Chapter 22A – Sound Waves A PowerPoint Presentation by

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Presentation transcript:

Chapter 22A – Sound Waves A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

Objectives: After completion of this module, you should be able to: Define sound and solve problems relating to its velocity in solids, liquids, and gases. Use boundary conditions to apply concepts relating to frequencies in open and closed pipes.

Source of sound: a tuning fork. Definition of Sound Sound is a longitudinal mechanical wave that travels through an elastic medium. Source of sound: a tuning fork. Many things vibrate in air, producing a sound wave.

Is there sound in the forest when a tree falls? Sound is a physical disturbance in an elastic medium. Based on our definition, there IS sound in the forest, whether a human is there to hear it or not! The elastic medium (air) is required!

Sound Requires a Medium The sound of a ringing bell diminishes as air leaves the jar. No sound exists without air molecules. Evacuated Bell Jar Batteries Vacuum pump

Sound as a pressure wave Graphing a Sound Wave. Sound as a pressure wave The sinusoidal variation of pressure with distance is a useful way to represent a sound wave graphically. Note the wavelengths l defined by the figure.

Factors That Determine the Speed of Sound. Longitudinal mechanical waves (sound) have a wave speed dependent on elasticity factors and density factors. Consider the following examples: A denser medium has greater inertia resulting in lower wave speeds. steel water A medium that is more elastic springs back quicker and results in faster speeds.

Speeds for different media Young’s modulus, Y Metal density, r Metal rod Bulk modulus, B Shear modulus, S Density, r Extended Solid Fluid Bulk modulus, B Fluid density, r

Example 1: Find the speed of sound in a steel rod. r = 7800 kg/m3 Y = 2.07 x 1011 Pa vs = ? v =5150 m/s

Speed of Sound in Air For the speed of sound in air, we find that:  = 1.4 for air R = 8.34 J/kg mol M = 29 kg/mol Note: Sound velocity increases with temperature T.

Example 2: What is the speed of sound in air when the temperature is 200C? Given: g = 1.4; R = 8.314 J/mol K; M = 29 g/mol T = 200 + 2730 = 293 K M = 29 x 10-3 kg/mol v = 343 m/s

Dependence on Temperature Note: v depends on Absolute T: Now v at 273 K is 331 m/s. g, R, M do not change, so a simpler formula might be: Alternatively, there is the approximation using 0C:

Solution 2: v = 331 m/s + (0.6)(270C); Example 3: What is the velocity of sound in air on a day when the temp-erature is equal to 270C? Solution 1: T = 270 + 2730 = 300 K; v = 347 m/s Solution 2: v = 331 m/s + (0.6)(270C); v = 347 m/s

Musical Instruments Sound waves in air are produced by the vibrations of a violin string. Characteristic frequencies are based on the length, mass, and tension of the wire.

Vibrating Air Columns Just as for a vibrating string, there are characteristic wavelengths and frequencies for longitudinal sound waves. Boundary conditions apply for pipes: Open pipe A The open end of a pipe must be a displacement antinode A. Closed pipe A N The closed end of a pipe must be a displacement node N.

Velocity and Wave Frequency. The period T is the time to move a distance of one wavelength. Therefore, the wave speed is: The frequency f is in s-1 or hertz (Hz). The velocity of any wave is the product of the frequency and the wavelength:

Possible Waves for Open Pipe Fundamental, n = 1 1st Overtone, n = 2 2nd Overtone, n = 3 3rd Overtone, n = 4 All harmonics are possible for open pipes:

Characteristic Frequencies for an Open Pipe. L Fundamental, n = 1 1st Overtone, n = 2 2nd Overtone, n = 3 3rd Overtone, n = 4 All harmonics are possible for open pipes:

Possible Waves for Closed Pipe. Fundamental, n = 1 1st Overtone, n = 3 2nd Overtone, n = 5 3rd Overtone, n = 7 Only the odd harmonics are allowed:

Possible Waves for Closed Pipe. Fundamental, n = 1 1st Overtone, n = 3 2nd Overtone, n = 5 3rd Overtone, n = 7 L Only the odd harmonics are allowed:

The second overtone occurs when n = 5: Example 4. What length of closed pipe is needed to resonate with a fundamental frequency of 256 Hz? What is the second overtone? Assume that the velocity of sound is 340 M/s. Closed pipe A N L = ? L = 33.2 cm The second overtone occurs when n = 5: f5 = 5f1 = 5(256 Hz) 2nd Ovt. = 1280 Hz

Summary of Formulas For Speed of Sound Solid rod Extended Solid Liquid Sound for any gas: Approximation Sound in Air:

Summary of Formulas (Cont.) For any wave: Characteristic frequencies for open and closed pipes: OPEN PIPE CLOSED PIPE

CONCLUSION: Chapter 22 Sound Waves