Joule and the Conservation of Energy

Transcription

1 Jule and the Cnservatin f Energy The unit Jule, abbreviated J, is shrthand fr NewtnCmeter (NCm) which is defined via the definitin f wrk: Wrk Frce (cs θ) Displacement where θ is the angle between the frce vectr and the displacement vectr. Nte that multiplying tw vectrs in this manner results in a scalar number r value. It is independent f the vectrs rientatin in space. It is the prduct f: the cmpnent f ne vectr alng the directin f the ther vectr, multiplied by the magnitude f the ther vectr. The " sign f wrk is determined by ( cs ) since the magnitude f a vectr,, is always psitive. Wrk is negative when the angle lies between 90E and 70E; a prime example is a frictin frce which always ppses the mtin ( 180E ). When frce and displacement are perpendicular t each ther (90E) the wrk dne is 0. Whenever wrk is dne n an bject, then ne r mre f the bject s mtinal, psitinal, r internal amunt f jules change; that is in general an bject can have: Jules f mtin Jules f psitin Jules internal t the bject called kinetic energy called ptential energy called internal energy. Nw there is a law f nature, discvered abut 1840 and always bserved t be true since then, which states that energy is cnserved. Jules cannt be created r destryed, they can nly be transferred r transfrmed; it is called the principle f cnservatin f energy. This law f energy cnservatin is cnnected t the time symmetry in nature. N exceptins t this law have ever been encuntered; it is thught always t be true. example: Suppse yu bserve an bject underging sme prcess in nature. Yu ntice that befre the prcess (initially) the bject has: mtinal jules Kinetic Energy KE initial 1 J psitinal jules Ptential Energy PE initial 15 J internal jules Internal Energy IE initial 4 J During this particular prcess the internal energy f the bject des nt change and the bject s psitinal energy reduces by 7 J. What is the final mtinal r kinetic energy f the bject? slutin: we write the cnservatin f energy: energy befre the prcess energy after the prcess Answer: J KE final + (15-7) + 4 J KE final 8 J In this example, we have assumed that the bject was left t itself and was nt given any additinal energy nr did it lse any energy; the bject was treated as an islated bject. C. Deurzen 1 /

2 General - Cncept f System In general, cnsider a system instead f nly an bject; it is an imaginary bx r sphere bundary, arund r cntaining the bject r regin f interest. Anything utside this bx r sphere des nt belng t the system f interest; it belngs t the rest f the universe. We keep track f the amunt f energy (Jules) entering (+) r leaving (!) the system. We speak f the system having an energy level E analgus t the water level in a well. Rest f the universe, nt in the system. External t the system. Energy depsit + Energy withdrawal! system E energy level The mst general frm f the cnservatin f energy law is: (system islated r nt) J in r ut Change in energy f the system )E Nte: 1. Energy is thus invlved in nature s changes during prcesses and it tracks Jules.. It is the net amunt f energy entering and leaving the system r transferred int the system frm the rest f the universe that determines hw much the energy level f the system changes. We dente this net amunt f energy here by J in r ut. When the energy enters the system J in r ut is a psitive number, )E is psitive and the system s energy level rises. When energy leaves the system J in r ut is a negative number, )E is negative and the system s energy level decreases. 3. Fr a system islated frm the rest f the universe (n interactin) J in r ut is zer. 4. Wrk is a measure f the energy transferred by a frce acting ver a distance. example: Suppse yu have a system whse initial energy level befre sme prcess is 78 J and during the prcess 14 J enter the system frm the rest f the universe, fr example electrmagnetic light phtns r sund r heat enters the system. What is the final energy level f the system after the prcess? slutin: We write the cnservatin f energy law: + 14 J E final! E initial + 14 J E final! 78 J Answer: E final 9 J ; the energy level f the system increased by 14 J frm 78 J t 9 J. / C. Deurzen

3 Wrk and Frce: Cnservative & Nn-Cnservative Frces Wrk is a measure f the energy transferred by frce acting ver a distance. Identify sme system: Nn-system Rest f the universe Energy exchange System has an energy level E E + Wrk Fr wrk dne by frces within a system: 0 Nte: Wrk dne by a system is at the expense f the system s energy; i.e. the wrk dne by a system always has the ppsite sign (±) f the change in energy f the system. A. Cnservative frces Wrk dne n the system by the rest f the universe Cnsider a system f tw bjects and their interactive cnservative frce. Fr example, a mass m free falls tward a secnd fixed mass M. Cnservative frces change the ptential energy (PE) f a system, which can cnvert it back if the prcess is reversed. 0 PE + Wrk 0 PE + F cns (cs0 ) s frce is parallel t and in the directin f the displacement s 0 m P + m Field (1) s where P is the change in the ptential. P Fr cnservative frces the field is the negative gradient f the ptential: Field s B. Nn-Cnservative frces Jules in r ut E system Wrk dne by the system n the rest f the universe Cnsider a system f tw bjects and an interactive nn-cnservative frce. Fr example, a mass m slides with frictin n a secnd mass M. Nn-Cnservative frces change the internal energy (IE) f a system, which can nt cnvert it back if the prcess is reversed. 0 IE + Wrk 0 IE + F nn-cns (cs180 ) s frictin ppses the mtin: cs180 1 IE Wrk nn-cns F nn-cns ( 1) s a psitive number > 0 Nn-Cnservative frces cause the internal energy f a system t increase. C. Deurzen 3 /

4 Example: Cnservative Frce A particle has mass.8 x 10 6 kg and carries electric charge.5 µculmb. Upn release, in a regin having electric field, this particle is bserved t have speed 45 m/s at pint A, and later arrives at pint B with speed 5 m/s. a) What is the ptential difference (vltage drp) between the tw pints? Cnsider the particle and the regin with the electric field as the system. After release n energy is put int the system; it is an islated system: 0 PE + KE 1 0 q V + m( vf vi ) q charge mmass m m( vf vi ).8x10 kg ( 5 45 ) V s Jule 6 q x Culmb Culmb b) Which f the tw pints A r B is at the higher electric ptential? Pint A is the initial pint and pint B is the final pint. Thus: V V final V initial V B V A 784 vlt V A V B vlt Answer: Pint A has higher electric ptential than pint B. c) What is the directin f the electric field between the tw pints A and B? Wrk n the particle by the field is at expense f the change in ptential energy: PE Fcnservative csθ s q V qefield csθ s q 784 vlt qe csθ s 0 ( ) field Hence (csθ 0) the frce is ppsite in directin t the displacement s, but with negative charge the field is in the directin f the displacement, frm pint A t pint B; it decelerates the negatively-charged particle. 4 /

5 Example: Nn-Cnservative Frce A child n a sled slides dwn a 1 hill withut changing its speed. The ttal mass f the sled and the child is 45 kg; the vertical drp is 4 m. Nte: the length f the hill is s h/sin1 4 m/ m. a) What is the change in internal energy f the system? Cnsider the sled/child and the hill with the gravity field as the system. After release n energy is put int the system; it is an islated system: 0 PE + KE + IE with KE 0 N IE PE (45 kg)(9.81 )(4 m) J kg when the ptential energy f the system decreases. Answer: the internal energy f the system increased by 19 kj. This energy cmes frm the decrease in gravitatinal ptential energy. There was n change in the translatinal kinetic energy KE f the system. b) What frictinal frce acts between the sled and the hill? Cnsider the wrk dne n the sled by the system s frictin frce. 0 IE + Wrknn cnservative IE Wrknn cnservative IE F frictin csθ s 0 nte : csθ 0 r with the frictin frce alng the incline cs180 1 IE J F frictin + 93 N (cs180 ) s ( 1) 00m Answer: the magnitude f the frictin frce is 93 N; its directin is ppsite the displacement vectr s, thus uphill. 5 /

6 Ntatin and Example Starting with we have E final E initial + J in r ut J in r ut )E system r r KE final + PE final + IE final KE initial + PE initial + IE initial + J in r ut KE final! KE initial + PE final! PE initial + IE finall! IE initial J in r ut r )KE + )PE + )IE J in r ut where we have used the cncept that the ttal energy f the system is made up f the system s kinetic plus ptential plus internal energy. Often we have a prcess during which the internal energy f the system des nt change, we then have )IE is zer and we speak f the mechanical energy ME r its change; in such a case we have r )KE + )PE + zer J in r ut r )ME + zer J in r ut where mechanical energy is defined as the sum f the kinetic and the ptential energy. example: slutin: At a carnival, yu manage t just barely ring the bell by striking a target with a 7-kg hammer which sends a 0.3-kg metal piece upward tward the 4.75-m high bell. With what speed must the hammer be mving when it strikes the target, if ¾ r 75 % f its kinetic energy is dissipated int heat, frictin, and sund, leaving ¼ r 5 % t d the mechanical wrk? Chse the system f interest t be the metal piece ( m0.3 kg ) and the hammer transfers energy int the system. The metal piece is at rest initially and als as it just strikes the bell, s the change in its kinetic energy is zer. The input jules are the kinetic energy f the hammer which is ½ M V as the hammer cmes t rest. We write the cnservatin f energy: J in r ut )KE + )PE + )IE ½ M V zer + mg)h + ¾ ( ½ M V ) r ¼ ( ½ M V ) zer + mg)h slve fr V when M 7 kg, m 0.3 kg, g 9.81 m/s )h 4.75 m Answer: V 4 m/s 6 /

7 Energy Expressins 1. Kinetic Energy 1 translatinal mass ( velcity ) 1 rtatinal rtatinal rtatinal inertia velcity relativistic Ttal Energy rest mass energy. Ptential Energy 3. Internal Energy G M m gravity r kqq electric r elastic 1 k x... r r mgh qv ( ) thermal mass thermal capacity temperature thermal 3 N kbltzmant fr a mnatmic ideal gas chemical heat energy released when a fuel is burned nuclear... m c Smetimes a little f the energy that we start with disappears int internal energy and we speak f energy lsses in the sense that the missing energy is n lnger readily available t d useful wrk. Fr example the transfrmatin f external wrk int thermal energy by mtin-impeding frces which causes increased agitatin f atmic mtin (increased kinetic energy); r the permanent defrmatin f bjects during cllisins which causes the reshaping f chemical mlecular bnds (increased ptential energy). 7 /

8 Energy and time The reader will have nticed that nwhere in the cnservatin f energy des the time interval duratin f the prcess appear. That is simply due t the system s energy stability ver time; withut sme energy-exchange prcess ccurring with the rest f the universe a system s energy level stays fixed and cnstant, there is n means fr it t change. S if we wish t knw the time duratin f a prcess, we must track the rate at which the energy change f the system ccurs. This rate is called pwer and is defined P / )E/)t. The unit f pwer is Jule per secnd which is called a Watt, abbreviated W. example: Suppse a system changes its energy by a ttal f 4 J and des s at the rate f 3 Jule every secnd r Watt. Hw lng des the )E prcess take? slutin: Frm the definitin f pwer )t )E/P ( 4 Jule ) / (3 Watt ) 8 s. Energy Efficiency The efficiency f any system, nrmally a machine, is defined as the rati f the energy (Jules) ut f a system t the ttal energy (Jules) put int the system, while the system remains at equilibrium. It is a pure rati and therefre has n units. Jin system efficiency efficiency E energy level Jut Jut t J J in in t Pwer Pwer ut in Nte: Efficiency als keeps accunt f the lss factr which is 1 - efficiency. example: Cnsider a mtr requiring 1800 W input while its utput shaft delivers.0 hrsepwer. mtr' s efficiency Pwer 0. hp x Watt/ hp ut r 83 % Pwer 1800Watt in The mtr s lss factr is r 17 % which means that the system itself dissipates 17% f its input r 0.17 x 1800 W which is 309 Watt r 309 Jule every secnd. 8 /

9 Energy Efficiency and Related Ratis 1. Mechanical engineers like t refer t mechanical frce advantage. efficiency Jut J in Frce Frce ut in displacement displacement ut in Frceut Frcein displacementin displacement ut AMA IMA where frces and their crrespnding displacements are taken t be parallel t each ther. The frce rati is called the Actual Mechanical Advantage. The displacement rati is called the Ideal Mechanical Advantage because it is the maximum pssible Mechanical Advantage which ccurs when there is zer energy dissipatin (zer transfrmed t IE) in the system, that is zer energy is absrbed by the system, that is when the efficiency is 1. A certain device has efficiency The input frce acts ver a displacement f 10 cm while the utput frce acts ver a distance f cm. What is the AMA f this device? slutin: IMA displacementin 10 cm displacement cm ut Frceut AMA efficiency IMA ( 080. ) 5 4 Frce in. Electrical engineers like t refer t the gain f a system r amplifier stage. Nte that electric pwer is current times ptential difference r vlts that is P IV. 5 rati r gain Pwerut Pwer in I I ut in V V ut in r Pwer Current Vltage gain gain gain A certain amplifier stage has input current 14 ampere while its vltage gain is 7 and its verall pwer gain is 4. What is the utput current f this amplifier? slutin: I ut Pwer I V 4 7 I 8 ampere. gain gain gain ut 14 ampere Nte that in rder t btain an energy r pwer gain frm a system, there must f curse be an energy surce inside the system ( fr example a battery ) as well as an energy sink. 9 /

10 Applicatin: Electric Energy & Electricity Laws Kirchhff s Laws cncerning the currents and ptential differences in electric circuits can be derived frm the principle f cnservatin f energy. Frm the definitins f current and ptential difference, it fllws that electric pwer is current times vltage, P I V. Cnsider an energy surce r pwer supply cnnected t an energy sink r electric lad. Let I be the current that flws between the surce and sink and V the ptential difference ( vlt) that is cmmn t bth the surce terminals and the sink terminals, as shwn. energy surce pwer supply I V energy sink electric lad We write the cnservatin f energy as: E surce t E sink which are pwers. The pwer ut f the surce which is transferred int the sink is P ttal I V. Thus the pwer supply exchanges J in r ut IVA)t energy with the sink part f system. Nw, when the electric lad cnsists f several separate lads, the ttal pwer dissipated in the sink will be I 1 V 1 + I V + I 3 V as the energy must g smewhere in the sink. The abve cnservatin f energy statement then becmes: IV I 1 V 1 + I V + I 3 V Case 1 - Series Circuit The electrical lad cnsists f several lads all f which take the same cmmn current I. Then the ttal pwer invlved is IV I 1 V 1 + I V + I 3 V where all currents are I. Since the cmmn current cancels, we btain V V 1 + V + V which states that arund any clsed lp the sum f the sink (dissipating) vltages is equal t the sum f the surce (supply) vltage(s); this is knwn as Kirchhff s Vltage Law ( KVL ) r the Circuit Lp Law, r the sum f the vltage drps is equal t the sum f the vltage rises. It can als be stated as: the sum f the vltages arund a cmplete and clsed lp is zer; where supply vltages have ne sign ( " ) and sink vltages have the ppsite sign. Case - Parallel Circuit The electrical lad cnsists f several lads all f which take the same cmmn vltage V. Then the ttal pwer invlved is IV I 1 V 1 + I V + I 3 V where all vltages are V. Since the cmmn vltage cancels, we btain I I 1 + I + I which states that at any junctin the sum f the utging currents is equal t the sum f the incming current(s); this is knwn as Kirchhff s Current Law ( KCL ) r the Circuit Junctin r Nde Law. It can als be stated as: the sum f the algebraic currents at a junctin r nde is zer; where incming currents have ne sign ( " ) and utging currents have the ppsite sign. t 10 /

11 Example: Kirchhff s Vltage and Current Law. Kirchhff s Vltage Law: arund any clsed electrical circuit lp, KVL: the sum f the vltage drps is equal t the sum f the vltage rises. Kirchhff s Current Law: at any junctin nde in an electrical circuit, KCL: the sum f the incming currents equals the sum f the utging currents. Apply these tw circuit laws t the circuit shwn and demnstrate cnservatin f energy. I 1 R 1 R 1 cntains 3 kω õ 1 A 6 kω õ I R I 3 R 3 R cntains 1.5 kω.5 kω Examine the 3 currents at nde A and apply Kirchhff s Current Law as stated abve. Similarly examine the tw lps and apply Kirchhff s Vltage Law as stated abve. Apply Ohm s rule ( VIR r IGV ) fr the vltage drps acrss the energy sinks. KCL at nde A : I 1 I + I 3 KVL fr lp with õ 1 : õ 1 I 1 R 1 + I R KVL fr lp with õ : õ + I 3 R 3 I R Given: õ 1 11 vlt õ 33 vlt G 3 1/6 ms Nte: R 1 ks R 4 ks R 3 6 ks With these given values, there results 3 equatins in the 3 unknwns. The reader shuld slve the mathematics fr the three unknwn currents: Answer: I 1 &0.5 ma I +3.0 ma I 3 &3.5 ma 11 /

12 Having slved fr the three unknwn currents, we knw everything that can be knwn abut each individual device in this electrical circuit. We fill in a table with the values pertaining t each device in the circuit: Device ID ptential difference V vlt current I ma resistance RV/I ks cnductance G I/V ms pwer IV mw R 1 &1 &0.5 1/ 0.5 R /4 36 R 3 &1 & / õ 1 11 &0.5 &5.5 õ Blded values are given r pre-determined values, all ther values are derived with the 3 currents. The ttal pwer cnsumed by the energy sinks is P ttal mw. The ttal pwer supplied by the energy surces is P ttal (-5.5) 110 mw. Thus the same net amunt f energy per secnd is leaving the supply surce(s) as is being cnsumed r entering the sinks, and we see that energy is indeed cnserved. Nte that I 1 turned ut t be a negative number, which really means the actual current is in the ppsite directin f the arrw f I 1, (arrw directin is psitive). The current in the lp cntaining the õ 1 energy surce is running cunter-clckwise! That current is entering the õ 1 energy surce which means õ 1 is being charged using the energy frm the secnd õ energy surce r pwer supply. õ is being depleted at the rate f millijules per secnd f which 110 are absrbed by the three energy sinks and 5.5 mw are being pumped int the first õ 1 pwer supply. 1 /

13 Applicatin: Pwer Transfer - Impedance Matching Cnsider a resistive lad R Lad attached t a pwer supply õmf which has internal resistance R supply. Investigate the amunt f pwer transferred frm the pwer supply int the lad by pltting the pwer absrbed int the lad versus the lad s resistance. Pwer int the Lad P Lad I Lad V Lad Where is the peak? PLad ILad RLad small R Lad large R Lad 0 Z R Lad R Lad Y 4 V Lad 0 I Lad 0 P Lad 0 P Lad 0 with KVL: Hw des the P Lad vary with the resistive size f the lad itself, that is with R Lad? Let X / R Lad and R s / R Supply then the pwer absrbed: dp dx Lad P Lad [ { ( s ) } { s } () 1] 3 E X R + X + R + X R Supply ( ) P E X R + X Lad R Lad E + R Lad R Lad s Find where the slpe is zer. dp dx Lad ( s + ) Rs X E 3 ( R + X) ( R + X) E R X X s s 3 0 when X RSupply R Lad R Supply is called Impedance Matching fr Maximum Pwer Transfer, at which the pwer int the lad is P Lad { E/(R Supply ) } R Lad ½I sc ½V c. Prblem: What maximum pwer can be transferred ut f a pwer supply with electr-mtive-frce õmf 1 vlt and internal resistance 18 hm? Slutin: P max ½ { I shrt-circuit 1 vlt/18 } ½ { V pen-circuit 1 vlt } P max ½ { A } ½ { 1 vlt } Watt 13 /

14 Applicatin: Thermal Energy Devices Devices which depend fr their successful peratin n a temperature difference can be readily understd using the principle f cnservatin f energy. Such thermal devices can be classified int tw general categries: heat engines and heat pumps. Suppse ne has a high-temperature reservir f thermal energy and als a lw-temperature reservir, between which a device perates by virtue f the temperature difference between the reservirs. The device must be thermally cnnected t bth reservirs ( T ht and T cld ) s that thermal energy ( Q ) flws thrugh the device, as shwn in the diagrams. Ht reservir Ht reservir Q ht Q ht Heat Engine W ut Heat Pump W in Q cld Q cld Cld reservir Cld reservir Cnservatin f energy: Q ht Q cld + Wut Q ht Q cld + Win efficiency heat engine Wut Q Q ht ht Q COP means Cefficient f Perfrmance Q cld ht Kelvin s abslute temperature means Q is prprtinal t T, which permits all three ratis t be written in terms f the temperatures f the reservirs; the maximum ratis are: COP COP refrigeratr heat pump Qcld Win Q Qht W Q cld heat engine refrigeratr heat pump ht Qcld Q Qht Q T T Tht efficiency COP COP T T T Nte, these results say nthing abut: the wrking substance, the material the heat device is made f, r hw a device is cnstructed; it is nly the abslute temperatures that matter. Carnt s Therem (184) says that n practical thermal device can exceed these ratis. in ht ht cld cld 14 /

15 Applicatin: Rtatinal Energy and Inertia Cnsider a cart n fur wheels (mass m 1 ) n a level track cnnected by a light string ver a pulley t a secnd suspended mass m (see diagram belw). Assume bth cart and pulley are frictinless, and ignre the mass f the string. Assume the string is nn-stretchable, then bth masses have the same acceleratin. The tensin frce F tensin in the string is the same n bth sides f the pulley, because we are neglecting the frictin in the pulley fr the mment. Nte: the suspended weight m g is the applied frce causing the entire system t accelerate. Analyze the mtin f this system frm the pint f view using: (a) frces and (b) energy. A. Frce analysis f the mtin f the system (cart plus suspended mass). The free-bdy diagrams fr the cart and suspended mass are illustrated t assist in analysis. m 1 cart Nrmal m m 1 F Tensin F Tensin m 1 g m Cart F net F tensin m 1 a 1 F tensin m g Suspended Mass F net m g - F tensin m a m g - F tensin Cmbining these tw equatins by substituting F tensin and accunting fr the frictin: ( ) mg F m+ m a frictin 1 (1) Nte, Equatin (1) states: the net frce (applied frce m g less the ppsing frictin) and the acceleratin a f the system are prprtinal with cmbined mass cnstant f (m 1 + m ). We may view the cmbined cart-plus-suspended mass as ne system f ttal mass m 1 + m. In this analysis, the cause f any rtatin f the pulley and the cart wheels has been ignred. 15 /

16 B. Energy analysis f the mtin f the system (cart plus suspended mass). As the system mves it gains kinetic energy f mtin at the expense f the ptential energy f psitin f the suspended weight m g. When the suspended weight falls thrugh vertical distance y its ptential energy decreases by m gy, and using the cnservatin f energy: mgy KE + KE + KE + IE cart pulley wheels translatin rtatin rtatin frictin mgy ( m1+ m) v + Ipulleyω + Iwheelsω + RFfrictinθ R radius f pulley r wheels. Fr nn-slip rtatinal mtin: v I ω and thus Iω v R R 1 I I mgy m1+ m + 4 v RF + + frictin R pulley R wheel θ We differentiate this energy equatin with respect t time and recgnize that: dy y system velcity v dt dv v system acceleratin a dt dθ θ rtatinal velcity ω dt 1 I I mgy m1+ m + 4 vv RF + + R pulley R wheel I I mg m1+ m + 4 a F + + R pulley R wheel frictin frictin θ v R () Nte: Equatin () is the same as (1) stating frce is prprtinal t acceleratin but with a different mass cnstant which incrprates the rtatinal inertia f the pulley and wheels. 16 /

17 Applicatin: Oscillatr Energy Harmnic Mtin with laminar damping. Damped, driven harmnic scillatr. Cnsider a damped scillatr frced t perate by a sinusidal driver frce F sin(ωt) Analyze the mtin f this system frm the pint f view using: (a) frces and (b) energy. A. Frce analysis f the mtin f the system (driver frce & damped scillatr). B. Energy analysis f a driven damped scillatr system (with energy surce). C. Shw that the quality f an scillatr frequency at resnance Q. full width at half maximum pwer A. Frce analysis f the mtin f the system (frce driving a damped scillatr). Shw x Asin(Tt+N) satisfies the equatin f mtin f a driven harmnic scillatr. Equatin f mtin: mx + cx + k x F sinωt c c m x m x k m x F sinωt τ + + m Substituting the first and secnd differentiatin f the prpsed slutin x Asin(Tt+N): ω F ( ω ω ) Asin( ωt+ ϕ) + Acs( ωt+ ϕ) sin( ωt) τ using: sin(tt+n) sintt csn + cstt sinn cs(tt+n) cstt csn - sintt sinn ( ) ( ) 1 k ω m ω ω ϕ ω ϕ ω ω ω ϕ ω F A t ϕ A ωt ωt τ + + cs sin sin sin cs τ cs sin m This equatin can nly be satisfied fr all time t if the cefficients f sintt and cstt are each zer, which establishes A and the phase angle N: m A F / m F / m ω ω ϕ ω + ω τ τ ( ) cs sinϕ ( ω ω ) sinϕ ω / τ tanϕ csϕ ω ω 17 /

18 where we used sin N + cs N / 1 in rder t btain: sinϕ ω / τ ω τ csϕ ω ω ( ω ω ) + ( ω ω ) ω + τ Thus under the influence f an external driving frce F F sin Tt the system respnse is: Psitin: x A sin ( Tt + N ) displacement always lags (N < 0) the driving frce Amplitude A F / m ( ω ω ) ω + τ ω / τ tan ϕ ω ω a) At lw driving frequency T << T Respnse cntrlled by spring (k) b) At resnance driving frequency T T Respnse cntrlled by damping (c) c) At high driving frequency T >> T Respnse cntrlled by inertial mass (m) A F m F / ϕ ω k 0 0 A F m F / / ω π ϕ ω / τ c A F / m F / ω ϕ π ω m Nte in particular the rati f the respnse at resnance t the respnse at zer frequency. It turns ut t be the quality f the scillatr at the resnance frequency: respnse at resnance ( ) A ω ω 1/ ω ωτ ω ( ω 0) respnse at zer frequency A / τ Q resnance 18 /

19 B. Energy analysis f a driven damped scillatr system (i.e. with energy surce). Derive the pwer absrptin P(T) fr a driven damped harmnic scillatr. Fr the mtin f a frced r driven scillatr we have: frce F F sin Tt ; psitin x A sin ( Tt + N ) ; velcity v TA cs ( Tt + N ) Nte the phase angle N f the psitin x relative t the driving frce F and an additinal 90E fr the velcity v which is the derivative f the psitin. The time averaged pwer is the time average f the frce times the velcity Pwer + F velcity, +, / time average Pwer + ( F sintt { TA cs( Tt + N ) }, cs( Tt + N ) cstt csn - sintt sinn Pwer F T A + sintt cstt csn & sintt sintt sinn, The time average f sintt cstt is zer, that is + sintt cstt, 0 and the time average f sin Tt is ne half, that is + sin Tt, ½ Pwer F T A ( 0 & ½ sinn ) and previusly: 1 Pwer F A F ω m F ( ω ω ) / m + ω τ ω/ τ ( ω ω ) + ω τ sinϕ ω / τ and ω ω + τ ( ω ) 1 Pwer m F τ m ( ωτ / ) ( ω ω ) + ω τ Nte the frequency dependence factr cntaining T, damping Jc/m, spring T k/m 1 m factr ( ω A) τ ( ω / τ) ( ω ω ) + ω τ 19 /

20 C. Shw that the quality f any scillatr frequency at resnance Q. full width at half maximum pwer The quality factr f an scillatr is defined such that the energy lss in an scillatr is described expnentially with a time decay cnstant f τ Q/ω. E / T π E E t Q with ω and P then P /π T t E τ The quality f an scillatr is a figure f merit which transiently describes the number f repetitins (cycles) an scillatr will vibrate back and frth befre cming t rest, after the driver energy surce is remved i.e. befre all its energy is dissipated via the scillatr s damping mechanism. The greater a quality factr, the lnger the scillatr vibrates, r the less external pwer is required in rder t keep the scillatr vibrating. Pwer Absrptin: Pwer is the time average f the frce times the velcity (see energy analysis): Near resnance (T.T ) the half-pwer pints ccur when: r T & T T/J T )T. T/J Thus the full width ( )T) f the resnance at half maximum pwer is equal t 1/J and the quality f an scillatr Q measures the sharpness f damping near resnance: P res ½ P res FWHM small damping ( ωτ / ) 1 Pwer F v P where P m F < > res res τ m ( ω ) ω + ( ω/ τ) frequency at resnance Quality Q ωτ ω ω full width at half maximum pwer large damping ω 1 ω frequency where: ω ω - ω 1 is the full width at half maximum pwer f the scillatr s resnance curve, and ω 1 and ω are the frequencies at which the pwer dissipatin in the scillatr has drpped t ne-half its resnance value: 1 F Pwerresnance m τ m The quality factr is a measure f the sharpness f an scillatr s resnance curve. 0 /

21 Numerical Example f Damped Harmnic Oscillatr Given scillatr data: mass m 1 gram Then: spring cnstant k 10 N/m relaxatin time J 0.5 secnd N 10 k ω m radian cycles m 10 kg secnd secnd free scillati n frequency ω ω τ 4 radian secnd Quality factr Q radian ωτ secnd 50 secnd The time fr the amplitude t damp t e & 1 f its initial value is: the value f the damping cnstant is: τ 0.5sec 1sec Given the driving frce: F 0.1 N sin ( 90 t ) thus F 0.1 N ; m 1 gram c τ 05. secnd gram secnd F 0.1 N N N m 1 gram gram kg and the driving frequency is T 90 rad/sec The respnse f the scillatr is: n & 5.4E Amplitude A 100 N / kg ( ) / cm tan ϕ The full width f the resnance curve between half-pwer pints is: FWHM )T T /Q 100 rad/sec / 50 rad/sec 0.3 Hz. 1 / F / m 100 N / kg the amplitude respnse at resnance is A( ω ω) 50 cm ω rad / τ 100 /sec 05. secnd F 01. N the amplitude at lw frequency is A( ω 0) 1 cm k 10 N / m The pwer absrptin at resnance: 1 3 Pres 10 kg( 100 N / kg) 05. sec. 5Watt C. Deurzen

22 Applicatin: Nuclear (Mass) Energy A nucleus can underg transfrmatin when sme agent (nuclear) frce changes the internal energy structure f the nucleus. During such a prcess, be it fissin r fusin, the system (nucleus) absrbs r thrws ff energy in the frm f: interactin, particles, and/r electrmagnetic radiatin energy. Accrding t Einstein (experimentally verified) the existence f mass implies the presence f internal energy; mass is anther frm f energy. Fr any (cmplex) system f mass m, its internal energy is E mc where c dentes the speed f light in vacuum ( m/s). During a nuclear transfrmatin, the change in internal energy due t a mass change is thus E mc. The binding energy f a cmpsite bject f mass m against separatin int parts is the mass difference multiplied by the speed f light squared: J in r ut E binding { (sum f masses f the separate parts) m } c. Electrmagnetic radiatin is viewable either as waves r as energy packets called phtns. Fr a wave, its speed c fλ where f is the frequency and λ is the wavelength f the wave. Wave and phtn are crrelated by: a phtn carries energy E hf (Einstein, 1905) and carries mmentum p h/ λ (DeBrglie, 194) where h is Planck s (1900) cnstant. example: The fissin f a uranium-35 atm thrws ff r releases 00 MeV f energy. What percent is this fissin energy f the atm s ttal internal mass energy? E 00 MeV 1 amu Answer: r 0.091% mc 35 amu 931 MeV If sme f the lst mass (energy) were detected in the frm f electr-magnetic radiatin having wavelength λ 0.18 pm (gamma radiatin), what is the energy f a detected phtn? Answer: µ µ λ 0.18 x10 m 1eV 19 hc ev m ev m x J 1 E hf 6.89 MeV 1.1x10 J 1 λ If 75 % f a reactin s ttal released energy is depsited and harnessed int useful utput energy, what is the pwer utput f a nuclear plant burning 600 kg f uranium per year? m kg kg 1 year 8 m 9 Answer: Pwer x x10 s 1. GWatt 7 year 3.16 x 10 s s s /