Question

Find the resonance frequency of the ear canal. Treat it as a half-open pipe of diameter \(8.0 \mathrm{~mm}\) and length \(25 \mathrm{~mm}\). Assume that the temperature inside the ear canal is body temperature \(\left(37^{\circ} \mathrm{C}\right)\).

Short answer

Answer: The resonance frequency of the ear canal is approximately 3445 Hz.

Step-by-step solution

Convert temperature to speed of sound

Using the given temperature, which is the body temperature of \(37^{\circ} \mathrm{C}\), we can find the speed of sound \(v\) in the medium by plugging the values into the equation: \(v = 331.4\sqrt{1+\dfrac{37}{273.15}}\) Calculate the value using a calculator: \(v \approx 331.4\sqrt{1.1355} \approx 344.5 \mathrm{~m/s}\)

Convert length and diameter to meters

We need to convert the given dimensions of ear canal, diameter (D) and length (L), into meters: \(D = 8.0 \mathrm{~mm} = 8.0 \times 10^{-3} \mathrm{~m}\) \(L = 25 \mathrm{~mm} = 25 \times 10^{-3} \mathrm{~m}\)

Compute the resonance frequency

Now that we have the speed of sound \(v\) and the length \(L\) of the half-open pipe, we can calculate the resonance frequency \(f\) using the formula: \(f = \dfrac{v}{4L}\) Substitute the values for \(v\) and \(L\): \(f = \dfrac{344.5}{4 \times 25 \times 10^{-3}}\) Calculate the value using a calculator: \(f \approx \dfrac{344.5}{0.1} \approx 3445 \mathrm{~Hz}\) The resonance frequency of the ear canal is approximately \(3445 \mathrm{~Hz}\).