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A \(3.00-\mathrm{kHz}\) tone is being produced by a speaker with a diameter of \(0.175 \mathrm{m} .\) The air temperature changes from 0 to \(29^{\circ} \mathrm{C} .\) Assuming air to be an ideal gas, find the change in the diffraction angle \(\theta\).

Short Answer

Expert verified
The change in diffraction angle is \(2.5^{\circ}\).

Step by step solution

01

- Understanding the Problem

We need to calculate the change in the diffraction angle \( \theta \) due to a temperature change. The frequency of the sound is given as \( 3.00 \mathrm{kHz} \) and the speaker's diameter as \( 0.175 \mathrm{m} \). The temperature changes from \( 0^{\circ}C \) to \( 29^{\circ}C \).
02

- Using the Speed of Sound Formula

The speed of sound \( v \) in air varies with temperature and is given by \( v = 331.4 + 0.6 \times T \) m/s, where \( T \) is the air temperature in degrees Celsius.
03

- Calculate Speed at 0°C

At \( 0^{\circ}C \), the speed of sound \( v_1 \) is \( v_1 = 331.4 + 0.6 \times 0 = 331.4 \) m/s.
04

- Calculate Speed at 29°C

At \( 29^{\circ}C \), \( v_2 = 331.4 + 0.6 \times 29 = 348.8 \) m/s.
05

- Calculating Wavelengths

The wavelength at each temperature is calculated using \( \lambda = \frac{v}{f} \), where \( f = 3000 \) Hz. Thus, \( \lambda_1 = \frac{331.4}{3000} \) and \( \lambda_2 = \frac{348.8}{3000} \).
06

- Determine Wavelengths

\( \lambda_1 = 0.1105 \) m and \( \lambda_2 = 0.1163 \) m.
07

- Diffraction Angle Formula

The diffraction angle \( \theta \) is given by \( \theta = \sin^{-1}\left(\frac{m \cdot \lambda}{D}\right) \), where \( m \) is the order of the minimum (usually 1 for the first minimum), \( \lambda \) is the wavelength, and \( D \) is the diameter of the speaker. For simplicity, we'll consider the first minimum \( m=1 \).
08

- Calculate \(\theta_1\) at 0°C

\( \theta_1 = \sin^{-1}\left(\frac{1 \cdot 0.1105}{0.175}\right) = \sin^{-1}(0.6314) \approx 39.2^{\circ} \).
09

- Calculate \(\theta_2\) at 29°C

\( \theta_2 = \sin^{-1}\left(\frac{1 \cdot 0.1163}{0.175}\right) = \sin^{-1}(0.6646) \approx 41.7^{\circ} \).
10

- Calculate Change in Diffraction Angle

The change in diffraction angle \( \Delta \theta \) is \( \theta_2 - \theta_1 = 41.7^{\circ} - 39.2^{\circ} = 2.5^{\circ} \).

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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 behaves in different environments. Essentially, sound is a wave that travels through a medium, typically air. Its speed depends significantly on several factors. One of the most influential factors is the temperature of the air. The speed of sound in air can be calculated using the formula:

\[ v = 331.4 + 0.6T \]

where \( v \) is the speed of sound in meters per second and \( T \) is the temperature in degrees Celsius.

In general, as the temperature increases, the speed of sound also increases. This is because warmer air results in faster-moving molecules, which helps sound waves propagate quicker through the medium.
  • At \( 0^{\circ}C \), the speed is \( 331.4 \) m/s.
  • At \( 29^{\circ}C \), it increases to \( 348.8 \) m/s.
These changes in speed due to temperature fluctuations are important when analyzing sound behavior through mediums.
Wavelength Calculation
Wavelength is another essential factor in understanding sound. Wavelength, represented by \( \lambda \), is the distance between successive crests of a wave. It can be calculated if you know the speed of sound and the frequency of the sound wave using the formula:

\[ \lambda = \frac{v}{f} \]

where \( v \) is the speed of sound and \( f \) is the frequency. In this particular exercise, our sound has a frequency of 3000 Hz.
  • At \( 0^{\circ}C \), when the speed of sound is \( 331.4 \) m/s, the wavelength \( \lambda_1 \) is \( 0.1105 \) m.
  • At \( 29^{\circ}C \), the speed is \( 348.8 \) m/s, and the wavelength \( \lambda_2 \) becomes \( 0.1163 \) m.
As you can see, temperature changes affect wavelength by altering the speed at which sound travels. The frequency remains constant, but because the speed changes, so does wavelength.
Temperature Effect on Sound
Temperature has a profound effect on sound propagation, as evident in both speed and wavelength changes. When air temperature rises, sound travels faster due to increased molecular energy and motion. This increase in speed, alongside a static frequency, results in a longer wavelength. In practical scenarios, such as when designing loudspeakers or understanding acoustics in various environments, these temperature-related changes must be considered.

The diffraction angle \( \theta \) of sound around obstacles also changes with temperature. The relationship used is:

\[ \theta = \sin^{-1}\left(\frac{m \cdot \lambda}{D}\right) \]

where \( m \) is the diffraction order, \( \lambda \) is the wavelength, and \( D \) is the obstacle diameter. As temperature increases, the wavelength increases, causing a change in the diffraction angle.
  • For example, when the air temperature increases from \( 0^{\circ}C \) to \( 29^{\circ}C \), the diffraction angle changes from about \( 39.2^{\circ} \) to \( 41.7^{\circ} \).
  • Such a change is vital for understanding sound behavior in varying conditions.
Overall, temperature is a key factor that affects how sound waves interact with their environment, impacting design and application in technology and daily life.

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Most popular questions from this chapter

One method for measuring the speed of sound uses standing waves. A cylindrical tube is open at both ends, and one end admits sound from a tuning fork. A movable plunger is inserted into the other end at a distance \(L\) from the end of the tube where the tuning fork is. For a fixed frequency, the plunger is moved until the smallest value of \(L\) is measured that allows a standing wave to be formed. Suppose that the tuning fork produces a \(485-\mathrm{Hz}\) tone, and that the smallest value observed for \(L\) is \(0.264 \mathrm{m} .\) What is the speed of sound in the gas in the tube?

A string has a linear density of \(8.5 \times 10^{-3} \mathrm{kg} / \mathrm{m}\) and is under a tension of \(280 \mathrm{N}\). The string is \(1.8 \mathrm{m}\) long, is fixed at both ends, and is vibrating in the standing wave pattern shown in the drawing. Determine the (a) speed, (b) wavelength, and (c) frequency of the traveling waves that make up the standing wave.

An organ pipe is open at both ends. It is producing sound at its third harmonic, the frequency of which is \(262 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the length of the pipe?

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube open at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about \(2.9 \mathrm{cm} .\) The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the fundamental frequency of the canal? (Interestingly, the fundamental frequency is in the frequency range where human hearing is most sensitive.)

The A string on a string bass vibrates at a fundamental frequency of \(55.0 \mathrm{Hz}\). If the string's tension were increased by a factor of four, what would be the new fundamental frequency?

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