PRINCIPLES OF PHYSICS I
PRINCIPLES OF PHYSICS I PHYS 2211
Popular in Course
Popular in Physics 2
This 29 page Class Notes was uploaded by Josiane Blick on Saturday October 3, 2015. The Class Notes belongs to PHYS 2211 at Armstrong Atlantic State University taught by William Baird in Fall. Since its upload, it has received 14 views. For similar materials see /class/217874/phys-2211-armstrong-atlantic-state-university in Physics 2 at Armstrong Atlantic State University.
Reviews for PRINCIPLES OF PHYSICS I
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 10/03/15
PHYS 2211 Notes follow and parts taken from Fundamentals at Physics 17 Edition Halliday ResnickI amp Walker College Physics Wilson amp Bn al and Physics 16 EditionI Cntnell amp Johnson Fluids amp Pressure We need a concept beyond force when we look at uids For a uid we can only apply a force over some area it s hard to push water around with a needle This force per unit area is called pressure just like stress Only the component of the force which is normal to the surface will increase the pressure 7 any component parallel to the area will do nothing to squeeze the uid We could then write iFcosG A A We re able to sense relatively small differences in pressure 7 when we drive up even a small mountain our ears tell us something s happening Similarly if you dive to the bottom of a pool you ll notice a pressure difference in your ears When we re talking about air the dependence of air pressure on altitude is more complicated than in the case of liquids For the atmosphere it s an exponential dependence which basically means that most of the Earth s air is close to the surface We will assume liquids are incompressible This isn t a bad approximation for most liquids but it s horrible for gases and we can t make it there Gases are very compressible 7 that s part of the de nition of a gas In reality liquids can be compressed by a very small amount under high pressures but it s not generally significant for us This means that we ll consider the density of the liquids to constant at all pressures The source of atmospheric pressure or the pressure at the bottom of the ocean is just gravity pulling on the material Since pressure is forcearea and force is mass gravity we can solve for the pressure at a given depth just by knowing the density of the material F m V Ah A A A A where we ve used the fact that the volume of the liquid sitting on our areaA is just equal to that area A times the height h of the column of uid If we re on Earth s surface and if we re not that g doesn t belong there we also have to add the atmospheric pressure to this because it s all around us In other words the pressure at the surface of the liquid isn t zero it s just normal atmospheric pressure We can rewrite our formula as ppopgh PHYS 2211 where p0 is atmospheric pressure at the Earth s surface about 101325 Pa or 147 lbsin2 One of the interesting things about a uid is that pressure applied to part of it is transmitted throughout the uid This m n0t mean that pressure is a constant throughout a liquid We still have the changes with depth shown above It does mean though that if we magically doubled Earth s atmospheric pressure that change by p0 would be felt at the top bottom sides and all other points of the liquid This idea is behind hydraulic assist devices of all kinds like your power brakes garage car lifts the jaws of life etc Basically we can use the fact that FA is constant except for height variations to turn a small force on a small area into a large force on a large area We can take a sealed tube of uid with a small piston on one end and a large piston on the other We push lightly on the small piston and the large piston pushes forcefully as shown below m Small force on small area gives pressure p Pressure p acting on much larger area produces much larger force Since the pressures are equal we see that the force on the big piston is equal to the ratio of the areas multiplied by the input force 1 small big big big small small big small Are we really getting something for free here No We re multiplying the force but there s no way that a tube of water can do any work on its own The work done by the large end will be the same as the work done on the small end The reason is that Work Force distance Let s see what a worked example looks like Assume the area of the small piston is 01 m2 and the area of the large piston is 1 m2 Ifthe force on the small piston is 50 N the formula above shows us that the force the large piston can exert isjust 1 m2 01 m2 50 N 500N The force does get multiplied How about the work done at the small end If the small piston moves forward by 02 meters the work done on it will be 50 N02 m 10 J That movement will push a volume of uid out of the small piston s cylinder into the larger part of the tube The volume moved is just V A d and we knowA and d already so we get V 01 m2 02m 002 m3 That same volume being pushed into the big tube is what pushes the larger cylinder out of the way again assuming the uid is incompressible How far does the big piston move We use the same volume formula to see what distance it has to travel to make room for this 002 m3 of water Now rearrange V A dto get d VA 002 m3 1 m2 giving a displacement of 002 m for the large piston How much work will it do Just use WFd and get 500 N002 m 10 J As we expected we re multiplying force but we re not multiplying work which would PHYS 2211 rk 39 39 the lever a great deal a small applled force beeorues a huge force but only over a small dstauee y gauge pressure Absolute pressure would be the pressure as eoruparedto avaeuuru In the formula above ppn pgh p ls the absolute pressure andpgls the atmosphene pressure Gauge pressure wouldbethe p b p 39 paorolm pamorpms pgalg At sealevel u tb F rtb lm Archimedes Prln ple amp Bugyanq welght e an lceberg ear welgh mllllons oftorrs or more amp sull oat but apenny wlll slnk 39 a foot b t u l force of gravlty far Also y faces eorue from7 How eould alarger or smaller force know what sldequot rt was supposedto be assoerated Wth7 The forces on the sldes ruustbalauee eael other Let39s look at what39s left upward and F ls dreeted downward Fm FrFr Pgth PghlAPghq hl1 PHYS 2211 Notice though that 112 7 h1A is just the volume of the block Also we re multiplying that volume by the density of the water We could then restate this in terms of the mass of the water that would ll the same volume as the block known as the displaced volume Fnet p g Vdisplaced m uid displaced g W uid displaced The net upward force is then just equal to the weight of the uid that s displaced If the cube has a side length of 10 cm its volume is then just 10 cm 10 cm 10 cm 1000 cm 1 liter or 0001 m3 Water has a density of1 g cm3 so that means we re displacing 1000 g or 1 kg of water 1 kg of water has a weight of 98 N so the upward force on the cube is 98 N Ifthe weight ofthe cube is less than 98 N it will bob to the surface If it s more than that it will sink to the bottom if released Assume the cube has a mass of 025 kg so its weight is 245 N How far out of the water will the cube oat It can t go all the way to the top and come completely out of the water because the buoyant force there is essentially zero not exactly zero because air is a uid also We don t usually see things oat in air because it s around 1000 times less dense than water There s a buoyant force due to the air on everything including you but it s generally so small as to not be noticeable Gravity would pull it back down into the water The balancing point occurs when the buoyant force is equal to the object s weight The volume of water which would have a mass of 245 N is 245 N 98 N 1000 cm3 This would be 250 cm3 Ifthe cube s face area is 100 cm2 this means the lower 25 cm of the cube is in the water and the rest of it is out of the water We see that average density is the important thing here Steel ships oat because their mass divided by the volume of the whole ship is less than the density of water Incidentally we can talk about densities by using something called speci c gravity This is essentially just the density of something usually a liquid compared to water If the speci c gravity of something is 135 m high 7 this is actually mercury it s 135 times as dense as water so one cm3 of it would have a mass of 135 g Mercury would de nitely sink in water Incidentally this is the idea behind antifreeze testers which are basically fat eyedroppers with lots of little spheres in them Each sphere has a different density and as the speci c gravity of your coolant increases more of them will oat The speci c gravity of the stuff in your radiator depends on the percentages of water and coolant in the radiator If it s all water the antifreeze tester will show that your coolant will freeze around 0 C A 5050 miX is ideal and should give you protection down to 410 C Fluid Dynamics amp Bemoulli s Equation We can use the principles of conservation of energy and conservation of mass to discover the way uids behave in motion This is in general a horribly complicated subject but if we make some assumptions we can greatly simplify it We will therefore consider the dynamics of an ideal uid sometimes jokingly called dry water We assume that the ow of the uid is steady which means all particles of the uid have the same velocity as they pass any given point This approximation is OK for low velocities Think of the ow of a slow stream If the velocity is large like rapids the ow becomes turbulent The ow should also be irrotational meaning there are no whirlpools in the stream Also if we dropped a small paddlewheel in the stream it wouldn t rotate as it went downstream We assume the uid is nonviscous meaning there s no internal PHYS 2211 friction Syrup and peanut butter are very viscous and water is not very viscous The realworld effect of viscosity is to slow the uid when it gets near the walls of the pipe since it s stopped Finally we assume the uid is incompressible meaning its density doesn t change We can use the conservation of mass to determine how the uid speeds up or slows down in response to changes in the size of the pipe carrying it We call the input end of the pipe 1 and the output end 2 The mass going into end 1 in a short time At depends on the pipe s size and the uid s velocity amp density We can write this mass entering the tube in that time as Aml pl AVI pl A1 V1 At At end 2 the mass leaving in time At is AmzzpzAszpzszzAt The mass added at end 1 in time At has to be the same as the mass leaving end 2 in time At Since the masses are equal we can set these two equations equal to one another We can also get rid of At which is the same on both sides Since we ve assumed there can be no change in density p1 and p2 are the same and can be cancelled We re left with the ow rate equation which says A1 V1 2A2 V2 For example if the area ofthe pipe at point 1 is 025 m2 and the uid velocity is 10 ms there and the pipe shrinks at point 2 down to an area of 005 ml the uid s velocity at that point is 025 m2 005 ml 10 ms 50 ms This effect is the reason why you use your thumb on the end ofa garden hose to get a faster stream of water The water leaves at a higher velocity but the volume delivered per minute doesn t change You won t fill a bucket any faster with your thumb over the end ofthe hose We can now use the conservation of energy to derive Bernoulli s equation which describes the ow of uids in a gravitational potential If we look at the work done onby the uid at each end of the pipe the difference will give us the net work WF1 x1F2 x2 p1A1V1fp2A2V2I By the ow equation A M A 2V2 The time is of course the same in both terms These three terms multiplied together are just the total volume which we can also write as the mass divided by the density We get Am W Z p1 p 2 i PHYS 2211 where Am is just the mass involved This net work is the same as the total of the changes in the kinetic and potential energies The kinetic energy change is as we ve seen before 1 AKeAmv v5 2 while the change in potential energy of the gravitational eld is AUZAmglyz y1 Equating W with the sum of these two terms we get Am 1 T121 1225A2nv VI Amgy2 y1 We get rid of the Am and move p over to get Bemoulli s equation 1 1 p1EPv5pgy1p2 Pv pgy2 We haven t said points 1 and 2 have to be anywhere in particular so this must be a constant everywhere in the pipe 1 p5pv2pgyC0nsz The units on each term would be energy except we have p instead of In so we really have energyvolume or Jm3 Under our original four assumptions if we know the value of this constant somewhere we know it everywhere in the uid A common example of this is to nd the speed of water exiting a hole at the bottom of a bucket For this purpose we ll assume that the atmospheric pressure is the same at the top of the bucket and the hole a very good approximation for a real bucket so p1 p2 The surface ofthe water will drop with a velocity we ll call 122 and we ll call the unknown velocity of the water leaving the hole V For this part we can use the ow rate equation to prove that A1121 A2122 where A 1 is the area of the hole and A 2 is the area of the bucket s open top For a small hole A2 is much greater than A1 The ow rate equation tells us that this means V2 must be much smaller than V1 We can therefore make the approximation that V2 0 which basically states that the lowering of the water level happens so slowly that the water m the bucket doesn t have any significant kinetic energy of course it s a different story outside the hole Taking all of this into account and representing the height of the water s surface by y2 and the height of the hole as y we then cancel p and get PHYS 2211 V122gy2y1 gtVI 2gy2y1 The picture below illustrates this w Vibrations amp Waves When a mass executes the same pattern of motion repeatedly like a mass bobbing on a spring or a pendulum swinging back and forth in a clock we call it simple harmonic motion SHM A force that produces simple harmonic motion has a form simpler than anything except a constant force like what gravity is approximately near the Earth s surface where the acceleration is just g and one example of this is called Hooke s Law This is the force exerted by an ideal spring 7 it s proportional to the extension or compression of the spring but it s in the opposite direction to re ect the fact that the spring opposes being stretched or compressed We can write it as F3 kx where k is the spring constant which we ve seen before and x is the amount by which the string is stretched or compressed relative to its rest length which we can write as x0 Sometimes you ll see this stated explicitly when the formula is written as FS k xxO The reason this force causes oscillations is that when the spring is extended and released the force above acts on the mass to accelerate it back towards the equilibrium position The force gets weaker but is still in the same direction as the mass approaches its original position When it reaches its initial position the force is zero but the mass now has some velocity and therefore shoots past the position x0 Once it gets past the original point now on the other side of it the force acts to drag it backwards The force will eventually slow it down until it stops and bring it back towards the equilibrium position again If there were no friction this would continue forever PHYS 2211 It will be convenient to describe the motion of the mass in terms of a few parameters The position of the mass at any time is called its displacement and the maximum of its displacement is called its amplitude A 01 Km for maximum x displacement It represents the distance between the equilibrium position and the furthest extension of the spring As the mass bobs back amp forth the time for it to make one complete cycle maximum positive displacement to equilibrium to maximum negative displacement to equilibrium and nally back to maximum positive displacement is called the period T Amplitude is measured in meters and period is measured in seconds as you would expect Sometimes the inverse of the period is also a useful thing to know This is called the frequency f or sometimes V and it is the number of cycles completed in one second You ve probably figured out by now that the relationship between frequency and period is just We already know that a stretched or compressed spring has a potential energy which is equal to Ukx x where xf and x0 are the spring s final and uncompressed positions The total energy will be the potential energy plus the kinetic energy remember that the mass is ying back and forth which we can write as Ezlmv2 lkx2 2 2 This is of course conserved in the absence of friction The energy just moves back amp forth between kinetic and potential terms never changing its total value We can use this to find the total value of the energy 7 we know that when the displacement is equal to the amplitude the mass is as far out as it s going to go by the definition of amplitude and is going to turn around Its velocity is momentarily zero so at that point the energy can be written as l E k 2 Knowing that this is the total energy 7 X m we can subst1tute 1t 1n the prevrous equation and find the velocity of the mass at any point x along its path We can write PHYS 2211 lkxi zlmv2 lkx2 gt v 2 2 2 i m We expect that v should depend on all of these things 7 the mass will certainly go faster if you extend the spring more initially or if it is small or if the spring is very stiff so that k is large The thing to notice here is that v is also a function of x which is changing constantly That also makes sense 7 the mass is stopped at the endpoints where xxm and it is moving the fastest at equilibrium because it s been accelerated in the same direction for a long time At equilibrium where x 0 the velocity is Equations of Motion A helpful fact about simple harmonic motion can be found by looking at circular motion If an object is moving in a circle and it casts a shadow in the same plane its shadow moves back and forth along a line in simple harmonic motion Kw If we call the position along this line x and the circle s radius is xm the two are connected by x xm Cos G If the angular velocity is held constant we know from our earlier work that 9 D twhere D is the angular velocity and tis the time Where we called this angular velocity earlier we can also call it the angular frequency since it represents the number of radians the object moves through per second Since there are 2 TE radians in a circle the number of cycles per second f the previous definition of frequency is related to the angular frequency by PHYS 22 ll 00 21 f We can also write this in terms of the period T rather than the frequency Ifwe do that we get xtxm Cos E T To generalize this a little further we can imagine that the x position may not be zero when we start the motion at t 0 We take this into account by adding a phase constant I which represents the position when time starts We can write the most general form of x as xt er C0sootq You can see that this motion will repeat every time t is a wholenumber multiple of T that is every period We can nd the period for simple harmonic motion by taking advantage of this circular reference system For the object moving in a circle the period is just the distance traveled divided by the velocity in this case the velocity of the object in the circle is at all times equal to the maximum velocity of the shadow which is in simple harmonic motion The reason the speed changes for the shadow but not the real object is that the component of velocity in the x direction changes as the ball moves around the circle The magnitude of its velocity is the same but the shadow only reveals the x component which grows and shrinks We know the maximum velocity for something in SHM and the distance around the circle is obviously just 2 TE xm so we get 21txm xmxkm k for the period An important result here is that the period depends only on the stiffness of the spring and the object s mass It does not depend on the amplitude so pulling the mass very far from equilibrium and letting it go will produce oscillations with the same period as a small displacement from equilibrium this is approximately true This is why a pendulum clock works so well 7 it does not depend on the pendulum moving exactly x centimeters from left to right every time A pendulum works in a way very similar to the mass on a spring When it is displaced from equilibrium there is a force which tries to move it back to its resting position When the pendulum returns to that position it has a velocity which carries it past rest and to the other side where the force acts on it again to drag it back down See the diagram below for the explanation of this PHYS 2211 Resting position Restoring force is mg Sin 0 In SHM the restoring force is proportional to the displacement Here the restoring force is proportional to the Sine of the angular displacement This means the equation of motion what is the angular displacement at a given time t after releasing the pendulum is in general very complicated and not solvable analytically we can t put it into a simple form Ifthe size of the angular displacement is small compared to one in radians 7 for our purposes less than about 15 20 the Sine of the angle is very close to the angle itself this only works in radians For example if we look at a displacement of 57 which is 01 radians we see that the Sine of 01 radians is 00998 or approximately 0 1 If we make the approximation known as the small angle approximation that the Sine of the angle is equal to the angle we get a restoring force approximately equal to m g 0 and we re back to a restoring force proportional to the displacement The formula which gives the position of a mass in SHM as a function of time uses the Cosine function to connect time and position In fact we could just as easily use Sine instead The initial conditions of the problem determine which one we should use For example if we use the Cosine function that says that the x position at time F0 is xm since the Cosine of zero is 1 If the mass is really at x xm at time F0 we need to add an angle to the Cosine argument to show that We would write instead of the formula above xzxm Cos E180 T The 180 angle here is again our phase constant or phase shift This is actually very much like the constants we saw in the kinematic equations For example one of those equations was x x0 Vt In this equation x0 is a constant which depends on where you start measuring distances If we start time when the mass is passing through the point x 0 we need for the formula to re ect that We could either use Sine since Sine of zero is zero or we could do the equivalent thing which is to make the phase shift 90 You can check for yourself that Sin 0 is equal to Cos0 90 We ll always get oscillation at the same frequency so the period is also the same but the position at time t depends on where the mass is when we start the clock The phase shift stands for the choice we made We can find the velocity and acceleration by looking at the formulas we already have for v and x If the phase shift in our problem is zero we get PHYS 2211 k gt7 4x xiC0S2DZ m k 2 2 xm x m we can simplify this more and get vti xmql Coszwt 5x Sinwt gt0xm Sinwt taking advantage of the fact that 0 4 k m Of course if we have a phase shift that makes us write x as a Sine instead of Cosine we ll get a Cosine term for the velocity This all boils down to the relationship between Sine and Cosine All we re really doing is using the old formula dxll vz The acceleration is easier to find since it s just the force divided by the mass Remember though that since the position is changing in time so is the force and therefore the acceleration so we ll see a t in our formula for acceleration We know that the force has the form F kx so we get 611 5 kx jxm C0s0l 02 xm C0s0l 02xl m m m 0139 am it dam dl dr This shows us that the acceleration is exactly out of phase with the displacement This is what we expect 7 if the displacement is positive it will take a negative acceleration to bring the mass back to equilibrium This oscillation back and forth could theoretically continue forever In reality though there is always some energy loss with each cycle This energy loss is known as damping and we ll look at it more closely soon PHYS 2211 Other examples of Simple Harmonic Motion We can also look at a mass on a string which is twisted The internal forces of the string will tend to make the mass rotate in such a way as to remove the twists As before by the time the string is untwisted the mass will have acquired an angular velocity and momentum and will not stop until the string has been twisted a certain amount in the m direction Then the process repeats Using our earlier work we know that it takes a torque to change an angular momentum This torque will be proportional to the amount of twist in the string measured as an angle 9 The constant of proportionality known as the torsion constant which depends on the stiffness of the string is usually written as K giving us TZ KG I T27 J where I is the moment of inertia of our mass We find that the period will be given by If we look more closely at the pendulum in this case the simple pendulum which is just a mass at the end of a massless string we find our restoring force gives a torque of C LmgSinG for a mass on a string of length L Using the angular form of Newton s second law and our small angle approximation from earlier we get X giving us T227 lg For a real pendulum known as the physical pendulum the moment of inertia is more complicated than that of a simple point mass We get I T227 7 mgh PHYS 2211 When damping is present we have damped harmonic motion instead of pure SHM Frequently the damping is small and we can ignore it for our purposes Including it makes the equations more complicated We can add to the equations of motion a velocitydependent force of the form Fd bv where the damping constant b must have units of forcevelocity or kgs Newton s second law gives bv kxma which we can solve to get ibtZm xlxm e cos039lq where For small values of b weak damping we can prove that the energy will decrease obviously we mean mechanical energy is being converted into heat energy via friction 7 total energy will of course be conserved as EZlkx e w39quot 2 We can correct for this energy loss by adding energy from an external source This is why a grandfather clock has hanging weights As they drop they retum lost energy to the pendulum to keep it swinging For some applications damping is desirable In your car you have springs to soften the ride when your car hits a pothole If you didn t have big dampers called shock absorbers your car would keep bouncing for quite some time and you wouldn t like the feel of it at all One of the tests for bad shock absorbers is to push each corner of your car up amp down a few times so that you get a large amplitude of motion and then stop The car should quit bouncing very quickly 7 usually within one cycle Waves Waves can generally be thought of as disturbances of a medium which carry energy Light and other forms of electromagnetic radiation are waves which don t require a medium We won t talk about these waves until next semester We ll typically be looking at motions which repeat and are PHYS 2211 therefore penode The partreles whlch make up the wave water forwaterwaves aufor sound waves roek for earthquake waves ete move baek and forth tn spaee as me goes on Thatmeans same spot rt wlll eomplete one full evele ofmonon tn one penod Penode tn spaee means that lf 39H aflxed dlstance known as awavelength Ifthe wavelength ls 10 em then eaeh peak of the wave wlll be separated by 10 em Wavelength 9 4 D Amplrtude A should glve us the waves speedln meters per seeond vlf Also the xraxl s A longrtudnal Imaglne A travel 1 Wu m wk 39 L L travels than everywhere else Sound waves tn atr are longltuddnal as thevrepresent pressure dlfferences lrghtlv ttn p mt That39s PHYS 2211 free it by sending a transverse wave down the hosewire Sending a longitudinal wave down the wire wouldn t do you any good You need to be able to move it perpendicularly to its length and that requires a transverse wave Some earthquake waves are transverse and these waves are also called shear waves because the wave is trying to shear the medium As we ve seen before liquids and gases don t support shear you ll never see a nail made out of water They do of course support compression so this is how sound travels in a liquid or a gas We can introduce other convenient ways to talk about the space and time variations in a wave 0 is the angular frequency of the wave with the same relation to the ordinary frequency f as before 0 2 T f It s sometimes convenient to talk about the number of waves per meter wave number symbolized by k instead of the number of meters per wave wavelength which is 7 We de ne the wave number as 31 7 The period and frequency keep the same relationship they ve always had f l T For a wave which travels in the positive direction along the x axis and oscillates along the y axis with angular frequency 0 and wavelength 7 we can write an equation that tells us what the position on the y axis is at any value of x and t We find that it is k yxlym sinkx 0l Looking at this formula we see that this wave has zero height along the y axis at x 0 t 0 Notice though that if we change position while still keeping time frozen we ll see a corresponding change in the y position until it reaches its maximum value which would be ym at x 7 2 from the origin If we keep moving to the right we find a zero again at x 7 a minimum ym at x 3 7 2 and zero again at x 2 7 We have a wave in space If we instead stay at the origin F0 and let time start we see that we have a minimum ym at t T 4 zero at t T 2 a maximum ym at t 3 T 2 and zero again at t lfort T We have a wave in time If we look at the whole picture we can see that this wave will travel from left to right negative x to positive x as time increases One way to understand this is to see that if t increases it will make the argument of the sine slightly more negative we can increase the value of x to counteract that A piece of this wave with a yvalue of 09823 ym for example therefore moves from left to right as time passes We can do this same analysis with any piece of the wave and we ll get the same answer Therefore the entire wave is moving from le to right If we change the sign of the Qt term we make the wave travel right to left Finally just as we saw in other oscillatory motion which is all that a wave really represents we have to include a phase constant to take the initial conditions of the problem into account Our full form for a wave traveling to the right is then PHYS 2211 yxlym sinkx 0l Waves on a String The velocity of a wave on a string depends on its mass per unit length because a km of dental oss may have the same mass as a meter of heavy rope 7 we need to know the mass of a given length and the tension in the string Since tension is measured in kg msz and mass per unit length is measured in kgm the form of the velocity is lt H tl where T is the tension and u is the masslength The energy of this wave will have kinetic and potential parts For a small piece of the string of length dx the mass will be dm u dx and the kinetic energy of the string which is a maximum when the string element has a y coordinate of zero will then be 12 dm vy2 where vy is 8y vy zgz coym cosk x cot The power transmitted by the wave is a combination of the time derivative of the kinetic energy and the time derivative of the potential energy These two energies are equal so we can nd the average power through LK Q 2 2 2 l 2 2 Pavg 2 dt JHdtw ymlcos kx Ming 2va ym where we have used the fact that the average of the square of cosine or sine over one full cycle 12 and Walt is just the wave speed v The Wave Equation amp Other Wave Properties What we ve seen so far of wave behavior has shown that there is a very close connection between the wave s behavior in time and its behavior in space That connection derived in your book gives us the wave equation and it is azyziazy 8x2 v2 BIZ PHYS 2211 mm y propomonahty constantxsjustthe wave39s valency squared Tms says thesum Below 7 quot andbluewaves all wnhmd wdnh m x axis their m m V H wave 5 much smaller 1 pm x A r V r r y 5111 1f you have a m n k r u m add PHYS 2211 th e MM H t spot and we ea11 tt destructive interference The waves have the ese ptetures are exam 1es oftmal cnnslnlctivvdeslnlctive interference amphtude frequency wavelength and ether the same phase or exaetly oppostte phases For m m V er hmquot We nd the resultant wave by addmg them duecdy y39xt ym sinkx 70 t sinch 7 0 MW Usmg mg tdehttttes we ear reduee thts to the shghuy slmplerform y39xt2ym 0054 2sinkxiat 2 PHYS 2211 and also notlee that thls amphde ls constmtfor aglven phase relanonshlp See what thls wlll glve tn th mmng eases whene 0 and 180 Standing Waves amp Resonance medla or anythlng else some othem getne eeted that39s how we see thtngs For waves ln stnngs we could have a 180 the wall there W111 be are ecnon beeause the hlll wtll txyto pull the wall up lnto lts shape The down the wall lt W111 do Justthat andthe wave W111 retum m the same shape as lt earne ls were aregular wave the hllls would be travellng down the rope and all potnts wouldmove at dlfferenttlmes Here the wall eneates awave movmg back down the rope canymg the same energy ealled nudes two anunodes ls one half of a wavelength or M2 PHYS 2211 Since the two waves are identical except for their direction of travel we can nd the resultant wave by adding the leftgoing and rightgoing waves y xtym sinkx cot sinkxcot We can again use trig identities to nd that this simpli es to y39xr 12 ym sinkxlcoswt Notice that the oscillations in time and space are now separated The amplitude changes in space only this is why the location of the nodes and antinodes does not shift with time and the position of any given element changes with time To see the difference between this and a traveling wave notice that any random point on the string will eventually go to the maximum possible amplitude above and later below zero at some time 7 all parts of the string will eventually visit i39ym For the standing wave the maximum range of motion for any given part of the string is less than iym except for the antinodes Various frequencies will cause this behavior 7 as the frequency increases the number of hills amp valleys that t between you amp the wall increases These special frequencies are called resonant frequencies or natural frequencies These resonant frequencies are found by examining the length of the rope If an integral number whole number of half wavelengths can t in the length of the string the resonant pattern will appear The connection between wavelength and string length is then Ln 7quot wheren123 s0 kn27L n The relation between frequency wave speed and wavelength means that the resonant frequencies are then v 2711 7 11 where n123 fquot M f1 The lowest frequency where n 1 is called the fundamental frequency and all other frequencies are integer multiples of it known as harmonics The velocity of a wave on a string depends on its mass per unit length because a km of dental oss may have the same mass as a meter of heavy rope 7 we need to know the mass of a given length and the tension in the string Using our earlier equation for v we nd the equation for resonant frequencies reduces to 21 PHYS 2211 T fn 1 7 where n 1 23 2M H Resonance is also found when you push someone on a swing at the righ time over amp over again to send them very high You can nd the resonant frequency of a wine glass by wetting your nger amp running it around the rim of the glass over amp over again In the past the wind has hit at least one bridge at just the right frequency resonance to cause it to swing back amp forth so violently that it collapsed Sound Sound waves in air are essentially all 39 quot J39 39 The t v r can t r r because gases and liquids don t support shear so when a mass of water is dragged past another mass of water there s essentially no interaction between the two masses These changes in pressure are detected by the eardrum and the tiny bones in your inner ear Humans can generally hear sounds which have frequencies from about 20 Hz to about 20000 Hz Whales elephants etc can hear some of the sounds below 20 Hz known as the infrasound region Dogs and other animals can hear above 20 kHz ultrasonic region Bats can hear up to about 100 kHz and they use this ability to nd the insects they eat They send out a very highpitched noise and listen for re ections which means they re really using sonar The frequency has to be very high because the bat needs a small wavelength to detect small insects Waves with a frequency of 30 kHz would have a wavelength of about a centimeter The same technique is used by doctors to examine people without exposing them to Xray radiation There is an ongoing debate about the effects of small doses of radiation on people We know that large doses can cause sickness or even death but determining the effects of small doses is much more dif cult There are two general theories 7 one is that below a certain threshold radiation has no effect on people You can think of this as working like a car wreck 7 in a collision with a brick wall at 80 mph perhaps 99 of people involved would be killed If we lower the speed the crash becomes more survivable but there is still a good chance for death even at 35 mph However if we drop the speed to 1 mph there is really no damage done to anyone If a million people were in collisions like this we would not expect it to cause any deaths Another opinion says that every little bit of radiation does damage and that while a large dose might kill 99 ofthe people exposed a very small dose may still kill 10 people out of 1000000 by causing a mutation that causes cancer As uncertain as this debate is in people it s even cloudier when applied to a fetus since the fetus cells are already dividing very rapidly and cells are thought to be more vulnerable during division For this reason no one wants to Xray a pregnant woman That s why pictures of a fetus are always made using ultrasound because there seems to be no way for low levels of sound to cause damage although they can cause heating or a skin burn if the technician is not careful 22 PHYS 2211 Speeds Sound moves faster in solids than in air in general because the molecules in a solid are connected to each other with stronger forces and therefore snap back into place more quickly when disturbed The velocity depends on Young s modulus for a solid or the Bulk modulus for a liquid or gas Also the density of the material is important as that describes how close the molecules are to one another The velocity can then be written vsolid 7 0r vliquid B In air the speed of sound depends on the temperature and other things we ll ignore like the humidity and we get vair33l06TCms where the 06 must clearly have units of m C s This gives us a speed of about 343 ms at room temperature in air For comparison sound moves at about 1500 ms in water and over 6000 ms in aluminum Traveling Sound Waves Sound is a longitudinal pressure wave That means that the molecules in the medium involved are alternately compressed and rare ed spread apart as the wave passes through The molecules aren t moving large distances when they transmit sound they mainly oscillate back and forth around an equilibrium position We can define the displacement of these molecules from this equilibrium position as follows s x I sm c0skx 0 I where Sm is the maximum displacement of the molecule amplitude and this displacement is along the direction of motion that s what a longitudinal wave means The other variables have the same meanings as in the case of transverse waves We could write a similar expression for the change in pressure at a given point in space and time as the wave passes ApxtApm sinkx xt where Apm represents the maximum change in pressure along the wave Notice that we have a sine function instead of cosine Places along the wave where the displacement is a maximum will be the places where the pressure change from the case without sound is zero and vice versa We can relate displacement to pressure change proven in your book using 23 PHYS 2211 Apm v p w Sm Sound Phenomena As with other wave disturbances we can have interference re ection refraction and diffraction with sound waves Re ection of sound can be found in echoes It s easy for you to experiment with this if you can find a large solid wall or a set of concrete bleachers If you clap your hands loudly you can hear the claps coming back to you as they bounce off of different objects Fireworks shows in cities are a great way to hear this happen if you can get to the roof of a building Refraction in sound is less common but it can happen if there is a layer of cool air underneath a layer of warm air As sound moves out and encounters the warm air where it speeds up it will bend at the interface and can come back down to ground enabling you to hear sounds from a greater distance than normal Diffraction is heard every time you hear a sound from around a comer 7 the wall is not sufficient to block all of the sound and the sound will bend around the comer Interference works the same way we mentioned before At some points two peaks will coincide and produce a much higher peak while at other points a peak and a trough will coincide and the resulting intensity will be greatly reduced Ifthere are two sources of sound and a wall producing an echo can be one of the sources the important thing to know is the phase of each wave at a point If the two waves are in phase we will get constructive interference and increased intensity Ifthe two waves are out of phase we ll get destructive interference and reduced intensity For the case of two speakers sending out the same sound at the same time all we need to know is the distance from our measuring device to speaker A and the distance from our device to speaker B In fact we really just need to know the difference in these distances If the difference corresponds to a whole number of wavelengths we will get constructive interference because the waves will be peaking at the same time and the valleys will also occur at the same time Ifthe difference is equal to an odd number of half wavelengths we get two waves 180 out of phase and we should get silence It has to be an odd number of half wavelengths because an even number of them would give a whole number of full wavelengths which gives constructive interference We can then write our condition for perfect constructive or destructive interference as AL 11 where n 0 24consl or n 1 3 5desl where AL is the path length difference source 1 to measuring device minus source 2 to measuring device and 7 is the wavelength Ask yourself why you might not have noticed this effect with your home stereo even if you listened to a pure note like a recording of atuning fork Hint why do you have two different outputs for your speakers Are they playing exactly the same sounds Having two notes which are almost the same frequency call them 0 and 0 causes the ear to hear beat frequencies Mathematically we add the two sounds together as we have before We ll stay fixed at a particular value of x to make things easier and we can therefore let the kx part of our 24 PHYS 2211 phase olrop out We get s39t2vm cosm39tcosmt ft le Sound Inten 39u The or Wm2 hght ampgravlty we ean ndlntensltv by usmg a slmple geometne argument Imagme apolnt equally formlng spheres Assume the mlcrophone has a square face Wth an area ofl em2 Look below at the power hlttlng the mlcrophone at dlfferentpolnts If all h u H has to eover alarger area Thls Wordlmenslonal drawmg ls mlsleadmg beeause when we move the In realltv 22 we moveol the mlcrophone 3 umes as far away the sounol would be spreaol over an area whlch ls 32 or mne umes as great Thls reoluees the lntensltv to 19 of what rt was We ean guess now that the general form of the lntensltv formula shouldlnclude afactorwhlch shows h w the area wlll ehange We39ve sald lntensltvls power dlvlded by area and we39ve sarolthe sounol expands ln spheres so we ean ndthe measunng devlce PHYS 2211 P P 1at a dlst r away om source 7 72 A 47E r This tells us that the intensity of sound falls off as the inverse square of the distance to the source This general result will return when you examine light later Basically intensity goes as lr2 because area increases as r We could also use relationships between power and displacement to write 1 7 vmzs2 29 m The range of intensities humans can hear and stand is enormous The threshold of hearing quietest sound you can detect has an intensity of about 103912 Wmz while the loudest sound you can stand threshold of pain has an intensity of around 1 Wmz Our ears don t respond linearly to sound In other words an intensity of 10396 Wm2 does not seem 10 times louder to us than an intensity of 10397 Wmz It would actually only seem about twice as loud For this reason it is common to use a logarithmic scale to describe sound intensity if you study astronomy you ll see a similar logarithmic scale is used because the eye s response to light is not linear either In astronomy this is called the magnitude system Logarithmic scales are a way to compress a wide range of information into a smaller range at the expense of detail For example if you want to make someone a map of how to get from your house to Atlanta you will probably want to have a great deal of detail near your house so that they can nd the streets that will take them to 116 Once on 116 you will need to omit most of the details except the big ones like the connection to 175 Once they are on 75 you might draw the entire length of I 75 in Georgia as just a straight line a few inches long Your scale is a compromise 7 you want to give the person a very detailed map to get himher out of town but there is no way or need for you to give a very detailed map of the whole trip from here to Atlanta The logarithmic scale for sound is a similar compromise A linear scale from 0 to the threshold of pain would make everyday sounds appear too tiny to distinguish from silence This logarithmic scale is the source of the decibel level we re familiar with To calculate decibels we first establish a reference level of sound called 10 which is the threshold of hearing or 103912 Wmz Intensities of sound are divided by this intensity and log of the result is then taken For example if we want to know the decibel level of a sound with an intensity of 10394 Wmz we would do this I 10 4Wm2 logilog 10 108 8 10 10 12Wm2 g B 26 PHYS 2211 The B is the abbreviation for bel One tenth of a bel is a decibel so this sound would be 80 decibels pretty loud This scale ts the range from nearsilence to painfully loud into 12 bels or 120 decibels We could invert the equation above and write 5 10dB log L 0 Doppler Effect An effect familiar to everyone is the change in pitch of a police or ambulance siren as the siren moves towards or away from you This is called the Doppler effect and it is caused by the bunching up or spreading out of sound waves as the source moves towards or away from us The same effect happens with light but is not noticed in ordinary situations because the size of the effect depends on the speed of the source compared to the speed of sound or light Since light is so much faster than sound we don t notice a color shift as a police car approaches us but if we had incredibly accurate measuring devices we could see it We get a similar effect if we move and the source is stationary The situations is not identical because there is a preferred reference frame provided by the air the sound travels through This is not the case for light which can travel without a medium and therefore has only one formula for Doppler shift where the relative velocity is the important thing For sound in air with a moving source and a stationary observer we get the formula below for the frequency heard by the observer V f39zf Viv s where is the frequency sent by the source v is the velocity of sound and V is the velocity of the source where a sign indicates that the source is moving away from the observer and a sign indicates a source moving towards the observer For example if an ambulance siren has a frequency of 1000 Hz and the ambulance is moving at 30 ms towards you you should hear a frequency of about 343343301000 1096 Hz After it passes you ll hear a frequency of about 343343301000 920 Hz If you re doing the moving the formula is slightly different The frequency you observe is given by vi v fl f D V where the now represents motion toward the source and means motion away from it Now if you re moving towards the ambulance at 30 ms you ll hear its 1000 Hz siren as 34330343 1000 1087 Hz and 343303431000 913 Hz as you move away 27 PHYS 2211 We eould Wnte arnore general formula as ata p u r arr wewrlllneara snnichnnm The boom long as ltmoves fasterthan the speed ofsounde bamer s W there39s notjust a slngle boom when the sound ls broken Thl ls the reason the Coneorde doesn39t y from NYto LA Lhe plane would l n w n m annoyrng or V r r n from breaklng the sound bamerwhlle over the heavllyrpopulated continental US Speeds of A plan e R l the Doppler effect for a movmg souree 1n thls plcture PHYS 2211 r w M m um uw quotH errele and you can see that it ls very far from me ongln The blue obseryer wlll be hlt by very many whlle the real frequency note Supersorue speeds wouldhave me erreles appeanrrg nutside prevlous erreles as below booms cone ls found by usmg the formula Sin 6 vv