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Chapter 17: Problem 7

Note: Use the following values as needed unless otherwise specified: the equilibrium density of air at \(20^{\circ} \mathrm{C}\) is \(\rho=1.20 \mathrm{kg} / \mathrm{m}^{3} .\) The speed of sound in air is \(v=\) \(343 \mathrm{m} / \mathrm{s} .\) Pressure variations \(\Delta P\) are measured relative to atmospheric pressure, \(1.013 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) Problem 70 in Chapter 2 can also be assigned with this section. A bat (Fig. P17.7) can detect very small objects, such as an insect whose length is approximately equal to one wavelength of the sound the bat makes. If a bat emits chirps at a frequency of \(60.0 \mathrm{kHz},\) and if the speed of sound in air is \(340 \mathrm{m} / \mathrm{s},\) what is the smallest insect the bat can detect? PICTURE CANT COPY

Short Answer

Expert verified
The size of the smallest insect that the bat can detect is around 5.67 millimeters.

Step by step solution

01

Understand the Given Variables

Here, we're asked to determine the size of the smallest insect a bat can detect. The bat emits chirps at a frequency of \(60.0 \mathrm{kHz} \) which is equal to \( 60000 \mathrm{Hz} \). The speed of sound is given as \( 340 \mathrm{m} / \mathrm{s} \). The given values must be used in the formula for the velocity of sound which is velocity = wavelength * frequency.
02

Formulate the Formula to Calculate Wavelength

We can rearrange the formula for the velocity of sound to calculate the wavelength. Hence, the formula would be wavelength = velocity / frequency.
03

Insert the Given Values into the Formula

Now let's plug in the given values into our formula. Hence, wavelength = \(340 \mathrm{m} / \mathrm{s}\) / \(60000 \mathrm{Hz}\).
04

Calculate the Wavelength

Doing the calculation, we get the wavelength to be approximately \(0.00567 \mathrm{m}\), or 5.67 mm. This represents the size of the smallest insect the bat can detect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Sound in Air
The speed of sound in air is a fundamental concept in physics, referring to how fast sound waves can travel through the atmospheric medium. At a temperature of approximately 20掳C, the speed is generally taken as 343 meters per second (m/s).

Sound is a mechanical wave that requires a medium to travel, being air in this case. Variations in temperature, humidity, and atmospheric pressure can affect the speed of sound in air, but for many calculations 鈥� like the exercises we approach in physics 鈥� we consider it under standard conditions.

The speed is essential for various practical applications. For example, by understanding how sound travels through air, architects can design better acoustics for buildings, meteorologists can use the sound of thunder to calculate the distance of lightning strikes, and it's indispensable for calibrating instruments that use sonar.
Frequency of Sound Waves
Frequency of sound waves refers to the number of complete oscillations or cycles that a sound wave completes in one second, measured in Hertz (Hz). To put it simply, it's how often the sound waves are hitting your ear in a second.

The frequency of a sound wave determines its pitch; higher frequencies result in higher pitches, while lower frequencies make for a lower pitch. These frequencies can range from very low (infrasound) to very high (ultrasound). Humans can generally hear frequencies between 20 Hz and 20 kHz, and anything above that is considered ultrasound, which is utilized by certain animals for communication and navigation and in various technological applications.
Ultrasound in Animal Navigation
Ultrasound serves as a sophisticated biological navigation mechanism for many animals, most notably in bats and marine mammals like dolphins and whales.

These animals emit high-frequency sound waves that are inaudible to humans. These sound waves bounce back upon hitting an object, a phenomenon called echolocation. By interpreting the echoes that return, animals can determine the size, shape, distance, speed, and even some characteristics of objects or prey around them.

In the case of bats, which can emit frequencies of up to 120 kHz or more, they can detect and manoeuvre around obstacles and hone in on tiny insects with precision. Their ability to process this information allows them to fly and hunt effectively in complete darkness, showcasing an excellent example of adaptation and the importance of ultrasound in the animal kingdom.

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Most popular questions from this chapter

A sinusoidal sound wave is described by the displacement wave function $$s(x, t)=(2.00 \mu \mathrm{m}) \cos \left[\left(15.7 \mathrm{m}^{-1}\right) x-\left(858 \mathrm{s}^{-1}\right)t\right]$$ (a) Find the amplitude, wavelength, and speed of this wave. (b) Determine the instantaneous displacement from equilibrium of the elements of air at the position \(x=0.0500 \mathrm{m}\) at \(t=3.00 \mathrm{ms} .\) (c) Determine the maximum speed of the element's oscillatory motion.

A fireworks rocket explodes at a height of 100 m above the ground. An observer on the ground directly under the explosion experiences an average sound intensity of \(7.00 \times 10^{-2} \mathrm{W} / \mathrm{m}^{2}\) for \(0.200 \mathrm{s}\). (a) What is the total sound energy of the explosion? (b) What is the sound level in decibels heard by the observer?

A bat, moving at \(5.00 \mathrm{m} / \mathrm{s},\) is chasing a flying insect (Fig. \(P 17.7) .\) If the bat emits a \(40.0 \mathrm{kHz}\) chirp and receives back an echo at \(40.4 \mathrm{kHz},\) at what speed is the insect moving toward or away from the bat? (Take the speed of sound in air to be \(v=340 \mathrm{m} / \mathrm{s} .\))

Prove that sound waves propagate with a speed given by Equation \(17.1 .\) Proceed as follows. In Figure \(17.3,\) consider a thin cylindrical layer of air in the cylinder, with face area \(A\) and thickness \(\Delta x\). Draw a free-body diagram of this thin layer. Show that \(\Sigma F_{x}=m a_{x}\) implies that \(-[\partial(\Delta P) / \partial x] A \Delta x=\) \(\rho A \Delta x\left(\partial^{2} s / \partial t^{2}\right) .\) By substituting \(\Delta P=-B(\partial s / \partial x),\) obtain the wave equation for sound, \((B / \rho)\left(\partial^{2} s / \partial x^{2}\right)=\left(\partial^{2} s / \partial t^{2}\right) .\) To a mathematical physicist, this equation demonstrates the existence of sound waves and determines their speed. As a physics student, you must take another step or two. Substitute into the wave equation the trial solution \(s(x, l)=\) \(s_{\max } \cos (k x-\omega t) .\) Show that this function satisfies the wave equation provided that \(\omega / k=\sqrt{B / \rho} .\) This result reveals that sound waves exist provided that they move with the speed \(v=f \lambda=(2 \pi f)(\lambda / 2 \pi)=\omega / k=\sqrt{B / \rho}\)

The intensity of a sound wave at a fixed distance from a speaker vibrating at a frequency \(f\) is \(I\) (a) Determine the intensity if the frequency is increased to \(f^{\prime}\) while a constant displacement amplitude is maintained. (b) Calculate the intensity if the frequency is reduced to \(f / 2\) and the displacement amplitude is doubled. GRAPH CANT COPY
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