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Q17.5-50PE

Expert-verified
Found in: Page 595

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# The ear canal resonates like a tube closed at one end. (See Figure $$17.39$$.) If ear canals range in length from $$1.80$$to$$2.60\;{\rm{cm}}$$in an average population, what is the range of fundamental resonant frequencies? Take air temperature to be$$37.0{\rm{^\circ C}}$$, which is the same as body temperature. How does this result correlate with the intensity versus frequency graph (Figure $$17.37$$ of the human ear?

The range of the fundamental frequencies is$${\rm{3}}{\rm{.39}}\;{\rm{kHz}}$$to$${\rm{4}}{\rm{.9}}\;{\rm{kHz}}$$

See the step by step solution

## Given Data

The temperature is$${{\rm{t}}_{\rm{1}}}{\rm{ = 37}}{\rm{.0^\circ C = }}\left( {{\rm{37 + 273}}} \right)\;{\rm{K = 310}}\;{\rm{K}}$$

The length of the canal ranges from $${{\rm{l}}_{\rm{1}}}{\rm{ = 1}}{\rm{.80}}\;{\rm{cm = 0}}{\rm{.018}}\;{\rm{m}}$$ to $${{\rm{l}}_{\rm{2}}}{\rm{ = 2}}{\rm{.60}}\;{\rm{cm = 0}}{\rm{.026}}\;{\rm{m}}$$

The diagram of Human Ear is shown below:

The human ear

The diagram of intensity Vs frequency is shown below:

The intensity-frequency graph

## Calculation of the ratio of the frequencies

For a closed tube, the resonance frequencies are,

$${{\rm{f}}_{\rm{n}}}{\rm{ = n}}\frac{{\rm{v}}}{{{\rm{4l}}}}{\rm{,}}\;{\rm{n = 1,3,}}\;{\rm{5,}}\;....$$

For an open tube, the resonance frequencies are,

$${{\rm{f}}_{\rm{n}}}{\rm{ = n}}\frac{{\rm{v}}}{{{\rm{2l}}}}{\rm{,}}\;{\rm{n = 1,2,}}\;{\rm{3,}}\;....$$

The speed of sound at$${t_1}$$ temperature is

$$\begin{array}{c}{\rm{v = 331}}\sqrt {\frac{{{{\rm{t}}_{\rm{1}}}}}{{{\rm{273}}}}} \\{\rm{ = 331}}\sqrt {\frac{{{\rm{310}}}}{{{\rm{273}}}}} \\{\rm{ = 352}}{\rm{.7}}\;{\rm{m/s}}\end{array}$$

The frequencies are,

$$\begin{array}{c}{{\rm{f}}_{\rm{1}}}{\rm{ = }}\frac{{{\rm{352}}{\rm{.72}}}}{{{\rm{4 \times 0}}{\rm{.018}}}}\\{\rm{ = 4898}}{\rm{.89}}\;{\rm{Hz}}\\{\rm{ = 4}}{\rm{.9}}\;{\rm{kHz}}\end{array}$$

$$\begin{array}{c}{{\rm{f}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{352}}{\rm{.72}}}}{{{\rm{4 \times 0}}{\rm{.026}}}}\\{\rm{ = 3391}}{\rm{.54}}\;{\rm{Hz}}\\{\rm{ = 3}}{\rm{.39}}\;{\rm{kHz}}\end{array}$$

Our ear can hear$${\rm{ - 20}}\;{\rm{dB}}$$ to$${\rm{10}}\;{\rm{dB}}$$ with$${\rm{0}}\;{\rm{phon}}$$, $${\rm{10}}\;{\rm{dB}}$$to$${\rm{40}}\;{\rm{dB}}$$with maximum of$${\rm{40}}\;{\rm{phon}}$$ and$${\rm{40}}\;{\rm{dB}}$$to$${\rm{60}}\;{\rm{dB}}$$with maximum of$${\rm{60}}\;{\rm{phon}}$$.