# Waves - Sound

1. Jul 9, 2010

### rizardon

1. The problem statement, all variables and given/known data
The auditory canal of the human ear extends about 2.5 cm from the outside ear to the eardrum
(a)Explain why the human ear is especially sensitive to sounds at frequencies around 3500Hz.

2. Relevant equations
Speed of sound v=344m/s

3. The attempt at a solution
I know the human can hear sound with frequency between 500Hz-3500Hz,but , I don't how to explain how and why we hear what we hear. It would be great if anyone could give me some clues.

2. Jul 9, 2010

### Tusike

I'm not exactly sure about this, but it gives the right solution, so I'll tell it anyway:)
The ear can be thought of as a pipe with one end open, the other closed. This means that resonance of the sound waves can happen every s=(2k+1)*lambda / 4 distances, where s is the length of the pipe. Now, substitute s = 0.025m and k=0, we get that lambda = 0.1m, and since f=c/lambda, f = 3440Hz. So because of the resonance, we would be especially sensitive to frequencies near that. (For k>0, we get frequencies so high that the human can't sense. Btw, I think we're ranged from 100Hz to 4000Hz, although I'm not sure of the lower boundry).

3. Jul 9, 2010

### rl.bhat

Ear is sensitive for the frequency which forms anti node at the opening of the ear and node at the ear drum. Length of the auditory canal is 2.5 cm. So the wavelength of the frequency is 4*2.5 cm.
Now velocity v = f*lambda.
Find f.

4. Jul 9, 2010

### Bartek

We can hear sound with frequency between 16Hz and 20000Hz!

Sound is a travelling wave which is an oscillation of pressure transmitted through an air. When that wave reaches eardrum, the eardrum begins to vibrate. We interpret this vibrations as a sound. No matter how - it is biology.

Imagine auditory canal as a pipe with eardrum at the one end and open at the second. You should calculate which is frequency of standing (stationary) sound wave in that pipe. Wave with that frequency will be strengthened.

Last edited: Jul 9, 2010