1 Rörelse 1.1 1.3.3 Kurvlinjär rörelse 1.1.1 Kraft Acceleration F = ma Bärkraft F = ρgV Coulomb F = qE Friktion F = µFn 1.1.2 Fjäder F = −kx Potential F = −grad(φ) Magnet F = q(v × B) Energi 2 d2 x dt2 d2 x dt2 b= 1.1.3 3 R Generellt W = S F · dl H Konservativa W = E(x2 ) − E(x1 ), F · dl = 0 1.1.4 3.4.1 + λ1 dx + λ2 x = c ⇒ Ae r dt λ2 − e + Singelspalt m-te minimum d sin(θ) = mλ 3.4.2 √ √ + λx = c ⇒ A sin( λ + φ) + B cos( λ + φ) + −1 2 λ1 t i(bt+φ) Interferensen och diffraktion c λ c λ2 λ2 1 4 Dubbelspalt interferens Konstruktiv: d sin(θ) = mλ Destruktiv: d sin(θ) = (m + 21 )λ För små θ ym = R mλ d 3.4.3 Multispalt m-te maximum Rak d sin(θ) = mλ Sned d sin(φ) + d sin(θ) = mλ 3.4.4 Ljus Cirkulär spalt första minimum D sin(θ1 ) = 1.22λ Synligt: 400nm till 750 nm Ljusets energi Ey = h f = hc λ (Plancks konstant h = 6, 62607 ∗ 10−34 ) 3.4.5 3.1 4 Generella vågor k= 2π λ Rayles kriterium Två objekt kan särskiljas om: θR = 1.22 λd Effekt Medel P = 1.1.5 ∆W ∆t Momentan dW dt =F·v Definitioner Elastisk: Energi och rörelsemängd konserveras Stängt system E1 = E2 Öppet E1 + Ein = E2 Inelastisk: Rörelsemängd konserveras Stängt system E1 = E2 − E f Öppet E1 + Ein = E2 − E f BKontrollera P alltid rörelsemängden Jämvikt: F = 0 Spänning (T) Dragande kraft i rep 1.1.6 Cirkulär rörelse ω = 2π f S = rθ[rad] θ̈ = ω̇ = α 1.2.1 Radiell komponent Acceleration arad = rω2 = Kraft Frad = mrω 1.2.2 2 Polarisering Malus lag för linjärt polariserat ljus: I = I0 η cos2 (θ) (η är förluster i filtret) ~ (Vinkel θ mellan genomsläppsriktningen och E) I Malus lag för opolarizerat ljus: I = 20 Polarisering på en yta: P polariserat: Parallell med planet av incident S polariserat: Ortogonal med planet av incident 3.2 Sfärisk spegel 1 2 1 s + s0 = R = 0 1 f Sfärisk refraktion nb nb −na na s + s0 = R 0 m = − nna ss m = − ss b Lins avstånd s från objekt, bild hamnar på avstånd s’ från linsen m är förstoringen s > 0 objekt på infallande sida s0 > 0 objekt på utgående sida m < 0 inverterad bild R = ∞ ⇔ f = 0 betyder plan Linsmakarens formel Tangentiell komponent Hastighet av en våg q v= tråd v= F µ q vätska v= q B ρ Y ρ 4.2 Effekt/Power Pav = 1 2 4.3 Stående våg √ µFω2 A2 y = 2Asin( 2πx λ ) För trådar: f1 = 4.4 1 2L q F µ Intensitet P A 4.4.1 β = 10log10 ( II ) 0 I0 = 10−12 [Wm−2 ] Interferens ∆φ = 2nπ ⇔ Konstruktiv interferens ∆φ = (2n + 1)π ⇔ Destruktiv interferens Intensitet vid interferens: √ Itot = I1 + I2 + 2 I1 I2 cos(∆φ12 ) Moment τ = Iα τ = Ftan r τ = r × F Tröghetsmoment Punktmassa I = mr2 2 Disk rotation runt egen axel I = m r2 Rotation distansen d från masscentrum Id = Imc + md2 Rörelsemängdsmoment P L = Iω Rörelsemängd bevaras L = 0 mi r Tyngdpunkt rtyngd punkt = m +m y(x, t) = A cos(kx − ωt) v = λf I= Fart v = rω Acceleration atan = rα Kraft Ftan = mrα 1.2.3 ω = 2π f Geometrisk optik 3.2.1 v2 r 4.1 solid Annat Rörelsemängd p = mv Position, Fart, Acceleration ẍ = v̇ = a 1.2 3.4 Diff-ekvationer Kinetisk Ek = 21 mv2 Elastisk E f R= 12 kx2 Gravitation Eg = mgh Kraft E = S F · dl Deformationsenergi: skillnad i kinetisk energi före och efter en stöt Arbete Drivna system F = −kx − bv + Fd (b dämpningsfaktor) Resonans ω = ωd Odämpat ⇒ A → ∞ Fmax A= q (k − mω2d )2 + b2 ω2d P i 1.2.4 W= R j Fasskillnad brytning (anti-reflex): ∆φ12 = τdθ Er = Pendel Fjädrar 1.3.1 F = −kx Odämpade system ω= q 2π λ0 nb 2t 1 2 2 Iω Svägning: x(t) = xm cos(ωt + φ) q q Simple: ω = Lg Fysisk: ω = mgd I k m Etot = E f + Ek = 12 kx2 + 12 mv2 Max utsträckning Etot = E f = 21 kx2 Jämnviktspunkt Etot = Ek = 12 mv2 1.3.2 Gränsytor Arbete och Energi 1.2.5 1.3 3.3 Dämpade system F = −kx − bv (b dämpningsfaktor) Etot = E f + Ek − E f = 12 kx2 + 21 mv2 − bv2 √ Dämpning: b < 2 km ⇔ underdämpad √ b = 2 √km ⇔ kritisk dämpad b > 2 km ⇔ över dämpad 4.5 Reflektionslagen: θi = θr Refraktionslagen: n1 sin(θ1 ) = n2 sin(θ2 ) n Kritisk vinkel: sin(θcrit ) = n2 1 Totalreflektion vid infallsvinkeln θcrit n Polariseringsvinkeln (Brewster): tan(θ p ) = n2 1 Reflekterade ljuset S polariserat vid infallsvinkel θ p Reflex i högre n ger fasskift på π 3.3.1 n= c v 3.3.2 Ir I0 Ir I0 Fasförskjutning ∆φ = ∆φ12 + k∆r + t∆ω + ∆R ∆φ12 Skillnad i utgångsfas ∆r Skillnad i gångväg (optisk + geometrisk) ∆ω Skillnad i fas ∆R Fasförskutning som resultat av reflektion. Ljus: Om n2 > n1 ⇒ ∆R = π annars ∆R = 0 Ljud: ∆R = π Tråd: Fast ∆R = π Fri ∆R = 0 Brytningsindex λn = λ n LO = Ln 4.6 Fresnels lag n1 cos θi − n2 cos θt n1 cos θi + n2 cos θt n1 cos θt − n2 cos θi = Rp = n1 cos θt + n2 cos θi = Rs = T s = 1 − Rs T p = 1 − Rp Dopplereffekt Ljud 2 2 v + vL f s , fL lyssnare, f s source v + vs Ljus q fR = f s c−v c+v ( fR Reciever, om relativ rörelse från v > 0, annars fL = v < 0, c = 3 ∗ 108 m/s) 5 Ljudvågor befinner sig a) om högtalarna sänder ut musik från två olika stereoanläggningar? eller b) ljudfarten i rummet är 344 m/s och högtalarna avger en ton med frekvensen 387 Hz samt är kopplade med lika långa sladdar till samma funktionsgenerator? v = 340[ms−1 ] dρ dp Tryckmodulen B = ρv2 = −V dV = ρ dV Kompressibilitet B−1 Övertrycket [Pa=Nm−2 ]: p(x) = BkA sin(kx ± ωt) = pmax sin(kx) 5.1 Akustisk impedans √ = ρv = qB Ir (Z2 − Z1 )2 R = = 2 Ii (Z2 + Z1 ) Z= 8.2.1 p(x) vy (x) 5.2 T= 4Z2 Z1 It = Ii (Z2 + Z1 )2 ! Intensitet I = p(x)vy (x) = BkωA2 sin2 (kx ± ωt) p2max √ Imedel = 21 BkωA2 = 12 ρBω2 A2 = 2|Z| 5.3 Stående ljudvåg Öppet rör fn = 6 nv 2L Stängt rör fm = nv 4L Enheter Frekvens Vinkelfrekvens Kraft Energi Effekt Intensitet Tryck Tröghetsmoment Rörelsemängd Rörelsemängdsmoment 7 f=[Hz] ω = [rad/s] N=[kgm/s2 ] J=[Nm] W=[J/s] I=[W/m2 ] p=[Pa]=[N/m2 ] I=[kgm2 ] P=[Ns] L=[Nms] Differentialekvationer Homogena lösningen: y”+ay’+by=0, karaktäristik ekvation r2 + ar + b = 0 Två reella rötter: y = Aer1 x + Ber2 x En reell dubbelrot: y = erd x (Ax + B) Två icke reella rötter: y = eαx (Acos(βx) + Bsin(βx)), , r = α ± βi Partikulära lösningen: Högerled: → Ansats: x+6 → ax+b x2 + 6x + 3 → ax2 + bx + c 4e3x → aekx sin(kx) → Asin(kx) + Bcos(kx) 8 Exempelsamling 8.1 Fråga interferens Två identiska högtalare till en stereoanläggning vid en utomhuskonsert är placerade på avståndet 6,00 m från varandra. Vid en testning av anläggningen skickas en signal med frekvensen 200 Hz till de båda högtalarna. a) Hur många minimum i ljudintensitet kommer en person som från början befinner sig långt bort framför den ena av högtalarna och sedan vandrarmot densamma att uppfatta? b) Hur långt från högtalaren befinner sig personen i dessa fallen? 8.1.1 Lösningsförslag Antag att betraktaren befinner sig på avståndet L framför den ena högtalaren. Pythagoras sats ger oss att vägskillnaden mellan vågorna från de båda högtalarna kan tecknas ∆L = L1 − L2 = √ L2 + d2 − L. Detta ger en fasskillnaden mellan vågorna kan v tecknas ∆Φ = k ∗ ∆L = 2π λ ∆L där våglängden λ = f = 1, 65m. √ 2π 2 ∆Φ = 0 då L = oändligheten och ∆Φ = 1,65 ( 0 + 62 − 0) = 7, 27π då L = 0. Interferensminimum (destruktiv interferens) fås då ∆Φ = (2p + 1)π och p är ett heltal. Följdaktligen kommer minimum att inträffa då ∆Φ = π, 3π, 5πoch7π (motsvarande att p = 0, 1, 2 och 3), vilket ger fyra minimum i ljudintensiteten när betraktaren förflyttar sig mot högtalaren. För att beräkna avstånden från högtalaren när dessa minimum inträffar, löser man ut L ur villkoret för destruktiv interferens. Avstånden från högtalaren erhålls nu genom insättning av de olika p som ger minimum. 8.2 Fråga akustik Två identiska högtalare som vardera avger den totala ljudeffekten 1 mW lika fördelat över alla riktningar i rummet står placerade på ett avstånd av 5,0 m från varandra. En person står på avståndet 3,0 m från den enda och 7,0 m från den andra. Vilken blir ljudintensitetsnivån (uttryckt i dB relativt I0 ) där personen Lösningsförslag a) Eftersom högtalarna är kopplade till olika stereoanläggningar finns det ingen bestämd fasrelation mellan de ljudvågor som sänds ut av högtalarna. Ljudet är således inkoherent och vi kan addera intensiteterna rakt av. Intensiterna från vardera av högtalarna i den punkt där personen befinner sig är I1 = P 2 = 8, 84 ∗ 106Wm2 I2 = P 4π(r2 )2 4π(r1 ) = 1, 62 ∗ 106Wm2 vilket ger att den I +I totala intensiteten är β = 10 log IIt ot = 10 log 1I 2 = 70, 2dB b) 0 0 Nu är ljudvågorna koherenta, vilket betyder att det nns en konstant fasskillnad mellan de vågor som når personen i rummet. På grund av denna fasskillnad kommer vågorna att interferera när de överlagras. I detta fallet är det således vågornas amplituder som ska adderas, inte dess intensiteter. Först måste dock fasskillnaden mellan vågorna beräknas på vanligt sätt: ∆Φ = 2π f 2π λ ∆L = v (L2 − L1 ) = 9π. Eftersom man får destruktiv interferens när fasskillnaden ∆Φ = (2p + 1)π. och p är ett heltal, kommer vågorna från högtalarna att vara i motfas när de når personen (faskillnad ). Den totala intensiteten måste nu beräknas utående från en vektorsummering av tryckamplituderna. I detta fallet är emellertid fasskillnaden varför man kan subtrahera am√ 2 t ot) = I + I 2 I I Insättning plituderna från varandra It ot = (P2|z| 1 2 1 2 av I1 och I2 ger Itot som ger β. 8.3 Fråga transmittans En ljudvåg med frekvensen 400 Hz faller in vinkelrätt från luft mot vatten. a) Beräkna förhållandet mellan den transmitterade vågens och den infallande vågens intensiteter, IIt , samt b) i beräkna förhållandet mellan den transmitterade vågens tryck och Pt den infallande vågens tryck, P . Ljudfarten i luft är 340 m/s och i i vatten är den 1500 m/s. Densiteterna är 1,29 kg/m3 respektive 1000 kg/m3. 8.3.1 Lösningsförslag a) När ljudvågen träffar vattenytan kommer en del av intensiteten att reflekteras och en del att transmitteras. Med hjälp av refektionskoeficienten R för en våg som träffar en yta och villkoret att intensiteten (effekten per yta) ska vara kontinuerlig i gränsytan z −z 2 (Ii = Ir +It ), får vi att It = Ii −Ir = Ii −R∗Ii = Ii [1− zl +zvv ) Där zl l och zv är de akustiska impedanserna för luft respektive vatten, ρl och vl är densiteten respektive ljudfarten i luft och ρ v och vv är motsvarande storheter för vatten. Det sökta förhållandet blir IIt b) i Intensiteterna hos den infallande, den reflekterade och den trans(∆Pi )2 2|Zi | (∆Pi )2 2ρl vl mitterade vågen kan allmänt tecknas som Ii = = samma för Ir och It . där ∆P är motsvarande övertryck orsakat av vågen. Om vi i dessa ekvationer dividerar den transmitterade vågens intensitet med den infallande vågens intensitet får 2ρl vl t 2 vi att IIt = ( ∆P ∆Pi ) ∗ 2ρv vv Eftersom den transmitterade vågen i med nödvändighet måste ha samma fas som den infallande, ska vi välja plustecknet framför rottecknet. Med insatta numeriska t värden fås det sökta förhållandet ∆P ∆P . i 9 Fråga dopplereffekten 24. Ett flygplan passerar med konstant hastighet och på låg höjd en observatör på ett flygfält en tidig vårdag då termometern visar + 5◦ C. Vilket fart ska planet ha för att frekvensen hos bullret ska tyckas dubbelt så högt före passagen som efter? 9.0.1 vinkelrätt infall av gult ljus (590 nm). Antag att man istället vinklar glasytan. Vid vilken infallsvinkel kommer man att se maximal reflektans för rött ljus med våglängden 650 nm? 9.1.1 Lösningsförslag 32. Enligt uppgiften ska skiktets tjocklek vara d = 34 , där givetvis ska tolkas som våglängden inuti antireflexskiktet, Vi får 3∗559 att d = 4∗1,38 = 320, 7nm. När glasytan istället vinklas har vi att interferensvillkoret vid snett infall kan skrivas 2ndcosθb = p, vilket med det minsta värdet på p ( = 1) ger med insatta värden att 650 = 42.74◦ Infallsvinkeln beräknas ur bryθb = arccos( 2∗1,38∗320,7 tningslagen sinθi = nsinθb − > θi = arcson(1, 38sinang42, 74) = ◦ 69.50 9.2 Fråga polarisation 40. Två polarisatorer är orienterade relativt varandra så att inget ljus transmitteras genom den andra polarisatorn (analysatorn). Mellan polarisatorerna placeras ytterligare en vridbar polarisator. Härled den formel som visar hur det transmitterade ljusets intensitet varierar med orienteringen av den mellersta polarisatorn. För vilka vinklar erhålls intensitetsmaximum? 9.2.1 Lösningsförslag 40. Eftersom polarisatorerna i ursprungsläget inte släpper igenom något ljus, betyder det att de är korsade, dvs den andra polarisatorns genomsläppsriktning ligger 90◦ förskjuten i förhållande till den första (inget av det linjärpolariserade ljuset från den första polarisatorn släpps igenom). När en tredje polarisator nu förs in mitt emellan dessa båda polarisatorer, kommer denna polarisators genomsläppsriktning att ligga en vinkel θ förskjuten i förhållande till den första polarisatorn. Enligt Malus lag kommer då den intensitet som lämnar den mittersta polarisatorn att vara I = I1 cos2 (90◦ − θ) = I0 sinθ . Vinkeln mellan genomsläppsriktningarna för denna mittersta polarisator samt den sista polarisatorn blir i detta fallet 90◦ θ. Malus lag ger nu att den intensitet som passerar den sista polarisatorn är 9.3 Fråga polarisering 42. Opolariserat ljus träffar en glasplatta under en infallsvinkel som gör att det reflekterade ljuset är linjärpolariserat. Hur många procent av den infallande intensiteten transmitteras in i plattan om plattan har brytningsindex 1,50 och omges av luft? 9.3.1 Lösningsförslag 42. Enligt uppgiften är det reflekterade ljuset linjärpolariserat. Detta kan endast ske ifall det opolariserade ljuset faller in under polarisationsvinkeln (Brewstervinkeln). Detta betyder att n las infallsvinkeln mot glasytan ges av: tanθi = ngust − > θi = l ◦ arctan( 1,50 1,00 ) = 56.31 samt att brytningsvinkeln ges ur 1, 00 ∗ sinθi = 1, 50 ∗ sinθb , vilket get θb = 33.69◦ (detta kan även i detta speciella fallet räknas fram utgående från att Brewstervinkeln inträffar när θi + θb = 90◦ ). Då vi har opolariserat ljus som faller in mot ytan, är det rimligt att i medeltal beskriva det som att hälften av ljusets intensitet ligger polariserad i infallsplanet och hälften ligger polariserad vinkelrätt mot infallsplanet. Summering av dessa båda bidrag tillsammans med att It = Ii − Ir för vardera komponenten ger att It,tot = It,s + It,p = sin(θi −θb ) 2 tan(θi −θb ) 1 1 2 = 0, 9260 ∗ I0 2 Ii (1 − ( sin(θ +θ ) ) ) + 2 Ii (1 − ( i 9.4 tan(θi +θb )) ) b Fråga linser Lösningsförslag Frekvensen hos ljudvågen ändras beroende på flygplanets rörelser, vilket visar att det rör sig om Dopplereffekt. Eftersom det endast är vågkällan som rör sig (observatören står still), gäller att frekvenserna då flygplanet närmar sig respektive avlägsnar sig v v kan tecknas fn = ( v−v ) f och fa = ( v+v ) f Villkoret i uppgiften s s att fn = 2 fa ger att v s = 13 v Ljudfarten v den aktuella dagen kan q exempelvis bestämmas ur v = γRT M ger att v s = 111, 7 m/s där C vi hämtat, γ = Cvp , allmänna gaskonstanten, R, samt molmassa för luft, M, ur tabell. 50. Ett sätt att komma över skriven information på olagligt sätt är att med kikare läsa den genom ett fönster. Antag att vi har en normal text med linjeavståndet 5 mm som vi vill kunna läsa på 50 m håll med hjälp av en kamera med teleskoptillsats. Första linsen har fokallängden 400 mm. 446,0 mm bakom denna placeras en lins med fokallängden 40 mm. Filmen placeras sedan där en skarp bild av det 50 m avlägsna objektet hamnar. Hur stort blir linjeavståndet i bilden? 9.4.1 Lösningsförslag Linsformeln ger för den första linsen att 1 50000mm 9.1 Fråga reflektion 32. Betrakta en glasyta (brytningsindex 1,60) som är belagd med ett antireflexskikt (brytningsindex 1,38) som inte är tunnast möjliga utan istället för en tjocklek som motsvarar λ4 har en tjocklek som motsvarar 3λ 4 . Skiktet ger minimal reflektion vid + 1 q1 = > q1 = 403, 2mm Bilden från den första linsen blir objekt till den andra linsen. Avståndet från mellanbilden till andra linsen är p2 = (446 − 403, 2)mm = 42, 8mm. Linsformen på an1 1 dra linsen ger att 42,8 + q1 = 40mm − > q2 = 617mm. Totala 1 400mm − 2 förstoringen i linssystemet blir M = M1 ∗ M2 = q1 p1 ∗ q2 p2 −>M= = 0, 12 vilket ger att linjeavståndet i bilden blir 0, 12 · 5 mm = 0,58 mm 403,2∗617 50000∗42,8 Tentafrågor Hej kompis, lycka till på Tentan! r Straight-line motion, average and instantaneous xvelocity: When a particle moves along a straight line, we describe its position with respect to an origin O by means of a coordinate such as x. The particle’s average x-velocity vav−x during a time interval ∆t = t2 − t1 is equal to its displacement ∆x = x2 − x1divided by ∆t . The instantaneous x-velocity vx at any time t is equal to the average x-velocity for the time interval from t to t + ∆t in the limit that ∆t goes to zero. Equivalently, vx is the derivative of the position function with respect to time. vav−x = x2 − x1 t2 − t − 1 ∆x vx = ∆t line, their components along that line are related by Eq. (3.33). More generally, these velocities are related by Eq. (3.36). t2 − t1 = (relative velocity along a line) ∆vx ∆t Straight-line motion with constant acceleration: When the x-acceleration is constant, four equations relate the position x and the x-velocity vx at any time t to the initial position x0 , the initial x-velocity v0x (both measured at time t = 0), and the x-acceleration ax . Constant x-acceleration only: 2 v0x + vx F̄ F̄ = 0 Mass, acceleration, and Newton’s second law: The inertial properties of a body are characterized by its mass. The acceleration of a body under the action of a given set of forces is directly proportional to the vector sum of the forces (the net force) and inversely proportional to the mass of the body. This relationship is Newton’s second law. Like Newton’s first law, this law is valid only in inertial frames of reference. The unit of force is defined in terms 1kg of unit of force is the newton (N), equal to 2 2 2 2 vx = v0x + 2ax (x − x0 ) x − x0 = ( X The net force on a body and Newton’s first law: Newton’s first law states that when the vector sum of all forces acting on a body (the net force) is zero, the body is in equilibrium and has zero acceleration. If the body is initially at rest, it remains at rest; if it is initially in motion, it continues to move with constant velocity. This law is valid only in inertial frames of reference. X ax t W = (relative velocity in space) P = object, B = moving air, A = ground observer Force as a vector: Force is a quantitative measure of the interaction between two bodies. It is a vector quantity. When several forces act on a body, the effect on its motion is the same as when a single force, equal to the vector sum (resultant) of the forces, acts on the body. vx = v0x + ax t 1 )t 2 m/s Freely falling bodies: Free fall is a case of motion with constant acceleration. The magnitude of the acceleration due to gravity is a positive quantity, g. The acceleration of a body in free fall is always downward. Straight-line motion with varying acceleration: When the acceleration is not constant but is a known function of time, we can find the velocity and position as functions of time by integrating the acceleration function. vx = v0x + x = x0 + Z t 0 Z t 0 vx dt v̄av = v̄ = vx = dx dt ā = t2 − t1 = ∆t dr̄ ∆r̄ lim = ∆→0 ∆t dt , vy = dy dt , vz = dz dt ∆v̄ dv̄ = lim ∆→0 ∆t dt Projectile motion: In projectile motion with no air resistance, ax = 0 and ay = -g. The coordinates and velocity components are simple functions of time, and the shape of the path is always a parabola. We usually choose the origin to be at the initial position of the projectile. x = (v0 cos α0 )t y = (v0 sin α0 )t − 1 X 2 gt 2 vx = v0 cos α0 , vy = v0 sin α0 − gt Uniform and nonuniform circular motion: When a particle moves in a circular path of radius R with constant speed v (uniform circular motion), its acceleration ā is directed toward the center of the circle and perpendicular to v̄ . The magnitude arad of the acceleration can be expressed in terms of v and R or in terms of R and the period T (the time for one revolution), where v = 2pR . T If the speed is not constant in circular motion (nonuniform circular motion), there is still a radial component of āgiven by Eq. (3.28) or (3.30), but there is also a component of ā parallel (tangential) to the path. This tangential component is equal to the rate of change of speed, dv . Forces dt in circular motion: In uniform circular motion, Acceleration in uniform circular motion: the acceleration vector is directed toward the center of the circle. The motion is governed by Newton’s second law x1 F cos φdl = = Z P 2 P1 F̄ = mā P1 F|| dl F̄ · dl̄ ∆W lim = F̄ · v̄ ∆→0 ∆t Gravitational potential energy and elastic potential energy: The work done on a particle by a constant gravitational force can be represented as a change in the gravitational potential energy Ugrav = mgy . This energy is a shared property of the particle and the earth. A potential energy is also associated with the elastic force Fx = −kx exerted by an ideal spring, where x is the amount of stretch or compression. The work done by this force can be rep- resented as a change in the elastic potential energy of the spring, Uel = 1 kx2 . 2 Wgrav = mgy1 − mgy2 = F̄AonB = − X F̄BonA Using Newton’s first law: When a body is in equilibrium in an inertial frame of reference—that is, either at rest or moving with constant velocity—the vector sum of forces acting on it must be zero (Newton’s first law). Freebody diagrams are essential in identifying the forces that act on the body being considered. Newton’s third law (action and reaction) is also fre- quently needed in equilibrium problems. The two forces in an action–reaction pair never act on the same body. (See The normal force exerted on a body by a surface is not always equal to the body’s weight. Using Newton’s second law: If the vector sum of forces on a body is not zero, the body accelerates. The acceleration is related to the net force by Newton’s second law. Just as for equilibrium problems, free-body diagrams are essential for solving problems involving Newton’s second law, and the normal force exerted on a body is not always equal to its weight. Magnitude of kinetic friction force: fk = µk n Magnitude of kinetic friction force : fs ≤ µs n Work done by a force: When a constant force F̄ acts on a particle that undergoes a straight-line displacement s̄, the work done by the force on the particle is defined to be the scalar product of F̄ and s̄. The unit of work in SI units is 1 joule = 1 newton-meter (1J = 1N m). Work is a scalar quantity; it can be positive or negative, but it has no direction in space . Kinetic energy: The kinetic energy K of a particle equals the amount of work required to accelerate the particle from rest to speed v. It is also equal to the amount of work the particle can do in the process of being brought to rest. Kinetic energy is a scalar that has no direction in as the units of work: (1J = 1N m = 1kg ṁ2 /s2 ). K = 1 mv 2 2 The work–energy theorem: When forces act on a parti- cle while it undergoes a displacement, the particle’s kinetic energy changes by an amount equal to the total work done on the particle by all the forces. This rela- tionship, called the work–energy theorem, is valid whether the forces are constant or varying and whether the particle F̄ (t2 − t1 ) = Z t X 2 t1 K1 + U1 = K2 + U2 K1 + U1 + Wother = K2 + U2 Conservative forces, nonconservative forces, and the law of conservation of energy: All forces are either conservative or nonconservative. A conservative force is one for which the work–kinetic energy relationship is completely reversible. The work of a conservative force can always be represented by a potential-energy function, but the work of a non- conservative force cannot. The work done by non- conservative forces manifests itself as changes in the internal energy of bodies. The sum of kinetic, potential, and internal energy is always conserved. ∆K + ∆U + ∆Uint = 0 Determining force from potential energy: For motion along a straight line, a conservative force is the negative derivative of its associated potential-energy function U. In three dimensions, the components of a conservative force are negative partial derivatives of U. Fx = − δx , Fy = − dU (x) dx δU δy , Fz = − δU δz Momentum of a particle: The momentum p̄ of a particle is a vector quantity equal to the product of the particle’s mass m and velocity v̄ . Newton’s second law says that the net force on a particle is equal to the rate of change of the particle’s momentum. p̄ = mv̄ X F̄ = dp̄ dt F̄ dt = p̄2 − p̄1 If . X F̄ = 0 → P̄ = const Collisions: In collisions of all kinds, the initial and final total momenta are equal. In an elastic collision between two bodies, the initial and final total kinetic energies are also equal, and the initial and final relative velocities have the same magnitude. In an inelastic two-body collision, the total kinetic energy is less after the collision than before. If the two bodies have the same final velocity, the collision is completely inelastic. Center of mass: The position vector of the center of mass of a system of particles, r̄cm , is a weighted average of the positions r¯1 , r¯2 , ... of the individual particles. The total momentum P̄ of a system equals its total mass M multiplied by the velocity of its center of mass, v̄cm . The center of mass moves as though all the mass M were concentrated at that point. If the net external force on the system is zero, the center-of-mass velocity v̄cm is constant. If the net external force is not zero, the center of mass accelerates as though it were a particle of mass M being acted on by the same net external force. P i mi r̄i P i mi P̄ = m1 v̄1 +m2 v̄2 +m3 v̄3 +... = M v̄cm When total mechanical energy is not conserved: When forces other than the gravitational and elastic forces do work on a particle, the work Wother done by these other forces equals the change in total mechanical energy (kinetic energy plus total potential energy). δU F̄ ∆t P̄ = p̄A + P̄B + ... = ma v̄a + mb v̄b + ... r̄cm = When total mechanical energy is conserved: The total potential energy U is the sum of the gravitational and elastic potential energy: U = Ugrav + Uel . If no forces other than the gravitational and elastic forces do work on a particle, the sum of kinetic and potential energy is conserved. This sum E = K + U is called the total mechanical energy. Fx (x) = − X Conservation of momentum: An internal force is a force exerted by one part of a system on another. An external force is a force exerted on any part of a system by something outside the system. If the net external force on a system is zero, the total momentum of the system P̄ (the vector sum of the momenta of the individual particles that make up the system) is constant, or conserved. Each component of total momentum is separately conserved. Ugrav,1 − Ugrav,2 = −∆Ugrav Uel,1 − U el, 2 = −∆Uel w = mg X J¯ = Z P 2 Power: Power is the time rate of doing work. The average power Pav is the amount of work ∆W done in time ∆t divided by that time. The instantaneous power is the limit of the average power as ∆t → 0. When a force F̄ acts on a particle moving with velocity Sv, the instantaneous power (the rate at which the force does work) is the scalar product of FS and Sv. Like work and kinetic energy, power is a scalar quantity. The SI unit of power is 1 watt = 1 joule/second (1W = 1J/s) P = J¯ = Fx dx 1 2 2 kx1 − kx2 = Wel = 2 2 (and sub components) Weight : X ∆r̄ P1 Z x2 Impulse and momentum: If a constant net force F̄ acts on a particle for a time interval ∆t from t1 to t2 , the impulse J¯ of the net force P is the product of the net force and the time interval. If F̄ varies with time, SJ is the integral of the net force over the time interval. In any case, the change in a particle’s momentum during a time interval equals the impulse of the net force that acted on the particle during that interval. The momentum of a particle equals the impulse that accelerated it from rest to its present speed. P 1 Newton’s third law and action–reaction pairs: Newton’s third law states that when two bodies interact, they exert forces on each other that at each instant are equal in magnitude and opposite in direction. These forces are called action and reaction forces. Each of these two forces acts on only one of the two bodies; they never act on the same body. ax dt r̄ = xî + y ĵ + z k̂ r¯2 − r¯1 W = Z P 2 v̄P /A = v̄P /B + v̄B/A R̄ = ∆vx dvx lim = ∆→0 ∆t dt x = x0 + v0x t + Work done by a varying force or on a curved path: When a force varies during a straight-line displacement, the work done by the force is given by an integral, Eq. (6.7). When a particle follows a curved path, the work done on it by a force F̄ is given by an integral that involves the angle f between the force and the displacement. This expression is valid even if the force magnitude and the angle f vary during the displacement. vP /A−x + vB/A−x Average and instantaneous x-acceleration: The average x-acceleration aav-x during a time interval ∆t is equal to the change in velocity ∆vx = v2x − v1x during that time interval divided by ∆t. The instantaneous xacceleration ax is the limit of aav−x as ∆t goes to zero, or the derivative of vx with respect to t. ax = Relative velocity: When a body P moves relative to a body (or reference frame) B, and B moves relative to A, we denote the velocity of P relative to B by v̄P /B , the velocity dx v2x − v1x Wtot = K2 − K1 = ∆K T2 to A by v̄B/A . If these velocities are all along the same lim = ∆→0 ∆t dt aav−x = R 4π 2 R of P relative to A by v̄P /A , and the velocity of B relative ∆x = arad = moves along a straight or curved path. It is applicable only to bodies that can be treated as particles. v2 arad = X F̄ext = M ācm Rotational kinematics: When a rigid body rotates about a stationary axis (usually called the z-axis), its position is described by an angular coordinate u. The angular velocity vz is the time derivative of u, and the angular acceleration az is the time derivative of vz or the second derivative of u. (See Examples 9.1 and 9.2.) If the angular acceleration is constant, then u, vz , and az are related by simple kinematic equations analogous to those for straight-line motion with constant linear acceleration. ωz = αz = ∆θ dθ lim = ∆→0 ∆t dt ∆ωz dωz d2 θ lim = = ∆→0 ∆t dt dt2 θ = θ0 + ω0z t + 1 2 2 αt Relating linear and angular kinematics: The angular speed ω of a rigid body is the magnitude of its angular velocity. The rate of change of v is a = dω/dt. For a particle in the body a distance r from the rotation axis, the speed v and the components of the acceleration ā are related to ω and a. v = rω atan = dv = r t dω = rα dt v2 arad = Wave functions and wave dynamics: X dL̄ τ̄ = dt X 2 = ω r r X X 2 2 2 I = m1 r1 + m2 r2 + ... = mi ri i K = 1 Iω 2 2 Calculating the moment of inertia: The parallelaxis theorem relates the moments of inertia of a rigid body of mass M about two parallel axes: an axis through the center of mass (moment of inertia Ic m) and a parallel axis a distance d from the first axis (moment of inertia IP ). If the body has a continuous mass distribution, the moment of inertia can be calculated by integration. τ = Fl τ̄ = r̄ × F̄ τz = Iαz W = Z theta 2 θ1 Wtot = 1 2 ω f = P = τz ωz Angular momentum: s L̄I ω̄ 1q Pav = Maxwell etc: 0 c = √ µF ω 2 A2 2 (average power, sinusoidal wave) k n = 2 r1 1 2π s 1 2 m 1 1 2 2 2 mvx + kx = kA = const 2 2 Angular SHM: Moment of intertia I and torsion constant κ s κ ,f = 1 ω = 2L = nf1 ω Physical pendulum: A physical pendulum is any body suspended from an axis of rotation. The angular frequency and period for small-amplitude oscillations are independent of amplitude, but depend on the mass m, distance d from the axis of rotation to the center of gravity, and moment of inertia I about the axis. , f1 = s mgd I v u u , T = 2π t 10−12 W/m2 x = Ae 2 0 ω = v u u k t m I nv 2L (open), fn = nv (openended) 4L Interference: ∆ϕ = 2πd λ x sin θ Waves and their properties: v = λf tan θp nb na = Thin lenses: The object–image relationship, given by Eq. (34.16), is the same for a thin lens as for a spherical mirror. Equation (34.19), the lensmaker’s equation, relates the focal length of a lens to its index of refraction and the radii of curvature of its surfaces. + 1 s0 = 1 f , 1 f = (n − 1)( 1 R1 − 1 ) R2 Camera: F ocallength f = Aperturediameter D 0 θ 25cm Simple magnifier: M = = θ f I mgd b2 4m2 law: p2 max 2 2 ψBω A = 2ψv Standing sound waves in pipes: 0 cos(ω t + φ − Brewster’s Definition of sound intensity level: fn = −(b(2m)t 2 I = Imax cos φ f-number = Intensity of a sinusoidal sound wave: 1p nb na Malus’s law: Imax is incident on a polarizing filter, used as an analyzer, the intensity I of the light transmitted through the analyzer depends on the angle φ between the polarization direction of the incident light and the polarizing axis of the analyzer. I0 Damped oscillations: LzA = mv0 cos α = LzB = mv cos θ Total internal reflection: sin θcrit = s Intensity and sound intensity level: The intensity I of a sound wave is the time average rate at which energy is transported by the wave, per unit area. For a sinusoidal wave, the intensity can be expressed in terms of the displacement amplitude A or the pressure amplitude pmax . The sound intensity level b of a sound wave is a logarithmic measure of its intensity. It is measured relative to I0 , an arbitrary intensity defined to be I n na sin θa = nb sin θb 1 pmax = BkA β = (10dB) log ω = v u 1 uF t 2L µ (string fixed at both ends) Sound waves: Sound consists of longitudinal waves in a medium. A sinusoidal sound wave is characterized by its frequency f and wavelength l (or angular frequency v and wave number k) and by its displacement amplitude A. The pressure amplitude pmax is directly proportional to the displacement amplitude, the wave number, and the bulk modulus B of the wave medium. The speed of a sound wave in a fluid depends on the bulk modulus B and density r. If the fluid is an ideal gas, the speed can be expressed in terms of the temperature T, molar mass M, and ratio of heat capacities g of the gas. The speed of longitudinal waves in a solid rod depends on the density and Young’s modulus Y. Sinousoidal sound wave: ω v u uL f = , T = 2π t L 2π g g v 2π Simple pendulum: A simple pendulum consists of a point mass m at the end of a massless string of length L. Its motion is approximately simple harmonic for sufficiently small amplitude; the angular frequency, frequency, and period then depend only on g and L, not on the mass or amplitude. s y(x, t) = (ASW sin kx) sin ωt k λ0 θr = θa When a sinusoidal wave is reflected from a fixed or free end of a stretched string, the incident and reflected waves combine to form a standing sinusoidal wave with nodes and antinodes. Adjacent nodes are spaced a distance λ > 2 apart, as are adjacent antinodes. When both ends of a string with length L are held fixed, standing waves can occur only when L is an integer multiple of λ > 2. Each frequency with its associated vibration pattern is called a normal mode. fn = n ,λ = Reflection and refraction: (Standing wave on a string, fixed at end x = 0 m c v y(x, t) + y1 (x, t) + y2 (x, t) x 1 0 µ0 Light and its properties: When light is transmitted from one material to another, the frequency of the light is unchanged, but the wavelength and wave speed can change. The index of refraction n of a material is the ratio of the speed of light in vacuum c to the speed v in the material. 2 r2 = fS v vs 8.85410−12 F /m, = µ0 = 4π10−7 N/A2 m Energy in simple harmonic motion: Energy is conserved in SHM. The total energy can be expressed in terms of the force constant k and amplitude A. rigid body rotating about axis of symmetry Rotational dynamics and angular momentum: The net external torque on a system is equal to the rate of change of its angular momentum. If the net external torque on a system is zero, the total angular momentum of the system is constant (conserved). v u uF v = t µ v + vS Shock wave: sin α = δt2 k = 2π L̄ = r̄ × p̄ = r̄ × mv̄ particle v2 Wave superposition: x = A cos ωt + φ τz dθ 1 2 2 Iω2 − Iω1 2 = v + vL fL = 1 δ 2 y(x, t) I2 = − m I W = τz (θ2 − θ1 ) = τz ∆θ (waves on a string) Wave power: 2π Fx Fx = −kx, ax = ω = Work done by a torque: A torque that acts on a rigid body as it rotates does work on that body. The work can be expressed as an integral of the torque. The work– energy theorem says that the total rotational work done on a rigid body is equal to the change in rotational kinetic energy. The power, or rate at which the torque does work, is the product of the torque and the angular velocity δx2 I1 Simple harmonic motion: If the restoring force Fx in periodic motion is directly proportional to the displacement x, the motion is called simple harmonic motion SHM. In many cases this condition is satisfied if the displacement from equilibrium is small. The angular frequency, frequency, and period in SHM do not depend on the amplitude, but only on the mass m and force constant k. The displacement, velocity, and acceleration in SHM are sinusoidal functions of time; the amplitude A and phase angle Φ of the oscillation are determined by the initial position and velocity of the body. E = Rotational dynamics: The rotational analog ofNewton’s second law says that the net torque acting on a body equals the product of the body’s moment of inertia and its angular acceleration. − t)] T 2 Torque: When a force FS acts on a body, the torque of that force with respect to a point O has a magnitude given by the product of the force magnitude F and the lever arm l. More generally, torque is a vector τ̄ equal to the vector product of r̄ (the position vector of the point at which the force acts) and F̄ . δ 2 y(x, t) f ω = 2πf = w x 1 ,T = T ω = IP = Icm + M d y(x, t) = A cos[ω( τ̄ = 0 1 f = , ω = 2πf = vk λ Fx,y,z = 0 about any point Periodic motion: Periodic motion is motion that repeats itself in a definite cycle. It occurs whenever a body has a stable equilibrium position and a restoring force that acts when it is displaced from equilibrium. Period T is the time for one cycle. Frequency f is the number of cycles per unit time. Angular frequency ω is 2π times the frequency. Moment of inertia and rotational kinetic energy: The moment of inertiaI of a body about a given axis is a measure of its rotational inertia: The greater the value of I, the more difficult it is to change the state of the body’s rotation. The moment of inertia can be expressed as a sum over the particles mi that make up the body, each of which is at its own perpendicular distance ri from the axis. The rotational kinetic energy of a rigid body rotating about a fixed axis depends on the angular speed v and the moment of inertia I for that rotation axis. 2π k = Conditions for equilibrium: For a rigid body to be in equilibrium, two conditions must be satisfied. First, the vector sum of forces must be zero. Second, the sum of torques about any point must be zero. The torque due to the weight of a body can be found by assuming the entire weight is concentrated at the center of gravity, which is at the same point as the center of mass if ḡ has the same value at all points. Doppler effect: The Doppler effect for sound is the frequency shift that occurs when there is motion of a source of sound, a listener, or both, relative to the medium. The source and listener frequencies fS and fL are related by the source and listener velocities vS and vL relative to the medium and to the speed of sound v. λ x sin θ λ = 2πx sin θ λ = 0, ±1, ±2, ...(inphase) = ± 1 2 ,± 3 2 Two source interference of light: d sin θ = mλ(m = 0, ±1, ±2, ...) (constructive interfer- ence), d sin θ = (m + 1 )λ (destructive interfer- 2 ence) Intensity in interference patterns: I = I0 cos2 φ 2 Fresnels: How large quotas of S versus P-polarized light is reflected Rs = n1 cos θi − n2 cos θt 2 Rp = n2 cos θi − n1 cos θt 2 n1 cos θi + n2 cos θt n2 cos θi + n1 cos θt , ...(outof phasebyhalf aphase) Bragg: 2d sin θ = nλ Extraswag Lite formler för oss glömska Cirkel: O = 2πr A = πr2 Klot: A = 4πr2 3 V = 4πr 3 Cylinder: A = 2πrh V = πr2 h Kon: 2 V = πr3 h A = 4πr2 Pyramid: V = Bh 3 Cirkelsektor: b = 360θ◦ 2πr A = br2 = 360v◦ πr2
