1 / 9

ONE Sound Projection & Absorption

ONE Sound Projection & Absorption

morna
Télécharger la présentation

ONE Sound Projection & Absorption

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. ONE Sound Projection & Absorption A man hears the sound of freeway noise 40’ away, at an intensity loudness of 70 decibels. The freeway is 90 feet from the exterior wall of a certain room. The only way sound from the freeway can be heard inside the room is through the exterior wall of the room. The wall is 12’ x 20’ and is made of materials that give it an NRC of 32. The room inside the wall is 20’ x 30’ x 12,’ and STC values for the room surfaces are; floor = .46, walls = .25, and ceiling = .76. Ignore doors and windows. Find the level of sound in the room from the freeway noise. 1 find the IE at the man: it involves loudness and energy. IL = 10 log [ IE/10-16] 70 = 10 log [ IE/10-16] ; 7 = log [ IE/10-16] 107 = IE/10-16 ; IE = 107 x 10-16 ; IE = 10-9

  2. 2 find IE outside at the wall: IE1 / IE2 = [ d1 / d2 ]2 10-9 / IE2 = [90 / 40]2 = 5.0625 ; IE2 = 10-9/ 5.0625 =.1975 x 10-9 the energy at the wall 3 Find the IL at the wall; IL = 10 log [.1975 x 10-9/ 10-16] IL = 10 log (.1975 x 10-9 x 1016] = 10 log [.1975 x 107] IL = 10 (7 - .7044) = 10 ( 6.2956) = 62.96 decibels 4 Find absorption of the room ; A = (600x.46) + (600x.76) + (1200x.25) = 276 + 456 + 300 A = 1032 5 NR = TL + 10 log [AR / S] = 32 + 10 log [1032 / 240] = NR = 32 + 10 log 4.3 = 32 + 10 x .635 = 32 + 6.35 = 38.35 db Noise in the room = 62.96 – 38.35 = 24.61 decibels

  3. TWO Loudness from multiple sounds On a dark and foggy night, a an stranded on a tiny desert island hears a ship’s horn at an intensity loudness of 70 decibels. Purely by coincidence, another ship on the opposite side of the island sounds it’s horn, which the man on the island also hears at 70 decibels. Then, at the same time, both ships sound their horns. At the location of the man on the island, find A IE of the first horn. B IE of both horns at the same time. C IL of one horn. D IL of both horns at the same time. IE of first horn:IL = 10 log [ IE/10-16] ; 70 = 10 log [IE/10-16] 7 = log [ IE/10-16] ; 107 = IE/10-16 ; IE = 107 x 10-16 IE of first horn = 10-9 watts / cm2

  4. IE of both horns at same time: 2 x IE of first horn IEboth = 2 x 109 IL of one horn: IL = 10 log [ 10-9/10-16] ; IL = 10log[ 107] IL = 10 x 7 = 70 but remember, that was given in the problem . . . IL of both horns: IL = 10 log [ 2 x 10-9 / 10-16 ] IL = 10 log [ 2 x 10-9 x 1016 ] = 10 log [ 2 x 107 ] IL = 10 ( 7 + .301) = 10 (7.301) = 73.01 decibels

  5. Review of the formulas: Loudness and Energy • 1 IL = 10 log IE / 10-16 Distance and Energy • 2 IE1 / IE2 = [ d2 / d1 ]2 Absorption of Energy on Surfaces • 3 NR = 10 log [ a2 / a1 ] Remember here sound involves reverberation Time in seconds = .05 x [volume of space / total absorption of surfaces within a space] Isolation of Energy by a Barrier • 4 NR = TL+10 log ( a R / S )

  6. In answer to some of the questions that have been asked: 1 I do not grade on a curve. I will not base your grade on what someone else has earned. The objective of any course is to present a scenario where any student has the opportunity to learn a material – not merely obtain a grade. 2 Nothing that occurs in another course, present, future, or past, will have any influence on how this course is graded. A grade is based on the average of 5 tests. But a penalty may be incurred on a calculated grade, simply because the University requires students to attend class – and has a specific limit to absence from class. 3 A test cannot be scheduled during what the university calls “period of no examinations,” which is the 5 class days that precede the period of final examinations. That is a specific University Rule.

  7. Final Examination Policies • In accordance with OP 34.10, class-related activities, with the exception of office hours, are prohibited on designated individual study days and during the final examination period. These dates are set aside for students to prepare for and take scheduled final examinations. During this period, review sessions are not to be scheduled, quizzes are not to be given, and no other class-related activities can be scheduled. • No substantial examinations other than bona fide makeup examinations may be given during the last class week or during the individual study day.  Courses in which lab examinations and design studio reviews are normally scheduled the week prior to finals are excluded from this policy. No extracurricular activities of any kind may be scheduled within the individual study day and the final examination period without written permission of the Provost's Office.

  8. 5 The University makes the schedule for final examinations, and requires that examinations be given at the assigned times on the specific days. Your test number five is scheduled for this room, Tuesday, May 5, at 8:30 AM. The period actually is scheduled to begin at 7:30 AM, but since this is not a comprehensive test over the course material, you may come at 8:30 AM. However, we must give up the room at 10:00 AM. For those who wish to have more time, I will be present at 7:30 AM. As for the possibility of changing the date and time, there simply is not another room available that can accommodate 110 or more students at another time.

More Related