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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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Most popular questions from this chapter

Two identical half-open pipes each have a fundamental frequency of \(500 .\) Hz. What percentage change in the length of one of the pipes will cause a beat frequency of \(10.0 \mathrm{~Hz}\) when they are sounded simultaneously?

A sound meter placed \(3 \mathrm{~m}\) from a speaker registers a sound level of \(80 \mathrm{~dB}\). If the volume on the speaker is then turned down so that the power is reduced by a factor of 25 , what will the sound meter read? a) \(3.2 \mathrm{~dB}\) c) \(32 \mathrm{~dB}\) e) \(66 \mathrm{~dB}\) b) \(11 \mathrm{~dB}\) d) \(55 \mathrm{~dB}\)

A half-open pipe is constructed to produce a fundamental frequency of \(262 \mathrm{~Hz}\) when the air temperature is \(22^{\circ} \mathrm{C} .\) It is used in an overheated building when the temperature is \(35^{\circ} \mathrm{C} .\) Neglecting thermal expansion in the pipe, what frequency will be heard?

A string of a violin produces 2 beats per second when sounded along with a standard fork of frequency \(400 . \mathrm{Hz}\). The beat frequency increases when the string is tightened. a) What was the frequency of the violin at first? b) What should be done to tune the violin?

A source traveling to the right at a speed of \(10.00 \mathrm{~m} / \mathrm{s}\) emits a sound wave at a frequency of \(100.0 \mathrm{~Hz}\). The sound wave bounces off of a reflector, which is traveling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). What is the frequency of the reflected sound wave detected by a listener back at the source?
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