Physics 1501: Lecture 30 Today ’ s Agenda

Physics 1501: Lecture 30Today’s Agenda • Homework #10 (due Friday Nov. 18) • Midterm 2: Nov. 16 • Honor’s student … • Topics • Doppler effect • Fluids:static conditions • Pressure • Pascal’s Principle (hydraulic lifts etc.) • Archimedes’ Principle

Recap & Useful Formulas: y  A x • General harmonic waves • Waves on a string tension mass / length

v tension F mass per unit length  • For ideal gas g = CP / CV Speed of Sound • For a pulse propagating along a string • In general, the speed of sound depends on • Elastic property • Inertial property

Speed of Sound • For a compressible fluid • Compressibility Bplays the role of tension F • The density  plays the role of  where Bulk modulus • For a solid rod where Y : Young’s modulus • Some numbers • Air (0oC) : 331 m/s • Air (100oC) : 386 m/s • Water (25oC) : 1 490 m/s • Aluminum : 5 100 m/s

Waves, Wavefronts, and Rays • If the power output of a source is constant, the total power of any wave front is constant. • The Intensity at any point depends on the type of wave.

Intensity of sounds • The amplitude of pressure wave depends on • Frequency  of harmonic sound wave • Speed of sound vand density of medium  of medium • Displacement amplitude s0 of element of medium • Intensity of a sound wave is • Proportional to (amplitude)2 • True in general (not only for sound) • Threshold of human hearing: I0 = 10-12 W/m2

Intensity of sounds • Sound intensity level is given in decibels (dB) • On a log-scale with 0 for threshold of human hearing • Threshold of human hearing: I0 = 10-12 W/m2 • Some examples: • Sound pressure intensity intensity • amplitude (Pa) (W/m2) level (dB) • Threshold of hearing 3  10-5 10-12 0 • Classroom 0.01 10-7 50 • City street 0.3 10-4 80 • Car without muffler 3 10-2 100 • Indoor concert 30 1 120 • Jet engine at 30 m. 100 10 130

Doppler effect • If the source of sound is moving • Toward the observer   seems smaller • Away from observer   seems larger • If the observer is moving • Toward the source   seems smaller • Away from source   seems larger • If both are moving • Ex: police car, train, etc.

Fluids : Chapter 15 • At ordinary temperature, matter exists in one of three states • Solid - has a shape and forms a surface • Liquid - has no shape but forms a surface • Gas - has no shape and forms no surface • What do we mean by “fluids”? • Fluids are “substances that flow”…. “substances that take the shape of the container” • Atoms and molecules are free to move. • No long range correlation between positions.

Fluids • What parameters do we use to describe fluids? • Density units : kg/m3 = 10-3 g/cm3 r(water) = 1.000 x103 kg/m3 = 1.000 g/cm3 r(ice) = 0.917 x103 kg/m3 = 0.917 g/cm3 r(air) = 1.29 kg/m3 = 1.29 x10-3 g/cm3 r(Hg) = 13.6 x103 kg/m3 = 13.6 g/cm3

n A Fluids • What parameters do we use to describe fluids? • Pressure units : 1 N/m2 = 1 Pa (Pascal) 1 bar = 105 Pa 1 mbar = 102 Pa 1 torr = 133.3 Pa 1atm = 1.013 x105 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in2 (=PSI) • Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. • Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as:

F L L F2 F1 V F V - V Some definitions • Elastic properties of solids : • Young’s modulus: measures the resistance of a solid to a change in its length. • Shear modulus: measures the resistance to motion of the planes of a solid sliding past each other. • Bulk modulus: measures the resistance of solids or liquids to changes in their volume. elasticity in length elasticity of shape (ex. pushing a book) volume elasticity

Pb Gas (STP) H2O Steel         Bulk modulus (Pa=N/m2) Fluids • Bulk Modulus • LIQUID: incompressible (density almost constant) • GAS: compressible (density depends a lot on pressure)

Consider an imaginary fluid volume (a cube, face area A) • The sum of all the forces on this volume must be ZERO as it is in equilibrium: F2 - F1 - mg = 0 Pressure vs. DepthIncompressible Fluids (liquids) • When the pressure is much less than the bulk modulus of the fluid, we treat the density as constant independent of pressure: incompressible fluid • For an incompressible fluid, the density is the same everywhere, but the pressure is NOT!

For a fluid in an open container pressure same at a given depth independent of the container • Fluid level is the same everywhere in a connected container, assuming no surface forces • Why is this so? Why does the pressure below the surface depend only on depth if it is in equilibrium? • Imagine a tube that would connect two regions at the same depth. Pressure vs. Depth • If the pressures were different, fluid would flow in thetube! • However, if fluid did flow, then the system was NOT in equilibrium since no equilibrium system will spontaneously leave equilibrium.

A) r1 < r2 B) r1 = r2 C) r1 > r2 Lecture 30, ACT 1Pressure • What happens with two fluids?? Consider a U tube containing liquids of density r1 and r2 as shown: • Compare the densities of the liquids: dI r2 r1

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. Pascal’s Principle • So far we have discovered (using Newton’s Laws): • Pressure depends on depth: Dp = rgDy • Pascal’s Principle addresses how a change in pressure is transmitted through a fluid. • Pascal’s Principle explains the working of hydraulic lifts • i.e. the application of a small force at one place can result in the creation of a large force in another. • Does this “hydraulic lever” violate conservation of energy? • Certainly hope not.. Let’s calculate.

Pascal’s Principle • Consider the system shown: • A downward force F1 is applied to the piston of area A1. • This force is transmitted through the liquid to create an upward force F2. • Pascal’s Principle says that increased pressure from F1 (F1/A1) is transmitted throughout the liquid. • F2 > F1 : Have we violated conservation of energy??

Pascal’s Principle • Consider F1 moving through a distance d1. • How large is the volume of the liquid displaced? • This volume determines the displacement of the large piston. • Therefore the work done by F1 equals the work done by F2 We have NOT obtained “something for nothing”.

M dA A1 A10 A) dA = (1/2)dB B) dA = dB C) dA = 2dB M dB A) dA = (1/2)dC B) dA = dC C) dA = 2dC A2 A10 • If A10 = 2´A20, compare dA and dC. dC M A1 A20 Lecture 30, ACT 2Hydraulics • Consider the systems shown to the right. • In each case, a block of mass M is placed on the piston of the large cylinder, resulting in a difference dI in the liquid levels. • If A2 = 2´A1, compare dA and dB.

Example Problems (1) • 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 104 meters)? Solution: d is the depth. The pressure increases one atmosphere for every 10m. This assumes that water is incompressible. P2 = P1 + rgd 2.02105 Pa = 1.01105 Pa + 103 kg/m3*9.8m/s2*d d = 10.3 m P2 = 1.01105 Pa + 103 kg/m3*9.8m/s2*104 m = 9.81107 Pa = 971 Atm For d = 104 m:

M1 h M2 Example Problems (2) • Two masses rest on two pistons in a U-shaped tube of water, as shown. If M1 = 3 kg and M2 = 4.5 kg, what is the height difference, h? The area of piston 1 is A1 = 200 cm2, and the area of piston 2 is A2 = 100 cm2. Solution: M1g/A1 + rgh = M2g/A2 h = (M2/A2-M1/A1)/r = 0.3 m 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.

Barometer Using Fluids to Measure Pressure • Use Barometer to measure Absolute Pressure • Top of tube evacuated (p=0) • Bottom of tube submerged into pool of mercury open to atmosphere (p=p0) • Pressure dependence on depth:

Manometer p0 p1 Dh Using Fluids to Measure Pressure • Use Manometer to measure Gauge Pressure • Measure pressure of volume (p1) relative to atmospheric pressure (º gauge pressure) • The height difference (h) measures the gauge pressure 1 atm = 760 mm (29.9 in) Hg = 10.3 m (33.8 ft) H20

W2? W1 W1 > W2 W1 < W2 W1 = W2 Archimedes’ Principle • Suppose we weigh an object in air (1) and in water (2). • How do these weights compare? • Why? • 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.

W2? W1 Archimedes: The buoyant force is equal to the weight of the liquid displaced. Archimedes’ Principle • The buoyant force is equal to the difference in the pressures times the area. • The buoyant force determines whether an object will sink or float. How does this work?

y F mg B Sink or Float? • The buoyant force is equal to the weight of the liquid that is displaced. • If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink. • We can calculate how much of a floating object will be submerged in the liquid: • Object is in equilibrium

y F mg B The Tip of the Iceberg • What fraction of an iceberg is submerged?

Pb styrofoam A) It sinks B) C) D) styrofoam Pb Lecture 30, ACT 3Buoyancy • 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. • If you turn the styrofoam+Pb upside down, what happens?

Cup II Cup I (D) can’t tell (C) the same (A) Cup I (B) Cup II ACT 3-AMore Fun With Buoyancy • Two cups are filled to the same level with water. One of the two cups has plastic balls floating in it. • Which cup weighs more?

oil water (B) move down (C) stay in same place (A) move up ACT 3-BEven More Fun With Buoyancy • A plastic ball floats in a cup of water with half of its volume submerged. Next some oil (roil < rball < rwater) is slowly added to the container until it just covers the ball. • Relative to the water level, the ball will:

Physics 1501: Lecture 30 Today ’ s Agenda