Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Resonance tube - By Aditya Abeysinghe by Aditya Abeysinghe... 1075 views
- Session 9 fossil energy part ii by Albania Energy As... 270 views
- Albpetrol status update in the era ... by Albania Energy As... 710 views
- Lecture 21 applications of moving... by Albania Energy As... 760 views
- Session 4 cycles and combustion by Albania Energy As... 1132 views
- The fiscal regime in Albania for up... by Albania Energy As... 883 views

3,182 views

Published on

Lecture 08 standing sound waves. resonance.

No Downloads

Total views

3,182

On SlideShare

0

From Embeds

0

Number of Embeds

3

Shares

0

Downloads

50

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Lecture 8 Standing sound waves. Resonance.
- 2. Interference with sound Superposition works exactly as it did for transversal waves. Additional complication: 3D waves! (next lecture)
- 3. Reflection of sound waves against a surface Consider a sound pulse (air moves to the right and back to initial position) traveling along a pipe toward a closed end: No displacement: s=0 A closed end is a “fixed end” Wave must be inverted (s becomes –s) Reflected pulse v Incoming pulse v s s s s v x x v
- 4. Reflection of sound at an open end v Pulse travels out into open air s Oscillation back from a larger slice moves more air into pipe… v …and increases pressure… v s v …and causes a wave to propagate back in (a reflection!) (and another wave is transmitted outside) Beyond the open end of the pipe, variations in the pressure must be much smaller than pressure variations (gauge pressure) in pipe. Just beyond the open end, p ≈0
- 5. Boundary conditions for sound Open end • gauge pressure = 0 • maximum (absolute) air displacement Closed end • air displacement = 0 • maximum (absolute) gauge pressure
- 6. Standing sound waves in pipe open at both ends A harmonic wave and its reflection on an open end: s1 (x ,t ) = smax sin(kx − ωt ) s2 (x ,t ) = smax sin(kx + ωt ) sall (x ,t ) = smax sin(kx − ωt ) + sin(kx + ωt ) a +b sin ( a ) + sin ( b ) = 2cos 2 a −b ÷sin ÷ 2 sall (x ,t ) = 2smax sin ( kx ) cos ( ωt ) At the openings: p~0 Large displacements Standing wave within pipe: does not travel, bounces back and forth. Amplitude will decrease as energy is transported out of the pipe
- 7. ACT: Pipe open at both ends This is the air displacement for a standing wave inside this tube. Sketch the gauge pressure vs position for this wave. Compare with your neighbor and discuss. p x p = 0 at open ends Maximum/minimum p at node
- 8. Higher harmonics DEMO: Organ pipes Each harmonic is a standing wave. λ 2 n = 1,2,3... 2L n n = 1,2,3... L =n λn = λ gets shorter, frequency increases 2 Visualize them: http://www.walter-fendt.de/ph11e/stlwaves.htm
- 9. ACT: Pipe closed at one end Open end: Max s, p = 0 Closed end: s = 0, max p s First harmonic or fundamental frequency: λ1 = 4L What is the standing wave for the next harmonic? A s = 0 at an open end? (No!) B And s max at a closed end? (No!) C λ2 = 4L 3 In general, λ L = nodd nodd = 1,3... 4
- 10. In-class example A tube with both ends open has a fundamental frequency f. What is the fundamental frequency of the same tube if one end is closed? A. B. C. D. E. 4f 2f f f/4 None of the above Close end = node Open end = antinode λ = 2L f′= λ ′ = 4L = 2λ v v f = = λ ′ 2λ 2
- 11. A little music DEMO: Xylophone When you blow air into a pipe, all the harmonics are present. Example: Blow into a tube of length 19.2 cm open at one end λ1 = 2L f1 = v 343 m/s = = 890 Hz 2L 2 ( 0.192 m ) Approx. A5 (La)
- 12. Resonance DEMO: Resonant slabs To produce a wave, we need to apply an external force (driving force). This driving force can be periodic with frequency fD. The amplitude of the perturbation is maximum when the frequency of the driving force is equal to one of the natural (or harmonic, or normal) frequencies of the system. Examples: Pendulum: resonance occurs when fD = 2π (A pendulum has only one normal frequency) g L String fixed at both ends: when fD = fn = F n µ 2L Pipe closed at one end: when fD = fn = v sound nodd 4L (L = length of string) for n = 1,2,... for nodd = 1,3,... (see lecture 6)

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment