Question

The frequency of one string 50 cm long with a diameter of 0.50 mm is 320 Hz. A second string under the same tension and made of the same material is 1.0 m long with a diameter of 0.25 mm. What is the frequency of the second string?

Answer 1

Given data:

* The length of the first string is,

$\begin{gathered} d_1=50\text{ cm} \\ d_1=0.5\text{ m} \end{gathered}$

* 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,

$f_1\propto\frac{1}{L_1d_1}\ldots\ldots(1)$

The frequency of the second string in terms of length and diameter is,

$f_2\propto\frac{1}{L_2d_2}\ldots\ldots....(2)$

By dividing (2) equation by (1) equation,

$\begin{gathered} \frac{f_2}{f_1}=\frac{\frac{1}{L_2d_2}}{\frac{1}{L_1d_1}} \\ \frac{f_2}{f_1}=\frac{L_1d_1}{L_2d_2} \end{gathered}$

Substituting the known values,

$\begin{gathered} \frac{f_2}{320}=\frac{0.5\times0.5}{1\times0.25} \\ \frac{f_2}{320}=\frac{0.25}{0.25} \\ \frac{f_2}{320}=1 \\ f_2=320\text{ Hz} \end{gathered}$

Thus, the frequency of the second string is 320 Hz (same as the frequency of the first string).