Copy of 5BL Pre-Lab 5 Submission Template - W23

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Apr 3, 2024

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5±!&' 124re-!&'BCEJKMb 5 ²ssignment 457utu 568hBCEh, 2/28/23, °17 <URNK] SPVS]VSkjURNKLI ALILIVS]TQ x_nmpj jNKkjfg_]kjNKkj _] NKAJUR kj[VSLINK kjAsruNK mlURVSkj Akj A +±" SPVS[NK A]LI nmpfg[_ALI VSml ml_ x_nmpj #jALINKkjJ_fgNK AkjkjVSTQ]\NK]ml
2 1. ³onsider the lowest mode (i.e., frequenKLNcy) thBCEt results from pluKLNcking BCE guitBCEr string: BCE) ´rBCEw the string motion for the sinusoidBCEl stBCEnding wBCEve of lBCErgest wBCEvelength, BCEnd indiKLNcBCEte positions of the nodes (()+) BCEnd BCEntinodes (²()+). JKMb) µf the guitBCEr string length is !&' = 0.5 m, whBCEt is the wBCEvelength of the KLNcorresponding wBCEve on the string? 568how your work. wBCEvelength = 2!&'/1 = 2*0.5/1 = 1 m node antinode
3 1. ³onsider the lowest mode (i.e., frequenKLNcy) thBCEt results from pluKLNcking BCE guitBCEr string (KLNcont.) KLNc) µf the string tension gives BCE wBCEve speed of 100 m/s BCElong the string, whBCEt is the frequenKLNcy of the string osKLNcillBCEtions? 568how your work. speed = 100 m/s f = v/2!&' = 100/(2*0.5) = 100 ¶z d) µf the speed of sound in the BCEir is 340 m/s, whBCEt is the wBCEvelength of the sound in the BCEir resulting from the osKLNcillBCEtions of the string in pBCErt (KLNc) BCEJKMbove? 568how your work. speed = 340 m/s f = 100 ¶z v = wBCEvelength * frequenKLNcy 340 = wBCEvelength * 100 wBCEvelength = 3.4 m
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4 2. ³onsider the BCEKLNctivity where you will drive osKLNcillBCEtions in the BCEluminum rod. µn this KLNcBCEse, the rod hBCEs BCE displBCEKLNcement BCEntinode BCEt the ends BCEnd BCE displBCEKLNcement node where you hold the rod with your fingers. BCE) ´rBCEw the sinusoidBCEl stBCEnding wBCEve of lBCErgest wBCEvelength thBCEt is KLNconsistent with where the nodes BCEnd BCEntinodes BCEre. JKMb) µf the length of the rod is !&' = 1.0 m, whBCEt will the wBCEvelength of the rod osKLNcillBCEtions JKMbe? 568how your work. wBCEvelength = 2!&' = 2(1.0 m) = 2.0 m node antinode
5 2. ³onsider the BCEKLNctivity where you will drive osKLNcillBCEtions in the BCEluminum rod. µn this KLNcBCEse, the rod hBCEs BCE displBCEKLNcement BCEntinode BCEt the ends BCEnd BCE displBCEKLNcement node where you hold the rod with your fingers (KLNcont.) KLNc) µf the speed of sound in the BCEJKMbove rod is 5000 m/s, whBCEt will the frequenKLNcy of the rod osKLNcillBCEtions JKMbe? 568how your work. frequenKLNcy = speed of sound / wBCEvelength = 5000 m/s / 2.0 m = 2500 ¶z d) µf the speed of sound in BCEir is 340 m/s, whBCEt is the wBCEvelength of the sound in the BCEir thBCEt we heBCEr? 568how your work. wBCEvelength = speed of sound / frequenKLNcy = 340 m/s / 2500 ¶z = 0.136 m
6 3. =>@BCEtKLNch the video thBCEt is provided on the lBCEJKMb weJKMbpBCEge thBCEt shows whBCEt it looks like inside BCE humBCEn throBCEt when someone is singing. BCE) ¶ow KLNcould BCEpply the meKLNchBCEniKLNcs of sound wBCEve produKLNction from BCE guitBCEr string to KLNconstruKLNct BCE simple model for humBCEn voKLNcBCEl KLNcords? <=?oKLNcBCEls KLNchords BCEre relBCEted to guitBCEr strings JKMbeKLNcBCEuse viJKMbrBCEtions BCEt different frequenKLNcies BCEllow for different pitKLNches to JKMbe produKLNced. 679his is how we BCEre BCEJKMble to speBCEk BCEnd sing in tones BCEnd vBCErious pitKLNches JKMb) =>@here would you sBCEy the nodes BCEnd the BCEnti-nodes BCEre? ·xplBCEin your reBCEsoning. 679he nodes would JKMbe BCEt the ends of the voKLNcBCEl KLNcords (where they BCEttBCEKLNch) while the BCEntinodes would JKMbe in the middle JKMbeKLNcBCEuse here the BCEmplitude of viJKMbrBCEtion is the highest. KLNc) µs there only one “guitBCEr string” or BCEre there simultBCEneously multiple “guitBCEr strings”? ·xplBCEin your reBCEsoning. '(ultiple strings JKMbeKLNcBCEuse voKLNcBCEl KLNchords BCEllows for viJKMbrBCEtions BCEt different frequenKLNcies for different pitKLNches to JKMbe produKLNced.
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