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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 the diffraction angle is approximately 4.3°.

Step by step solution

01

Understanding the Relationship between Temperature and Sound Speed

The speed of sound in air varies with temperature. The formula for the speed of sound \( v \) in air at a temperature \( T \) in degrees Celsius is given by: \[ v = 331.4 + 0.6T \] where \( v \) is in meters per second. First, we'll use this formula to find the speed of sound at 0°C and 29°C.
02

Calculate Speed of Sound at 0°C

Using the formula \( v = 331.4 + 0.6T \), substitute \( T = 0 \): \[ v_{0} = 331.4 + 0.6 imes 0 = 331.4 \, \text{m/s} \]
03

Calculate Speed of Sound at 29°C

Again, use the formula \( v = 331.4 + 0.6T \) and substitute \( T = 29 \): \[ v_{29} = 331.4 + 0.6 imes 29 = 348.8 \, \text{m/s} \]
04

Understand Diffraction Angle Formula

The diffraction angle \( \theta \) for a circular opening is given by \( \sin\theta = \frac{1.22 \lambda}{d} \), where \( d \) is the diameter of the speaker and \( \lambda \) is the wavelength of the sound. The wavelength \( \lambda \) can be found using the relationship \( \lambda = \frac{v}{f} \) where \( f \) is the frequency.
05

Calculate Wavelength at 0°C

Using \( \lambda = \frac{v}{f} \) and substituting \( v_0 = 331.4 \, \text{m/s} \) and \( f = 3000 \, \text{Hz} \): \[ \lambda_{0} = \frac{331.4}{3000} = 0.1105 \, \text{m} \]
06

Calculate Wavelength at 29°C

Similarly, find \( \lambda \) for \( v_{29} = 348.8 \, \text{m/s} \): \[ \lambda_{29} = \frac{348.8}{3000} = 0.1163 \, \text{m} \]
07

Calculate Diffraction Angle at 0°C

Using \( \sin\theta = \frac{1.22 \lambda}{d} \) with \( \lambda_0 = 0.1105 \, \text{m} \) and \( d = 0.175 \, \text{m} \): \[ \sin\theta_{0} = \frac{1.22 \times 0.1105}{0.175} = 0.7703 \] \( \theta_{0} = \arcsin(0.7703) \approx 50.3^\circ \)
08

Calculate Diffraction Angle at 29°C

Now use \( \lambda_{29} = 0.1163 \, \text{m} \): \[ \sin\theta_{29} = \frac{1.22 \times 0.1163}{0.175} = 0.8105 \] \( \theta_{29} = \arcsin(0.8105) \approx 54.6^\circ \)
09

Find the Change in Diffraction Angle

Finally, calculate the change in the diffraction angle \( \Delta \theta \): \[ \Delta \theta = \theta_{29} - \theta_{0} = 54.6^\circ - 50.3^\circ = 4.3^\circ \]

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Key Concepts

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

Speed of Sound and Temperature
Temperature has a direct effect on the speed of sound in air. As the air temperature increases, the speed of sound also increases. This relationship is modeled by the equation: \[ v = 331.4 + 0.6T \] where \( v \) represents the speed of sound in meters per second and \( T \) is the air temperature in degrees Celsius. This equation illustrates that for every degree Celsius the temperature rises, the speed of sound increases by 0.6 m/s.
To visualize this, consider air at two different temperatures: 0°C and 29°C. At 0°C, the speed of sound is 331.4 m/s, while at 29°C, it rises to 348.8 m/s. This increment is due to the kinetic energy increase in air molecules as temperature rises, allowing sound waves to travel faster.
Wavelength Calculation
The wavelength \( \lambda \) of a sound wave depends on the speed of sound and the frequency of the wave. It can be calculated using the formula: \[ \lambda = \frac{v}{f} \] where \( v \) is the speed of sound and \( f \) is the frequency.
For example, with a frequency of 3000 Hz:
  • At 0°C, use the speed \( v = 331.4 \, \text{m/s} \). So, \( \lambda_0 = \frac{331.4}{3000} = 0.1105 \, \text{m} \).
  • At 29°C, with \( v = 348.8 \, \text{m/s} \). Therefore, \( \lambda_{29} = \frac{348.8}{3000} = 0.1163 \, \text{m} \).
This shows that as the speed of sound increases with temperature, so does the wavelength, assuming the frequency remains the same. Longer wavelengths at higher temperatures imply that sound can diffract and spread more around obstacles.
Diffraction Angle Formula
The diffraction angle \( \theta \) describes how sound waves bend around obstacles. Its calculation for circular openings involves the formula: \[ \sin\theta = \frac{1.22 \lambda}{d} \] Here, \( \lambda \) is the wavelength of the sound wave, and \( d \) is the diameter of the speaker.
Using this formula, we can find the diffraction angle at different temperatures:
  • At 0°C, with \( \lambda_0 = 0.1105 \, \text{m} \) and \( d = 0.175 \, \text{m} \), \( \sin\theta_{0} = \frac{1.22 \times 0.1105}{0.175} \approx 0.7703 \) translating to \( \theta_0 \approx 50.3^\circ \).
  • At 29°C, \( \lambda_{29} = 0.1163 \, \text{m} \) results in \( \sin\theta_{29} = \frac{1.22 \times 0.1163}{0.175} \approx 0.8105 \) which implies \( \theta_{29} \approx 54.6^\circ \).
As air temperature increases, the diffraction angle increases too. This change allows sound waves to bend more prominently around the speaker edge.
Ideal Gas Assumption
The ideal gas assumption simplifies the study of gases, including air, under different conditions. Air is treated as an ideal gas to use the straightforward relationship between temperature, pressure, and volume found in the ideal gas law: \[ PV = nRT \] where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
This assumption helps in calculating the speed of sound, where higher temperatures generally lead to faster sound speeds as the gas molecules move quickly and transmit sound more efficiently. In our scenario, treating air as an ideal gas allows us to apply the formula for the speed of sound as a function of temperature. Even though air isn't perfectly ideal, this model provides good approximations in many everyday conditions.

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Most popular questions from this chapter

Multiple-Concept Example 3 reviews the concepts that are important in this problem. The entrance to a large lecture room consists of two side-by-side doors, one hinged on the left and the other hinged on the right. Each door is 0.700 m wide. Sound of frequency 607 Hz is coming through the entrance from within the room. The speed of sound is 343 m/s. What is the diffraction angle of the sound after it passes through the doorway when (a) one door is open and (b) both doors are open?

The approach to solving this problem is similar to that taken in Multiple- Concept Example 4. On a cello, the string with the largest linear density \(\left(1.56 \times 10^{-2} kg / m\right)\) is the C string. This string produces a fundamental frequency of 65.4 \(\mathrm{Hz}\) and has a length of 0.800 \(m\) between the two fixed ends. Find the tension in the string.

Suppose that the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends. The musical notes and corresponding fundamental frequencies of two of these strings are G (196.0 Hz) and E (659.3 Hz). The linear density of the E string is \(3.47 \times 10^{-4} kg/m\) What is the linear density of the G string?

Two out-of-tune flutes play the same note. One produces a tone that has a frequency of 262 Hz, while the other produces 266 Hz. When a tuning fork is sounded together with the 262-Hz tone, a beat frequency of 1 Hz is produced. When the same tuning fork is sounded together with the 266-Hz tone, a beat frequency of 3 Hz is produced. What is the frequency of the tuning fork?

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-Hz tone, and that the smallest value observed for \(L\) is 0.264 m. What is the speed of sound in the gas in the tube?

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