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

Transcription

1 Chapter 22A Sound Waes A PowerPoint Presentation by Paul E. Tippens, Proessor o Physics Southern Polytechnic State Uniersity 2007

2 Objecties: Ater completion o this module, you should be able to: Deine sound and sole problems relating to its elocity in solids, liquids, and gases. Use boundary conditions to apply concepts relating to requencies in open and closed pipes.

3 Deinition o Sound Sound is is a longitudinal mechanical wae that traels through an elastic medium. Many things ibrate in air, producing a sound wae. Source o sound: a tuning ork.

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

5 Sound Requires a Medium The sound o a ringing bell diminishes as air leaes the jar. No sound exists without air molecules. Batteries Vacuum pump Eacuated Bell Jar

6 Graphing a Sound Wae. Sound as a pressure wae The sinusoidal ariation o pressure with distance is a useul way to represent a sound wae graphically. Note the waelengths deined by the igure.

7 Factors That Determine the Speed o Sound. Longitudinal mechanical waes (sound( sound) ) hae a wae speed dependent on elasticity actors and density actors. Consider the ollowing examples: A denser medium has greater inertia resulting in lower wae speeds. A medium that is more elastic springs back quicker and results in aster speeds. steel water

8 Speeds or dierent media Y Metal rod Young s s modulus, Y Metal density, B 4 3 S Extended Solid Bulk modulus, B Shear modulus, S Density, B Fluid Bulk modulus, B Fluid density,

9 Example 1: Find the speed o sound in a steel rod. = 7800 kg/m 3 Y = 2.07 x Pa s =? Y x 10 Pa =5150 m/s 7800 kg/m 3

10 Speed o Sound in Air For the speed o sound in air, we ind that: B P and P RT M = 1.4 or air R = 8.34 J/kg mol M = 29 kg/mol B P RT M Note: Sound elocity increases with temperature T.

11 Example 2: What is the speed o sound in air when the temperature is 20 0 C? Gien: = 1.4; R = J/mol K; M = 29 g/mol T = = 293 K M = 29 x 10-3 kg/mol RT M (1.4)(8.314 J/mol K)(293 K) x 10 kg/mol = 343 m/s

12 Dependence on Temperature Note: depends on Absolute T: T RT M Now at 273 K is 331 m/s., R, M do not change, so a simpler ormula might be: T 331 m/s 273 K Alternatiely, there is the approximation using 0 C: m/s 331 m/s 0.6 tc C 0

13 Example 3: What is the elocity o sound in air on a day when the temp- erature is equal to 27 0 C? Solution 1: T 331 m/s 273 K T = = 300 K; = 347 m/s 331 m/s 300 K 273 K Solution 2: =331 m/s + (0.6)(27 0 C); = 347 m/s

14 Musical Instruments Sound waes in air are produced by the ibrations o a iolin string. Characteristic requencies are based on the length, mass, and tension o the wire.

15 Vibrating Air Columns Just as or a ibrating string, there are characteristic waelengths and requencies or longitudinal sound waes. Boundary conditions apply or pipes: The open end o a pipe must be a displacement antinode A. A Open pipe A The closed end o a pipe must be a displacement node N. N Closed pipe A

16 Velocity and Wae Frequency. The period T is the time to moe a distance o one waelength. Thereore, the wae speed is: 1 but T so T The requency is in s -1 or hertz (Hz). The elocity o any wae is the product o the requency and the waelength:

17 Possible Waes or Open Pipe Fundamental, n = 1 L 2L 1 1st Oertone, n = 2 2L 2 2nd Oertone, n = 3 2L 3 3rd Oertone, n = 4 2L 4 All harmonics are possible or open pipes: n 2L n n 1, 2, 3, 4...

18 Characteristic Frequencies or an Open Pipe. Fundamental, n = 1 1st Oertone, n = 2 2nd Oertone, n = 3 3rd Oertone, n = 4 L 1 2L 2 2L 3 2L 4 2L All harmonics are possible or open pipes: n n 2L n1, 2, 3, 4...

19 Possible Waes or Closed Pipe. Fundamental, n = 1 L 1 4L 1 1st Oertone, n = 3 1 4L 3 2nd Oertone, n = 5 1 4L 5 3rd Oertone, n = 7 1 4L 7 Only the odd harmonics are allowed: n 4L n n 1, 3, 5, 7...

20 Possible Waes or Closed Pipe. Fundamental, n = 1 L 1 1 4L 1st Oertone, n = L 2nd Oertone, n = L 3rd Oertone, n = L Only the odd harmonics are allowed: n n 4L n1, 3, 5, 7...

21 Example 4. What length o closed pipe is needed to resonate with a undamental requency o 256 Hz?? What is the second oertone?? Assume that the elocity o sound is 340 M/s. N Closed pipe L =? A n n 4L n 1, 3, 5, (1) 340 m/s ; L 4L 4 4(256 Hz) L = 33.2 cm 1 The second oertone occurs when n = 5: 5 = 5 1 = 5(256 Hz) 2nd Ot. = 1280 Hz

22 Summary o Formulas For Speed o Sound Solid rod Y Extended Solid 4 B 3 S Liquid B Sound or any gas: RT M Approximation Sound in Air: m/s 331 m/s 0.6 tc C 0

23 Summary o Formulas (Cont.) For any wae: Characteristic requencies or open and closed pipes: n OPEN PIPE CLOSED PIPE n n 1, 2, 3, L n 4 1, 3, 5, 7... n n L

24 CONCLUSION: Chapter 22 Sound Waes