Question

A string 0.50 m long is stretched under a tension of 2.0 x 102 N and its fundamental frequency is 400 Hz. If the length if the string is shortened to 0.35 m and the tension is increased to 4.0 x 102 N, what is the new fundamental frequency?

Answer 1

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,

$f_1=\frac{\sqrt[]{\frac{T_1}{\mu}}}{2L_1}$

Where μ is the mass per unit length of the string.

On rearranging the above equation,

$\begin{gathered} 4L^2_1f^2_1=\frac{T_1}{\mu} \\ \Rightarrow\mu=\frac{T_1}{4L^2_1f^2_1} \end{gathered}$

On substituting the known values,

$\begin{gathered} \mu=\frac{2.0\times10^2}{4\times0.50^2\times400^2} \\ =1.25\times10^{-3}\text{ kg/m} \end{gathered}$

The fundamental frequency after the string was shortened is given by,

$f_2=\frac{\sqrt[]{\frac{T_2}{\mu}}}{2L_2}$

On substituting the known values,

$\begin{gathered} f_2=\frac{\sqrt[]{\frac{4\times10^2}{1.25\times10^{-3}}}}{2\times0.35} \\ =808.1\text{ Hz} \end{gathered}$

Thus the fundamental frequency after the string is shortened and the tension is increased is 808.1 Hz