A child has an ear canal that is 1.3cm long. At what sound frequencies in the audible range will the child have increased hearing sensitivity?
There will be enhanced sensitivity whenever the sound frequency corresponds to a fundamental or harmonic of an organ pipe with one open end and one closed end, 1.3 cm long. This will excite resonant oscillations within the ear
i appreciated your work... but how do i get answer in Hz.
There will be resonance whenever the 1.3 cm ear canal length equals an odd number of quarter wavelengths (1, 3, 5 etc.). For the fundamental resonant frequency, wavelength = 4x1.3 = 5.2 cm. Divide the sound speed (about 30,000 cm/s) by that wavelength for the frequency in Hz.
Check my thinking. I may have the fundamental resonant wavelength wrong.
I didn't quite follow ...
To determine the sound frequencies at which a child would have increased hearing sensitivity in the audible range, we need to consider the length of the ear canal.
The human ear is capable of perceiving sound frequencies between approximately 20 Hz and 20,000 Hz (or 20 kHz). However, certain frequencies within this range may be amplified due to the resonant properties of the ear canal.
The length of the ear canal determines its resonant frequency. The formula for calculating the resonant frequency of a cylindrical tube, such as the ear canal, is given by:
Resonant frequency (f) = (Speed of sound in air) / (2 * length of the tube)
The speed of sound in air is approximately 343 meters per second at room temperature.
To calculate the resonant frequency for the given ear canal length of 1.3 cm, we first need to convert the length to meters:
Length of the ear canal = 1.3 cm = 0.013 meters
Now, substitute the values into the formula:
f = 343 / (2 * 0.013)
f ≈ 13192 Hz (or 13.2 kHz)
Therefore, considering the resonant frequency of the child's ear canal, the child may have increased hearing sensitivity around 13.2 kHz.