17.8

Concepts & Calculations

Diffraction is the bending of a traveling wave around an obstacle or around the edges of an opening and is one of the consequences of the principle of linear superposition. As Equations 17.1 and 17.2 indicate, the extent of diffraction when a sound wave passes through an opening depends on the ratio of the wavelength of the sound to the width or diameter of the opening. Example 8 compares diffraction in two different media and reviews some of the fundamental properties of sound waves.

| Concepts & Calculations Example 8 | | | Diffraction in Two Different Media

A sound wave with a frequency of 15 kHz emerges through a circular opening that has a diameter of 0.20 m. Find the diffraction angle when the sound travels in air at a speed of 343 m/s and in water at a speed of 1482 m/s. Concept Questions and Answers The diffraction angle for a circular opening is given by Equation 17.2 as , where is the wavelength of the sound and D is the diameter of the opening. How is the wavelength related to the frequency of the sound?

Answer According to Equation 16.1, the wavelength is given by , where v is the speed of sound and f is the frequency.

Is the wavelength of the sound in air greater than, smaller than, or equal to the wavelength in water?

Answer According to Equation 16.1, the wavelength is proportional to the speed v for a given value of the frequency f. Since sound travels at a slower speed in air than in water, the wavelength in air is smaller than the wavelength in water.

Is the diffraction angle in air greater than, smaller than, or equal to the diffraction angle in water?

Answer The extent of diffraction is determined by , the ratio of the wavelength to the diameter of the opening. Smaller ratios lead to less diffraction or smaller diffraction angles. The wavelength in air is smaller than in water, and the diameter of the opening is the same in both cases. Therefore, the ratio is smaller in air than in water, and the diffraction angle in air is smaller than the diffraction angle in water.

Solution Using (Equation 17.2) and (Equation 16.1), we have

Applying this result for air and for water, we find

As expected, the diffraction angle in air is smaller.
The next example deals with standing waves of sound in a gas. One of the factors that affect the formation of standing waves is the speed at which the individual waves travel. This example reviews how the speed of sound depends on the properties of the gas.

| Concepts & Calculations Example 9 | | | Standing Waves of Sound

Two tubes of gas are identical and are open at only one end. One tube contains neon (Ne) and the other krypton (Kr). Both are monatomic gases, have the same temperature, and may be assumed to be ideal gases. The fundamental frequency of the tube containing neon is 481Hz. What is the fundamental frequency of the tube containing krypton?
Concept Questions and Answers For a gas-filled tube open at only one end, the fundamental frequency (n = 1) is given by Equation 17.5 as f1 = v/(4L), where v is the speed of sound and L is the length of the tube. How is the speed related to the properties of the gas?

Answer According to Equation 16.5, the speed is given by , where is the ratio of the specific heat capacities at constant pressure and constant volume, k is Boltzmann’s constant, T is the Kelvin temperature, and m is the mass of an atom of the gas.

All of the factors that affect the speed of sound are the same except the atomic mass. The periodic table located on the inside of the back cover gives the atomic masses of neon and krypton as 20.180 u and 83.80 u, respectively. Is the speed of sound in krypton greater than, smaller than, or equal to the speed of sound in neon?

Answer According to Equation 16.5 the speed of sound is , so the speed is inversely proportional to the square root of the mass m of an atom. Thus, the speed is smaller when the mass is greater. Since krypton has the greater mass, the speed of sound in krypton is smaller than in neon.

Is the fundamental frequency of the tube containing krypton greater than, smaller than, or equal to that of the tube containing neon?

Answer The fundamental frequency is given by Equation 17.5 as . Since the speed of sound in krypton is smaller than in neon, the fundamental frequency of the krypton-filled tube is smaller than that of the neon-filled tube.

Solution Using (Equation 17.5) and (Equation 16.5), we have

Applying this result to both tubes and taking the ratio of the frequencies, we obtain

Solving for f1, Kr gives

As expected, the fundamental frequency for the krypton-filled tube is smaller than that for the neon-filled tube.

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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