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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

In this problem, you will estimate the smallest kinetic energy of vibration that the human ear can detect. Suppose that a harmonic sound wave at the threshold of hearing $\left(I=1.0 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\right)\( is incident on the eardrum. The speed of sound is \)340 \mathrm{m} / \mathrm{s}\( and the density of air is \)1.3 \mathrm{kg} / \mathrm{m}^{3} .$ (a) What is the maximum speed of an element of air in the sound wave? [Hint: See Eq. \((10-21) .]\) (b) Assume the eardrum vibrates with displacement \(s_{0}\) at angular frequency \(\omega ;\) its maximum speed is then equal to the maximum speed of an air element. The mass of the eardrum is approximately \(0.1 \mathrm{g} .\) What is the average kinetic energy of the eardrum? (c) The average kinetic energy of the eardrum due to collisions with air molecules in the absence of a sound wave is about \(10^{-20} \mathrm{J}\) Compare your answer with (b) and discuss.
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