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Chapter 15: Problem 2

Audible Sound. Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 \(\mathrm{Hz}\) to about 20.0 \(\mathrm{kHz}\) . (a) If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound and a blue dot for the short-wavelength sound, how far apart wonld the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meterstick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 \(\mathrm{m} / \mathrm{s}\) . How far apart would the dots be in each set? Could you readily measure their separation with a meterstick?

Short Answer

Expert verified
In air, red dots are 17.15m apart, blue dots 0.01715m apart. In water, red dots are 74m, blue dots 0.074m apart. Red dots are measurable in both mediums; blue dots are difficult to measure in air.

Step by step solution

01

Understanding Wave Concepts

To find the distance between the wave dots (which represent the wavelength), we need to use the formula for wavelength: \( \lambda = \frac{v}{f} \), where \( \lambda \) is the wavelength, \( v \) is the speed of sound, and \( f \) is the frequency of the sound.
02

Calculating Wavelength in Air for Long Wavelength Sound

For air, the speed of sound is approximately 343 m/s. For the low-frequency sound, \( f = 20.0 \) Hz. Using the formula, \( \lambda_{\text{low}} = \frac{343 \text{ m/s}}{20.0 \text{ Hz}} = 17.15 \text{ m} \). The red dots would be 17.15 meters apart.
03

Calculating Wavelength in Air for Short Wavelength Sound

For the high-frequency sound, \( f = 20.0 \text{ kHz} = 20000 \text{ Hz} \). Using the formula, \( \lambda_{\text{high}} = \frac{343 \text{ m/s}}{20000 \text{ Hz}} = 0.01715 \text{ m} \). The blue dots would be 0.01715 meters apart.
04

Determining Measurability in Air

The distance of 17.15 meters is large and easily measurable with a meterstick. However, 0.01715 meters (or 1.715 cm) is quite small, making it difficult to measure precisely with a standard meterstick.
05

Calculating Wavelength in Water for Long Wavelength Sound

In water, the speed of sound is 1480 m/s. For low-frequency sound, \( f = 20.0 \) Hz. Using the formula, \( \lambda_{\text{low, water}} = \frac{1480 \text{ m/s}}{20.0 \text{ Hz}} = 74 \text{ m} \). The red dots would be 74 meters apart.
06

Calculating Wavelength in Water for Short Wavelength Sound

For high-frequency sound, \( f = 20000 \text{ Hz} \). Using the formula, \( \lambda_{\text{high, water}} = \frac{1480 \text{ m/s}}{20000 \text{ Hz}} = 0.074 \text{ m} \). The blue dots would be 0.074 meters apart.
07

Determining Measurability in Water

The distance of 74 meters is large and easily measurable with a meterstick. The distance 0.074 meters (or 7.4 cm) is small but still measurable with more precision compared to air.

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Key Concepts

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

Frequency Range
Sound waves encompass a broad range of frequencies, and our human ears capture only a part of that spectrum. We typically hear sounds ranging from 20 Hz to 20 kHz. This is considered the audible frequency range.

Lower frequencies, like those at 20 Hz, represent deep bass sounds. These sounds come with longer wavelengths because they oscillate fewer times per second. On the other hand, higher frequencies, like those at 20 kHz, are the realm of high-pitched sounds. These shorter wavelengths result from more oscillations per second.
  • The 20 Hz frequency is often felt more than heard, especially in environments where large spaces allow for the propagation of these waves.
  • The 20 kHz frequency might be challenging for some ears to detect, as hearing sensitivity diminishes with age.
Understanding these frequencies helps us to explore how sound behaves in different mediums and environments.
Wavelength Calculation
Wavelength is crucial in understanding how sound travels. It is the distance a single wave covers during one cycle. Calculating wavelength uses the formula \( \lambda = \frac{v}{f} \), where \( \lambda \) is the wavelength, \( v \) is the velocity, and \( f \) is the frequency.

Consider sound traveling in air, where the speed of sound is approximately 343 m/s.
  • For a low-frequency of 20 Hz, \( \lambda = \frac{343 \text{ m/s}}{20 \text{ Hz}} = 17.15 \text{ m} \). This long wavelength means wave peaks are far apart.
  • For a high-frequency of 20 kHz, \( \lambda = \frac{343 \text{ m/s}}{20000 \text{ Hz}} = 0.01715 \text{ m} \), showing the peaks are very close together.
In water, where sound travels faster at 1480 m/s:
  • The wavelength for 20 Hz stretches further to 74 m.
  • For 20 kHz, it equals \( 0.074 \text{ m} \) due to increased speed.
Calculating wavelengths in different mediums helps in applications ranging from sonar to acoustics.
Speed of Sound
The speed of sound refers to how fast sound waves travel through a medium. This speed varies based on the medium鈥檚 properties like density and elasticity. In general, sound moves faster in solids, followed by liquids, and slowest in gases.

For example, in air under normal conditions (around 20掳C), the speed of sound is about 343 m/s. However, this speed changes with temperature and pressure shifts. Despite being slower in air compared to other mediums, the speed allows for a clear calculation of how sound travels over distances.

Meanwhile, sound zips through water at around 1480 m/s, more than four times faster than in air. This speed is advantageous in marine environments, facilitating explorations and communications via sonar.
  • The speed increase in water results from tightly packed molecules, creating more pathways for vibrations.
  • In designing sound equipment or investigating environmental acoustics, understanding the speed of sound in different mediums is essential.
So, knowing how sound behaves across various materials is crucial for technological and scientific advancements.

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Most popular questions from this chapter

A thin, 75.0 -cm wire has a mass of 16.5 g. One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire. (a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 \(\mathrm{cm}\) makes 875 vibrations per second? (b) How fast would this wave travel?

A simple harmonic oscillator at the point \(x=0\) generates a wave on a rope. The oscillator operates at a frequency of 40.0 \(\mathrm{Hz}\) and with an amplitude of 3.00 \(\mathrm{cm}\) . The rope has a linear mass density of 50.0 \(\mathrm{g} / \mathrm{m}\) and is stretched with a tension of 5.00 \(\mathrm{N}\) . (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function \(y(x, t)\) for the wave. Assume that the oscillator has its maximum upward displacement at time \(t=0\) . (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.

A jet plane at take-off can produce sound of intensity 10.0 \(\mathrm{W} / \mathrm{m}^{2}\) at 30.0 \(\mathrm{m}\) away. But you prefer the tranguil sound of normal conversation, which is 1.0\(\mu \mathrm{W} / \mathrm{m}^{2}\) . Assume that the plane behaves like a point source of sound. (a) What is the closest distance you should live from the airport runway to preserve your peace of mind? (b) What intensity from the jet does your friend experience if she lives twice as far from the runway as you do? (c) What power of sound does the jet produce at take-off?

Transverse waves on a string have wave speed 8.00 \(\mathrm{m} / \mathrm{s}\) , amplitude \(0.0700 \mathrm{m},\) and wavelength 0.320 \(\mathrm{m}\) . The waves travel in the \(-x\) -direction, and at \(t=0\) the \(x=0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves, (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at \(x=0.360 \mathrm{m}\) at time \(t=0.150 \mathrm{s}\) s. (d) How much time must elapse from the instant in part (c) until the particle at \(x=0.360 \mathrm{m}\) next has maximum upward displacement?

A transverse sine wave with an amplitude of 2.50 \(\mathrm{mm}\) and a wavelength of 1.80 \(\mathrm{m}\) travels from left to right along a long, horizontal, stretched string with a speed of 36.0 \(\mathrm{m} / \mathrm{s}\) . Take the origin at the left end of the undisturbed string. At time \(t=0\) the left end of the string has its maximum upward displacement, (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\)for a particle 1.35 \(\mathrm{m}\) to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverse velocity of a particle 1.35 \(\mathrm{m}\) to the right of the origin at time \(t=0.0625 \mathrm{s}\) .
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