Chapter 15, Problem 57A

To determine

To calculate:The approximate value of the lowest resonance frequency.

Answer to Problem 57A

The approximate value of lowest resonance frequency is $2.9\text{ kHz}$ .

Explanation of Solution

Given info:

Length of the auditory canal, $L\text{=3}\text{.0 cm}$

The auditory canal is a closed pipe.

Formula used:

For a closed pipe, the first resonance wavelength

  ${\lambda }_1=4L$

Here, ${\lambda }_1$ represent lowest resonance wavelength and $L$ represents length of the pipe.

  $f_1=\frac{v}{{\lambda }_1}=\frac{v}{4L}$

$f_1$ represents lowest resonance frequency and $v$ represents the speed of sound in air.

Calculation:

As per the problem,

Length of the auditory canal ( $L$ ) = $\text{3}\text{.0 cm= 0}\text{.03 m}$

Speed of sound in air ( $v$ ) = $343\text{ m/s}$

Use the formula

  $f_1=\frac{v}{4L}$

Substitute the values

  $\begin{array}{l}{f}_1=\frac{343\text{ m/s}}{4\cdot (0.03\text{ m)}}\\ =2858.33\text{ Hz}\\ =2.9\text{ kHz [approximate value]}\end{array}$

Conclusion:

Thus, approximate value of lowest resonance frequency is $29\text{ kHz}$ .