PHYSICS 151 Notes for Online Lecture 4.1

Transcription

1 PHYSICS 5 Nte r Online ecture 4. Peridicity Peridic ean that ethin repeat itel. r exaple, eery twenty-ur hur, the ae a cplete rtatin. Heartbeat are an exaple peridic behair. I yu l at heartbeat n an electrcardira, they ae a reular pattern. he pattern that the heart bey i rather cplicated. In thi ectin, we re in t be dealin with a peciic type peridic tin called iple harnic tin Harnic ean that the tin can be decribed uin ine and cine. Siple ean that the tin can be decribed uin a inle requency. A a n a prin (hrizntal r ertical) i a x 0 d exaple iple 0 harnic tin (r SHM r hrt). he tin the prin i repeated er and er. et tart with a hrizntal prin, retin n a rictinle table. We pic a reerence pint - x n the a r exaple, the center the a. he pitin the center the a when the prin i untretched i called the equilibriu x x pint (x 0). Nw I pull the a an arbitrary ditance x t the riht. he prin exert a rce in the directin ppite the diplaceent (t the let in thi cae). he rce i ien by He aw: x where i the prin cntant and ha unit N/. I I pull the prin t the riht, the prin exert a rce t the let. Alternately, I can puh the prin in a ditance x. Nw the prin exert a rce tward the riht. Reeber that He law nly wr when the diplaceent are all. I yu ae a ery lare diplaceent, He law den t apply anyre and nne what I abut t tell yu will apply either. A pecial characteritic iple harnic tin i that the acceleratin i directly prprtinal t the diplaceent. We can tart with Newtn' ecnd law a, and then inert He' law r the rce n the prin.

2 a x a a x Any yte in which the acceleratin i prprtinal t the diplaceent will exhibit iple harnic tin. hi can be teted experientally. Plt. x n a raph and tae the lpe the reultin traiht line. I yu d thi and yu dn t et a traiht line, it ean that the prin can t be decribed by He law. Siple Harnic Mtin Vcabulary diplaceent ti e When I pull the a n a prin and releae it, the a exhibit a peridic tin the pitin the prin cntantly repeat itel. I I were t plt the diplaceent the a a a unctin tie, it wuld l ethin lie thi: I yu want t ind ut where the a i at any pint in tie, yu llw the x-axi ut t the tie yu're intereted in and they e up t the cure t ee where the a' pitin i. We can deine a nuber characteritic iple harnic tin. r exaple, the aplitude i the axiu diplaceent the a. he ybl r aplitude i x. hi i a ditance, the unit huld be eter. he tie it tae r the a t ae ne cplete cycle that i, t r tretched t cpreed and bac aain i called the perid, which we repreent by. Reind yurel that the "picture" the wae i a picture the a a a unctin tie. It' nt a napht the wae itel. he requency i the nuber cycle that are cpleted in ne ecnd. he requency i ien by I the a tae 3.0 t cplete a cycle, the requency i / (/). We hae a pecial nae r the unit requency, which i the Hertz (Hz).

3 Hz Quantity Sybl Deinitin Unit Perid tie r ne cycle requency nuber cycle per ecnd / Hz Aplitude x axiu diplaceent diplace ent a plitude ti e peri d Decribin SHM uin ine r cine he raph the wae we hae been diarain can be expreed a a ine r cine wae. In eneral, we can write any SHM a a ine wae r a cine wae. Yu nw r tri that the ine and cine wae hae perid π. he aruent the tri unctin ha t be ultiplied by a actr uch that the perid yur wae i a ultiple π. he cale actr turn ut t be t/. When the tie i equal t ne perid, yu want yur wae t be bac where it tarted. At t, the aruent i equal t π. xt ( ) x c( π?) xt () x c t π H G I K J I yu tae a cine wae and hit it by ne quarter a cycle (90 deree r π/ radian), yu ind that the reult i a ine wae. π t π πt xt () xc xin

4 Hw d yu nw which i which? he anwer i that yu hae t iure ut hw the wae tart. r exaple, at t 0, the ine unctin will alway be zer, reardle the alue ea. he wae belw in blue ut be a ine wae becaue it tart at zer. diplaceent Cine wae tie Sine wae We can al ind the elcity and the acceleratin the a a a unctin tie. I xt () x c then t () in and at () a c πt H G I K J πt H G I K J πt H G I K J Nte that ur cntraint that x and a ut be prprtinal t each ther i atiied by thee exprein.

5 Ex. : he tin an cillatr a 0. i ien by: xt () (. 050)c 09. t b where x i in and t i in a) ind the aplitude b) ind the perid c) ind the requency cillatin d) ind the pitin the a at t 0, 0.75,.5, 3.0 and 6.0 We irt hae t put thi in the ae r -a xt () x c hi ie u a) x 0.50 b) he aruent in the cine unctin i πt c) / / Hz ie π t/ t π H G I K J t xt () ( 0.5) c π 3.0. he perid ut therere be 3.0. c HG π t 30. I K J X(t) π / π π 0.5 Ex. : A a at the end a hrizntal prin ha pitin 0 when t 0. he aplitude i 0.5 and the cycle tart by in t the riht irt. he a ae.0 cplete cillatin each ecnd. What i the equatin r the pitin a a unctin tie? Slutin: he unctin will be either a ine r a cine. Hw d we nw which t pic? We re tld that the pitin at t 0 i x 0. Cpare the cine and in unctin.

6 unctin t 0 alue I HG K J c(0) c π t in π t I HG K J in(0) 0 S anytie that the a tart r x 0, yu will hae a in unctin. I the a tart r it aplitude alue, x 0, yu need t hae a cine unctin. Since we re tartin r 0, we need t ue a in unctin. πt xt () x in We are tld that the yte cplete tw cillatin eery ecnd. hi i the requency, H G I K J / he perid,, i ien by / 0.5 he aplitude i ien t u a x 0.5. Puttin thee in ur equatin, we hae: πt xt () b. in H G I 05. in t. K J b05 b4π 05 Why were we tld that the cillatin tarted tward the riht? S that we wuld nw whether we needed a pitie r a neatie in ut rnt. When the a tart at zer, it can either pitie r neatie in diplaceent. I we tae t the riht a pitie, the equatin will nt need a neatie in. I the a were in t the let, we wuld hae a neatie in ut rnt. he ertical prin What i the prin yu hae i hun ertically intead hrizntally? De what we jut dicered till hld? Ex. 3: A prin prin cntant 5 N/ ha a a 0.5 hun r it. Hw ar de the prin tretch when the a i placed n it? When the a i n the prin, it pull the prin dwn, but then it jut han there. We can draw a ree-bdy diara r the a. he acceleratin i zer, and the nly rce actin are raity dwn and the rce the prin up. x0 x x eq

7 Σ 0 x 0 eq xeq (. 05) xeq 98. 5N/ x eq hi i where the eect raity ce in- it hit the equilibriu pitin the prin. Once thi ha been accunted r - by tain the ptential enery t be zer when the a i at x. Graity ha n eect n the SH tin at all. et l at the prin when it diplaced a ditance x. Draw the ree-bdy diara. (x-x ) x0 x x eq x x + x eq he net rce i We und in part a that x eq /. ( x+ x ) HG HG x+ x+ he nly rce cauin the SHM i the prin! eq I K J I K J x+ x S analyzin SHM in the ertical and the hrizntal directin i the ae, except that the equilibriu pitin hit ut be accunted r.

8 Cneratin Enery r SHM A the prin i tretched r cpreed, enery i cnerted r the tin the a and prin t enery tred in the cil and bac aain. he elatic ptential enery due t a prin (and ther tretchy thin lie rubber band) i: PE el x where, unlie raitatinal ptential enery, we tae the zer t be the equilibriu pitin the prin (i.e. x 0 crrepnd t the pint at which there i zer ptential enery). 0 x - x 0 We can write the ttal echanical enery r a prin a: E KE+ PE E + x At the axiu diplaceent (x x ), the a i entarily tandin till. he ttal enery i then: 0 x 0 x E + x E x0 All ptential enery! When x 0, the a ha a elcity, which i the axiu elcity that the a can hae. he ttal enery i then: 0 E + x E Becaue the ttal enery i cntant at eery place aln the tin, All inetic enery

9 x x x here i ne ther relatinhip that we will need t ue (which can be deried by cniderin SHM i the prjectin circular tin). π et' reiew the deinitin and relatinhip we hae x aplitude axiu diplaceent - ccur when 0 (A i al ued r aplitude) axiu elcity ccur when x 0 perid () requency (Hz /) prin cntant (N/) Relatinhip: π x x π Ex. 4: he tin an cillatr a 0. i ien by: xt () (. 050)c 09. t where x i in and t i in. Nte that thi i the ae equatin a Exaple 3-. b e) ind the prin cntant ) ind the ttal enery ) ind the axiu elcity We irt hae t put thi in the ae r a xt () x c t π H G I K J hi ie u t xt () ( 0.5) c π 3.0

10 a) prin cntant: We ntice irt that the perid i 3.0, b) tal enery π ( π ) ( π ) ( ) ( ) 0. π ( 3.0 ) N E x N E c088. hb050. E. x0 J c) Maxiu elcity: he axiu elcity ccur when x 0, the enery i entirely inetic E E (. x0 J) Yu ry It! A a at the end a hrizntal prin i pulled bac t a ditance 0.5. At t 0, the a i releaed and ae 3.0 cplete cillatin each ecnd. ind: a) the elcity when the a pae the equilibriu pint b) the elcity when the a i 0.0 r equilibriu c) the ttal echanical enery the yte Knwn: A Hz a) he quantity we are lin r i. In exainin the equatin r elcity and pitin, we und that x

11 Unrtunately, we dn't nw, but we can ind r Nw put thi in ur exprein r π π π I HG K J π π b x d i π π x x πx π( 30. Hz)(. 05) 8. + x + π Uin cneratin enery: x x x π. x b b b π c h c π(. h b. x Stp t ee i thi ae ene. he elcity ut be le than, which it i. c) tal enery

12 E b c h E E 0. J Pendula A iple pendulu cnit an bject upended r a trin. he tin the pendulu winin bac and rth can al be decribed by iple harnic tin (under certain cnditin). θ θ c(θ) in(θ) let irt draw a ree-bdy diara r the bb n the pendulu. We can decpe the weiht,, int a cpnent aln the directin the trin and a cpnent perpendicular t the trin. he cpnent the weiht in the directin the trin will be: c(θ) he cpnent the weiht perpendicular t the trin i - in(θ) Write Σ r 0. We hae t write ne equatin r each directin. Aln the trin In the directin aln the trin, there i n acceleratin, tain tward the pit a pitie, Perpendicular t the trin here i, hweer, acceleratin in the directin perpendicular t the trin due t the unbalanced rce:

13 Σ r aln thetrin 0 c( θ) 0 c( θ) Σ r perpendicular t the trin in( θ) a One the thin we learned in the lat lecture, i that bject that under iple harnic tin llwin ethin that l lie He law, with the rce prprtinal t the diplaceent. hi equatin den t l lie He law. θ θ θ I θ i ery all, we can ae an apprxiatin. I we let be the diplaceent the pendulu bb, will be an arc. I θ i all, we can write that inθ We can then write in( θ) that the prin cntant r thi prble i

14 where the eectie prin cntant i ien by " " H G I K J We can ue the ae relatinhip we deried r the a n a prin t ind the iilar quantitie r the pendulu Sprin Pendulu π π π π Nte that, r the pendulu, all thee reult are independent the a the pendulu. Enery Cncern r the Pendulu ae h 0 t be the lwet part the pendulu win. he heiht at any pint i then Θ c(θ) h -c(θ) At the tp the win, the elcity i zer, there i n inetic enery and the ttal enery i entirely ptential E PE h E (-c(θ)) At the btt the win, the ptential enery i zer and all the enery i inetic, - c(θ) h 0 here E where i the axiu elcity.

15 r cneratin enery, the ttal enery at thee tw pint ut be equal. E c( θ) b b c( θ) c( θ) b Ex. 5: A pendulu lenth 0.50 ae.0 cillatin in 30.0 ecnd. What i the alue? nuber cillatin.0 requency 0.7 Hz tie 30.0 π bπ bπ π b b c π 07. ( 05. ) 967. h

16 Ex. 6: A pendulu with a requency 6 cillatin per ecnd i taen t the n, where raity i /6 th the raity earth. What will the pendulu requency be n the n? Slutin: he pendulu lenth i cntant. π b 4 π bπ bπ bπ 4π Mn Mn 4 Mn π b hee tw ut be equal, 4π 4π b b b b bmn b Mn Mn Mn Mn Mn Mn Mn Mn 6Hz 5. Hz 6 Mn Mn b b 6 6 Mn Mn