1 / 46

March 25, 2014 Musical Instruments and Geometric Optics

March 25, 2014 Musical Instruments and Geometric Optics. This is how the picture was taken. Two plane mirrors Were facing each other with The candle in between. Announcements & Reminders. To the teacher: Turn on the recording! To students: About equipment needed for L17. L149, Problem 1.

satin
Télécharger la présentation

March 25, 2014 Musical Instruments and Geometric Optics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. March 25, 2014Musical InstrumentsandGeometric Optics

  2. This is how the picture was taken. Two plane mirrors Were facing each other with The candle in between.

  3. Announcements & Reminders • To the teacher: Turn on the recording! • To students: • About equipment needed for L17

  4. L149, Problem 1 B = 2683 Hz f = B/2p = 427.0 Hz Frequency printed on tuning fork = 426. 6 Hz

  5. L149, Problem 2 n1 = 256 Hz n2 = 2x256 Hz n6 = 6x256 Hz B = 1608 Hz f = B/2p = 255.9 Hz Frequency printed on tuning fork = 256 Hz

  6. L149, Problem 3 B = 9498 Hz f = B/2p = 1512 Hz • vsound = 344 m/s at 21 oC • = v/f = 0.23 m The wavelength is the length of the pipe. This requires the waveform shown to the right. A N A N A n = 2

  7. L149, Problem 4 B = 1.502E4 Hz fn= B/2p = 2391 Hz n = 2391 / 756 Hz = 3.16 This must be the 3rd harmonic. From the last problem, f2 = 1512 Hz. Thus, f1 = f2/2 = 756 Hz. n = 6

  8. L149, Problem 5 B = 2682 Hz fn= B/2p = 427 Hz The frequency is 0.56f1. This is approximately an octave lower and may be produced by closing the end of the pipe as seen in P211. From the last problem, f1 = 756 Hz. Why isn’t the frequency exactly an octave lower than the fundamental of the open pipe? The effective lengths of the open and closed pipes may be different. For a closed pipe with a fundamental of 427 Hz, the wavelength is l = (344 m/s)/(427 Hz) = 0.806 m. For the fundamental of a closed pipe Lc = l/4. In this case, that gives Lc = 0.203 m. A smaller length would make sense, since closing the end prevents the waveform from extending beyond the end of the pipe.

  9. Harmonics of a Longer Toy Flute

  10. f1 = 546 Hz

  11. f2 = 1087 Hz (2f1 = 1092 Hz)

  12. n = 2 n = 4 n = 6 n = 8

  13. Waveforms and Frequency Spectra of Various Instruments

  14. waveform by Justin Luo

  15. waveform by Ashwin Monian

  16. waveform by Hunter Denham

  17. waveform by Jennie Cunningham

  18. waveform by Hayden Rudd

  19. waveform by Jesse Sykes

  20. waveform by Will King

  21. waveform by Will King

  22. waveform by Charles Zhao

  23. waveform by Arjun Adhia

  24. waveform by Drew Smith

  25. waveform by Trey Faddis

  26. waveform by Param Sidhu

  27. waveform by Hunter Denham

  28. waveform by Kayla Howes

  29. waveform by Michael Schroeder

  30. waveform by Jesse Sykes

  31. Mirror Diagnostic Questions

  32. The mirror is grayed out so that you can’t see whether it’s plane, convex, or concave. The object is the black, upright arrow, and the image is the gray, inverted arrow. Is the mirror plane or curved?

  33. The mirror is grayed out so that you can’t see whether it’s plane, convex, or concave. The object is the black, upright arrow, and the image is the gray, inverted arrow. The mirror is curved. (The object and image are different sizes.) Is the image real or virtual?

  34. The mirror is grayed out so that you can’t see whether it’s plane, convex, or concave. The object is the black, upright arrow, and the image is the gray, inverted arrow. The mirror is curved. (The object and image are different sizes.) The image is real. (Light rays actually converge at the position of the image.) Is the mirror convex or concave?

  35. The mirror is grayed out so that you can’t see whether it’s plane, convex, or concave. The object is the black, upright arrow, and the image is the gray, inverted arrow. The mirror is curved. The image is real. The mirror is concave. (A convex mirror won’t produce a real image of a real object.)

  36. Everything is revealed here. o = object distance i = image distance f = focal length M = magnification

  37. Where should the object be placed so that the image is inverted and the same size as the object?

  38. Where should the object be placed so that the image is inverted and the same size as the object? The object must be placed at the center of curvature of the mirror. This is twice the focal length.

  39. Is the image real or virtual? What kind of mirror is it?

  40. The image is virtual. (It’s behind the mirror.) The mirror is concave. (The image is larger than the object.)

  41. A convex mirror produces virtual images smaller than the object. Hence the phrase, “Objects in the mirror are closer than they appear.”

  42. Is the image real or virtual? What kind of mirror is it?

  43. The image is virtual. The mirror is plane.

  44. Is the image real or virtual? What is the image distance? What is the magnification?

  45. The image is real. The image distance is found using 1/i = 1/f – 1/o. The object distance and focal length are read from the scale. o = 12.0 cm and f = 10.0 cm. This gives i = 60.0 cm. The magnification is –i/o = -5.00. This also equals the ratio of image height to object height. The negative sign indicates the inversion of the image.

  46. The image is real. Compare I = 60.0 cm found in the previous slide to 59.9 cm above. Compare M = -5.00 found in the previous slide to -4.99 above.

More Related