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Chapter 16: Problem 5

BIO UItrasound and Infrasound. (a) Whale communication. Blue whales apparently communicate with each other using sound of frequency 17 \(\mathrm{Hz}\) , which can be heard nearly 1000 \(\mathrm{km}\) away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1531 \(\mathrm{m} / \mathrm{s} ?\) (b) Dolphin clicks. One type of sound that dolphins emit is a sharp click of wavelength 1.5 \(\mathrm{cm}\) in the ocean. What is the frequency of such clicks? (c) Dog whistles. One brand of dog whistles claims a frequency of 25 \(\mathrm{kH} z\) for its product. What is the wavelength of this sound? (d) Bats. While bats emit a wide variety of sounds, one type emits pulses of sound having a frequency between 39 \(\mathrm{kHz}\) and 78 \(\mathrm{kHz}\) . What is the range of wavelengths of this sound? (e) Sonograms. Ultrasound is used to view the interior of the body, much as \(\mathrm{x}\) rays are utilized. For sharp imagery, the wave length of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is 1.0 \(\mathrm{mm}\) across if the speed of sound in the tissue is 1550 \(\mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
(a) 90.06 m, (b) 102066.67 Hz, (c) 1.372 cm, (d) 0.00440 m to 0.00879 m, (e) 6.2 MHz

Step by step solution

01

Understanding the Problem

Read carefully through the entire problem to identify the questions being asked and the given values for each part of the exercise. The key quantities to focus on are frequency, wavelength, and the speed of sound, as these will be necessary for calculations.
02

Part (a) Whale Communication: Using the Speed and Frequency to find Wavelength

We use the equation for wave speed, \( v = f \lambda \), where \( v \) is the speed of sound in seawater (1531 \( \text{m/s}\)), \( f \) is the frequency (17 \( \text{Hz}\)), and \( \lambda \) is the wavelength. Rearrange the equation to solve for \( \lambda \):\[ \lambda = \frac{v}{f} = \frac{1531}{17} \approx 90.06 \text{ m} \].
03

Part (b) Dolphin Clicks: Using Wavelength to find Frequency

Given that the wavelength \( \lambda = 1.5 \text{ cm} = 0.015 \text{ m} \), and the speed of sound in water \( v = 1531 \text{ m/s} \), use the wave speed equation, \( v = f \lambda \), to solve for the frequency \( f \): \[ f = \frac{v}{\lambda} = \frac{1531}{0.015} \approx 102066.67 \text{ Hz} \].
04

Part (c) Dog Whistle: Finding Wavelength using Frequency

We use the wave speed equation, \( v = f \lambda \), where \( v = 343 \text{ m/s} \) (speed of sound in air), \( f = 25 \text{ kHz} = 25000 \text{ Hz} \). Solve for wavelength \( \lambda \):\[ \lambda = \frac{v}{f} = \frac{343}{25000} = 0.01372 \text{ m} = 1.372 \text{ cm} \].
05

Part (d) Bats: Range of Wavelengths from Frequency Range

For frequencies \( f_1 = 39 \text{ kHz} \) and \( f_2 = 78 \text{ kHz} \), use the speed of sound in air, \( v = 343 \text{ m/s} \), to find the range of wavelengths. Calculate:For \( f_1 = 39 \text{ kHz} \):\[ \lambda_1 = \frac{343}{39000} \approx 0.00879 \text{ m} \].For \( f_2 = 78 \text{ kHz} \):\[ \lambda_2 = \frac{343}{78000} \approx 0.00440 \text{ m} \].Thus, the range is approximately 0.00440 m to 0.00879 m.
06

Part (e) Sonograms: Frequency Required for Imaging a Tumor

To have a wavelength one-fourth the size of a 1.0 mm object, \( \lambda \approx \frac{1.0}{4} \text{ mm} = 0.25 \text{ mm} = 0.00025 \text{ m} \). Use \( v = 1550 \text{ m/s} \) for the speed of sound in tissue to solve for \( f \): \[ f = \frac{v}{\lambda} = \frac{1550}{0.00025} = 6200000 \text{ Hz} = 6.2 \text{ MHz} \].

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Key Concepts

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

Frequency
Frequency in wave physics refers to how often the wave oscillates or completes a cycle in one second. It's measured in hertz (Hz). A higher frequency indicates more oscillations per second, while a lower frequency means fewer. This is crucial in understanding how different sounds or signals behave in various media.

To calculate frequency, you can use the formula:\[ f = \frac{v}{\lambda} \]where:
  • \( f \) is the frequency.
  • \( v \) is the speed of sound in the medium.
  • \( \lambda \) is the wavelength.
For instance, if the wavelength in seawater is known, and the speed of sound in that medium is 1531 \(\text{m/s}\), you can easily find the frequency by rearranging the formula. Frequency is vital because different animals and devices utilize specific frequencies for communication and detection.
Wavelength
Wavelength is the distance between two consecutive points that are in phase on a wave. It is typically measured in meters (m) and is an essential aspect of wave physics and acoustics. Wavelength gives us insight into the "spread out" nature of waves and can vary greatly between different types of waves.

The relationship between wavelength, frequency, and the speed of sound is given by the formula \( v = f \lambda \). This relationship indicates that:
  • For a given speed, a higher frequency results in a shorter wavelength.
  • Conversely, a lower frequency corresponds to a longer wavelength.
For example, in the problem of whale communication, given a frequency of 17 Hz and a speed of sound of 1531 m/s in seawater, the wavelength can be found using \( \lambda = \frac{v}{f} \). Understanding wavelength helps in designing equipment in both human industries and understanding animal behaviors.
Speed of Sound
The speed of sound is how fast sound waves travel through a medium and is influenced by factors like temperature, medium type, and pressure. It is typically measured in meters per second (m/s). In air, the speed of sound is approximately 343 m/s, but it differs in other media, such as seawater or different tissues.

The formula to find the speed of sound is \( v = f \lambda \), which relates it to wavelength and frequency.
  • In media like seawater, the speed of sound is often higher than in air due to its denser nature.
  • Knowing the speed of sound is crucial for solving problems like those involving dolphin clicks or ultrasound used in medical imaging.
For specific applications, such as sonar in marine environments or ultrasound in medical fields, understanding variations in the speed of sound allows for precise calculations and solutions tailored to the specific medium and situation.

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Most popular questions from this chapter

CP Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F .\) (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { beat }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 Hz. One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?

Example 16.1 (Section 16.1\()\) showed that for sound waves in air with frequency 1000 \(\mathrm{Hz}\) , a displacement amplitude of \(1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2}\) Pa. Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s} .\) For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain whyyour answer is much less than \(1.2 \times 10^{-8} \mathrm{m}\) .

Extrasolar Planets. In the not-too-distant future, it should be possible to detect the presence of planets moving around other stars by measuring the Doppler shift in the infrared light they emit. If a planet is going around its star at 50.00 \(\mathrm{km} / \mathrm{s}\) while emitting infrared light of frequency \(3.330 \times 10^{14} \mathrm{Hz},\) what frequency light will be received from this planet when it is moving directly away from us? (Note: Infrared light is light having wavelengths longer than those of visible light.)

singing in the Shower. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you can hear them if you are inside the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda_{n}=2 L / n\) and the frequencies are given by \(f_{n}=n v / 2 L=n f_{1},\) where \(n=1,2,3, \ldots,\) (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 \(\mathrm{m}\) tall. Are these frequencies audible?

Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from A. What is the closest you can be to \(B\) and be at a point of destructive interference?
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