PHYSICS 151 Notes for Online Lecture #23

Transcription

1 PHYSICS 5 Ntes fr Online Lecture #3 Peridicity Peridic eans that sething repeats itself. r exaple, eery twenty-fur hurs, the Earth aes a cplete rtatin. Heartbeats are an exaple f peridic behair. If yu l at heartbeats n an electrcardigra, they ae a regular pattern. he pattern that the heart beys is rather cplicated. In this sectin, we re ging t be dealing with a specific type f peridic tin called siple harnic tin Harnic eans that the tin can be described using sines and csines. Siple eans that the tin can be described using a single frequency. A ass n a spring (hrizntal r ertical) is a x 0 gd exaple f siple 0 harnic tin (r SHM fr shrt). he tin f the spring is repeated er and er. Let s start with a hrizntal spring, resting n a frictinless table. We pic a reference pint - x n the ass fr exaple, the center f the ass. he psitin f the center f the ass when the spring is unstretched is called the equilibriu x x pint (x 0). Nw I pull the ass an arbitrary distance x t the right. he spring exerts a frce in the directin ppsite the displaceent (t the left in this case). he frce is gien by He s Law: x where is the spring cnstant and has units f N/. If I pull the spring t the right, the spring exerts a frce t the left. Alternately, I can push the spring in a distance x. Nw the spring exerts a frce tward the right. Reeber that He s law nly wrs when the displaceents are sall. If yu ae a ery large displaceent, He s law desn t apply anyre and nne f what I abut t tell yu will apply either. A special characteristic f siple harnic tin is that the acceleratin is directly prprtinal t the displaceent. We can start with Newtn's secnd law a, and then insert He's law fr the frce n the spring. Lecture 3 Page

2 a x a a x Any syste in which the acceleratin is prprtinal t the displaceent will exhibit siple harnic tin. his can be tested experientally. Plt s. x n a graph and tae the slpe f the resulting straight line. If yu d this and yu dn t get a straight line, it eans that the spring can t be described by He s law. Siple Harnic Mtin Vcabulary displaceent ti e When I pull the ass n a spring and release it, the ass exhibits a peridic tin the psitin f the spring cnstantly repeats itself. If I were t plt the displaceent f the ass as a functin f tie, it wuld l sething lie this: If yu want t find ut where the ass is at any pint in tie, yu fllw the x-axis ut t the tie yu're interested in and they e up t the cure t see where the ass's psitin is. We can define a nuber f characteristics f siple harnic tin. r exaple, the aplitude is the axiu displaceent f the ass. he sybl fr aplitude is x. his is a distance, s the units shuld be eters. he tie it taes fr the ass t ae ne cplete cycle that is, t g fr stretched t cpressed and bac again is called the perid, which we represent by. Reind yurself that the "picture" f the wae is a picture f the ass as a functin f tie. It's nt a snapsht f the wae itself. he frequency is the nuber f cycles that are cpleted in ne secnd. he frequency is gien by f If the ass taes 3.0 s t cplete a cycle, the frequency is / (/s). We hae a special nae fr the unit f frequency, which is the Hertz (Hz). Lecture 3 Page

3 Hz s Quantity Sybl Definitin Units Perid tie fr ne cycle s requency f nuber f cycles per secnd /s Hz Aplitude x axiu displaceent displace ent a plitude ti e peri d Describing SHM using sines r csines he graph f the wae we hae been diagraing can be expressed as a sine r csine wae. In general, we can write any SHM as a sine wae r a csine wae. Yu nw fr trig that the sine and csine waes hae perids f π. he arguent f the trig functin has t be ultiplied by a factr such that the perid f yur wae is a ultiple f π. he scale factr turns ut t be t/. When the tie is equal t ne perid, yu want yur wae t be bac where it started. At t, the arguent is equal t π. xt ( ) x cs( π?) xt () x cs t π H G I K J If yu tae a csine wae and shift it by ne quarter f a cycle (90 degrees r π/ radians), yu find that the result is a sine wae. π t π πt xt () xcs xsin Lecture 3 Page 3

4 Hw d yu nw which is which? he answer is that yu hae t figure ut hw the wae starts. r exaple, at t 0, the sine functin will always be zer, regardless f the alue f ega. he wae belw in blue ust be a sine wae because it starts at zer. displaceent Csine wae tie Sine wae We can als find the elcity and the acceleratin f the ass as a functin f tie. If xt () x cs then t () sin and at () a cs πt H G I K J πt H G I K J πt H G I K J Nte that ur cnstraint that x and a ust be prprtinal t each ther is satisfied by these expressins. Lecture 3 Page 4

5 Ex. 3-: he tin f an scillatr f ass 0. g is gien by: xt () (. 050)cs 09. t b g where x is in and t is in s a) ind the aplitude b) ind the perid c) ind the frequency f scillatin d) ind the psitin f the ass at t 0 s, 0.75 s,.5 s, 3.0 s and 6.0 s We first hae t put this in the sae fr -as xt () x cs his gie us a) x 0.50 b) he arguent in the csine functin is πt c) f / /3.0 s 0.33 Hz ie π t/ t π H G I K J t xt () ( 0.5) cs π 3.0. he perid ust therefre be 3.0 s. cs HG π t 30. I K J X(t) s π / s π π 0.5 Ex. 3-: A 0.50-g ass at the end f a hrizntal spring has psitin 0 when t 0. he aplitude is 0.5 and the cycle starts by ing t the right first. he ass aes.0 cplete scillatins each secnd. What is the equatin fr the psitin as a functin f tie? Slutin: he functin will be either a sine r a csine. Hw d we nw which t pic? We re tld that the psitin at t 0 is x 0. Cpare the csine and sin functins. Lecture 3 Page 5

6 functin t 0 alue I HG K J cs(0) cs π t sin π t I HG K J sin(0) 0 S anytie that the ass starts fr x 0, yu will hae a sin functin. If the ass starts fr its aplitude alue, x 0, yu need t hae a csine functin. Since we re starting fr 0, we need t use a sin functin. πt xt () x sin We are tld that the syste cpletes tw scillatins eery secnd. his is the frequency, f H G I K J f /s he perid,, is gien by /f 0.5 s he aplitude is gien t us as x 0.5. Putting these in ur equatin, we hae: πt xt () b. sin H G I 05 g. sin t. K J b05 g b4π g 05 Why were we tld that the scillatins started tward the right? S that we wuld nw whether we needed a psitie r a negatie sign ut frnt. When the ass starts at zer, it can g either psitie r negatie in displaceent. If we tae t the right as psitie, the equatin will nt need a negatie sign. If the ass were ging t the left, we wuld hae a negatie sign ut frnt. he ertical spring What if the spring yu hae is hung ertically instead f hrizntally? Des what we just discered still hld? Ex. 3-3: A spring f spring cnstant 5 N/ has a ass f 0.5 g hung fr it. Hw far des the spring stretch when the ass is placed n it? When the ass is n the spring, it pulls the spring dwn, but then it just hangs there. We can draw a free-bdy diagra fr the ass. he acceleratin is zer, and the nly frces acting are graity dwn and the frce f the spring up. x0 Lecture 3 Page 6 x x eq

7 Σ 0 g x 0 eq xeq g (. 05g) xeq 98. 5N/ s x eq his is where the effect f graity ce in- it shifts the equilibriu psitin f the spring. Once this has been accunted fr - by taing the ptential energy t be zer when the ass is at x. Graity has n effect n the SH tin at all. Let s l at the spring when it s displaced a distance x. Draw the free-bdy diagra. (x-x ) x0 x x eq g x x + x eq he net frce is We fund in part a that x eq g/. ( x+ x ) g x+ x+ he nly frce causing the SHM is the spring! HG HG eq g g I K J I K J x+ g g x S analyzing SHM in the ertical and the hrizntal directins is the sae, except that the equilibriu psitin shift ust be accunted fr. g g Lecture 3 Page 7

8 Cnseratin f Energy fr SHM As the spring is stretched r cpressed, energy is cnerted fr the tin f the ass and spring t energy stred in the cils and bac again. he elastic ptential energy due t a spring (and ther stretchy things lie rubber bands) is: PE el x where, unlie graitatinal ptential energy, we tae the zer t be the equilibriu psitin f the spring (i.e. x 0 crrespnds t the pint at which there is zer ptential energy). 0 x - x 0 We can write the ttal echanical energy fr a spring as: E KE+ PE E + x At the axiu displaceent (x x ), the ass is entarily standing still. he ttal energy is then: 0 x 0 x E + x E x0 All ptential energy! When x 0, the ass has a elcity, which is the axiu elcity that the ass can hae. he ttal energy is then: 0 E + x E Because the ttal energy is cnstant at eery place alng the tin, All inetic energy Lecture 3 Page 8

9 x x x here is ne ther relatinship that we will need t use (which can be deried by cnsidering SHM is the prjectin f circular tin). π Let's reiew the definitins and relatinships we hae x aplitude axiu displaceent - ccurs when 0 (A is als used fr aplitude) axiu elcity ccurs when x 0 perid (s) f frequency (Hz /s) spring cnstant (N/) Relatinships: π f x x f π Ex. 3-4: he tin f an scillatr f ass 0. g is gien by: xt () (. 050)cs 09. t where x is in and t is in s. Nte that this is the sae equatin as Exaple 3-. b g e) ind the spring cnstant f) ind the ttal energy g) ind the axiu elcity We first hae t put this in the sae fr as xt () x cs t π H G I K J his gies us t xt () ( 0.5) cs π 3.0 Lecture 3 Page 9

10 a) spring cnstant: We ntice first that the perid is 3.0 s, s b) tal energy π ( π ) ( π ) ( ) ( ) 0.g π ( 3.0 s) g N s E x N E c088. hb050. g E. x0 J c) Maxiu elcity: he axiu elcity ccurs when x 0, s the energy is entirely inetic E E (. x0 J) 0. g 05. s. s Yu ry It! A 0.50-g ass at the end f a hrizntal spring is pulled bac t a distance f 0.5. At t 0, the ass is released and aes 3.0 cplete scillatins each secnd. ind: a) the elcity when the ass passes the equilibriu pint b) the elcity when the ass is 0.0 fr equilibriu c) the ttal echanical energy f the syste Knwn: A g f 3.0 Hz a) he quantity we are ling fr is. In exaining the equatins fr elcity and psitin, we fund that x Lecture 3 Page 0

11 Unfrtunately, we dn't nw, but we can find fr Nw put this in ur expressin fr π π f πf I HG f K J π πf b x g d i πf πf x x πfx π( 30. Hz)(. 05) 8. s + x + πf Using cnseratin f energy: x f x x π. x b b b π c sh c π(. sh b. g s f x Stp t see if this aes sense. he elcity ust be less than, which it is. c) tal energy g g g Lecture 3 Page

12 E b gc sh E 05. g 8. E 0. J Lecture 3 Page