Chapter 12, Problem 38P

To determine

To Find: The frequency of standing waves in the ear canal.

Explanation of Solution

Given:

Length of human ear canal $=2.5\text{ cm }=0.025\text{ m}$

Speed of the sound wave $=343\text{m/s}$

Formula used:

The frequency is given as,

  $f=\frac{nv}{4L}$

Where, $L$ is the string length and $v$

is the speed of the sound

Calculation:

Since one end is open and other is closed, it is to be considered as closed pipe. So, for closed pipe the standing frequency can be determined by the following equation.

  $f=\frac{nv}{4L}$

  $v=343\text{ m/s}$

Now, substitute the values in the equation f

  $f=\frac{343}{4\times 0.025}\text{ Hz}\text{.}$

  $f=3430\text{ Hz}\text{.}$

The frequency is 3430 Hz.

The resonant frequency is given by $f=\frac{nv}{4L}$ for odd values of n.

Therefore, the first few frequencies are given as,

  $f_1=3430\text{Hz},f_3=10300\text{Hz}\text{and}{f}_5=17200\text{Hz}$ and so on.

The given graph shows the relation between the frequency and sound level.

  

In the graph, most sensitive frequencies lie between $3,000\text{ Hz}$ and $4,000\text{ Hz}$ which correspond to the ear canal’s fundamental resonant frequency. The frequency $10,000\text{ Hz}$ is relatively sensitive region corresponding to the first overtone resonant frequency of the ear canal because sensitivity is decreases above the $4,000\text{ Hz}$ , but it appears flatten out around $10,000\text{ Hz}$ . This indicates higher sensitivity near $10,000\text{ Hz}$ frequency.

Conclusion:

The first few frequencies of the standing wave are $f_1=3430\text{Hz},f_3=10300\text{Hz}$ and $f_5=17200\text{Hz}$ and so on. Hence, the frequency $10,000\text{ Hz}$ is relatively sensitive region corresponds to the first overtone resonant frequency of the ear canal