πx 4πR and that of the entire sphere is therefore the mass

Transcription

1 Answers to test yoursel questions Topi 9 9 imple harmoni motion They are not simple harmoni beause as shown in the textboo the restoring ore whereas opposite to, is not proportional to the isplaement away rom the equilibrium position I howeer the amplitue o osillations is small the ore oes beome approximately proportional to the isplaement an the osillations are then approximately simple harmoni a We notie that x os π ωt x sinωt an so the phase is π b At t the equation says that x x osφ The next time x assumes this alue is at a time gien by x os( ωt φ) x osφ Thus we must sole the equation os( ωt φ) osφ This means that the angles π ωt φ an φ ier by π an so solutions are ωt φ φ π T a At t we hae y os( ) mm ω b At t s we use the alulator (in raian moe) to in y os( ) mm os( t) t os 9 t 99 s Use ± ω x x We now that ω s Thereore, 6 ± x x 9 x ± mm a The equation is simply y os( π t) os( π t) b The eloity is thereore π sin( π t) an the aeleration is a ( π ) os( π t) At t s we ealuate (in raian moe) y os( π ) m, π sin( π ) m s an a ( π ) os( π ) 6 m s The angular requeny is ω π π 6 9π The maximum eloity is ω A 9π m s an the maximum aeleration is ω π A 9 ( ) m s 6 a The equation o the string may be rewritten as y ( 6 sin( πx) ) os( π t) rom whih we eue that the requeny o all points is Hz an that the phase o all points is zero b From a the amplitue is A 6 sin( π x) an so is ierent or ierent points on the string The maximum amplitue is obtaine when sin( πx ), ie the maximum amplitue is 6 mm The isplaement is always zero at the ens o the string, in partiular at the right en where x L, the length o the string The isplaement is zero all the time when 6 sin( πx ) ie when πx π x m L e When x m the amplitue is 6 sin( πx ) 6 sin( π ) mm a The area is approximately m (the exat alue is m) b This is the isplaement rom when the eloity is zero to when it is zero again ie rom one extreme position until the other ie twie the amplitue t The perio is s an so the equation or isplaement is x π sin t sin ( π ) PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE ANWE TO TET YOUELF QUETION 9

2 We nee to graph the equation a ω π π x where ω s The slope woul be ω s T or just sine we are plotting m on the horizontal axis a/m s x/m 9 a The eining relation or HM is that a ω x whih implies that a graph o aeleration ersus isplaement is a straight line through the origin with negatie slope just as the gien graph b The slope o the graph gies ω The measure slope is s an so ω s Thus the π perio is T 6 s The maximum eloity is ω A 9 m s The maximum net ore is ma mω A N e The total energy is ET mω A J a The ores on the mass when the plate is at the top are shown below: N mg The net ore is mg N ma ine we hae simple harmoni motion a ω x π x in magnitue, an the largest aeleration is obtaine when x A, the amplitue o the osillation The requeny is Hz The g ritial point is when N Ie g π A an so A 9 99 m The amplitue must π π not exee this alue b At the lowest point: N mg ma mπ A N mg mπ A N 9 π 99 N N a The olume within the sphere o raius x is πx π an that o the entire sphere is thereore the mass enlose is the ration M x Mx b F G m GMmx x ANWE TO TET YOUELF QUETION 9 PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE

3 GMmx The aeleration o the mass is gien by ma GM a x whih is the onition or HM with ω GM T e T π π ω π GM 6 ( 6 ) s min From graitation we now that m GMm GM T π T π as in GM a When extene by an amount x the ore pulling ba on the boy is x an so ma x a m x an so ω 9 s giing a perio o m π π T s ω 9 b With the springs onnete this way, an the mass pulle to the sie by small amount one spring will be ompresse an the other extene Hene the net ore on the mass will still be x so the perio will not hange a At the top the woman s total energy is graitational potential energy equal to mgh where h is the height measure rom the lowest position that we see At the lowest position all the graitational potential energy has been onerte into elasti energy x an so mgh x ine h x we hae that mgh ( h ) We must now sole or the height h: 6 h ( h ) 6h ( h h ) h 9h h 9h The physially meaningul solution is h m b The ores on the woman at the position in (a) are her weight ertially ownwars an the tension in the spring upwars Hene the net ore is Fnet T mg x mg ( ) 6 N hene Fnet a m s m 6 Let x be the extension o the spring at some arbitrary position o the woman Then the net ore on her is Fnet T mg x mg irete upwars ie opposite to the iretion o x o ma ( x mg) The aeleration is not proportional to the isplaement so it loos we o not hae HM But we must measure isplaement rom an equilibrium position This is when the extension o the spring is x an x mg In other wors all the isplaement to be y x x Then ma ( ( y x ) mg) y x mg y sine x mg Hene we o hae the onition or HM An so a m y so that ω π ω 9 s an inally T s m ω he will ome to rest when the tension in the spring equals her weight ie when mg 6 x mg x m Hene the istane rom the top is m e It has been onerte to other orms o energy mainly thermal energy in the air an at the point o support o the spring PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE ANWE TO TET YOUELF QUETION 9

4 9 ingle-slit iration The iration angle is θ ra an so the angular with is ouble this ie 666 ra or b 6 The iration angle is θ ra an so the angular with is ouble this ie ra The b linear with is thereore θ m 6 a The iration angle is about θ ra an so b b b In i we hae a smaller with an so a larger iration angle In ii there will be no hange sine both waelength an with hale Notie howeer that i we were to pay attention to the ertial axis sale, with a smaller slit with less light woul go through so in both ases the intensity woul be less Intensity 6 θ/ra 9 Intererene 9 6 The separation is gien by the boolet ormula s mm The two lashlights are not oherent This means that the phase ierene between them eeps hanging with time (ery ast, on a time sale o nanoseons) Thus, whateer intererene pattern is proue at any moment in time, a ierent pattern will be proue a nanoseon later Thereore all we an obsere is an aerage o the rapily hanging patterns on the sreen, ie no intererene at all 9 sinθ n 6 an sin θ ( n ) Thus n 6 ( n ) 6n n n a The separation o the bright ringes is s m From s we get s m b The waelength in water woul be less (by a ator o ) an so the istane woul also be less a We must hae sin an so 9 sin b From s, the istane between the bright ringes woul ouble i hales a Use sinθ n with b n n θ mm an 6 nm to get: ANWE TO TET YOUELF QUETION 9 PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE

5 There will be a phase hange o π at both reletion points an so the onition or estrutie intererene (or normal iniene) is where n is the reratie inex o the oating n This gies 6 n nm The least thiness is obtaine or an is nm The relete light must show onstrutie intererene There is a phase hange only at the top reletion so the onition or onstrutie intererene is where n is the reratie inex o the ilm Then n n 6 nm Possible alues o are then nm, nm et a Coherent light means light where the phase ierene between any two points on the same ross setion o the beam is onstant Monohromati light means light o the same waelength b The irst maximum is obsere at sinθ θ sin ra an this gies the ollowing graph Intensity θ/ra The angle in b woul now hale to ra giing the ollowing graph Intensity 6 a ee, or example, Figure 99 in the ourseboo b i The intensity inreases, the maxima beome thinner an there are seonary maxima ii the intensity o the maxima stays the same but their separation inreases θ/ra PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE ANWE TO TET YOUELF QUETION 9

6 9 esolution The angular separation o the points is θ A 6 ra The iration angle is 9 6 θ ra The objets will not be resole sine θ A < θ a The angular separation is θ A 9 an the iration angle is θ ra For resolution we nee θa θ, ie an so m b It woul erease sine the iration angle woul get smaller 9 a The iration angle is θ ra an this is the smallest angular separation that an be resole b With θ θa s we get s m a The iration angle is θ ra an this is the smallest angular separation that an be resole 6 b The angular separation o the two stars is 6 6 < θ 6 so the stars annot be resole The iration angle is θ ra The angular separation o two points on a iameter o Anromea is > θ 6 so the telesope sees Anromea as an extene objet The iration angle is θ ra When this is about equal to the angular separation o the earth an the moon, ie θ A, the objets will be resole This means m a The iration angle is θ ra b It is ree rom atmospheri isturbanes suh as light pollution, turbulene in the air et a From sinθ n, we get using the aerage waelength o the two lines: m sin b For resolution: mn N 9 m 9 9 a From 9 nm mn b Inreasing m an N both erease an so improe resolution Howeer the intensity o the light ereases with inreasing m an so it is preerable to inrease N instea 9 The oppler eet Note: Tae the spee o soun in still air to be m s 6 soure moing stationary obserer a b 6 ANWE TO TET YOUELF QUETION 9 PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE

7 a This is a ase o a soure moing towars the obserer an so 66 Hz b i 6 m ii 6 m 66 a This is a ase o a soure moing away rom the obserer an so Hz b i m ii m 9 a This is a ase o an obserer approahing a stationary soure an so the releant ormula is an so Hz b i 66 m ii 66 m 9 9 a This is a ase o an obserer moing away rom a stationary soure an so the releant ormula is an so Hz b i 6 m ii m 6 An obserer on the approahing ar will measure a higher requeny ( ) than that emitte ( ) beause we hae a ase o the oppler eet with an approahing soure The wae will then be relete with requeny The ar is now ating as an approahing soure The requeny reeie ba at the soure ( ) will be higher than that emitte rom the ar This is the ase o a ouble oppler eet The requeny reeie by the reeing obserer is (obserer moing away) The wae is relete bawars The moing obserer now ats as the soure o the waes an the requeny emitte by this soure is thereore, the original soure now ats as a stationary obserer an so the requeny it reeies is now Hene ine 96 we hae m s PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE ANWE TO TET YOUELF QUETION 9

8 ANWE TO TET YOUELF QUETION 9 PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE Hint: You an put the equation iretly into the OLVE o your graphis alulator (with x ) an get the answer immeiately without any o the teious algebra aboe The requeny reeie by the stationary obserer is (soure moing towars) The wae is relete bawars The stationary obserer now ats as the soure o the waes an the requeny emitte by this soure is The original soure now ats as a moing obserer an so the requeny it reeies is now Hene ine we hae 6 6 m s As ar as the obserer is onerne the eloity o the soure is an the spee o the wae is o using the ormula o the stationary obserer an an approahing soure we hae a The requeny emitte is The obserer is moing away so he reeies a requeny This requeny is relete rom the objet whih now ats as a reeing soure The requeny reeie ba at the original soure is then b I is small then Hene i 6 6 m s ii Beause there is a range o spees or the bloo ells an the ultrasoun is not inient normally on the ells m s

9 a m s b The spee erie in a is just the omponent o eloity along the line o sight, not the total eloity π π The spee o a point on the un s equator is 9 m s T 6 6 The emitte requeny is Hz The shits are then 9 Hz 9 a There is no shit sine the eloity is at right angles to the iretion o obseration The stars are neither approahing or moing away rom the obserer at that time b The spees o the stars are m s 6 6 m s an PHYIC FO THE IB IPLOMA CAMBIGE UNIVEITY PE ANWE TO TET YOUELF QUETION 9 9