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Chapter 12: Problem 1

Bats emit ultrasonic waves with a frequency as high as $1.0 \times 10^{5} \mathrm{Hz} .$ What is the wavelength of such a wave in air of temperature \(15^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The wavelength of the ultrasonic wave is approximately \(3.4\,\text{mm}\).

Step by step solution

01

Calculate the speed of sound in air at the given temperature

To calculate the speed of sound in the air at 15°C, we use the formula \(v = 331.4 + 0.6 \cdot T,\) where \(v\) is the speed of sound and \(T\) is the temperature in Celsius. Plugging in the given temperature, we have: \(v = 331.4 + 0.6 \cdot 15 = 331.4 + 9\) \(v = 340.4 \,\text{m/s}\) The speed of sound in air at 15°C is 340.4 m/s.
02

Calculate the wavelength of the ultrasonic wave

The ultrasonic wave frequency \(f\) is given as \(1.0 \times 10^5 \,\text{Hz}\). To calculate the wavelength \(\lambda\), we can use the speed of sound \(v\) and the frequency \(f\) in the following formula: \(\lambda = \frac{v}{f}\) Plugging in the speed of sound and frequency, we have: \(\lambda = \frac{340.4}{1.0 \times 10^5} = 3.404 \times 10^{-3} \,\text{m}\) The wavelength of the ultrasonic wave in air at 15°C is \(3.404 \times 10^{-3}\,\text{m}\) or approximately \(3.4\,\text{mm}\).

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Most popular questions from this chapter

An auditorium has organ pipes at the front and at the rear of the hall. Two identical pipes, one at the front and one at the back, have fundamental frequencies of \(264.0 \mathrm{Hz}\) at \(20.0^{\circ} \mathrm{C} .\) During a performance, the organ pipes at the back of the hall are at $25.0^{\circ} \mathrm{C},\( while those at the front are still at \)20.0^{\circ} \mathrm{C} .$ What is the beat frequency when the two pipes sound simultaneously?
Derive Eq. \((12-4)\) as: (a) Starting with Eq. \((12-3),\) substitute \(T=T_{\mathrm{C}}+273.15 .\) (b) Apply the binomial approximation to the square root (see Appendix A.5) and simplify.
A musician plays a string on a guitar that has a fundamental frequency of \(330.0 \mathrm{Hz}\). The string is \(65.5 \mathrm{cm}\) long and has a mass of \(0.300 \mathrm{g} .\) (a) What is the tension in the string? (b) At what speed do the waves travel on the string? (c) While the guitar string is still being plucked, another musician plays a slide whistle that is closed at one end and open at the other. He starts at a very high frequency and slowly lowers the frequency until beats, with a frequency of \(5 \mathrm{Hz}\), are heard with the guitar. What is the fundamental frequency of the slide whistle with the slide in this position? (d) How long is the open tube in the slide whistle for this frequency?
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.
A sound wave arriving at your ear is transferred to the fluid in the cochlea. If the intensity in the fluid is 0.80 times that in air and the frequency is the same as for the wave in air, what will be the ratio of the pressure amplitude of the wave in air to that in the fluid? Approximate the fluid as having the same values of density and speed of sound as water.
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