Respuesta :
Given,
The initial length of the string, L₁=0.50 m
The tension on the string, T₁=2.0×10² N
The initial fundamental frequency of the string, f₁=400 Hz
The length of the string after it was shortened, L₂=0.35 m
The increased tension on the string, T₂=4.0×10² N
The fundamental frequency of the string before it was shortened is given by,
[tex]f_1=\frac{\sqrt[]{\frac{T_1}{\mu}}}{2L_1}[/tex]Where μ is the mass per unit length of the string.
On rearranging the above equation,
[tex]\begin{gathered} 4L^2_1f^2_1=\frac{T_1}{\mu} \\ \Rightarrow\mu=\frac{T_1}{4L^2_1f^2_1} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} \mu=\frac{2.0\times10^2}{4\times0.50^2\times400^2} \\ =1.25\times10^{-3}\text{ kg/m} \end{gathered}[/tex]The fundamental frequency after the string was shortened is given by,
[tex]f_2=\frac{\sqrt[]{\frac{T_2}{\mu}}}{2L_2}[/tex]On substituting the known values,
[tex]\begin{gathered} f_2=\frac{\sqrt[]{\frac{4\times10^2}{1.25\times10^{-3}}}}{2\times0.35} \\ =808.1\text{ Hz} \end{gathered}[/tex]Thus the fundamental frequency after the string is shortened and the tension is increased is 808.1 Hz
