Answers to Coursebook questions Chapter 4.1

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1 Carige Physis or the IB Diploa nswers to Courseook questions Chapter 41 1 n osillation is any otion in whih the isplaeent o a partile ro a ixe point keeps hanging iretion an there is a perioiity in the otion, ie the otion repeats in soe way In siple haroni otion, the isplaeent ro an equiliriu position an the aeleration are proportional to an opposite to eah other In an SHM osillation the perio is inepenent o the aplitue an the aeleration is opposite an proportional to the isplaeent ro equiliriu It is an osillation, sine we ay eine the isplaeent o the partile ro the ile point an in that ase the isplaeent hanges iretion an the otion repeats The otion is not siple haroni, however, sine there is no aeleration that is proportional (an opposite) to the isplaeent 4 It is an osillation, sine the otion repeats The otion is not siple haroni, however, sine the aeleration is onstant an is not proportional (an opposite) to the isplaeent 5 We know that a ass at the en o a spring when isplae will peror siple haroni osillations with a perio given y T = π Hene, easuring the spring k onstant an the perio o osillations allows eterination o the ass 6 They are not siple haroni eause, as shown (see page 01 in Physis or the IB Diploa), the restoring ore, although opposite to, is not proportional to the isplaeent away ro the equiliriu position I, however, the aplitue o osillations is sall, the ore oes eoe approxiately proportional to the isplaeent an the osillations are then approxiately siple haroni 7 t t = 0 the equation says that x= osφ The next tie x assues this value is at a tie given y os( T + φ) = osφ Thus we ust solve the equation os( T + φ) = osφ This eans that the angles π T + φ an φ ier y π an so solutions are T + φ = φ + π T = 8 a t t = 0 we have y = 50os(0) = 50 t t = 1 s we use the alulator (in raian oe) to in y = 50os( 1) = 7 0 = 50os( t) t = os ( ) = 198 t = 099 s 5 Use v=± x We know that = 0 s Thereore, x 5 x 90 x 400 =± = =± Copyright Carige University Press 011 ll rights reserve Page 1 o 10

2 Carige Physis or the IB Diploa 9 a The equation is siply y= 80os(π 14 t) = 80os(8π t) The veloity is thereore v= 80 8πsin(8π t) an the aeleration is a= 80 (8π) os(8π t) t t = 005 s we evaluate (in raian oe): y = 80os(8π 005) = 47, v a = 80 8π sin(8π 005) = 57 s an = 80 (8π) os(8π 005) = 6 10 s 10 The angular requeny is = π = π 460 = 90π The axiu veloity is = 90π = 14 s an 4 the axiu aeleration is = (90π) = 4 10 s y = 60sin(π x)os(π 50 t), ro whih we eue that the requeny o all points is 50 Hz an that the phase o all points is zero 11 a The equation o the string ay e rewritten as ( ) Fro a the aplitue is = 60sin(π x) an so is ierent or ierent points on the string The axiu aplitue is otaine when sin(π x ) = 1, ie the axiu aplitue is 60 The isplaeent is always zero at the ens o the string, in partiular at the right en, where x = L, the length o the string The isplaeent is zero all the tie when 60sin(π x ) = 0, ie when πx= π x= 10 e When L x = = 075 the aplitue is 60sin(π x ) = 60sin(075π) = t the aeleration is axiu in agnitue an is irete to the right t B it is irete to the right an has saller agnitue than at t C we have the equiliriu position an so the aeleration is zero t D the aeleration is to the let an has the least agnitue opare to an B Net ore: sine F = a whatever is true or aeleration is also true or net ore 1 a t axiu isplaeent we will have the axiu aeleration, ie 4500 aax = The aplitue is = 45 an = π = π s Thereore, aax = π = s 60 s the piston oves past its equiliriu position it has the axiu spee an 4500 so vax = π = 1 s 60 F ax aax N = = Copyright Carige University Press 011 ll rights reserve Page o 10

3 Carige Physis or the IB Diploa 14 a The area is approxiately 050 (the exat value is 051 ) (ieally the graph shoul have een given on graph paper an we woul then ount squares) This is the isplaeent ro when the veloity is zero to when it is zero again, ie ro one extree position until the other, ie twie the aplitue The perio is 04 s an so the equation or isplaeent is πt x = 05sin( ) = 05sin ( 5πt ) 04 π π 15 We nee to graph the equation a= x, where = = = 157 s T 05 The slope woul e = 158 s or just 158 sine we are plotting on the horizontal axis This gives the graph in the answers (see page 799 in Physis or the IB Diploa) 16 a The eining relation or SHM is that a= x whih iplies that a graph o aeleration versus isplaeent is a straight line through the origin with negative slope just as the given graph The slope o the graph gives 15 The easure slope is = 15 s an so = 15 = 87 s 010 π Thus the perio is T = = 16 s 87 The axiu veloity is The axiu net ore is = = 09 s a = = = 0 N e The total energy is J ET = = = Copyright Carige University Press 011 ll rights reserve Page o 10

4 Carige Physis or the IB Diploa 17 a The ores on the ass when the plate is at the top are shown elow: N g The net ore is g N = a Sine we have siple haroni otion a= x= 4π x in agnitue, an the largest aeleration is otaine when x =, the aplitue o the osillation The requeny is 50 Hz The ritial g 98 point is when N = 0, ie g = 4π an so = = = π 4π 5 The aplitue ust not exee this value t the lowest point: 4π N = g+ 4π N = + N = 5 N N g = a = π a We have siple haroni osillations aout a ean value o sale reaing o 70 kg whih is the ass o the passenger The perio is 10 s t the top point o the osillations the reaing is 50 kg an at the otto it is 90 kg t the lowest point (see Q17) we have that 4π N g = + 4π = g + T 70 4 π = = 71 (This passenger has proaly regrette uying a tiket on this partiular ruise) Copyright Carige University Press 011 ll rights reserve Page 4 o 10

5 Carige Physis or the IB Diploa 4πx 19 a The volue within the sphere o raius x is an that o the entire 4πR x sphere is ; thereore the ass enlose is the ration M R Mx GMx F = G R = x R GMx GM The aeleration o the ass is given y a = a x R = R, whih is GM the onition or SHM with = R T = π R π = GM e 6 (64 10 ) T = π = 5085 s = 85 in Fro gravitation we know that π v = = = T = π v GM GM R R R R R T GM as in 0 a The tension in eah string is T Taking oponents o the two tension ores we see that the horizontal oponents T osθ anel out, leaving a ownwar net ro the vertial oponents o T sinθ T sinθ Fro Fnet = a we get a = T sinθ g an hene a = g Sine the tension is very large an the ass is sall, the irst ter is oun to oinate, an so we ay neglet the aeleration o gravity in the expression T sinθ aove Hene, approxiately, a = x x 4T Now sinθ tanθ = = an inally a= x L L L 4T This eans we will have siple haroni osillations with = L The perio is given y π L π = 4 T 1 The perio o osillation oes not epen on the aeleration o gravity elerating ay e thought o as hanging the aeleration o gravity (inreasing it or upwars aelerations an ereasing it or ownwar aelerations) Hene the perio will e unaete Copyright Carige University Press 011 ll rights reserve Page 5 o 10

6 Carige Physis or the IB Diploa The isplaeent is in general y= os( t+ φ) an so the veloity is v= sin( t+ φ) So the initial isplaeent is x0 = osφ an the initial veloity is v0 = sinφ In other wors, x0 osφ = v0 sinφ = But os φ sin φ 1 + =, ie = 1 = x0 + x v v a In this ase the putty will stik to the ass eause o ritional ores etween the, an so soe energy will e lost reuing the aplitue o osillation The ass will inrease, inreasing the perio o osillations When the lok is oentarily at rest the putty will rest on it without any ritional ores ating an so the energy o the spring ass syste will e unaete Hene the aplitue o osillation will also e unaete Sine the ass has inrease, however, the perio will inrease as well 4 a The aplitue is 060 The requeny is π = 680 = Hz 1 1 an the perio is T 094 s 108 The total energy is ET 1 = Now (680) (060) 540 J E T = = = π = 680 s an so The elasti potential energy is (680) (015) EP = x = = 0650 J The kineti energy is thereore E K = = 475 J 5 a When extene y an aount x, the ore pulling ak on the oy is kx an so k k 10 a = kx a = x, an so = = = 1095 s, giving a 0 π π perio o T = = = 057 s 1095 With the springs onnete this way, an the ass pulle to the sie y a sall aount, one spring will e opresse an the other extene Hene the net ore on the ass will still e kx so the perio will not hange Copyright Carige University Press 011 ll rights reserve Page 6 o 10

7 Carige Physis or the IB Diploa 6 a t the top the woan s total energy is gravitational potential energy equal to gh, where h is the height easure ro the lowest position that we seek t the lowest position all the gravitational potential energy has een onverte 1 into elasti energy kx an so 1 gh = kx 1 Sine h= 15 + x we have that ( 15) gh = k h We ust now solve or the height h : h= 0 ( h5) 600h= 110( h 0h+ 5) 110h 900h = 0 11h 90h+ 475 = 0 The physially eaningul solution is h = 7 e The ores on the woan at the position in a are her weight vertially ownwars an the tension in the spring upwars Hene the net ore is F = T g = kx g = 0 (7 5) 600 = 084 N ; hene net F net 084 a = = = 47 s 60 Let x e the extension o the spring at soe aritrary position o the woan Then the net ore on her is Fnet = T g = kx g irete upwars, ie opposite to the iretion o x So a = ( kx g) The aeleration is not proportional to the isplaeent, so it looks as i we o not have SHM But we ust easure isplaeent ro an equiliriu position This is when the extension o the spring is x 0 an kx0 = g In other wors all the isplaeent to e y = x x0 Then a = ( k( y + x0) g) = ky kx0 + g = ky sine k kx0 = g Hene, we o have the onition or SHM n so a = y so that k π = = 191s an inally T = = 8 s She will oe to rest when the tension in the spring equals her weight, ie when g kx0 = g x0 = = = 7 Hene the istane ro the top is k = 177 Sine she will peror any osillations eore oing to rest, the osillations are unerape It has een onverte to other ors o energy, ainly theral energy in the air an at the point o support o the spring Copyright Carige University Press 011 ll rights reserve Page 7 o 10

8 Carige Physis or the IB Diploa 7 n SHM osillation is haraterise y the equation a= x With aping there will e, in aition, a ore opposing the otion, whih is usually proportional to the spee o the partile Depening on the agnitue o this ore we an have three istint ases: i ii iii very sall opposing ore in whih ase the osillations ie out slowly speii larger ore whih akes the partile return to its equiliriu position as ast as possile ut without osillations Finally, any ore larger than the ore in ii, in whih ase the partile again returns to its equiliriu position without osillations ut in a tie longer than that in ii 8 Desirale: i ii Shok asorers in ars It is esirale eause without aping every tie the ar hits a up the ar (an its passengers) woul suer unoortale osillations up an own Sales that are use in weighing things I the osillations o the pointer were not ape, the pointer woul osillate or a long tie eore settling own to the true reaing o the ass o the ojet eing weighe Unesirale: i ii The osillations in your granather s lok hanging on the wall The osillations o a hil in a swing when the ather is too lazy to push! n a The ratio o suessive aplitues oes appear to e onstant at aout 06 n The ration o energy lost is n n n = n + 1 The ratio o suessive aplitues woul now e saller than 06 an the n perio woul get very slightly longer 0 See Figures 117, 118 an 119 (pages 07 an 08 in Physis or the IB Diploa) 1 They shoul ieally e ritially ape so that the passengers o not go on osillating or a long tie ater the ar hits a up in the roa as they woul in the ase o unerape osillations They shoul not e overape sine in that ase the ar woul return to its noral (an sae) onition or riving in a longer tie s hinte at in Q1, overape shok asorers ight ake the passenger eel the up less ut it woul take a longer tie to ring the ar into its noral riving onition This is not sae as the rakes will take a longer istane to ring the ar to a stop in this ase Copyright Carige University Press 011 ll rights reserve Page 8 o 10

9 Carige Physis or the IB Diploa See Figures 117, 118 an 119 (pages 07 an 08 in Physis or the IB Diploa) 4 a The angular requeny is k = an the requeny is oun ro = π The isplaeent is in general y= os( t+ φ) an so the veloity is v= sin( t+ φ) Hene x os( t + φ) = v sin( t + φ) = But os ( t φ) sin (( t φ) =, ie x v + = 1 e The graph is an ellipse entre at the origin The area o the ellipse in is π With aping, the energy o the syste will e reue an the aplitue will erease Hene the area o the ellipse will erease This will happen as the graph in eoes an elliptial spiral that loses onto the origin 5 Free osillations are osillations in whih no other ore, other than the restoring ore that is opposite an proportional to isplaeent, at on the syste In the ase o ore osillations there are other aitional ores ating on the syste 6 ny struture has its own natural requeny o osillation When the struture is sujete to an external perioi ore the aplitue o the resulting osillations will e the largest when the natural requeny o the syste an that requeny o the external ore are the sae This is alle a state o resonane n exaple o resonane is the irowave oven, in whih irowaves o requeny equal to that o the virating water oleules are irete at oo The water oleules osillate with the greatest aplitue in this ase, whih eans that the teperature o the water oleules in the oo rises an the oo is ware 7 In ase the perioi arhing gets the rige in resonane with angerous largeaplitue osillations that ight reak the rige Copyright Carige University Press 011 ll rights reserve Page 9 o 10

10 Carige Physis or the IB Diploa 8 a 1 k = π The net ore on the ass is the tension in the spring, kx an so where L 0 is the natural length o the spring πr (π R ) Now, v = = πr an so kx T L0 + x = We ust solve this equation to in the extension x : 4π ( L + x) = kx 4π L x( k 4π ) 0 0 = 4π L0 x = k 4π v kx L + x = 0 The enoinator in is 1 k k 4π = k 4π = 0, whih is ipossile 4π We have a ase o resonane where a isplaeent eoes unontrollaly large This is an iealise situation, though, where aping has een opletely neglete In a real situation with rition an aping this woul not happen Copyright Carige University Press 011 ll rights reserve Page 10 o 10