P h y s i c s F a c t s h e e t

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1 P h y s i c s F a c s h e e Sepembe 2001 Numbe 20 Simple Hamonic Moion Basic Conceps This Facshee will:! eplain wha is mean by simple hamonic moion! eplain how o use he equaions fo simple hamonic moion! descibe he enegy ansfes in simple hamonic moion! show how o epesen simple hamonic moion gaphically! eplain he link beween simple hamonic moion and cicula moion Lae Facshees will deal wih specific eamples of simple hamonic moion, such as he simple pendulum and masssping sysem, and damped oscillaoy moion. Wha is simple hamonic moion? Simple hamonic moion (SHM) is one fom of oscillaoy moion. SHM occus when he esulan foce acing on a body has paicula popeies: Equaions fo SHM The definiion of SHM above can be epessed in he fom of an equaion: F = k Noe ha he minus sign appeas because he foce is dieced back owads he fied poin. Fo he sake of convenience, his is moe usually wien: F = mω 2 The eason why his fom is moe convenien will become appaen sholy. By using F = ma, his leads o he equaion a = ω 2 This equaion leads o he following equaions (bu you do no need o know how unless you ae sudying SHM in A-level Mahs!): A body pefoms SHM if i is aced upon by a foce! he magniude of which is popoional o he disance of he body fom a fied poin! he diecion of which is always owads ha fied poin. v 2 = ω 2 ( 2 2 ) whee v = speed = displacemen = ampliude = maimum displacemen A simple eample of SHM can be obseved by aaching a mass o a sping, hen pulling he mass down and eleasing i. I will bob up and down his moion is SHM. In his case, he fied poin in he above definiion is he equilibium posiion of he mass whee i was befoe i was pulled down. The esulan foce acing on he mass is composed of is weigh and he ension in he sping. This will always be ying o pull he mass back o he equilibium posiion, and he fuhe away fom equilibium he mass is, he songe his foce will be. = cosω whee = ime This equaion assumes ha he paicle sas a a poin of maimum displacemen in he case of he sping eample, his would mean i sas fom he pulled down posiion If we saed iming fom he equilibium posiion hen he displacemen would be given by = sinω v = ωsinω Again, his assumes he paicle sas a he poin of maimum displacemen. if i sas fom equilibium, we d have v = ωcosω Some helpful mahs 1. cosω means cos(ω ) so you have o wok ou ω fis, hen find is cosine. 2. Radians ae anohe way of measuing angles, ahe han degees. Angula velociies (see also Facshee 19 Cicula Moion) ae measued in adians pe second (ad s -1 ). When you ae using funcions like cosω, you will need o wok in adians. The bes way o do his is o pu you calculao ino adians mode, ene he value you have and wok ou is cosine as nomal.! On a sandad scienific calculao, you can can change in in and and ou ou of of adians mode mode using using he he buon buon labelled labelled DRG DRG (whee (whee D = D degees = degees and R and = adians). R = adians). Some Some gaphical gaphical calculaos calculaos wok wok in his in way his way oo; oo; on ohes on ohes you need you need o go o hough go hough he seup he seup menu. menu.! If you have o conve beween degees and adians (which you usually won ): # To change degees o adians, muliply by π/180. # To change adians o degees, muliply by 180/π 3. sine and cosine epea hemselves evey 360 o (= 2π adians). This is because 360 o is a full cicle, so if you add 360 o o an angle in degees, you ge back o whee you saed. We can say ha sine and cosine have a peiod of epeiion of 360 o o 2π adians 1

2 Physics Facshee We can deduce some moe esuls fom hese equaions: Maimum speed If we look a he speed equaion v 2 = ω 2 ( 2 2 ), we can see ha he bigge 2 is, he smalle v 2 is and vice vesa. The smalles possible value of 2 is 0. Puing 2 = 0 in gives us v 2 = ω 2 2, hence: v ma = ω; his occus when = 0 Maimum acceleaion Since a = ω 2, he acceleaion will be geaes in magniude when is geaes in magniude, hence maimum magniude of a = ω 2 Acceleaion a any ime Using a = ω 2 ogehe wih = cosω, we have a = ω 2 cosω Peiod and fequency We know = cosω. Since cosine epeas evey 2π adians (see mahs bo on page 1), he displacemen will fis eun o is inial value when ω = 2π. The ime equied fo his is he peiod of he moion, T. Fom above, we have T = 2 π ω The fequency of he moion is he numbe of oscillaions pe second. 1 ω f = = T 2π So ω = 2πf I is woh noing ha he peiod and fequency do no depend on he ampliude of he moion. Eam Hin: You may need o know how o deive he epessions fo maimum speed and maimum acceleaion. Ohe equaions can be lean bu check you fomula shee, o make sue you do no wase ime leaning equaions you will be given. The equaions ae summaised in Table 1 below Table 1. SHM equaions soed by ype acceleaion equaions speed equaions displacemen equaions peiod/fequency equaions a = ω 2 v 2 = ω 2 ( 2 2 ) Τ = 2 π ω a = -ω 2 cosω v = -ωsinω = cosω ω = 2πf a ma = ω 2 v ma = ω ma = Using he equaions A fis sigh, hee may seem o be a bewildeing numbe of equaions o choose fom. Hee ae some saegies o ensue you use he coec one(s): Focus on ω. If you ae given he peiod o fequency, use hese o find ω befoe doing anyhing else. Similaly, if you ae asked o find he peiod o fequency, find ω fis. Wie down wha you know and wha you wan. Then choose he equaion wih jus hese symbols in i. 2 You won need o use he equaions fo he maimum value of displacemen, velociy o acceleaion unless maimum values ae specifically menioned. If ime is no menioned anywhee, you ae pobably going o be using v 2 = ω 2 ( 2 2 ). You may also find he following useful: The ime equied fo he body o go beween a maimum value of displacemen o he equilibium posiion is one quae of he peiod The ime equied fo he body o go beween one maimum of displacemen o he ohe (i.e. he wo ends of he moion) is half of he peiod The way o appoach poblems is bes seen fom woked eamples: Eample 1. An objec is oscillaing wih simple hamonic moion. Is maimum displacemen fom is equilibium posiion is 0.2m. The peiod of he moion is 0.1 s. Find is speed when i is 0.06m fom is equilibium posiion Since we ae given T, ou fis sep is o calculae ω. 2π 2 π T = ω = = 62.8 ads -1 ω Τ We now know: (= ma displacemen) = 0.2m ω = 62.8 ad s -1 =0.06 m We wan: v =? Since, ω, and v ae involved (and no ), we use v 2 = ω 2 ( 2 2 ) v 2 = ( ) v 2 = v = 12.0 ms -1 Eample 2. A mass is moving wih simple hamonic moion; is displacemen was a a maimum of 1.1m when = 0. Is maimum speed is 0.33ms -1. Find: a) is fequency; b) is speed afe 2.0 seconds. a) Since we ae asked fo fequency, we need o find ω fis So we have: = 1.1 m v ma = 0.33 ms -1 ω =? So we need o use he equaion wih hese hee lees in i: v ma = ω 0.33 = ω1.1 ω = = 0.30 ads -1 Now we have ω, we find f using ω = 2πf 0.3 = 2πf f = 0.30 (2π) Hz b) We have = 2.0 s; ω = 0.30 ads -1 ; = 1.1 m; v =? So we mus use: v = -ωsinω = sin( ) = ms -1 Eam Hin: 1. Make sue you do befoe finding he sine 2. Ensue you calculao is in adians mode!

3 Physics Facshee Eample 3. A heavy body is pefoming simple hamonic moion. Is displacemen is a is maimum value of 0.40 m when = 0. I fis eaches a poin 0.20 m fom is equilibium poin afe 3.0 s. a) Find he peiod of he moion. b) Eplain why he wods fis eaches ae impoan fo you calculaion in a). c) Find he body s displacemen when is speed is ms -1 a) Since we ae asked fo he peiod, we fis need o find ω We have: = 0.40 m, = 0.20 m, = 3.0 s, ω =? So we use = cosω 0.20= 0.40cos(ω 3.0) To solve his so of equaion, we mus ge he pa wih he cos on is own fis: = cos(3ω) 0.50 = cos(3ω) Now we mus use cos -1 (using INV COS on he calculao) o find he angle in adians whose cos is 0.5: 3ω = cos -1 (0.50) = 1.05 ω = 0.35 ad s -1 Typical Eam Quesion A body pefoms SHM wih a peiod 3 seconds. Timing sas a one of he eemes of displacemen of he body. Deemine he ne hee imes when: (a) displacemen is a an eeme of he moion; [3] (b) velociy is zeo; [3] (c) acceleaion is zeo. [3] (a) i will be a he ohe eeme half a cycle lae, so he imes ae 1.5 s$, 3.0 s $, 4.5 s $ (b) Velociy is zeo when ω 2 ( 2 2 ) = 0 so when = ± $ So imes ae 1.5 s, 3.0 s and 4.5 s as in (a). $$ (c) Acceleaion = -ω 2. So acceleaion is zeo when body is a equilibium posiion.$ This is midway beween he imes i is a he eemes. So we have 0.75 s, 2.25 s, 4.75 s $$ Typical Eam Quesion The body in he diagam pefoms simple hamonic moion beween he poins shown as doed lines. The peiod of he moion is 2s. 20cm 2 π Now we can find T = =18 s ω b) Since SHM is epeiive, hee will be ohe imes when he body eaches his displacemen. c) We have ω = 0.35 ad s -1, = 0.40 m, v = ms -1, =? So we use v 2 = ω 2 ( 2 2 ) = ( ) = 0.122( ) = = = = = 0.37 m Eample 4. A body pefoms simple hamonic moion. Is moion is imed fom a poin of maimum displacemen. Two seconds lae, i eaches he equilibium posiion fo he fis ime. Is maimum acceleaion is 0.60 ms -2. Find: a) is peiod; b) is maimum speed. a) Iniially, we do no seem o have enough infomaion o use any equaion. Bu we do know ha i akes wo seconds o move fom a poin of maimum displacemen o he equilibium posiion which coesponds o a quae of he peiod. So he peiod is 4 2 = 8 seconds 2 π 2π b) Fo any fuhe calculaions, we will need ω = = Τ 8 =0.785 ad s -1 we know: a ma = 0.60 ms -2 So use a ma =ω = = = m We need v ma = ω = =0.76 ms -1 Calculae he following quaniies: (a) ampliude; [1] (b) maimum acceleaion; [2] (c) maimum speed. [2] (a) ampliude = disance fom equilibium o ma displacemen p The disance shown is wice ha so ampliude = 10cm = 0.10m$ 2 π (b) We need ω = =π Τ a ma = ω 2 =π 2 (0.10)$ = 0.99 ms -2 $ (c) v ma = ω = 0.10π $= 0.31 ms -1 $ Eam Hin: - Many candidaes lose maks hough omiing o change hei answes o SI unis in he above eample, if he ampliude had no been changed ino mees, he values fo maimum acceleaion and speed could have been inconsisen. Enegy Tansfes in SHM idenical ligh helical spings In SHM, wih no eenal foces (so no damping o focing ), he oal enegy of he oscillaing sysem emains consan. Alhough oal enegy emains consan, enegy is ansfeed beween kineic enegy and poenial enegy when he body is a he equilibium posiion, i is moving a is fases, so is kineic enegy is maimum, and when i is a he poins of maimum displacemen, is speed is zeo so is kineic enegy is also zeo. These enegy ansfes ae bes epesened gaphically (see ne secion). Noe: we could have obained his by dividing a ma by ω. 3

4 Physics Facshee Gaphical Repesenaion of SHM Displacemen, speed and acceleaion agains ime The gaphs of hese gaphs, as migh be epeced fom hei especive equaions, poduce a sandad sine wave shape: displacemen peiod, T ampliude, Enegy Gaphs Enegy agains displacemen We will conside oal enegy, kineic enegy and poenial enegy. Since oal enegy is consan, his gaph is simply a hoizonal line. Kineic enegy, howeve, is moe ineesing: We know ha v 2 = ω 2 ( 2 2 ) So kineic enegy = ½ mv 2 = ½ mω 2 ( 2 2 ) This leads o he gaph below: k.e. ½ mω speed v ω -ω acceleaion - 0 We can combine kineic enegy, poenial enegy and oal enegy on one gaph: enegy oal enegy ½ m 2 ω 2 poenial enegy kineic enegy ω 2 a 0 displacemen Noe ha he kineic enegy and poenial enegy a any ime add up o he oal enegy, which is consan. ω 2 Poins o noe The speed gaph is he gadien of he displacemen gaph, and hey have a phase diffeence of π/2 (= 90 o o a quae of a peiod). The acceleaion gaph is he gadien of he speed gaph, and is π/2 ou of phase wih he speed, and π ou of phase wih displacemen. Each of hese gaphs has he same peiod. The acceleaion is always opposie in sign o he displacemen Acceleaion agains Displacemen a ω 2 Enegy agains ime Since we know v = -ωsinω, we can deduce k.e. = ½mv 2 = ½ m 2 ω 2 sin 2 ω This poduces he following gaph: k.e. T ½ m 2 ω 2 Noe ha he kineic enegy goes hough wo cycles duing one peiod of he oscillaion. The oal enegy is, again, a consan, poducing a hoizonal saigh line gaph. The poenial enegy gaph is an upside down vesion of he kineic enegy: - 0 ½ m 2 ω 2 p.e. -ω 2 4

5 Physics Facshee Cicula Moion and SHM To see he link beween cicula moion and SHM, conside an objec pefoming hoizonal cicula moion a consan speed in fon of a sceen, wih he plane of he moion pependicula o he sceen. Ligh is hen dieced ono he objec, so ha he objec s shadow falls ono he sceen. The diagam below shows a view looking down on he appaaus. Typical Eam Quesion A lage faigound wheel, which oaes a a consan ae, cass a shadow on o a neaby building. A a ime when he sun's ays sike he building hoizonally, a boy measues he speed of he shadow of one of he cas on he wheel as i passes diffeen floos of he building. A A* sceen Shadows LIGHT shadow Big wheel B B* Rays of sunligh The shadow s moion will be in a saigh line, beween poins A* and B*. In fac, his moion is simple hamonic; o show his we will need o inoduce some angles and simple igonomey. P A θ C We will measue he displacemen of he shadow fom he poin O, which is midway beween poins A and B. I is level wih he cene of he cicle, C. This coesponds o he equilibium poin in SHM. We will assume he objec sas is cicula moion a poin A. This means is shadow will sa a poin A*. This coesponds o ou assumpion ha SHM sas a a poin of maimum displacemen. A* O B* A a floo which is level wih he cene of he wheel, he speed of he shadow is 0.17 ms -1. A a floo 10m highe, he speed is 0.16 ms -1. Calculae: (a) he ime i akes o complee one oaion; [6] (b) he diamee of he wheel. [2] (a) Shadow is pojecion of cicula moion SHM Level wih cene of wheel speed is maimum. So 0.17 = ω$ When = 10 m, v = 0.16ms -1. So using v 2 = ω 2 ( 2 2 ): = ω 2 ( )$ = ω 2 2 ω $ Bu ω 2 2 = (ω) 2 = So = ω 2 $ 100ω 2 = = ω 2 = ω = ad s -1 $ T = 2π/ω = s$ (b) This is = ω = 0.17/ω = 29.8 m$ Diamee = 60 m (2SF)$ The angle he line CP makes an angle θ wih he line CA a ime. The paicle is moving wih a consan angula velociy ω. We now need o find an epession fo he displacemen,, of he shadow a ime. By igonomey, = cosθ, whee = adius of he cicle. Since he paicle has a consan angula velociy, θ = ω So a ime, displacemen of shadow is given by = cosω This is heefoe SHM. Any SHM can have cicula moion linked wih i in his way i is known as associaed cicula moion. The SHM is someimes descibed as a pojecion of he cicula moion. Noe ha ω in he SHM coesponds o he angula velociy in he associaed cicula moion Eam Hin: - This only woks if he cicula moion is a consan angula velociy so in many cases i will no apply o veical cicula moion. 5 Quesions 1. Eplain wha is mean by simple hamonic moion. 2. Wie down epessions fo he maimum speed and acceleaion of a paicle caying ou SHM. 3. Skech gaphs o illusae displacemen, speed and acceleaion agains ime fo a paicle caying ou SHM. Sae he elaionship beween hese gaphs. 4. Skech a gaph o show how kineic enegy vaies wih displacemen o a paicle caying ou SHM. Include on you skech he maimum value of he kineic enegy. 5. Eplain why a pojecion of cicula moion will no poduce simple hamonic moion if he angula velociy is no consan. 6. A paicle caying ou SHM has a peiod of 2.0 s and a maimum speed of 9.4 ms -1. Given ha iming sas when he paicle s displacemen is a a maimum, find he fis 3 imes when i is a a disance of 1.5 m fom is equilibium posiion. 7. A body caies ou SHM. When is displacemen fom is equilibium posiion is 0.10 m, is speed is 5.0 ms -1. When is displacemen fom equilibium is 0.30 m, is speed is 2.0 ms -1. Calculae is maimum displacemen.

6 Physics Facshee Eam Wokshop This is a ypical poo suden s answe o an eam quesion. The commens eplain wha is wong wih he answes and how hey can be impoved. The eamine s answe is given below. The spheical objec shown in he diagam below is known o pefom simple hamonic moion beween poins A and B. The objec appeas saionay when viewed wih a sobe ligh a 21 Hz and 28 Hz bu a no fequencies in beween. A Deemine: 4cm 20cm 9cm 10cm (a) he fequency of he moion. [1] 28Hz % 0/1 The suden clealy did no undesand wha was happening wih he sobe fequencies, bu giving some answe, ahe han no answe, was sensible, since i allows he suden o coninue wih he quesion and hence gain some maks. (b) he maimum speed of he objec. [2] ω = 2πf = 56π $ v=ω = 56π 20 = 3520 ms -1 (3SF) % 1/2 The suden gains he mak fo using his/he value of fequency o find ω, bu hen uses boh he wong value fo (emembe ampliude is half he peak-o-peak value his is commonly eamined!) and he wong unis (cm ahe han m) (c) he acceleaion a posiion. [2] ω 2 = 56π $= 31000ms -2 (3SF) 1/2 Again he suden gains cedi fo using his/he own value of fequency and hee, s/he has emembeed o wok in SI unis. Bu he suden has fogoen ha acceleaion is a veco i mus have a diecion! The wods owads he cene o he use of a minus sign would have compleed he answe. Also, alhough he suden did no make his eo in calculaion, i was unwise o wie 56π 2 when (56π) 2 is mean, since i may lead o making he misake of squaing jus he π. (d) he speed a poin y. [2] v 2 = ω 2 ( 2 2 ) $ v 2 = 56π 2 ( ) v = ms -1 % 1/2 Since he suden has shown a suiable mehod, one mak can be awaded. Howeve, s/he has fogoen o ake he squae oo! The size of he answe should have aleed him/he o somehing wong. y B Answes 1 4 can be found in he e 5. If he angula velociy of he cicula moion is no consan, hen he angle hough which he body uns will no be diecly popoional o ime. Accodingly, he displacemen of he pojecion will no be popoional o he cosω, when ω is consan. 6. T = 2 s ω = 2π/T = π ad s -1 v ma = ω = 3π = 3.0 m = 3.0cosπ Disance of 1.5m fom equilibium = ±1.5 m = 1.5 m 1.5 = 3.0cosπ 0.50 = cosπ = π = 0.33 s = -1.5 m -1.5 = 3cosπ = cosπ = π = 0.67 s We know he body akes half a peiod (= 1s) o avel fom one eeme displacemen o he ohe. So o avel fom = -1.5 m o he negaive eeme akes ( ) = 0.33 s I hen akes a fuhe 0.33s o eun o = -1.5 m So he fis 3 imes ae: 0.33s; 0.67s; 1.33s 7. v 2 = ω 2 ( 2 2 ) 5 2 = ω 2 ( ) & 2 2 = ω 2 ( ) ' & ': 5 ω ( -0.1 ) ( -0.1 ) = = ω ( -0.3 ) ( -0.3 ) 2 25 ( 0.01) == 2 4 ( 0.09) 25( ) = 4( ) = = = = 0.32 m (2SF) When you wie down you final answe you ae epeced o use he same numbe of significan figues as he daa ha you wee given. In mid calculaion i doesn eally mae if you use one moe significan figue because i is you mehod ahe han you mid-way esul ha is being maked. The same hing applies o unis. You may choose o leave ou unis in he middle of calculaions bu you mus include hem wih you final answe. Eamine s Answe (a) The fequency mus be a common faco of 21 and 28Hz i.e. 7Hz. $ (b) ω = 2πf = 14π. $ v ma = ω = π = 4.4ms -1 $ (c) a = ω 2 = 14 2 π = 77ms -2 $ owads he cene $ (d) v 2 = ω 2 ( 2 2 )$ v 2 =14 2 π 2 ((0.1) 2 (0.09) 2 ) v = 1.9ms -1 $ 6 This Facshee was eseached and wien by Cah Bown. Cuiculum Pess, Uni 305B The Big Peg, 120 Vyse See, Bimingham B18 6NF. Physics Facshees may be copied fee of chage by eaching saff o sudens, povided ha hei school is a egiseed subscibe. They may be newoked fo use wihin he school. No pa of hese Facshees may be epoduced, soed in a eieval sysem o ansmied in any ohe fom o by any ohe means wihou he pio pemission of he publishe. ISSN

PHYS PRACTICE EXAM 2

PHYS PRACTICE EXAM 2 PHYS 1800 PRACTICE EXAM Pa I Muliple Choice Quesions [ ps each] Diecions: Cicle he one alenaive ha bes complees he saemen o answes he quesion. Unless ohewise saed, assume ideal condiions (no ai esisance,