/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 A tuning fork oscillates at \(28... [FREE SOLUTION] | 91影视

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Chapter 23: Problem 7

A tuning fork oscillates at \(284 \mathrm{~Hz}\) in air. Compute the wavelength of the tone emitted at \(25^{\circ} \mathrm{C}\). Remembering that the speed of sound increases by \(0.61 \mathrm{~m} / \mathrm{s}\) for each \(1{ }^{\circ} \mathrm{C}\) increase in temperature, at \(25^{\circ} \mathrm{C}\), $$ v=331 \mathrm{~m} / \mathrm{s}+(0.61)(25) \mathrm{m} / \mathrm{s}=346 \mathrm{~m} / \mathrm{s} $$ Using \(\lambda=v T=v / f\) $$ \lambda=\frac{v}{f}=\frac{346 \mathrm{~m} / \mathrm{s}}{284 \mathrm{~s}^{-1}}=1.22 \mathrm{~m} $$

Short Answer

Expert verified
The wavelength is 1.22 meters.

Step by step solution

01

Understand the given information

We are given the frequency of a tuning fork as \(284 \, \text{Hz}\) and the temperature as \(25^{\circ} \text{C}\). We need to compute the wavelength of the sound emitted by the tuning fork.
02

Calculate the speed of sound

The speed of sound in air at \(0^{\circ} \text{C}\) is approximately \(331 \, \text{m/s}\). It increases by \(0.61 \, \text{m/s}\) for each \(1^{\circ} \text{C}\) increase in temperature. Therefore, at \(25^{\circ} \text{C}\), the speed of sound \(v\) is:\[v = 331 \, \text{m/s} + (0.61 \, \text{m/s} \times 25) = 346 \, \text{m/s}\]
03

Calculate the wavelength

The wavelength \(\lambda\) can be found using the equation \(\lambda = \frac{v}{f}\), where \(v\) is the speed of sound and \(f\) is the frequency.\[\lambda = \frac{346 \, \text{m/s}}{284 \, \text{Hz}} = 1.22 \, \text{m}\]

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Key Concepts

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

Wavelength Calculation
Understanding how to calculate the wavelength of a sound wave is essential when studying acoustics. The wavelength (\( \lambda \)) is the distance between two consecutive points in phase on a wave, such as from crest to crest or trough to trough. For sound waves, this distance can be affected by the speed of sound in the medium and the frequency of the sound.
The formula used for wavelength calculation is:
  • \( \lambda = \frac{v}{f} \)
Where:
  • \( \lambda \) is the wavelength,
  • \( v \) is the speed of sound,
  • \( f \) is the frequency.
In our exercise, with the speed of sound calculated to be 346 m/s and the tuning fork frequency at 284 Hz, the wavelength can be determined. Plugging the values into the formula: \[\lambda = \frac{346 \, \text{m/s}}{284 \, \text{Hz}} = 1.22 \, \text{m}\]This result indicates that the sound wave emitted by the tuning fork will have a wavelength of approximately 1.22 meters under the given conditions.
Tuning Fork Frequency
The tuning fork frequency is foundational for determining the characteristics of the sound it produces. Frequency (\( f \)) refers to the number of oscillations or vibrations that occur in one second, measured in Hertz (Hz). When a tuning fork vibrates, it creates a sound wave that matches its frequency.
Tuning forks, like the one mentioned in the exercise, are often used in musical settings for tuning instruments because they produce a clear, sustained tone at a specific frequency. In our case, the tuning fork oscillates at 284 Hz, meaning it completes 284 cycles every second. This frequency is crucial for calculating the wavelength of the sound it produces. When combined with the speed of sound, the frequency directly influences the wave's wavelength and how sound is perceived in different environments.
Temperature Effects on Sound Speed
The speed of sound in air is influenced by temperature. As the temperature of the air increases, the molecules move more rapidly, allowing sound waves to travel faster. This is because the speed of sound is proportional to the square root of the temperature. In a standard environment, the speed of sound at 0掳C is about 331 m/s.
For each degree Celsius rise in temperature, the speed of sound increases by approximately 0.61 m/s. This relationship is especially important in scientific calculations and practical applications, such as in music or meteorology.
In our exercise, with the air temperature at 25掳C, we calculate the speed of sound as follows:
  • Start with the speed at 0掳C: 331 m/s,
  • Calculate the increase: 0.61 m/s for each degree.
Thus, the speed of sound at 25掳C is: \[v = 331 \, \text{m/s} + (0.61 \, \text{m/s} \times 25) = 346 \, \text{m/s}\]Understanding this concept helps explain why sound travels faster in warm environments and impacts how we hear sounds in different climate conditions.

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Most popular questions from this chapter

Three seconds after a gun is fired, the person who fired the gun hears an echo. How far away was the surface that reflected the sound of the shot? Use \(340 \mathrm{~m} / \mathrm{s}\) for the speed of sound.

An organ pipe is tuned to emit a frequency of \(196.00 \mathrm{~Hz}\). When it and the G string of a violin are sounded together, ten beats are heard in a time of exactly \(8 \mathrm{~s}\). The beats become slower as the violin string is slowly tightened. What was the original frequency of the violin string?

An organ pipe whose length is held constant resonates at a frequency of \(224.0 \mathrm{~Hz}\) when the air temperature is \(15^{\circ} \mathrm{C}\). What will be its resonant frequency when the air temperature is \(24^{\circ} \mathrm{C}\) ? The resonant wavelength must have the same value at each temperature because it depends only on the length of the pipe. (Its nodes and antinodes must fit properly within the pipe.) But \(\lambda=v / f\), and so \(v / f\) must be the same at the two temperatures. Consequently, $$ \frac{v_{1}}{224 \mathrm{~Hz}}=\frac{v_{2}}{f_{2}} \quad \text { or } \quad f_{2}=(224 \mathrm{~Hz})\left(\frac{v_{2}}{v_{1}}\right) $$ At temperatures near room temperature, \(v=\left(331+0.61 T_{c}\right) \mathrm{m} / \mathrm{s}\), where \(T_{c}\) is the Celsius temperature. Then $$ f_{2}=(224.0 \mathrm{~Hz})\left[\frac{331+(0.61)(24)}{331+(0.61)(15)}\right]=0.228 \mathrm{kHz} $$

A tuning fork having a frequency of \(400 \mathrm{~Hz}\) (shown in Fig. 23-2) is moved away from an observer and toward a flat wall with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the apparent frequency \((a)\) of the unreflected sound waves coming directly to the observer, and \((b)\) of the sound waves coming to the observer after reflection? ( \(c\) ) How many beats per second are heard? Assume the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\) (a) The fork, the source, is receding from the observer in the positive direction and so we use \(+v_{s} .\) It doesn't matter what the sign associated with \(v_{o}\) is since \(v_{e}=0\). $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(400 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+0}{340 \mathrm{~m} / \mathrm{s}+2.0 \mathrm{~m} / \mathrm{s}}=397.7 \mathrm{~Hz}=398 \mathrm{~Hz} $$ The source is moving away from the observer and the frequency is properly shifted down from \(400 \mathrm{~Hz}\) to \(398 \mathrm{~Hz}\). (b) Think of the wall as a source that reflects sound of the same frequency as that which impinges upon it. The wave crests reaching the wall are closer together than normally because the fork is moving toward the wall. Therefore, the wall will appear as a stationary source emitting sound of a higher frequency than \(400 \mathrm{~Hz}\). due to the \(2.0-\mathrm{m} / \mathrm{s}\) motion of the fork. Alternatively we can think of the reflected wave as if it came from a source (the wall) moving at \(2.0 \mathrm{~m} / \mathrm{s}\) toward the observer. Hence, we enter \(-v_{x}:\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(400 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+0}{340 \mathrm{~m} / \mathrm{s}-2.0 \mathrm{~m} / \mathrm{s}}=402.4 \mathrm{~Hz}=402 \mathrm{~Hz} $$ and the frequency is properly shifted up. (c) Beats per second \(=\) Difference between frequencies \(=(402.4-397.7) \mathrm{Hz}=4.7\) beats per second

A locomotive moving at \(30.0 \mathrm{~m} / \mathrm{s}\) approaches and passes a person standing beside the track. Its whistle is emitting a note of frequency \(2.00 \mathrm{kHz}\). What frequency will the person hear \((a)\) as the train approaches and (b) as it recedes? The speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).
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