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Chapter 17: Problem 7

A bat (Fig. Pl7.7) 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?

Short Answer

Expert verified
The smallest insect a bat can detect is approximately equal to the wavelength of the sound, which is about 5.67 mm.

Step by step solution

01

Determine the Wavelength

The wavelength \(\lambda\) of the sound can be calculated using the relationship between speed \(v\), frequency \(f\), and wavelength, which is given by the formula \(v = f\lambda\). We have the speed of sound (\(v = 340\) m/s) and the frequency of the chirps (\(f = 60.0\) kHz), which must be converted to Hz by multiplying by \(10^3\).
02

Calculate the Wavelength

Using the formula \(v = f\lambda\), solve for \(\lambda\): \(\lambda = \frac{v}{f}\). Plug in the values \(v = 340\) m/s and \(f = 60.0 \times 10^3\) Hz to find the wavelength.
03

Find the Length of the Smallest Object

Since the length of the smallest object a bat can detect is approximately equal to one wavelength, the length is the same as the wavelength calculated in the previous step.

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Key Concepts

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

Speed of Sound
Understanding the speed of sound is crucial when studying how sound waves travel through different mediums. The speed of sound refers to how fast a sound wave travels through a medium, such as air, water, or solid materials. In the given exercise, we consider the speed of sound in air to be 340 meters per second (m/s). This value can vary depending on factors such as temperature, humidity, and altitude. Generally, sound travels faster in warmer air and slower in colder air. Additionally, sound travels significantly faster in solids than in liquids, and faster in liquids than in gases, because particles are closer together in solids than in liquids, and closer in liquids than in gases, allowing sound waves to be transmitted more quickly.
Frequency and Wavelength Relationship
The frequency and wavelength of a sound wave are intimately connected by the wave equation, which in this context is represented as \(v = f\lambda\). Frequency (\(f\)) is the number of waves that pass a point in one second, measured in hertz (Hz), whereas wavelength (\(\lambda\)) is the distance between consecutive corresponding points of the same phase on the wave, such as crest to crest or trough to trough. For a constant speed of sound, as the frequency increases, the wavelength decreases, and vice versa.

This inverse relationship is crucial for calculating the wavelength when you know the frequency and the speed of sound. By rearranging the wave equation to \(\lambda = \frac{v}{f}\), we can compute the wavelength if the frequency and the speed of sound are known. In the exercise, we implemented this calculation to determine the wavelength of a bat's chirp when its frequency (60.0 kHz) and the speed of sound in air were given.
Acoustic Detection in Bats
Bats are extraordinary creatures that utilize echolocation to navigate and hunt in the dark. Echolocation is a biological sonar system where bats emit high-frequency sound waves that bounce off objects in their environment and return to the bats as echoes. The bats then process these echoes to construct a detailed acoustic image of their surroundings. This allows bats to detect, locate, and even classify their prey with remarkable precision.

Bats commonly use frequencies in the range of 20 kHz to 200 kHz, which are well beyond the range of human hearing (20 Hz - 20 kHz). The high frequency of these sounds is ideal for detecting small objects. The exercise mentioned demonstrates how a bat's ability to detect an insect is related to the wavelength of the sound it emits. A wavelength that matches the insect's size makes the insect more detectable due to the sound waves reflecting off the insect's body and returning to the bat. This exemplifies how a bat's biology is finely tuned to its ecological niche, relying on the interplay between sound wave frequency, wavelength, and the speed of sound.

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Most popular questions from this chapter

Calculate the sound level in decibels of a sound wave that has an intensity of 4.00\(\mu \mathrm{W} / \mathrm{m}^{2}\) .

A block with a speaker bolted to it is connected to a spring having spring constant \(k=20.0 \mathrm{N} / \mathrm{m}\) as in Figure \(\mathrm{P} 17.40\) . The total mass of the block and speaker is 5.00 \(\mathrm{kg}\) , and the amplitude of this unit's motion is 0.500 \(\mathrm{m}\) . (a) If the speaker emits sound waves of frequency 440 \(\mathrm{Hz}\) , determine the highest and lowest frequencies heard by the person to the right of the speaker. (b) If the maximum sound level heard by the person is 60.0 \(\mathrm{dB}\) when he is closest to the speaker, 1.00 \(\mathrm{m}\) away, what is the minimum sound level heard by the observer? Assume that the speed of sound is 343 \(\mathrm{m} / \mathrm{s}\) .

Assume that a loudspeaker broadcasts sound equally in all directions and produces sound with a level of 103 \(\mathrm{dB}\) at a distance of 1.60 \(\mathrm{m}\) from its center. (a) Find its sound power output. (b) If the salesperson claims to be giving you 150 \(\mathrm{W}\) per channel, he is referring to the electrical power input to the speaker. Find the efficiency of the speaker-that is, the fraction of input power that is converted into useful output power.

The smallest wavelength possible for a sound wave in air is on the order of the separation distance between air molecules. Find the order of magnitude of the highest-frequency sound wave possible in air, assuming a wave speed of 343 \(\mathrm{m} / \mathrm{s}\) , density \(1.20 \mathrm{kg} / \mathrm{m}^{3},\) and an average molecular mass of \(4.82 \times 10^{-26} \mathrm{kg}\) .

A train whistle \((f=400 \mathrm{Hz})\) sounds higher or lower in frequency depending on whether it approaches or recedes. (a) Prove that the difference in frequency between the approaching and receding train whistle is $$ \Delta f=\frac{2 u / v}{1-u^{2} / v^{2}} f $$ where \(u\) is the speed of the train and \(v\) is the speed of sound. (b) Calculate this difference for a train moving at a speed of 130 \(\mathrm{km} / \mathrm{h}\) . Take the speed of sound in air to be 340 \(\mathrm{m} / \mathrm{s}\) .
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