2. Physics 1501: Lecture 30, Pg 1 Physics 1501: Lecture 30 Today’s Agenda l Homework #10 (due Friday Nov. 18) l Midterm 2: Nov. 16 l Honor’s student … l Topics çDoppler effect çFluids:static conditions çPressure çPascal’s Principle (hydraulic lifts etc.) çArchimedes’ Principle

3. Physics 1501: Lecture 30, Pg 2 Recap & Useful Formulas: y x A l Waves on a string l General harmonic waves tension mass / length

4. Physics 1501: Lecture 30, Pg 3 Speed of Sound l For a pulse propagating along a string l In general, the speed of sound depends on çElastic property çInertial property v tension F mass per unit length  l For ideal gas  C P / C V

5. Physics 1501: Lecture 30, Pg 4 Speed of Sound l For a compressible fluid çCompressibility B plays the role of tension F çThe density  plays the role of  Bulk modulus where Y : Young’s modulus where l For a solid rod l Some numbers çAir (0 o C): 331 m/s çAir (100 o C): 386 m/s çWater (25 o C): 1 490 m/s çAluminum: 5 100 m/s

6. Physics 1501: Lecture 30, Pg 5 Waves, Wavefronts, and Rays l If the power output of a source is constant, the total power of any wave front is constant. l The Intensity at any point depends on the type of wave.

7. Physics 1501: Lecture 30, Pg 6 Intensity of sounds l The amplitude of pressure wave depends on çFrequency  of harmonic sound wave çSpeed of sound v and density of medium  of medium çDisplacement amplitude s 0 of element of medium l Intensity of a sound wave is çProportional to (amplitude) 2 çTrue in general (not only for sound) l Threshold of human hearing: I 0 = 10 -12 W/m 2

8. Physics 1501: Lecture 30, Pg 7 Intensity of sounds l Sound intensity level is given in decibels (dB) çOn a log-scale with 0 for threshold of human hearing çThreshold of human hearing: I 0 = 10 -12 W/m 2 l Some examples: Soundpressureintensity intensity amplitude (Pa)(W/m 2 )level (dB) Threshold of hearing3  10 -5 10 -12 0 Classroom0.0110 -7 50 City street0.310 -4 80 Car without muffler310 -2 100 Indoor concert301120 Jet engine at 30 m.10010130

9. Physics 1501: Lecture 30, Pg 8 Doppler effect l If the source of sound is moving çToward the observer  seems smaller çAway from observer  seems larger l If the observer is moving çToward the source  seems smaller çAway from source  seems larger l If both are moving l Ex: police car, train, etc.

10. Physics 1501: Lecture 30, Pg 9 Fluids : Chapter 15 l At ordinary temperature, matter exists in one of three states çSolid - has a shape and forms a surface çLiquid - has no shape but forms a surface çGas - has no shape and forms no surface l What do we mean by “fluids”? çFluids are “substances that flow”…. “substances that take the shape of the container” çAtoms and molecules are free to move. çNo long range correlation between positions.

11. Physics 1501: Lecture 30, Pg 10 Fluids l What parameters do we use to describe fluids? çDensity units : kg/m 3 = 10 -3 g/cm 3  (water) = 1.000 x10 3 kg/m 3 = 1.000 g/cm 3  (ice) = 0.917 x10 3 kg/m 3 = 0.917 g/cm 3  (air) = 1.29 kg/m 3 = 1.29 x10 -3 g/cm 3  (Hg) = 13.6 x10 3 kg/m 3 = 13.6 g/cm 3

12. Physics 1501: Lecture 30, Pg 11 Fluids A n l Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. çForce (a vector) in a fluid can be expressed in terms of pressure (a scalar) as: units : 1 N/m 2 = 1 Pa (Pascal) 1 bar = 10 5 Pa 1 mbar = 10 2 Pa 1 torr = 133.3 Pa 1atm = 1.013 x10 5 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in 2 (=PSI) l What parameters do we use to describe fluids? çPressure

13. Physics 1501: Lecture 30, Pg 12 çYoung’s modulus: measures the resistance of a solid to a change in its length. çShear modulus: measures the resistance to motion of the planes of a solid sliding past each other. çBulk modulus: measures the resistance of solids or liquids to changes in their volume. Some definitions L LL F F1F1 F2F2 V V -  V F l Elastic properties of solids : elasticity in length elasticity of shape (ex. pushing a book) volume elasticity

14. Physics 1501: Lecture 30, Pg 13 Fluids l Bulk Modulus çLIQUID: incompressible (density almost constant) çGAS: compressible (density depends a lot on pressure)               Bulk modulus (Pa=N/m 2 ) H2OH2O SteelGas (STP) Pb

15. Physics 1501: Lecture 30, Pg 14 l When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid l For an incompressible fluid, the density is the same everywhere, but the pressure is NOT! Pressure vs. Depth Incompressible Fluids (liquids) l Consider an imaginary fluid volume (a cube, face area A) çThe sum of all the forces on this volume must be ZERO as it is in equilibrium: F 2 - F 1 - mg = 0

16. Physics 1501: Lecture 30, Pg 15 FIf the pressures were different, fluid would flow in the tube! FHowever, if fluid did flow, then the system was NOT in equilibrium since no equilibrium system will spontaneously leave equilibrium. Pressure vs. Depth l For a fluid in an open container pressure same at a given depth independent of the container l Fluid level is the same everywhere in a connected container, assuming no surface forces l Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium? FImagine a tube that would connect two regions at the same depth.

17. Physics 1501: Lecture 30, Pg 16 Lecture 30, ACT 1 Pressure What happens with two fluids?? Consider a U tube containing liquids of density  1 and  2 as shown: çCompare the densities of the liquids: A)  1 <  2 B)  1 =  2 C)  1 >  2 11 22 dIdI

18. Physics 1501: Lecture 30, Pg 17 Pascal’s Principle l So far we have discovered (using Newton’s Laws):  Pressure depends on depth:  p =  g  y l Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. l Pascal’s Principle explains the working of hydraulic lifts çi.e. the application of a small force at one place can result in the creation of a large force in another. çDoes this “hydraulic lever” violate conservation of energy? »Certainly hope not.. Let’s calculate.

19. Physics 1501: Lecture 30, Pg 18 Pascal’s Principle l Consider the system shown: çA downward force F 1 is applied to the piston of area A 1. çThis force is transmitted through the liquid to create an upward force F 2. çPascal’s Principle says that increased pressure from F 1 (F 1 /A 1 ) is transmitted throughout the liquid. l F 2 > F 1 : Have we violated conservation of energy??

20. Physics 1501: Lecture 30, Pg 19 Pascal’s Principle l Consider F 1 moving through a distance d 1. çHow large is the volume of the liquid displaced? l Therefore the work done by F 1 equals the work done by F 2 We have NOT obtained “something for nothing”. çThis volume determines the displacement of the large piston.

21. Physics 1501: Lecture 30, Pg 20 A1A1 A 10 A2A2 A1A1 A 20 M M M dCdC dBdB dAdA A) d A = (1/2) d B B) d A = d B C) d A = 2 d B »If A 10 = 2  A 20, compare d A and d C. A) d A = (1/2) d C B) d A = d C C) d A = 2 d C Lecture 30, ACT 2 Hydraulics l Consider the systems shown to the right. çIn each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference d I in the liquid levels. »If A 2 = 2  A 1, compare d A and d B.

22. Physics 1501: Lecture 30, Pg 21 l At what depth is the water pressure two atmospheres? It is one atmosphere at the surface. What is the pressure at the bottom of the deepest oceanic trench (about 10 4 meters)? P 2 = P 1 +  gd 2.02  10 5 Pa = 1.01  10 5 Pa + 10 3 kg/m 3 *9.8m/s 2 *d d = 10.3 m P 2 = 1.01  10 5 Pa + 10 3 kg/m 3 *9.8m/s 2 *10 4 m = 9.81  10 7 Pa = 971 Atm Solution: d is the depth. The pressure increases one atmosphere for every 10m. This assumes that water is incompressible. For d = 10 4 m: Example Problems (1)

23. Physics 1501: Lecture 30, Pg 22 l Two masses rest on two pistons in a U-shaped tube of water, as shown. If M 1 = 3 kg and M 2 = 4.5 kg, what is the height difference, h? The area of piston 1 is A 1 = 200 cm 2, and the area of piston 2 is A 2 = 100 cm 2. Example Problems (2) Solution: M1M1 M2M2 h The pressure difference due to the two masses must be balanced by the water pressure due to h. Note that g does not appear in the answer. M 1 g/A 1 +  gh = M 2 g/A 2 h = (M 2 /A 2 -M 1 /A 1 )/  = 0.3 m

24. Physics 1501: Lecture 30, Pg 23 Using Fluids to Measure Pressure Barometer l Use Barometer to measure Absolute Pressure çTop of tube evacuated (p=0) çBottom of tube submerged into pool of mercury open to atmosphere (p=p 0 ) çPressure dependence on depth:

25. Physics 1501: Lecture 30, Pg 24 Using Fluids to Measure Pressure 1 atm = 760 mm (29.9 in) Hg = 10.3 m (33.8 ft) H 2 0 p0p0 hh Manometer p1p1 l Use Manometer to measure Gauge Pressure  Measure pressure of volume (p 1 ) relative to atmospheric pressure (  gauge pressure) çThe height difference (  h) measures the gauge pressure

26. Physics 1501: Lecture 30, Pg 25 Archimedes’ Principle l Suppose we weigh an object in air (1) and in water (2). çHow do these weights compare? W2?W2? W1W1 W 1 < W 2 W 1 = W 2 W 1 > W 2 çWhy? »Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object.

27. Physics 1501: Lecture 30, Pg 26 l The buoyant force is equal to the difference in the pressures times the area. W2?W2? W1W1 Archimedes : The buoyant force is equal to the weight of the liquid displaced. l The buoyant force determines whether an object will sink or float. How does this work? Archimedes’ Principle

28. Physics 1501: Lecture 30, Pg 27 Sink or Float? l The buoyant force is equal to the weight of the liquid that is displaced. l If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink. Fmg B y l We can calculate how much of a floating object will be submerged in the liquid: çObject is in equilibrium

29. Physics 1501: Lecture 30, Pg 28 The Tip of the Iceberg l What fraction of an iceberg is submerged? Fmg B y

30. Physics 1501: Lecture 30, Pg 29 Lecture 30, ACT 3 Buoyancy l A lead weight is fastened to a large styrofoam block and the combination floats on water with the water level with the top of the styrofoam block as shown. çIf you turn the styrofoam+Pb upside down, what happens? styrofoam Pb A) It sinks C)B)D) styrofoam Pb

31. Physics 1501: Lecture 30, Pg 30 ACT 3-A More Fun With Buoyancy l Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it. çWhich cup weighs more? Cup I Cup II (A) Cup I (B) Cup II (C) the same (D) can’t tell

32. Physics 1501: Lecture 30, Pg 31 ACT 3-B Even More Fun With Buoyancy A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (  oil <  ball <  water ) is slowly added to the container until it just covers the ball. çRelative to the water level, the ball will: water oil (A) move up (B) move down (C) stay in same place