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

Short Answer

Expert verified
(a) 90.06 m, (b) 102 kHz, (c) 13.7 mm, (d) 4.4 to 8.8 mm, (e) 6.2 MHz.

Step by step solution

01

Understand the Formula for Wave Calculation

The relationship between speed, frequency, and wavelength is given by the formula \( v = f \times \lambda \), where \( v \) is the speed of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength.
02

Solve for Wavelength in Seawater

For part (a), we are given the frequency \( f = 17 \text{ Hz} \) and the speed of sound in seawater \( v = 1531 \text{ m/s} \). Using the formula, the wavelength \( \lambda \) can be calculated as:\[ \lambda = \frac{v}{f} = \frac{1531 \text{ m/s}}{17 \text{ Hz}} \approx 90.06 \text{ m} \]
03

Solve for Frequency of Dolphin Clicks

For part (b), the wavelength \( \lambda = 1.5 \text{ cm} = 0.015 \text{ m} \). Using the speed of sound in seawater \( v = 1531 \text{ m/s} \), solve for frequency \( f \):\[ f = \frac{v}{\lambda} = \frac{1531 \text{ m/s}}{0.015 \text{ m}} \approx 102067 \text{ Hz} \text{ or } 102 \text{ kHz} \]
04

Solve for Wavelength of Dog Whistle Sound

For part (c), the frequency \( f = 25 \text{ kHz} = 25000 \text{ Hz} \). The speed of sound in air is approximately 343 m/s. Using the formula to find wavelength:\[ \lambda = \frac{343 \text{ m/s}}{25000 \text{ Hz}} = 0.01372 \text{ m} \text{ or } 13.7 \text{ mm} \]
05

Determine Bat Sound Wavelength Range

For part (d), given the frequency range \( 39 \text{ kHz} \) to \( 78 \text{ kHz} \) in air (assuming speed of sound in air is 343 m/s), calculate the wavelength for both frequencies:- For \( 39 \text{ kHz} \): \[ \lambda = \frac{343 \text{ m/s}}{39000 \text{ Hz}} \approx 0.00879 \text{ m} \text{ or } 8.8 \text{ mm} \] - For \( 78 \text{ kHz} \): \[ \lambda = \frac{343 \text{ m/s}}{78000 \text{ Hz}} \approx 0.00440 \text{ m} \text{ or } 4.4 \text{ mm} \] The range is from 4.4 mm to 8.8 mm.
06

Solve for Frequency Needed for Sonogram

For part (e), with a wavelength of one-fourth the size of the tumor, \( \lambda = \frac{1 \text{ mm}}{4} = 0.25 \text{ mm} = 0.00025 \text{ m} \). Given \( v = 1550 \text{ m/s} \) in tissue, calculate the frequency:\[ f = \frac{v}{\lambda} = \frac{1550 \text{ m/s}}{0.00025 \text{ m}} = 6200000 \text{ Hz} \text{ or } 6.2 \text{ MHz} \]

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Key Concepts

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

Frequency and Wavelength Relationship
The relationship between frequency and wavelength is a fundamental concept in the study of waves. All types of waves, whether sound, light, or radio, display this relationship. The formula used to define this connection is \( v = f \times \lambda \), where \( v \) is the speed of the wave, \( f \) is the frequency, and \( \lambda \) represents the wavelength.

This formula exemplifies how the speed of a wave is the product of its frequency and wavelength. If you know any two of these variables, you can easily calculate the third. For example, higher frequency waves will have shorter wavelengths if the speed remains constant. Conversely, lower frequency waves will have longer wavelengths.

To understand through example, consider a blue whale's sound, which has a frequency of 17 Hz. When sounded through seawater, where the speed of sound is 1531 m/s, the wavelength of this sound can be calculated as \( \lambda = \frac{1531 \, \text{m/s}}{17 \, \text{Hz}} \approx 90.06 \, \text{m} \). This indicates that the wavelength associated with these low-frequency sounds is quite large, allowing them to travel tremendous distances in water.
Speed of Sound
The speed of sound is an important concept in acoustics and varies by medium.

  • In air, under standard conditions, it is approximately 343 m/s.
  • In seawater, it is between 1500 and 1600 m/s due to water's higher density compared to air.
  • In different human tissues, used commonly for medical imaging, it varies but is often around 1540 m/s.
Sound travels faster in denser mediums because the molecules can transmit vibrations more quickly. This is why you experience sound arriving almost instantly across water or through solids compared to air.

Let's look further into examples. For dolphin clicks in the ocean, we know the speed of sound is 1531 m/s. Given a click's wavelength of 1.5 cm, or 0.015 m, the frequency can be calculated as \( f = \frac{1531 \, \text{m/s}}{0.015 \, \text{m}} \approx 102 \, \text{kHz} \). On the other hand, sound in air like that from a dog whistle at 25 kHz, traveling at 343 m/s, results in a wavelength of 13.7 mm.
Ultrasound Imaging
Ultrasound imaging leverages sound at frequencies above human hearing (>20 kHz) to visualize interior body structures. The principle relies on echoes from sound waves as they reflect off different tissues.

For imagery to be sharp, sound waves require wavelengths of one-fourth or less than the objects being imaged. Considering a tumor that is 1 mm across, the required wavelength should be around 0.25 mm for a clear image. Given a tissue sound speed of 1550 m/s, the needed frequency is \( f = \frac{1550 \, \text{m/s}}{0.00025 \, \text{m}} = 6.2 \, \text{MHz} \).

Ultrasound waves penetrate soft tissues and reflect on different density boundaries, providing invaluable, detailed images in real-time. This non-destructive technique visualizes organs, blood flow, and abnormalities like tumors without the risks associated with ionizing radiation.
Animal Communication Frequencies
Animals utilize diverse sounds to communicate, with frequencies often beyond human perception.

  • Whales: Communicate through very low frequencies like 17 Hz, which travel long distances undersea.
  • Dolphins: Use high-frequency clicks, around 102 kHz, for echolocation, helping them navigate and hunt.
  • Bats: Emit ultrasonic sounds between 39 kHz to 78 kHz for echolocation, allowing them to "see" with sound.
  • Dogs: Detect ultrasonic sounds beyond human hearing, hence dog whistles at 25 kHz are inaudible to us but not to dogs.
For bats, the range of their sound wavelengths is from 8.8 mm for 39 kHz to 4.4 mm for 78 kHz. These sounds help them detect even the smallest insects in complete darkness, showcasing the evolutionary advantages of frequency diversity in auditory communication.

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Most popular questions from this chapter

Horseshoe bats (genus \(Rhinolophus\)) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the bat its name is a depression around the nostrils that acts like a focusing mirror, so that the bat emits sound in a narrow beam like a flashlight.) A \(Rhinolophus\) flying at speed \(v_{bat}\) emits sound of frequency \(f_{bat}\); the sound it hears reflected from an insect flying toward it has a higher frequency \(f_{refl}\). (a) Show that the speed of the insect is $$vinsect = v\Bigg[\frac{f_{refl}(v - v_{bat}) - f_{bat}(v + v_{bat})}{f_{refl}(v - v_{bat}) + f_{bat}(v + v_{bat})}\Bigg] $$ where \(v\) is the speed of sound. (b) If \(f_{bat} =\) 80.7 kHz, \(f_{refl} =\) 83.5 kHz, and \(v_{bat} =\) 3.9 m/s, calculate the speed of the insect.

A jet plane flies overhead at Mach 1.70 and at a constant altitude of 1250 m. (a) What is the angle a of the shock-wave cone? (b) How much time after the plane passes directly overhead do you hear the sonic boom? Neglect the variation of the speed of sound with altitude.

A bat flies toward a wall, emitting a steady sound of frequency 1.70 kHz. This bat hears its own sound plus the sound reflected by the wall. How fast should the bat fly in order to hear a beat frequency of 8.00 Hz?

You are trying to overhear a juicy conversation, but from your distance of 15.0 m, it sounds like only an average whisper of 20.0 dB. How close should you move to the chatterboxes for the sound level to be 60.0 dB?

In some applications of ultrasound, such as its use on cranial tissues, large reflections from the surrounding bones can produce standing waves. This is of concern because the large pressure amplitude in an antinode can damage tissues. For a frequency of 1.0 MHz, what is the distance between antinodes in tissue? (a) 0.38 mm; (b) 0.75 mm; (c) 1.5 mm; (d) 3.0 mm.

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