New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here


by: Favian Lemke


Marketplace > Georgia State University > Physics 2 > PHYS 1111K > INTRODUCTORY PHYSICS I
Favian Lemke
GPA 3.68


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Physics 2

This 185 page Class Notes was uploaded by Favian Lemke on Monday September 21, 2015. The Class Notes belongs to PHYS 1111K at Georgia State University taught by Staff in Fall. Since its upload, it has received 9 views. For similar materials see /class/209853/phys-1111k-georgia-state-university in Physics 2 at Georgia State University.

Similar to PHYS 1111K at GSU




Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/21/15
Web Sites Class httpwwwcharagsueduteamPHYSllll MWF 900am C80 Klnematlcs In 1 39D httpedugenwileyc0medugenclasscls523 70 MWF 10 00am h ttp edu gen wiley comedu genclasscls5 23 72 MWF 5 3 0pm h ttp edu gen wiley comedu genclasscls5 23 73 Definitions Displacement to Kinematics deals with the concepts that are a t 1 t needed to describe motion a Dynamics deals with the effect that forces X Disp39acement A have on motion lAi 2 31 320 2 displacement 3quot Together kinematics and dynamics form the branch of physics known as Mechanics V V Average SpEEd the distance traveled divided by the time required to cover the defined distance Distance Average speed 2 Elapsed t1me SI units for speed meters per second ms Example Jogger How far does a jogger run in 15 hours 5400 s if his average speed is 222 ms D Average speed 2 Stance Elapsed time Distance 2 Average speedElapsed time 222 ms 5400 s 12000 m Average VQIOCity is the displacement divided by the elapsed time Displacement Average ve1001ty Elapsed t1me lti i io Ai t to At SI units for velocity magnitude meters per second ms Speed and VeIOCIty Start Finish 72 25 A32 1609 m a r 439695 s 1010 5 Finish Start Thrust SSE 761035 pr n Berti 42 39 Ai 1609m httpwwwpeterrennclaranetsrhtm b Andy Green in the car ThrustSSC set a world record of 341 1 ms in 1997 To establish such a record the driver makes two runs through the course one in each direction to nullify wind effects From the data determine the average velocity for each run Speed amp Velocity Calculations 05 I 147405 3395m s t s a o Start Finish h A 1609 m a instant of time Instantaneous Velocity The instantaneous velocity indicates how fast the car moves and the direction of motion at each A 1609 14cgss 0i05 i 7 di x m 2 342 7 r V I 4 695s mS Fi Aldo At d1 39 A 1609 m b Acceleration Acceleration cont C ey A 90 kmh The notion of acceleration emerges when a change in A velocity is combined with the time during which the rpj0j change occurs Qquot a 0 quot15 Ar10 s 39o lt1quot 11 gt V 90 kinh A 20 s l V gt t 18 kmh 0 I Average Acceleration 390 1 w I U 1 3 1 39 39 Vquot In 141 TquotI he V gt of r 4 1 gt gt O DEFINITION OF AVERAGE ACCELERATION V V0AV 3 t to At Example Average Acceleration 0 vlt gt 39 3 w Tn g N F gt 39 I Determine the average acceleration ofthe plane vo0ms V260kmh t 0s 1295 0 V VO 260kmh Okmh 9Okmh zz 29s Os 39 s Negative Acceleration V Va 13ms 28ms t to 125 95 5 011152 Deceleration cont 50 msz Uquot 28 ms Ar10 5 up a L gt F 23 ms In 7 A20S My 9L gt V18mls Vector Symbol Confusion by ii i 6 6 a t Q t Q It is customary to dispense with the use of boldface symbols overdrawn with arrows for the displacement velocity and acceleration vectors We will however continue to convey the directions with a plus or minus sign Constant Acceleration Equations Let the object be at the origin when the x0 0 lg 0 clock starts Constant Acceleration Eqns cont 61 E9 61 t0 atv v vv at u The Five Kinematic Variables 1 displacement x 2 acceleration constant a 3 final velocity at time t v 4 initial velocity v0 5 elapsed time t The Displacement Equation vgat v0vtv0v0 att x votat2 n DisplacementAcceleration Example 1005 lAJr 805 i 20m 2 E 0450 I gt cgner 7 gt xvg a 60ms80 s20msz80 s2 110m Jet Catapult Example E Find the jet s displacement Assume constant acceleration l v00ms a31mS2 x v62ms Jet Catapult Algebraic Soln v v a 0 v v0 Constant Acceleration Jet Catapult Numeric Soln Variables and Equations 11 1 displacement X V 2 V0 at 3 r quot WW Ham52 mam 2 acceleration constant a 39 xvovlt 3 final velocity at time t V i 4 initial velocity v0 V2 V02 zax 2 2 2 2 v v 62ms Oms x 0 62 m 5 elapsed time t 2 2a 231ms2 xVoHtat Acceleration Problem Methodology Executing the HOWt0 1 Make a drawing A spacecraft is traveling with a velocity of 3250 ms Suddenly the retrorockets are fired and the spacecraft 2 Decide which directions are to be called posi ive and begins to Slow down with an acceleration Whose negative magnitude Is 100 ms2 What Is the velocrty of the 3 Write down the values that are given for any of the five spacecraft when the displacement of the craft is 215 kinematic variables km relative to the point where the retrorockets began firin 4 Verify thatthe information contains values for at leastthree of 9 he five kinematic variables Select the appropriate equation X a I V0 1 5 When the mo Ion Is dIVIded Into segments rememberthatthe final velocity of one segment is the initial velocity forthe next 215000 m 100 ms2 3250 6 Keep in mind that there may be two possible answers to a ms kinematics problem Drawing Coordinates Provided Data 7 A V39lEKEle T lyeozzsam m 2 rnm 5 39 inmm quotquot i 77quot N Calculating the Solution X a 1 V0 2 215000m 100 ms2 3250 ms 2 2 2 v i3250ms2 2100ms2215000 m 2500 ms FreeFalling Bodies In the absence of air resistance it is found that all bodies at the same location above the Earth fall vertically with the same acceleration lfthe distance ofthe fall is small compared to the radius of the Earth then the acceleration remains essentially constant throughout the descent This idealized motion is called freefall and the acceleration of a freely falling body is called the acceleration due to gravity g 2980ms2 or 322fts2 The Effect of Air Resistance g980ms2 Air filled Evacuated tube tube a b Falling Object Example During an astronomy lab 3 student drops an eyepiece from the top of the Urban Life Building After 300s of free fall what is the displacement y of the stone v0 0 ms Ilt gt J I KC 00 O 3 U N O 3 U 300 s Example Solution y a v v0 1 980 ms2 0 ms 300 s 1 2 y v0t3at 0 ms300 s 980 ms2 300 s2 2 441m Graphical Solution Velocity E R C o 5 395 o 0 8 T8 m 4 1 l O l 2 3 4 Time t s Slope 8m 4ms At 25 Graphical Solutions Velocity cont Instantaneous Velocity Positive gt Zero velocity lt Negative velocity E velocity 800 Tangent line Ax 26 m 1200 at 8 Position x m p O O Posnlon X m 400 0 l o zoo 30 530 sou moo 12 00 141m woo woo O 50 100 150 200 250 Tini6ls Time 1 S Graphical Solution Acceleration 36 2 g 24 D a 393 12 3 gt O O l 2 3 4 5 Time 1 5 Av 12ms Slope 6ms2 At 25 J Phys 1111K Lecture 04 CampJ Ch4 Forces and Newton s Laws of Motion 39 What is Force wwalalnascoIufunlmnl A push or a pull Contact forces arise from physical contact Actionatadistance forces do not require contact and include gravity and electrical forces Graphical Representation of Force Arrows are used to represent forces The length of the arrow is proportional to the magnitude of the force V 15N 5N Examples of Forces Contact Force ActionAtADistance Frictional Force Force Tensional Force Gravitational Force Normal Force Electrical Force Air Resistance Magnetic Force Applied Force Strong or Weak spring Force Nuclear force Mass is a measure of the amount of stuff contained in an object Inertia and Mass Inertia is the natural tendency of an object to remain at rest or in motion at a constant speed along a straight line The mass of an object is a quantitative measure of inertia 8 Unit of Mass kilogram kg Isaac Newton mu waqumups at strand i ukNhsmWPmDW aYNewmn mm Newton s First Law An object continues in a state of rest or in a state of motion at a constant speed along a straight line unless compelled to change that state by a net force The net force is the vector sum of all of the forces acting on an object Net Force The net force on an object is the vector sum of all forces acting on that object The SI unit of force is the Newton N Individual Forces Net Force 4N I 10N D75 Net Force 3N Individual Forces 4N Net Force cont N Gt F0 rce cont Mathematically the net force is written as gt 2F where the Greek letter sigma denotes the vector sum Newton s Second Law When a net external force acts on an object of mass m the acceleration that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass The direction of the acceleration is the same as the direction of the net force 52 Zi zm m SI Unit of Force kg kg m N This combination of units is called a newton N Units for Newton s Second Law Table 41 Units 0139 Mass Acceleration and Force System Mass Acceleration Fm ce SI kilogram kg metersecondl I ll52 newton N 68 gram g centimetersecom2 cm2 dym dyn BE slug sl RamsecondZ RSZ pound 1h A FreeBodyDiagram is a diagram that represents the object and the forces that act on it 3quot I l I g 2539 N 2T5 N 439 a a I 5 1 395 N DI paging farce 560 at B Freely13dI diagram of the car Net Force E 3 I I 215 N t 25 N i 7 39 7 1 j 4 i i 395 N v 7 l 39 I Op posing force 561 N I m b Free timinI diagram of the car The net force in this case is 275N 395N 560N11ON and is directed along the x axis of the coordinate system Calculate the Acceleration If the mass of the car is 1850 kg then by Newton s second law the acceleration is F 11 a Z 0059ms2 m 1850 kg Force Vector Components The direction of force and acceleration vectors can be taken into account by using X and y components EFm is equivalent to Zy m Z m Newton s Third Law Whenever one body exerts a force on a second body the second body exerts an oppositely directed force of equal magnitude on the first body Suppose that the magnitude of the force is 36 N If the mass of the spacecraft is 11000 kg and the mass ofthe astronaut is 92 kg what are the accelerations Solution On the spacecraft 2 F 13 On the astronaut 2 F 2 l3 00033 ms2 11000 kg S a 3 36N ms 5 13 36N A mA 92kg 039ms2 Law of Universal Gravitation Every particle in the universe exerts an attractive force on every other particle A particle is a piece of matter small enough in size to be regarded as a mathematical point The force that each exerts on the other is directed along the line joining hiltHWWWthSiESutahedUquenanIEIIEWAZEIMEWhtml Gravitational Force For two particles that have masses m1 and m2 and are separated by a distance r the force has a magnitude given by mm F G 2 I G6673gtlt10 Nm2kg2 k A r Gravitational Example Assume m1 is a 12 kg bicycle and m2 is a 25 kg child and they are 12 m apart In F A Fgt 1712 1 r 1 F 2 GM 1quot 667x10 11Nm2kg22521 12 m 14gtlt10 8N Astronomical Example M a 7M F MM F F ME Data The Earth orbit 149600000 km 100 E from Sun diameter 127563 km mass 5972624 kg The Moon Lbit 384400 km from Earth diameter 3476 km w 735622 kg h tpwwwsoarviewscomcapear wearhrmlhhn Nm2 Kg2 mEarth 5972 x 102 1 Kg mMoon 735 x 1022 Kg r 3844 x 108m G mEarth mMoon n11 G 667 x 10 11 F r2 Out15 198137 x 1020 N Weight The weight of an object on or above the earth is the gravitational force that the earth exerts on the object The weight always acts downwards toward the center of the earth On or above another astronomical body the weight is the gravitational force exerted on the object by that body SI Unit of Weight newton N httpphysicsnistgovcuuUnitschecklisthtml When the word quotweightquot is used the intended meaning is 39 V 5 clear In science and technology weight is a force for Hu I I asthma o1 wh1ch the SI un1t 1s the newton 1n commerce and everyday ank 1hnmm use weight is usually a synonym for mass for which the SI unit is the kilogram Weight Mass Relationship Object of mass m MEm WG 2 r Wmg gGMf I Mass of earth ME am on Earth MSL 598x1024 kg 667 10 N 2 k 2 X m g 638gtlt106m2 980 ms2 Normal Force Definition The normal force is one component of the force that a surface exerts on an object with which it is in contact namely the component that is perpendicular to the surface Normal Force Calculation Fb HDL45NO Kim N EWHIN UNEO 3 Pk4N Apparent Weight The apparent weight of an object is the reading of the scale It is equal to the normal force the man exerts on the scale Apparent Weight Details ZFy FN mg ma FN mgma I true apparent weight weight Frictional Forces When an object is in contact with a surface there is a force acting on that object The component of this force that is parallel to the surface is called the frictional force contact points Static Friction No movement a When the two surfaces are not sliding across one another the friction is called static friction When movement just begins 6 Magnitude of Static Friction The magnitude of the static frictional force can have any value from zero up to a maximum value fs SfSW jfsMAX SFN O lt lt 1 is called the coefficient of staticfriction S Contact Area Note that the magnitude of the frictional force does not depend on the contact area of the surfaces 7 Static vs Kinetic Friction Static friction opposes the impending relative motion between two objects Kinetic friction opposes the relative sliding motion motions that actually does occur fk kFN O lt lt 1 is called the coefficient of kinetic friction S Coefficient of Friction Table 42 Approximate Valuns uf 11c Coef cients ufmmmn fur uriuus Surfaces Coef cient of Static Coef cient 0 Materials Friction La Kinenc Frictiun Mk Glass on glass dry 094 04 Ice on ice clean 0 0 01 002 Rubber on dry concmc 10 03 Rubber on wet concrete 07 05 Stccl on ICC 01 005 Stccl on steel dry hard steel 078 042 Te on on Te on 004 004 Wood on wood 035 03 shortly Kinetic Friction uDX4O ms 7 vxOms i gti W m M 1m Freebody diagram for the sied and rider The sled comes to a halt because the kinetic frictional force opposes its motion and causes the sled to slow down Example at 444 N imam ms 7 atWanVs f 3 t A k gt Ega VJ quot W m in Hz Freebody d4agram for the sied and rider Suppose the coefficient of kinetic friction is 005 and the total mass is 40kg What is the kinetic frictional force fk IukFN Il lkmg 00540kg980ms2 20 N Tension Cables and ropes transmit forces through tension Rope and Pulley A massless rope will transmit tension undiminished from one end to the other Ifthe rope passes around a massless frictionless pulley the tension will be transmitted to the other end ofthe rope undiminished Equilibrium An object is in equilibrium when it has zero acceleration 2360 2130 Problem Methodology Select an objects to which the equations of equilibrium are to be applied Draw a freebody diagram for each object chosen above Include only forces acting on the object not forces the object exerts on its environment Choose a set of X y axes for each object and resolve all forces in the freebody diagram into components that point along these axes Apply the equations and solve for the unknown quantities FootInTraction Example U 38 S 35 5 1Cn w I I x F 35 H 3 1 39 3 IN v 35 35 Tlcn i X 1 g Tquot gt1 5 2 2 kg T1 sin35 T2 sin35 0 Tlcos35 T2 cos35 F 0 Engine Hoist Example m x w 7391 11 cos 100quot r 5m 10 0 I r2 cus aoolt m Freebody diagram to the ring Problem Setup Force X component ycomponent 1 T1 SinlOOo T1 coleO 92 T2 sin 800 T2 cos800 W 0 W W3150N Problem Solution ZFX T1 sin100 T2 sin800 0 ZFy T1 ooleO T2 0058000 WO The first equation gives T1 sin1000 Substitution into the second gives sin 800 TzooleOo T200580O WO sml00o Problem Answers W T2 o Sn 80D coleO 005800 sml00 T2 582N T1 33Ogtlt103N Acceleration amp Equilibrium When an object is accelerating it is not in equilibrium Partial Equilibrium ll Dnn 1300 E W Am lies hndy diagvam ia39 livHankei The acceleration is along the X axis so ay0 Problem Setup Force xcomponent ycomponent 1 T1 008300quot T1 sin 300 2 T2 cos300 T2 sin 300 I3 D O I R 0 Problem Solution ZFy I sin300 T2 sin300 0 2zz ZFX T1 oos300 T2 oos300D R max Problem Answers T max 2 D 153gtlt105 N ZCOS300 What is Thermodynamics Thermodynamics is the branch of physics that is built upon the fundamental laws that heat and work obey Phys 1111K Lecture 15 The collection of objects on which attention is being focused is called the system while everything else C80 Thermodynamics in the environment is called the surroundings Surroundings while walls that do not permit heat flow are called l Walls that permit heat flow are called diathermal walls adiabatic walls L System To understand thermodynamics it is necessary to describe the state ofa system l 7 l lwbrk l H at Thermal Equilibrium Zeroth Law of Thermodynamics Two systems are said to be in thermal equilibrium if there is no heatflow between then when they are brought into contact Adiabatic walls Two systems indIVIdually In Temperature is the Indicator ofthermal a thermal e a I I equilibrium in the sense that there is no q Ad39abailc Walls net flow of heat between two systems System are in thermal lquot in thermal contact that have the same 7 temperature 39 eqUIlibrium With each other Diathermal walls silver b Heat and Work Suppose that a system gains heat Q and that is the only effect occurring Consistent with the law of conservation ofenergy the internal energy ofthe system changes AUsz UizQ Heatis positive when the system gains heatand negative when the system loses heat If a system does work Won its surroundings and there is no heat flow conservation of energy indicates that the internal energy ofthe system will decrease AUsz Uiz W Work is positive when it is done by the system and negative when it is done on the system First Law of Thermodynamics The internal energy of a system changes due to heat and work AUsz UizQ W Heat is positive when the system gains heat and negative when the system loses eat Work is positive when it is done by the system and negative when it is done on the system Positive amp Negative Work Example In part a of figure the system gains 1500J of heat and 2200J of work is done by the system on its surroundings In part b the system also gains 1500J of heat but 2200J of work is done on the system In each case determine the change in internal energy of the system a AUQ W 1500 J 2200 J 700 J b AUQ W 1500 J 2200 J 3700 J 1 Work Heat Li Surroundings Surroundings System l1 Ideal Gas Example The temperature ofthree moles of a monatomic ideal gas is reduced from 540Kto 350K as 5500J of heat flows into the gas Find a the change in internal energy and b the work done by the gas AUsz UiQW UnRT 8 AU 2nRTf nRTi 30 mo1831 JmoiK350 K 540 K 7100 J b W Q AU 5500 J 7100J12600 J Quasistatic Processes A quasistatic process is one that occurs slowly enough that a uniform temperature and pressure exist throughout all regions of the system at all times isobaric constant pressure isochoric constant volume isothermal constant temperature adiabatic no transfer of heat Isobaric Process An isobaric process is one that occurs at constant pressure W Fs PAs PAV Isobaric process W PAV PltVf Isobaric Example Example One gram of water is placed in the cylinder and the pressure is maintained at 2 0x105Pa The temperature the water is raised by 31 C The water is in the liquid phase and expands by the small amount of 10x108m3 33951 Find the work done and the change in internal energy W PAV 20x105Pa10gtlt10 8m3 00020 Q chT 00010 kg4186 JkgCquot 31C 130J AUQ W130J 0 020J130J Work Graphcall y Speaking WPAVPltVf K j Pressure lt lt Volume Isochoric Process Pf 2 e A J isochoric constant volume quot1quot Jw PI l AU ZQ W 2Q Volume a 7 4 quoti 5 7 17 2 W O a Area Under the Graph Example 4 Work and the Area Under a PressureVolume Graph Determine the work for the process in which the pressure volume and temp erature of a gas are changed along the straight line in the figure Pressure The area under a pressurevolume graph is the work for any kind of process 20x105 Pi r gtl 10x10 4 m3 X Low temperature YHigh temperature I I II I I l I I I I I I lnlllmp Graphical Solution 20 x 105 P r gtl 10 x 1041 m3 Since the volume increases the work is positive t Estimate that there are 89 colored squares in the drawing W8920gtlt105Pa10x10 4m3 E XLow 180 J g temperature YHigh 9 temperature I I l i l7 1 l I l l I l l l l l Volume Isothermal Process Isothermal expansion or compression of an ideal gas Pressure Metal cylinder Volume 17 WISOtherma39 Example Adiabatic Processes Two moles ofthe monatomic gas argon expand isothermally at 298Kfrom and initial volume of0025m3 to a final volume of 0050m3 Assuming that adiabatic 0 tranSfer 0f heat argon is an ideal gas find a the work done by the gas b the change in internal energy of the gas and c the heat supplied to the gas Adiabatic V expansion or 3 f compression of W nR T T a W quotRT 1117 a mona omic 2 l f x ideal gas 3 2 0 mol8 3 1molK298K1n m 3400 J 0 025 m3 iquot b AUnRTf nRz 0 A t w la a 10 5 expansion or P fV g compression of 39 0 AU Q W a monatomic p ideal gas V V Q W J y cPCV I Volwe f quecific Heat Specific Heats for Gases For gases it is necessary to distinguish between the molar specific heat To relate heat and temperature change in solids and liquids we capacities which apply to the conditions of constant pressure and constant used Q chT volume CV CP specific heat U nRT ca ac y 3 l 5 Qconstantpressure W in Tf nRTf The amount ofa gas is conveniently expressed in moles so we write the V following analogous express39on rst law of W PM thermodynamics Q CnAT constant pressure mOIar spec39flc fora monatomic C heat capacity ideal gas Specific Heats cont nRT Qconstantvolume W 7 i 0 I f j first law of thermodynamics constant pressure 3 fora monatomic CV 2 3R ideal gas 5 monatomic y Cp ER 2 ideal as 9 CV R 3 anyideal gas C C ZR Second Law of Thermodynamics The second law is a statement about the natural tendency of heat to ow from hot to cold whereas the first law deals with energy conservation and focuses on both heat and work THE SECOND LAW OF THERMODYNAMICS THE HEAT FLOW STATEMENT Heat flows spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction Hot reservoir Heat Engine L QHl A heat engine is any device that uses heat to perform work It has three essential features 1 Heat is supplied to the engine at a relatively Engine 7 high temperature from a place called the hot reservoir 2 Part ofthe input heat is used to perform QC 1 work by the working substance of the engine Cold reservoir 3 The remainder ofthe input heat is rejected i to a place called the cold reservoir lQHl magnitude of input heat chl magnitude of rejected heat W magnitude of the work done Hot reservoir A Efficiency TU The efficiency ofa heat engine is defined as QH l the ratio ofthe work done to the input heat lW u 62 Engine gt IQHl lfthere are no other losses then QC 1 IQHI Cold reservoir i E ezl lQ Cl IQHI Automobile Efficiency Carnot s Principle 43 Example Ah automobile engine has an efficiency of 220 and produces 2510 J of work How much heat is rejected by the engine A reversible process is one in which both the system and the IWI environment can be returned to exactly the states they were in before the rocess occurred QHWQC e IQ lQH P Hi 6 chQHW CARNOT S PR NC PLE AN ALTERNATIVE STATEMENT OF THE SECOND LAW OF THERMODYNAMICS IQCI 1 No irreversible engine operating between two reservoirs at constant temperatures 9 9 can have a greater efficiency than a reversible engine operating between the same temperatures Furthermore all reversible engines operating between the same temperatures have the same efficiency 2510J 4 8900J 0220 Carnot Engine Tropical Sea Heat Engine The Carnot en ine is useful as an idealized Hat reserve Example model 9 temperature TH Water near the surface of a tropical ocean has a temperature of 298 2 K whereas the water 700 meters beneath the surface has a temperature of 280 2 A Ofthe heat input originates from a single QH K t has been proposed that the warm water be used as the hot reservoir and temperature and a the rejected heat goes the cool water as the cold reservoir of a heat engine Find the maximum into a cold reservoir at a single temperature poss39ble emc39ency for SUCh and engme39 W Since the effICIency can only depend on En ine T the reservoir temperatures the ratio of g ecam01 1 C heats can only depend on those temperatures TH i Mi QC T 280 2 K Q T I awn 1 C1 0060 TH 2982 K IQ T Cold reservoir e 21 C 1C temperature TC Refrigeration Processes Refrigerators air conditioners and heat pumps are devices that make heat flow from cold to hot This is called the refrigeration process iH t f 39sf rilbi39r temperature 71 Engine 39 Cold reservoir temperature Tc Refrigeration Process gt Qc l l 7 Cold reservoir temperature To Engine Process Air Conditioner QH Hot outdoors QC Cool room W Work done by electrical energy 39 efrigerator Cold reservoir Inslde refrigerator Hot reservoir outsnde refrigerator l QHQCW Is it Possible Conceptual Example Second Law of Thermodynamics Is it possible to cool your kitchen by leaving the refrigerator door open or to cool your room by putting a window air conditioner on the floor by the bed QH Hot outdoors Air 397 condjamprlx r QC Cool room gera Cold reservoir H i inside reingeratorl 39 W Work done by electrical energy I rquot Hot reservoir oursrde relrigeralor QHQCW ggi l i Engines vs Refrigerators Hot39reservoir temperature m I Cold reservoir temperature Tc Refrigeration Process Refrigerator or air conditioner Coef cient of performance Hot reserVOIr temperature Tn QH 1 Engine gt QC 1 i Cold reservoir temperatureTc Engine Process cl lWl Heat Pump 3533quot electrical energy The heatpump uses work to make heat from the wintry outdoors flow QH Q0 W 77 into the house Warm house if ll QC Cold outdoors Heat Pump Example An ideal or Carnot heat pump is used to heat a house at 294 K How much work must the pump do to deliver 3350 J of heat into the house on a day when the outdoor temperature is 273 K IQCI TC TC IQHI TH E 39QC39 lQH39E heatpump l Coef cient of performance WH 3 TC lWllQHl1E W QH1 l 3350 J1 240 J TH j 294 K Entropy In general irreversible processes cause us to lose some but not necessarily all of the ability to do work This partial loss can be expressed in terms of a concept called entropy Carnot Q engine 2 Igt Tc TH change entropy AS 2 reversible I Entropy cont Entropy like internal energy is a function ofthe state of the system Consider the entropy change of a Carnot engine The entropy of the hot reservoir decreases and the entropy of the cold reservoir increases TC TH Reversible processes do not alter the entropy of the universe 2nd Law of Thermo Using Entropy Entropy like internal energy is a function of the state of the system Consider the entropy change of a Carnot engine The entropy of the Reversible processes do not alter the entropy of the universe Avaiable Energy Example 51 hot reservoir decreases and the entropy of the cold reservoir increases i Hn resemir r o K W 12001 Example ucizem 1 Suppose that 1200 J of heat is used as input for an engine under two different conditions as shown on the 51380 I right Determine the maximum amount of work that un can be obtained for each case C 39 Hot reservoir Km TH350K T gin 1 QH39V39IZUOJ C ecarnot 1 e Cam w 680 J Q engine QC 52m Cold reservolr Ta 150 K ll Available Energy Solution The maximum amount of work will be achieved when the 1972qu engine is a Carnot Engine where 397 caldresenmir I Tc 150K T 150 K a 1 C1 077 8 TH 650 K am i W emotQH 0 771200 J 920 J 3233 mg l 0H1200J ecamot 21 T C 1 0 5 Camel quot368 TH 350 K W ecmtlQA 0 571200 J 680 J Och if gym I The irreversible process of heat through the copper rod causes some energy to become unavailable In More on Entropy Wunavailable TOAS universe d Puddle of water AS decrease 10 Third Law of Thermodynamics 17 THE TH RD LAW OF THERMODYNAM CS Us not posswme to owerthe temperature of any system to absomte Zero m a mute numberof 5 eps Phys 1111K Lecture 07 CampJ Ch7 IMPULSE AND MOMENTUM Impulse Force to tf Time I There are many situations when the 41 force on an object is not constant Impulse Definition DEFINITION OF IMPULSE The impulse of a force is the product of the average force and the time interval during which the force acts 3 FAt Impulse is a vector quantity and has the same direction as the average force newton seconds N s Force Linear Momentum Definition DEFINITION OF LINEAR MOMENTUM The linear momentum of an object is the product of the object s mass times its velocity p WIV Linear momentum is a vector quantity and has the same direction as the velocity kilogram meter second kg ms Baseball Impulse ImpulseMomentum Theorem When a net force acts on an object the impulse of this force is equal to the change in the momentum of the object ZEA1 mvf mvo final momentum initial momentum Rainstorm Example Rain comes down with a velocity of 15 ms and hits the roof of a car The mass of rain per second that strikes the roof of the car is 0060 kgs Assuming that rain comes to rest upon striking the car find the average force exerted by the rain on the roof Raindrop 0 vol 6f 0 ms 2 FAt me mV0 Ramstorm cont Rainodrop Neglecting the weight of 701 33 0 ms the raindrops the net force 1 on a raindrop is simply the r 7 force on the raindrop due to the roof EN me m70 E A jvo t F 0060 kgs 15 ms 090 N Conservation Laws WORKENERGY THEOREM lttCONSERVATION OF IMPULSEMOMENTUM THEOREM e Apply the impulsemomentum theorem to the collision between two objects ENERGY midair Midair Collision Internal forces Forces that objects within v c the system exert on each other quot39 External forces Forces exerted on objects v by agents external to the system OBJ ECT 1 t mlvfl m1V01 OBJECT 2 v f lui Alter collision W2 F21 t mZVfZ m2V02 Equation Derivation W1 12At mlvf1mlvol W2 E21At mzvfz mzvoz a1 W2 F12 F21Atmilnquot 712 7f2ml Equation Derivation cont The internal forces cancel out W1W2At E f sum of average external forcesAl E P It l Lcr collision Conservation of Momentum sum of average external forcesAl Pf P 0 If the sum of the external forces is zero then PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM The total linear momentum of an isolated system is constant conserved An isolated system is one for which the sum of the average external forces acting on the system is zero Net External Forces Illustrated PRINCIPLE OF CONSERVATION OF L NEAR MOMENTUM The total linear momentum of an isolated system is constant conserved An isolated system is one for which the sum of the average external forces acting on the system is zero In the top picture the net external force on the system is zero In the bottom picture the net external force on the system is not zero Ice Skaters Example Starting from rest two skaters 39 push off against each other on ice where friction is negligible quott9 O he IS a 54kg woman and 3 l i one is a 88 kg man The woman moves away with a speed of New 25 ms Find the recoil velocity of the man 7 bl After Ice Skaters Jul 2130 mlvf1m2vf2 0 Z 54 kg25ms 15ms 88kg f2 lb Aflcr Howto Solve Momentum Problems Applying the Principle of Conservation of Linear Momentum 1 Decide which objects are included in the system 2 Relative to the system identify the internal and external forces 3 Verify that the system is isolated 4 Set he final momentum ofthe system equal to its initial momentum Remember that momentum is a vector Collision Types The total linear momentum is conserved when two objects collide provided they constitute an isolated system Elastic collision One in which the total kinetic energy ofthe system afterthe collision is equal to the total kine ic energy before the collision Inelastic collision One in which the total kinetic energy ofthe system afterthe collision is not equal to he total kinetic energy before the collision if the objects stick together after colliding the collision is said to be completely inelas ic lln Inelastic collision 4 lc Completely inelastic collision Bullet Problem The mass of the block of wood is 250kg and the mass of the bullet is 00100kg The block swings to a maximum height of 0650 m above the initial position quot quot39z Vol Find the initial speed of the bullet In A 0 0650 m 4 in1 1quot Z l V Solving the Bullet Problem Apply conservation of momentum to the collision mlvfl m2vf2 mlvol m2v02 Bullet Problem Solution Hi vf 2l980ms2l0650 m T m1 m2 vf mlvo1 71 quot2 m2 m1 m2 vf U f 01 ml W M I o 550 m mgh zimv2 0550 m 01 m1 39 39 2 ltmm2gtghf mlmZv 2 l gh 1 00W 25 v f g g 2 vf ZWZW v01 001 kg J1 2i980ms l0650 m896ms Collisions in 2D 2D Collision Problem Setup q um 0900 ms n no 0900 mls I m10150kg TV mo 150kg sooni m so ordj D l V l 9 l J 1 77777777777777 quotL i W1 M Mr v13 lif r H gt H Y 0 121quot r quot75quot quot02 0540 quot 13 35 o wk 053 ms 02 0540 ms 350 m 053 quot90 260 kg l l 1112 0260 kg l U2 12 o 700 ms mlvflx m2vf2x mlvolx m2V02x mlvfly mzvfzy mlvoly mzvozy 12Body System Center of Mass Center of Mass Displacement The center of mass is a point that represents the average location for A x the total mass of a system an xl Ax2 m x m x cm 2 1 1 2 2 Ax mle1 mzsz mlv1 m2v2 m1 m2 cm 1111 m2 cm 1111 m2 Center of Mass Velocity More About The Skaters 3 BEFORE F quot 2m m1V1 mzvz 0 Van ch u m1 m2 m1 m2 t w i few 2 7 all In an isolated system the total linear momentum does not change 1 7 therefore the velocity of the center of mass does not change AFTER 88k 15m s 54k 25m s gx gt gx 0002z0 a 88 kg 54 kg Phys 1111K Lecture 08 CampJ Ch8 Rotational Kinematics n yAxis of rotation A In the simplest kind of rotation a points on a rigid object move on X circular paths around an axis of rotation I K Aquot E I A i 39 E i39a m 27 Angular Displacement The angle through which the object rotates is called the angular displacement AXIs of rotation I I I 747 I 39 A9 i 39 E I 4 6A A6 6 6 7 Reference 0 39 line Radial line Angular Displacement Definition When a rigid body rotates about a xed axis the angular displacement is the angle swept out by a line passing through any point on the body and intersecting the axis of rotation perpendicularly Axis of rotation 4 4 9 rls I 90 Reference line i By convention the angular displacement is positive ifit is counterclockwise and negative if it is I clockwise Radial line SI Unit of Angular Displacement radian rad Rotational Geometry 6 in radians Ar016ngth i Radius r 39 P g 39 Reference 7 gt line For a full revolution 27W L 27139 rad 360 6 2 39rad r Example Satellite Separation Synchronous satellites are put into an orbit whose radius is 423X107m If the angular separation of the two 239 satellites is 200 degrees nd the arc length that separates them 6 in radians Ar016ngth i Rad1us r 200 ght ml 00349 rad K360 deg s r0 423 x107m00349 rad 148gtlt105m 920 miles E Total Solar Eclipse Concept The diameter ofthe sun is about 400 times greater than hat ofthe moon By Ami r coincidence the sun is also about 400 times farther from I u f39w39 he earth than is the moon For an observer on the earth compare the angle subtended by the moon to the angle 7 quot subtended by the sun and explain why this result leads to a total solar eclipse H Total Solar Eclipse Geometry Nmoon Arc length i 6 in radians Rad1us r g g Change of Angular Displacement Average Angular Velocity T e Axis of Averageangular Vdocity AngulardISplacement rotat on I Elapsed t1me A66 6 60 M y Reference a he t t At How do we describe the rate E Radial he at which the angular displacement I is changing SI Unit of Angular Velocity radian per second rads E E High Bar Example Instantaneous Angular Velocity A gymnast on a high bar swings through two revolutions in a time of 190 s A wzllmwzhm Find the average angular velocnty AHO AHO At of the gymnast l l l l A6 200 rev 2 rad lreV 2 126 rad 126 a rad 663 rads 1903 Angular Acceleration Changing angular velocity means that an angular acceleration is occurring DEFINITION OF AVERAGE ANGULAR ACCELERATION Chan ein an ular velocit Average angular acceleratlon Elapsedtime 0 600 Am a t to At SI Unit of Angular acceleration radian per second squared rad2 Linear Movement Review Five kinematic variables 1 displacement X 2 acceleration constant a 3 final velocity at time t v 2 2 4 initial velocity v0 5 elapsed time t Jet Engine Example As seen from the front of the engine the fan blades are rotating with an angular speed of110 rads As the A plane takes off the angular velocity of the blades reaches 330 rads in a time of 14 s Find the angular acceleration assuming it to be constant 330 rads 1 lOrads 16rads2 14s Rotational Movement Equations r y I 39 39 ANGULARACCELERATION ANGULAR VELOCITY L a 00 at N6wowtltT TME ANGULAR DISPLACEMENT 2 2 a 00 2056 6w0tgat2 Q Table 82 Symbols Used in Rotational and Linear Kinematics Comparison Table 81 The Equations of Kinematics for Rotational and Linear Motion Rotational Linear Motion Quantity Motion 9 Displacement x 00 Initial velocity v0 0 Final velocity 17 a Acceleration 0 1 Time f Rotational Motion a constant Linear Motion 0 constant w a at 9 gm wt 9 wot m z w2 1102 2010 84 86 87 88 v on at x 5m U x trot Alatz 2 002 2ax 24 27 28 29 Bartending Example The blades are whirling with an angular velocity of 375 rads when the puree is pushed in quotL39s When the blend button is pushed the blades accelerate and reach a greater angular velocity after the blades have rotateg through an angular displacement of 440 39 rad Axis of rotation butB 4 The angular acceleration has a constant value of 1740 rad2 Find the final angular velocity of the blades l 1 if 5 Q Rotational Problem Steps 1 Make a drawing 2 Decide which directions are to be called positive and negative 3 Write down the values that are given for any of the five kinematic variables 4 Verify that the information contains values for at least three of the five kinematic variables Select the appropriate equation 5 When the motion is divided into segments rememberthat the final angular velocity of one segment is the initial velocity for the next 6 Keep in mind hat there may be two possible answers to a kinematics problem Bartending Solution 9 or w wo t 440 rad 1740 375 rEtClS2 rads 02 of 2059 p wzqiw 2a6l agwaTQJ 37 5 rads2 21740 rads2 440rad 52i2 rads E E Tangential Speed vs Velocity Tangential Acceleration VT 2 tangential velocity VT tangentia1 speed rwlrw0 aT 2 Mt a in rad52 Stationary skater St Ii VT 7a a 1n radS keigfry or pivot or pivot E H Helicopter Example Centripetal Acceleration i G A helicopter blade has an angular speed of 650 revs and an angular acceleration of 130 rev2 VT2 no2 For point 1 on the blade find ac 7 the magnitude of a the 300 m lial r02 win rads tangential speed and b the tangential acceleration a 650 408 rads 5 MW I VT ra 300m408rads122ms HI I a 130 r6112 rad 817mds2 lreV aT ra 300 m817rads2 245ms2 E Discus Thrower Example Starting from rest the thrower accelerates the discus to a final angular speed of 150 rads in a time of 0270 s before releasing it u During the acceleration the discus moves in a circular arc of radius 0810 m Find the magnitude of the total acceleration E Discus Solution a0 m2 0810 m150rads2 zlgzmsz aT rar 60600 039810mMj a I 0270s 450mSZ a Ja ac2 182msz450msz 187ms2 Rolling Motion The tangential speed of a point on the outer edge of the tire is equal to the speed of the car overthe ground Linear velocity V Sports car Example Starting from rest the car accelerates for 200 s with a constant linear acceleration of 0800 msZ The radius of the tires is 0330 m Lmemmci y39v What is the angle through which each wheel has rotated Sport scar Solution 2 M242rad 2 r 0330m 9 a w wo t 242rads2 Orads 2005 l9wot 0n2 9 242radsz2200 s2 484 rad Angular Velocity Vector RightHand Rule Grasp the axis of rotation with your right hand so that your fingers circle the axis in he same sense as the rotation Your extended thumb points along the axis in the direction of the angular velocity Right hand Right hand Vectors Magnitude and Direction N W E S 7 75 miles 230 N OfW r 75 miles a 157 Polar Coordinates Cartesian Coordinates 69 miles y 29 miles Notation Book Boldface for vectors R normal font for scalar R My lectures and solutions Arrow for vectors R or R no arrow for scalar R R 75 miles 23 N0fW is in II ta Wl CD CD A B Ia19 21 Ae180 gt gt A B21 B x I Dyaw his mum 1 mm mum uwsnmm Vania 4a Draw vccvm ham m a A m up m cm harveclnr a u quot615 me 125mmquot m 5mg 1cmYm Scam chzhu PK 9 C eii ABQ RI 6A Scaie 1 Gm 2 WEE Components of a Vector CX C c058 Cy C Sim CZ CXZ CyZ tan 8 CCX 139 8 tan391CyCX 393 CxCy v 25 ms 40 below x axis v 25 ms 6 220 339 V l X y v vx 25 msc0s 22 19 ms vy 25 mssin 220 16 ms m Carry only direction Magnitude 1 No dimensions no units Unit vectors separate the direction from the components gt CCx Cyfz 17vx vyf2 17 19ms 16msf2 Vector quotHm with f VH3 quot V312 w W v 4 v Vector Addition with Components i l gt N 1 1x172x 1yF2y J 1151 E X 55 RxAxBxCfDx Ry AyByCyDy A pilot leaves Peachtree Dekalb Airport and flies 240 miles in a direction 25 North of East He then changes course and flies 150 miles due South How far and in what direction must another pilot leaving PDK fly to reach the same spot A 240 miles 25 North of East B 150 miles due South R A B Use a coordinate system where East is positive xaxis and North is positive y axis AX 240 ml cos 25 218 mi Ay 240 ml sin 25 101 mi BX 750 ml cos 270 0 mi By 150 ml sin 270 150 mi RfA Bf2mmHOW ampm gagmwm4wmn4mm Rmyg wa3mi 6 tanquotRRX 13 347 R223 mi347 The second pilot must fly 223 miles in a direction 13 South of East Temperature Scales quot I Celsius H Fahrenheit scale scale Temperatures are reported in degrees Celsius or degrees Fahrenheit Phys 1111K Lecture 12 5 c 39 A o CampJ Ch12 Temperature amp Heat A timid Temperatures changed on the other hand are reported in Celsius degrees or Fahrenheit degrees mm i I J ice and water Bulb Kelvin Thermometers Thermometers make use of the change in some physical property with temperature A property that changes with temperature is called a thermometric property Kelvin K Celsius C KeIVIn temperature Steam point 37315 10000 Ice point 27315 000 5 l l a Hot Junction 3 l l 7 Copper 1 7 a I V a I I Object WY T Voltmeter Copper Conslantan One kelvin I l equals one Celsius degree 45 Reference iunctmn Icepaint bth 0 C r Absolute u m 0 7 27315 zero Linear Thermal Expansion Thermal Expansion of Solid NORMAL SOLIDS The length of an object changes when its temperature changes 1 Temperature TO L Temperature T0 1 I I I AL 2 aLOAT I l I l I I l I l coefficient of l Temperature To AT I I I linear expansion r I I I K f JW 39 I L0 AL I 1 o 1 L0 A L Common UnIt for the Coef CIent of Linear ExpanSIon CO C Table 121 Coefficients of Thermal Expansion for Solids and Liquidsa Coel cicnl ofThernml Expansion C l 39 m Substance Linear a Volume B p solids A concrete sidewalk is constructed between Aluminum 23 X l0quot 60 X 1039 7 Brass 19x 10b 57 x 1076 two bUIldlngs on a day when the Concrete l2 x 1039 36 x 10 temperature is 25 C As the temperature PP 397 X 390 quot 3 X 10 rises to 38 C the slabs expand but no Glass common 85 X 1039 26 X lquot I Glumpym 33 X 106 99 X 07 space Is prOVI e or errna expanSIon Gold 14 x 10 I 42 X 10 DetermIne the dIstance y In part b of the Iron or steel 12 x 10b 36 x 10 drawing Lead 29 x 10 87 x 10 Nickel 13 x 10 h 39 x10 lt1 Quarlzu used 050 x IO quot 15 x 10 quot AL 2 aL AT Silver 19X 10 h 57 x10quot 0 1 Liquids 12x10 6 l3 0 ml3 C 0 00047 In Benzene i 1240 X 10 Carbon tetrachlorith i 240 X 0396 Ethyl alcohol 7 llZO X 1076 Gasoline 7 950 X 0quot 2 2 x Mercury m X w y 3 00047 m 3 00000 m 0 053 m Methyl alcohol 7 l200 gtlt 10quot A Water i 207 x l0 0 39 quotThe values liIrnand Bpmzsin lmllemperlnure ncllr 20 DC mal Since liquids do nnl lune xed shapes llle coull Icienl nl linellr expansion is nnl de ned I39nl39 lllem bl 7 Beam Another Example The beam is mounted between two concrete supports when the temperature is 23 C What compressional stress Concrete Concrete must the concrete supports apply to support support each end of the beam if they are to keep the beam from expanding when the temperature rises to 42 C AL aL AT S 63 557 39 quoti Eleclnclly rrgt 5 quot 39 d O b quot 1 F AL r Xr 7 39 Stress Y YaAT e39 29 or A L 39 l 0 g L Blmelallic swucn cold 39 7 Blmelalllc swntch not Aquot Contacts dosed 5 7 Cunlacls separated gt J Coffee quotstreng1h adjustment knob o 1 o 391 0 6 C 2 LHealmgco mCollee pot39on39 InColfee pot 39oll Conceptual Example Volume Thermal Expansion The figure shows eight square tiles that are arranged to form a square pattern with a hold in the center If the tiles are heated what happens to the size of the The VOIUme Of an ObjeCt Changes When its temperature Changes hole Expanded hole H le i l 39 coefficient of 39 39 P volume expansion Il l heated u Unhealed b Heated c A hole in a piece of solid material expands when heated and contracts when cooled just as if it were filled with the material that surrounds it Common Unit for the Coef cient of Volume Expansion i Z 1 Co Auto Radiator Example A small plastic container called he coolant resenoir catches he radiator fluid that overflows when an automobile engine becomes hot The radiator is made of copper and the coolant has an expansion coefficient of 40x104C 1 fthe radiator is filled to its 15quart capacity when the engine is cold 6 C how much overflow wi spill into the reservoir when the coolant reaches its operating temperature 92 C Ame 4 10x10 4 0 1 15 quarts86 C 0 53 quarts AVMW 51x10 6 0 1 X15 quarts86 C 0 066 quarts AVSPUl O 53 quarts O 066 quarts O 46 quarts Coolant reservoir Radiator H20 Expansion Maximum density at 4 C 10000 9999 l i To street Ice 4 6 Density kgm3 9998 9997 connection 7 fig quot 9996 r 0 2 8 10 4 m 7 Temperature C L 1 N V 7 Water at Waker at normal very high pressure pyessure Closed faucet What Is Heat DEFINITION OF HEAT Heat is energy that flows from a higher temperature object to a lowertemperature object because of a difference in temperatures SI Unit of Heat joule J The heat that flows from hot to cold originates in the internal energy of the hot substance If is 1101 earncf 1 0 sa y flmf 1 substance allllfaills 1011 u B2 Heat flow 7 Causing Temperature Change SOLIDS AND LIQUIDS HEAT SUPPL ED OR REMOVED N CHANGING THE TEMPERATURE OF A SUBSTANCE The heat that must be supplied or removed to change the temperature of a substance is Q chT specific heat capacity Common Unit for Specific Heat Capacity Jkg C What About Gases GASES volume is held constant OTHER UNITS 1 kcal 4186 joules 1 cal 4186 joules The value of the specific heat of a gas depends on Whether the pressure or This distinction is not important for solids Table 122 Specific Heat Capacitiesquot ofSame Solids and Liquids Speci c Heal Capacity L39 h Substance lkg39C l Solids Aluminum 900 X it Copper Glass x40 Human body 3500 37 C average lcchlsnt 200X10 iron or steel 452 Lead 128 Silver 235 Liquillx Bcnzcnc I740 Ethyl alcohol Z450 Glycerin 24 Mercury 3 9 WalerllS Cl 486 Fxccpi as noted the values arc for 25 quotC and l aim nl39prcwire JJogger Example In a halfhour a 65kg jogger can generate 80x105J of heat This heat is removed from the body by a variety of means including the body s own temperatureregulating mechanisms lfthe heat were not removed how much would the body temperature increase Q chT 80X105J Aug 35C mc 65kg3500JlkgC l Measuring Heat CALORIMETRY hermumeter 4 V lfthere is no heat loss to the surroundings ll the heat lost by the hotter object equals the A heat gained by the cooler ones aiunmeter l up u l nsulalmg i 7 l onlainer Jnkncwn r naleriai Measuring Heat Capacity Solution 6 Example The calorimeter is made of 015 kg of aliminum and contains 020 kg of water mCAT mCAT Initially the water and cup have the same temperature of 180 C A 0040 kg A1 mass of unknown material is heated to a temperature of 970 C and then added to the water After thermal equilibrium is reached the temperature of the water the cup and the material is 220 C Ignoring the small amount of 3 Thermometer Water mCATHTLk 0WTl quot heat gained by the thermometer nd the speci c heat capacity ofthe unknown r 7 7 material h i i C mCATAl mCATwater quot 1 unknown Note Vquot unknown If there is no heat loss to the surroundings 39 Zalorlme er the heatost by the hotter object equals the ND t r 9 00X102 Jkg Com 15 kg4 0 Co 4186 Jkg Com 20 kg4 0 Co heat gamed by the cooler ones naming 0 040 kg75 0 c ontainer J n known if natarial quotquot 1300JkgC e During a phase change the temperature ofthe mixture does not change provided the system is in thermal equilibrium Phases of Matter Temp vs Heat amp APhase PUOO Water boils JL Water vapor 100 warms Up Water warms up Us a densing Temperature C i ezi ng l J i Send I Mewm Wm 30 Heat Latent Heat HEAT SUPPLIED OR REMOVED IN CHANGING THE PHASE Table 123 Latent Heatsa of Fusion and Vaporization OF A SU BSTANCE Latent Heat Latent Heat of Melting Point of Fusion Li Boiling Point Vaporization L The heat that must be supplied or removed to change the phase substance no kg 00 Wkg of a mass m of a substance is Ammonia 778 332 X 10quot 334 137 X 105 Benzene 55 126 X 10 801 394 X 105 Copper 1083 207 X 10quot 2566 473 X 10quot Q Ethyl alcohol ll44 MIR X 10quot 783 R55 X 105 Gold 1063 628 X 10quot 2808 172 X 105 k Lead 3273 232 X Ill4 1750 359 X Hquot latent heat Mercury 4139 114 gtlt to 3566 296 x 105 Nitrogen 200 257 X 10quot l 5tX 200 X 105 Oxygen 2Ixx 139 x 104 ix30 213 x 105 Water 00 335 X l0 1000 220 X 105 SI Units of Latent Heat Jkg 3The values pertain to i am pressure Lemonade Stand Physics Vapor PEssure V 7 7 777 Example x 5 Ice at 0 C is placed in a Styrofoam cup containing 032 kg of lemonade at 27 C The specific heat capacity of lemonade is virtually the same as that of water After the ice and lemonade reach and equilibrium temperature how much ice still remains Assume that i f f mass of the cup is so small that it absorbs a negligible amount of heat U 39 ML f L GMAT lemonade Exggmd 133123 ice 2 quot rVquotCo39nstanttemperature 539 v 5 Constant temperature 1 Lf 39 heated sand 1 39 heated sand a b 41861 k C 032k 27 C 0 C g gX 2 011 kg The pressure of vapor that coexists in equilibrium with the liquid is 5 335X10 Jkg called the equilibrium vaporpressure of the liquid 4x10539 100cc Nozzle iror gtlt1Ci5 Pa 3 3 x 105 7 g I 39 f Highrpressure i r 053 gtlt1O5 Pa propellant I Spray 5 2 x 105 v 39 83 DC vapor r l I 101x105 u 39 1 I Tube 053 x105 II quot 39 r 1 I n 0 50 o 50 53 100 150 r 7 i 3 QM Liquid propellant v propellant Temperature C 1 plus pmde 5 product Only when the temperature and vapor pressure correspond to a point on the curved line do the liquid and vapor phases coexist in equilibrium ml in r NW ewe Air is a mixture of gases As is the case for liquidvapor The total pressure is the sum of the partialpressures of the component equilibrium a solid can be in gases equilibrium with its liquid phase only at specific conditions of The partial pressure of water vapor depends on weather conditions It temperature and pressure Tequotme can be as low as zero or as high as the vapor pressure of water at the W given temperature 5 Normal meltingfreezing a WWW H20 To provide an indication of how much water vapor is in the air weather g forecasters usually give the relative humidity 101 x 105 iiiiii W l 1 Percent 7 Partial pressure of water vapor 100 l l Equilibrium vapor pressure of water at existing temperature Temperature C Relative Humidity Example One day he partial pressure of water vapor is 20X 103 Pa Using the vaporization curve determine the relative humidity ifthe temperature is 32 C Percent Partial pressure of water vapor 100 Equilibrium vapor pressure of water at existing temperature 2 0 gtlt103 Pa Relatrve hurmdrty 3gtlt 100 42 4 8x10 Pa 3 48 x 10 Q E D 25 x 103 Pressure Pa Dew Point O 2 Temperature C DeWpoint Ifthe air is gradually cooled while maintaining the moisture content constant the relative humiditv will rise until it reaches 100 This temperame at which the moisture content in the air wall saturate the air is called the dew point Ifthe air is cooled further some of the moisture will condense 10 en in o o A o Relative Humidity 7w N o o 30 60 Temperature OF Dew point of 25 C http hyperphysics Dwaer gsu eduhbasekinetcrelhum html 40 The temperature at which the relative humidity is 100 is called the dew point Dehumidified air Dehumidifier Cold coils Humid air Circulating refrigerant Receptacle Phys 1111K Lecture 16 CampJ Ch16 Waves and Sound What is a Wave 1 A wave is a traveling disturbance 2 A wave carries energy from place to place JWWWWWWWWW WWMWM 39 4 WMWWWWWWW gt w lllllllllll Transverse Waves L rWNWmmvmvmmmmmmmmwmm 39 WAMWMMWWWWWWWM gt WWW Longitudinal Waves i quot MMWWWMMWNMNMNWMWMMWVMW 1 1 Compressed region 1 u 1 1 w 1 WWWWWWWWWMWWMMWMM 2 VMMWMWMWWMWN U l l 1 1 1 Water Waves Direction of wave travel gt if Water particle Transverse moves on component circular path Longitudinal component Water waves are partially transverse and partially longitudinal Periodic Waves Periodic waves consist of cycles or patterns that are produced over and over again by the source In the figures every segment of the slinky vibrates in simple harmonic motion provided the end of the slinky is moved in simple harmonic motion iNNNMWWWWWWWMWM i willmwvwmmmllwvwwv 7 39 I f f i gt U WW WWW 39PR NVVMWWNWWWMWMMNVWWVN Stretched region Compressed region bl If WillWilli ra MNXMMWWWWWVWWWA r 39 H I 139 Periodic Wave Properties Vertical L DUSIHCIH Vertical 39u39r39aue ler39igll i A l at mm L Pennu T prlsitlon of the Slinky T pol nt 0 1 gt I l A the r l SI lnliy 39 l I l V Distance V V Time l Undlsturbed position mi At a particular time m m a partially location In the drawing one cycle is shaded in color The amplitude A is the maximum excursion of a particle of the medium from the particles undisturbed position The wavelength is the horizontal length of one cycle of the wave The period is the time required for one complete cycle The frequencyis related to the period and has units of Hz or 34 l fT Period Wavelength Frequency The time for one car to pass is the x 39 period T Velocity V gt k Wavelength 14quot A 2 A v T f Radio Example Example AM and FM radio waves are transverse waves consisting of electric and magnetic field disturbances traveling at a speed of 300x108ms A station broadcasts AM radio waves whose frequency is 1230x103Hz and an FM radio wave whose frequency is 91 9x106Hz Find the distance between adjacent crests in each wave xi 8 vft AM 1212300x103m32244m T f 1230gtltIO Hz 8 v FM 1213OOX10 ms 2326m f 919gtlt106Hz KN Wave Speed The speed at which the wave moves to the right depends on how quickly one particle of the string is accelerated upward in response to the net pulling force tension F linear density Electric Guitar Example Example Transverse waves travel on each string of an electric guitar after the string is plucked The length of each string between its two fixed ends is 0628 m and the mass is 0208 g for the highest pitched E string and 332 g for the lowest pitched E string Each string is under a tension of 226 N Find the speeds of the waves on the two strings H High E Transverse A vibration of the string v i N mS mL 0208x10393kg0628 m Low E v i 226N 207ms mL 332x10393kg0628m Wave Speed vs Particle Speed Conceptual Example Is the speed of a transverse wave on a string the same as the speed at which a particle on the string moves I particle String particle U wave Undisturbed position of string Loudspeaker Example m The frequency ofmotion is l 0 KHz and the amplitude is 0 mm a What Is the maximum speed of the diaphragm Where in the motion does this maximum speed occu V vx 7vT sin 19 7Amsinwt eve Vzux a vm Am A27rf 020x10395mX27r10x103Hz 13ms The maximum speed occurs midway between the ends of its motion Amplitude Frequency Phase What is the displacement y at time tof a particle located at x l o w ll H mm bl l i ll bl q 2 Trx Wave umn39mi toward 39 y A Sm 21139 i A ll g 2 7H WavenmrinnInward x y A sin 217 A LONGITUDINAL WAVES Sound Waves gt lllllllWllWlth VSllnky The distance between adjacent condensattons IS Condensation Normal alr equal to the wavelength of the pressure sound wave m l 4 3 WWquot l l l gquot l lvllllllllllllllllllll iquotStinky Wavelength l Wd Raretaction Normal alr pressure Condensation Eb Sound Vibrations Individual air molecules are not carried along with the wave Vibration of an individual air molecule Pitch Frequency THE FREQUENCY OF A SOUND WAVE The frequency is the number of cycles per second A sound with a single frequency is called a pure tone The brain interprets the frequency in terms of the subjective quality called pitch Sound Pressure Waves THE PRESSURE AMPLITUDE OF A SOUND WAVE Loudness is an attribute of V a sound that depends primarily High Mensa on the pressure amplitude miitude A of the wave l jL Distance Low Rarefaction Air pressure Atmospheric pressure Sound Speed Sound travels through gases liquids and solids at considerably different speeds able 161 Speed of Sound in Gases Liquids and Solids Substance Gases Air 0 1 A1 20 C Carbon dioxide 0 0 Oxygen 0 quotC Helium 0 C Liquids Chlorofonn 20 3 Ethyl alcohol 20 C Mercury 20 C Fresh water 20 C Seawat r D Solids Copper Glass Pyrex Lead Stccl Spccd ms Sound Speed in a Gas In a gas it is only when molecules collide that the condensations and rarefactions of a sound wave can move from place to place 3kT This actually overestimates Vrms m Speed of Sound in an Ideal Gas V WT k138gtlt10 23JK m Monatomic gas Ideal Gas y 5 or 3 More accurate equation for 7 5 Speed of Sound In an Ideal Gas 1 Diatomic gas Lightning amp Thunder Conceptual Example There is a rule of thumb for estimating how far away a thunderstorm is After you see a flash of lighting count off the seconds until the thunder is heard Divide the number of seconds by ve The result gives the approximate distance in miles to the thunderstorm Why does this rule work 10 mile 16 gtlt1O3 m Table 111 Mass Densitiesa s of Common Substance S ub dnuc o I s A I umil l um 133 Concrete ce Imn meal Lead Quartz Silver Wood yellow pine Liquids Blood whole 37 C7 1000 Ethyl alcohol Mercury Oil hydmulic Waler 4 Cl 33 Il Carbon dloxide Unless olherwisc iioled densilies are given a o c and l alm pressure LIQUIDS Mass Density p kgm 2700 V 2 ad 8470 p 2200 8890 Table 103 Values forthe Bulk 3520 Modulus of Solid and Liquid 19 300 Materials 917 7860 Bulk Modulus B 11300 Material Nin3lgtngt 266 Solids 125530 Aluminum 71 X 1039 Brass 67 X 1039 Copper l3 X 0quot 800 Diamond 443 X 0quot 13 600 Land 42 x 1039 5 00 Nylon 6 x lo 0 X 10 Osmium 462 x 0 Pyrex glass 26 X 1039 129 1798 Steel 14 X 10quot M79 Liquids 00399 Ethanol 89 X 0quot 125 oil 17 x l0quot 43 Water 22 X loquot SOLID BARS Table 101 Values for the Young39s Modulus of Solid Materials Young s Modulus Y Nm2 Malcrial Aluminum 69 X 10 Bone Cnrnprtlssinn 94 X I0quot Tension I 6 X 10 Brass 90 X 10 Brick 14 x 10 Copper 1 X 10quot Mohalr 29 X 10 Nylon 37 X IOU Pyrex glass 62 X 10 Steel 20 X Il Te on 37 X IO Tilzlnium 12 X lt Tungslcn 36 X 0 Power amp Intensity Sound waves carry energy that can be used to do work The amount of energy transported per second is called the power of the wave The sound intensity is defined as the power that passes perpendicularly through a surface divided by the area ofthat surface Sound Intensity Example Example 12x105W of sound power passed through the surfaces labeled 1 and 2 The areas ofthese surfaces are 40m2 and 12m2 Determine the sound intensity at each surface P 12x10 5 2W 30gtlt10 5 Wm2 A1 40m 5 i 12x10 2W 10gtlt10 5 Wm2 A 12m Sound Intensity Sensitivity For a 1000 Hz tone the smallest sound intensity that the human ear can detect is about 1x10 12Wm2 This intensity is called the threshold of Sound source at hearing On the other extreme continuous memfsphe39e exposure to intensities greater than 1Wm2 can be painful If the source emits sound uniformly in all directions the intensity depends on the distance from the source in a simple way power of sound source P I 4751 K 2 area of sphere Reflected Sound Waves Conceptual Example Suppose the person singing in the shower produces a sound power P Sound reflects from the surrounding shower stall At a distance r in front of the person does the equation for the intensity of sound emitted uniformly in all directions underestimate overestimate or give the correct sound intensny Reflected sound if Direct sound Reflected sound Decibel The decibel dB is a measurement unit used when comparing two sound intensities Because ofthe way in which the human hearing mechanism responds to intensity it is appropriate to use a logarithmic scale called the intensity level 1 6 10 dBlogI Ia 100gtlt10 12 wm2 Note that log10 so when the intensity of the sound is equal to the threshold of hearing the intensity level is zero Sound Intensity amp dB s 10 dBlog i In 100gtlt10quotZ wmZ 10 Table 162 Typiltal Sound Intensities and Intensity Levels Relative to the Threshold of Hearing Intenslly Imensity I Wmz Level dB Threshold ofhearing 10 x 10 0 Rustling Icavcs 10 x 10 10 ispcl I 0 x 10quot 20 Normal convclsation mctCI39 32 x 10 65 lnsidccarin cirymf c 10 x 0quot 80 Car without muf er 10 x 0 1 100 Live rock concen 10 120 Threshold of pain 10 130 Sound Intensity Example Example Audio system 1 produces a sound intensity level of 900 dB and system 2 produces an intensity level of 930 dB Determine the ratio of intensities 200 watts so D Hill g4 ET 7 20 watts Sound Intensity Solution 6 10 dBlogII 0 61 10 dBlog11 1 62 10 dB1ogU2 z 61 10 dB10g12 10dB10gj 1IOdB10g1210lOdB10g 0 1110 1 30 dB 10 dB10g J Doppler Effect Condensations Larger wavelength Truck at rest in Truck moving 7 Wavelength Smaller wavelength The Doppler effect is the change in frequency or pitch of the sound detected by an observer because the sound source and the observer have different velocities with respect to the medium of sound propagation Doppler Shift MOVING SOURCE Stationary observer VST V V V Truck at rest a Stationary observer 7 pvs 1 1 f0 fs 7 Doppler Equations source moving toward a stationary observer source moving away from a stationary observer f0 4 4 1 1vSv J Train Horn Example 6 Example A highspeed train is traveling at a speed of 447 ms when the engineer sounds the 415Hz warning horn The speed of sound is 343 ms What are the frequency and wavelength of the sound as perceived by a person standing at the crossing when the train is a approaching and b leaving the crossing 1 1 f0 2 fs f0 f3 1 1 vs v Vs V approaching 39eaVing 1 J2 477HZ f0 HZ 343ms 343ms Doppler Shift With Observer Moving MOVING OBSERVER Stationary V V SOUI CE ff0f10 A 0 s s A fs If V0 Moving observer v Observer moving Observer moving towards stationary away from source stationary source f0 lquot 0 f0 fsl quot 0 V V General Doppler Equation GENERAL CASE Numerator plus sign applies when observer moves towards the source Denominator minus sign applies when source moves towards the observer Medical Applications of Sound By scanning ultrasonic waves across the body and detecting the echoes from various locations it is possible to obtain an image Medical Applications of Sound Ultrasonic sound waves cause the tip of the probe to vibrate at 23 kHz and shatter sections of the tumor that it touches Medical Applications of Sound Transmitter Receiver When the sound is reflected from the red blood cells its frequency is changed in a kind of Doppler effect because the cells are moving What Does This Mean Intensity level dB Frequency Hz Phys 1111K Lecture 03 CampJ Ch3 Kinematics in 2D Displacement in 2D r0 1n1t1a1 posnlon F 2 nal position A F F0 displacement Average Velocity in 2D Average velocity is the displacement divided by the elapsed time r ro Ar 6 t to At Instantaneous Velocity in 2D The instantaneous velocity indicates how fast the car moves and the direction of motion at each instant oftime e A V 2 11m At gt0 A Instantaneous Velocity cont Average Acceleration I 1 j v 40 AV I a I t t At a A V 2 11m At gt0 5 quot 9 Our Kinematic Equations Constant Acceleration X axis v 2 v0 at x ltv0vt v Vigil Lil l Xi V V 2ax V 2V0xaxt xvoxvxt xv0tat2 xzvoxtaxt2 v v02x2ax Constant Acceleration Yaxis Constant Acceleration 2D y y vy voy ayt v 3 t g iiiIll l 2 xxx A y y voyt 2 ayt 1 7 1 4 I v quot all v it l l y y 2 0 y i x 0y x 2 2 The X part of the motion occurs exactly as it would if the v V y V 0y 261 yy y part did not occur at all and vice versa 2D Example Acceleration Problem Methodology In the Xdirection the spacecraft has an initial velocity component 1 Make a drawing of 22 mls and an acceleration of 24 mls2 In the y direction the analogous quantities are 14 mls and an accelera ion of 12 mls2 239 Derelde Whlch d39rectlons are to be called pos39tlve and Find a x and vX b y and vy and c the final velocity ofthe 993 9 39 spacecraft at time 70 s V 3 Write down the values that are given for any ofthe ve kinematic variables associated with each direction 4 Verify that the information contains values for at least three of the kinema ic variables Do this forxand y Select the appropriate equation 5 When the motion is divided into segments remember that the nal velocity of one segment is the ini ial velocity for the next 139 6 Keep in mind hat there may be two possible answers to a A 39 quot kinematics problem Movmg Spacecraft Example In the x direction the spacecraft has an initial velocity component of 22 ms and an acceleration of 24 msi In the y direction the analogous quantities are 14 ms and an acceleration of 12 msz Find a x and vX b y and Vy and c the final velocity of the spacecraft at time 70 s XI ax IVXI V ltl 240m52 22ms 70s lavlvl y V VW t I 120 m52l 14 ms 70s Solve for x and vX X ax vxlvaxltl 7 240mszl 22ms l705 x vat ch1 2 zzms70 s24msz 70 s2 740 m vX vax axt 22ms 24ms2 X10 s 190 ms Solve for y and vy 1 yl a Ivyl Vay 7 14msl705 V 7 120 msz y vgyl ayl 2 14ms70 s12mSZX70 s2 2 390 m vy v0y ayt 14ms 12ms2 70 s 98ms Combine the Components v vy 98 ms 6 vx190 ms v l190 ms2 98 ms2 210 ms 0tan 198190 27 Does It Make Sense y v 210 ms vy 98 ms 11 o 5 II 6 a a 9 gt x 190 ms Gravity in 2D Applications Underthe influence ofgravity alone an object near the surface of the Earth will accelerate downwards at 980m82 ay 980ms2 a 0 x Q vx vox constant Projectile Example The airplane is moving horizontally with a constant velocity a u of 115 ms at an altitude of 1050m Determine the time required forthe care package to hit the ground What are the magnitude and direction of the nal velocity of the care package Was Projectile Solution 1050 m 980 ms2 Yaxis Solution y ay Vy Voy 1050 m 980 ms2 0 ms 2 yv0ytayt y at t 2y mm ay 980ms 2 y 65 Velocity Solution Concepts A B um115 r l H v 139 if i I m w Velocity Solution Setup A 1 E gt We mm quotimam y ay ly V0 1 1050 m 980 ms2 0 ms 146 s Numeric Solution y ay vy Voy t 1050 m 980 ms2 399 0 ms 146 s vy voy ayt 0 980ms2146 s 143ms lVl Jv vj 1432 1152 21835 a184ms 6 tan 1V y tan X 51190 a 5120 vx 115 Relative Velocity Vm 90 ms gt Wm 20 ms 3 r I is l39 39 fl 7 O gt 739 110 ms Groundvbased Observer VPG VPT VTG Relative Velocity Example ml across a river that is 1800m wide The engine ofa boat drives it quotm quotquot E gt The velocity ofthe boat relative to the water is 40ms directed perpendicular to the current The velocity ofthe water relative to the b shore is 20ms a What is the Current velocity ofthe boat relative to the Ira shore b How long does it take for the boat to cross the river x 7 r 7 M V VBS 4ng VV2VS x4Oms2 2Oms2 45ms Travel Time Solution 1800m 4SOs 40ms K Phys 1111K Lecture 17 CampJ Ch17 The Principle of Linear Superposition and Interference Phenomena KN Pulse Mergers l lt w umml i Ollllllllllll39 u Overlap begins When the pulses merge the Slinky assumes a shape that is the sum of the shapes of the individual pulses e a Illlllllllllllllllll O llllllllllllllllllll in Total overlap the Slinky has twice the height Oi either pul 4 W W gt y 2 llllllllllllllll lll39 m The receding pulses Pulse Mergers II Wl Z We uuuwnw 0 uummm l m Ovevlap begins When the pulses merge the Slinky gt assumes a shape that is the sum of Ir 2 the shapes of the individual pulses nummmmmm3 mmmWmmummuuuv 4 e 391 U2 Totat overlap I may uumm mmv 2N r The recedmg pulses Principle of Linear Superposition When two or more waves are present simultaneously at the same place the resultant disturbance is the sum of the disturbances from the individual waves gt lW1 i ll I gt S I quotquotquot s 19 lllllllllll ulllllllll i1i llllllllllll39 Z Iii lu Overlap begins lii Overlap begins I gt E 7 mmnuwmm quotlmummmmmw illlllllllllllllllllll llllllllllllflllllllllllllllllllllll39 4 quot7quot In Total overlap the Slinky has Mice the height at either pulse llii Total overlap W Wuiiiiiiiiiiiiiiiw iiii39 quotWI2 quot quot quot mmquot lW try The receding pulses vi The receding Pulses Constructive Interference When two waves always meet condensationtocondensation and rarefactionto rarefaction they are said to be exactly in phase and to exhibit constructive interference Constructive interference Pressure Destructive Interference When two waves always meet condensationtorarefaction they are said to be exactly out of phase and to exhibit destructive interference Recewer rs Destrucnve Inler erence 0 Pressure 7 430 K A quotT 1Q me 31 1 o Qajk r5 I Noise Canceling Headphones Coherent Waves Ifthe wave patters do not shift relative to one another as time passes the sources are said to be coherent For two wave sources vibrating in phase a difference in path lengths that is zero or an integer number 1 2 3 of wavelengths leads to constructive interference a difference in path lengths that is a halfinteger number 72 1 72 2 V2 of wavelengths leads to destructive interference Sound or No Sound 1 Example Two inphase loudspeakers A and B are separated by 320 m A listener is stationed 2 40 m at C which is 240 m in front of speaker B Both speakers are playing identical 214Hz C tones and the speed of sound is 343 ms f 3 Does the listener hear a loud sound or no sound Calculate the path length difference 320 m2 240 m2 240 m 160 m Cl ltth I th l mS acuae ewaveeng f 160m Because the path length difference is equal to an integer 1 number of wavelengths there is constructive interference which means there is a loud sound OutOf Phase Example Conceptual Example To make a speaker operate two wires must be connected between the speaker and the amplifier To ensure that the diaphragms of the two speakers vibrate in phase it is necessary to make these connections in exactly the same way If the wires for one speaker are not connected just as they are for the other the diaphragms will vibrate out of phase Suppose in the figures next slide the connections are made so that the speaker diaphragms vibrate out of phase everything else remaining the same In each case what kind of interference would result in the overlap point InPhase vs OutOf Phase 39 f L 3 Destmclwe Interference 1 9 g Pressqu A T C V Q quota f I39I39quot 4 X JA J J e A Diffraction The bending of a wave around an obstacle or the edges of an opening is called diffraction b Without diffraction Single Slit Diffraction Room Vibrating air molecule located in the doorway single slit rst minimum Sill 6 Circular Opening Diffraction The high frequencies are beamed forward inside this cone 39 This person hears primarily the low frequencies This person hears the high and low frequencies xi Circular opening rst minimum Sill 6 B Beats iictive Destructive Constructivel Destructive I ponstr Small piece L of putty Two overlapping waves with slightly different frequencies gives rise to the phenomena of beats Beat Frequency Constructive i D Constructive es ruc ive Destructive 1 of putty Small piece The beat frequency is the difference between the two sound frequencies Transverse Waves Transverse standing wave patters lst Harmonic Fchlrznzv u Hun amemal rm 2nd halmumc Frequency 7 2 us uvenonel Mn 3m havmnmc F r 7 3 ran my 11 12nd marlanew Transverse Wave Reflections ILV In re ecting from the wall a fonNardtraveling halfcycle becomes a backwardtraveling halfcycle that is inverted Unless the timing is right the newly formed and reflected cycles tend to offset one another Repeated reinforcement between newly created and re ected cycles causes a large amplitude standing wave to develop Standing Wave Frequencies v String fixed atboth ends fr 2 n 1 23 4 Rock n Roll Physics Longitudinal Standing Waves A longitudinal standing wave pattern on a slinky N A N bA N TA N TA t Standing Wave Frequencies 7 fgt Fvequency Flute Example Example When all the holes are closed on one type of V flute the lowest note it can sound is middle quot C 2616 Hz If the speed of sound is 343 ms and the flute is assumed to be a cylinder open at both ends determine the distance L lmi n1234 L z zmzo smn 2f 22616 Hz More Standing Waves 3 Frequency 3 Tube open at one end fnn j n135


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Allison Fischer University of Alabama

"I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.