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Chapter 12: Problem 19

Humans can hear sounds with frequencies up to about \(20.0 \mathrm{kHz},\) but dogs can hear frequencies up to about \(40.0 \mathrm{kHz} .\) Dog whistles are made to emit sounds that dogs can hear but humans cannot. If the part of a dog whistle that actually produces the high frequency is made of a tube open at both ends, what is the longest possible length for the tube?

Short Answer

Expert verified
Answer: The longest possible length of the tube is approximately 4.3 mm.

Step by step solution

01

Write down the formula for the fundamental frequency of a tube open at both ends

The formula for the fundamental frequency (\(f_1\)) of a tube open at both ends is: \[f_1 = \frac{v}{2L}\] where: - \(f_1\) is the fundamental frequency, - \(v\) is the speed of sound, - \(L\) is the length of the tube.
02

Substitute the maximum frequency a dog can hear for the fundamental frequency

The maximum frequency a dog can hear is \(40.0 \mathrm{kHz}\). Substitute this value for \(f_1\) in the formula: \[\frac{v}{2L} = 40,000 \,\text{Hz}\]
03

Find the speed of sound in the air

To solve the equation for the length of the tube, we need to know the speed of sound in the air. Consider the speed of sound in air at room temperature (\(20^\circ C\)) to be \(v = 343 \,\text{m/s}\).
04

Solve the equation for the length of the tube

Now substitute the speed of sound into the equation and solve for the length of the tube: \[\frac{343}{2L} = 40,000\] First, multiply both sides of the equation by 2: \[343 = 80,000L\] Now, divide both sides by 80,000: \[L = \frac{343}{80,000} \,\text{m}\]
05

Calculate the longest possible length of the tube

Finally, calculate the value of \(L\): \[L = \frac{343}{80,000} = 0.0042875 \,\text{m}\] The longest possible length for the tube that can produce a frequency only dogs can hear is approximately \(0.0043 \,\text{m}\) or \(4.3 \,\text{mm}\).

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Most popular questions from this chapter

A source and an observer are each traveling at 0.50 times the speed of sound. The source emits sound waves at \(1.0 \mathrm{kHz}\). Find the observed frequency if (a) the source and observer are moving towand each other, (b) the source and observer are moving away from each other; (c) the source and observer are moving in the same direction.
A source of sound waves of frequency \(1.0 \mathrm{kHz}\) is traveling through the air at 0.50 times the speed of sound. (a) Find the frequency of the sound received by a stationary observer if the source moves toward her. (b) Repeat if the source moves away from her instead.
Doppler ultrasound is used to measure the speed of blood flow (see Problem 42). The reflected sound interferes with the emitted sound, producing beats. If the speed of red blood cells is \(0.10 \mathrm{m} / \mathrm{s},\) the ultrasound frequency used is \(5.0 \mathrm{MHz},\) and the speed of sound in blood is \(1570 \mathrm{m} / \mathrm{s},\) what is the beat frequency?
An ambulance traveling at \(44 \mathrm{m} / \mathrm{s}\) approaches a car heading in the same direction at a speed of \(28 \mathrm{m} / \mathrm{s}\). The ambulance driver has a siren sounding at \(550 \mathrm{Hz}\). At what frequency does the driver of the car hear the siren?
A 30.0 -cm-long string has a mass of \(0.230 \mathrm{g}\) and is vibrating at its next-to-lowest natural frequency \(f_{2} .\) The tension in the string is \(7.00 \mathrm{N} .\) (a) What is \(f_{2} ?\) (b) What are the frequency and wavelength of the sound in the surrounding air if the speed of sound is $350 \mathrm{m} / \mathrm{s} ?$
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