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Chapter 16: Problem 51

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

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

Step by step solution

01

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: \(v \approx 331.4\sqrt{1.1355} \approx 344.5 \mathrm{~m/s}\)
02

Convert length and diameter to meters

We need to convert the given dimensions of ear canal, diameter (D) and length (L), into meters: \(D = 8.0 \mathrm{~mm} = 8.0 \times 10^{-3} \mathrm{~m}\) \(L = 25 \mathrm{~mm} = 25 \times 10^{-3} \mathrm{~m}\)
03

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 \(3445 \mathrm{~Hz}\).

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