## Related questions with answers

The human ear canal is approximately 2.5 cm long. It is open to the outside and is closed at the other end by the eardrum. Estimate the frequencies (in the audible range) of the standing waves in the ear canal.

Solutions

VerifiedResonant frequencies are defined by:

$\begin{align*} f_n&=\frac{nv}{2l} \end{align*}$

where n=1,2,3,...

We can start solution with expression for frequency of an closed tube:

$\begin{align*} f_n&=\frac{nv}{4l} \tag{where $n$=1, 3, 5, $\dots$} \\ f_1&=\frac{1 \cdot v}{4l} \tag{fundamental frequency} \\ f_1&=\frac{343}{4 \cdot 2.5 \cdot 10^{-2}} \tag{substitute} \\ f_1&=\boxed{3430 \text{ Hz}} \\ f_3&=\frac{3 \cdot v}{4l} \tag{first overtone} \\ f_3&=\frac{3 \cdot 343}{4 \cdot 2.5 \cdot 10^{-2}} \tag{substitute} \\ f_3&=\boxed{10290 \text{ Hz}} \\ f_5&=\frac{5 \cdot v}{4l} \tag{first overtone} \\ f_5&=\frac{5 \cdot 343}{4 \cdot 2.5 \cdot 10^{-2}} \tag{substitute} \\ f_5&=\boxed{17150 \text{ Hz}} \\ f_7&=\frac{7 \cdot v}{4l} \tag{first overtone} \\ f_7&=\frac{7 \cdot 343}{4 \cdot 2.5 \cdot 10^{-2}} \tag{substitute} \\ f_7&=24010 \text{ Hz} \end{align*}$

$f_7$ is out of audible range, so that is not our solution. From graph we can see that the most sensitive frequencies are between 3000 Hz and 4000 Hz. In our case that is fundamental frequency. The sensitivity decreases in region until it reaches 10000 Hz where it increases rapidly again. That phenomenon can be explained with overtones.

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers with Modern Physics

4th Edition•ISBN: 9780131495081 (8 more)Douglas C Giancoli#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Introduction to Quantum Mechanics

3rd Edition•ISBN: 9781107189638Darrell F. Schroeter, David J. Griffiths## More related questions

1/4

1/7