\ f_2=320\text{ Hz} \end{gathered}[/tex]Thus, the frequency of the second string is 320 Hz (same as the frequency of the first string).
