Understanding Energy Conservation and Oscillatory Systems

1. Lecture 20 Goals: • Chapter 14 • Understand and use energy conservation in oscillatory systems. • Understand the basic ideas of damping and resonance. • Chapter 15 • Understand pressure in liquids and gases • Use Archimedes’ principle to understand buoyancy • Understand the equation of continuity • Use an ideal-fluid model to study fluid flow. • Investigate the elastic deformation of solids and liquids • Assignment • HW8, Due Wednesday, Apr. 8th • Tuesday: Read all of Chapter 15

2. SHM So Far • The most general solution isx(t) = A cos(t + ) where A = amplitude  = (angular) frequency = 2pf = 2p/T  = phase constant Velocity: v(t) = -A sin(t + ) Acceleration: a(t) = -2A cos(t + ) Spring constant Inertia Hooke’s Law Spring: Simple Pendulum:

3. T = 2/ A x(t) time /w /w /w A SHM So Far • The most general solution isx(t) = A cos(t + ) where A = amplitude  = (angular) frequency = 2pf = 2p/T  = phase constant Here  = 0 Velocity: v(t) = -A sin(t + ) Acceleration: a(t) = -2A cos(t + )

4. SHM So Far For SHM without friction • The frequency does notdepend on the amplitude ! • The oscillation occurs around the equilibrium point where the force is zero! • Mechanical Energy is constant, it transfers between potential and kinetic energies.

5. The shaker cart • You stand inside a small cart attached to a heavy-duty spring, the spring is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you. • At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground. • What effect does jettisoning the sandbag at the equilibrium position have on the amplitude of your oscillation? A. It increases the amplitude. B. It decreases the amplitude. C. It has no effect on the amplitude. Hint: At equilibrium, both the cart and the bag are moving at their maximum speed. By dropping the bag at this point, energy (specifically the kinetic energy of the bag) is lost from the spring-cart system. Thus, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must decrease

6. The shaker cart • Instead of dropping the sandbag as you pass through equilibrium, you decide to drop the sandbag when the cart is at its maximum distance from equilibrium. • What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation? A. It increases the amplitude. B. It decreases the amplitude. C. It has no effect on the amplitude. Hint: Dropping the bag at maximum distance from equilibrium, both the cart and the bag are at rest. By dropping the bag at this point, no energy is lost from the spring-cart system. Therefore, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must remain constant.

7. The shaker cart • What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the maximum speed of the cart? A. It increases the maximum speed. B. It decreases the maximum speed. C. It has no effect on the maximum speed. Hint: Dropping the bag at maximum distance from equilibrium, both the cart and the bag are at rest. By dropping the bag at this point, no energy is lost from the spring-cart system. Therefore, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must remain constant.

8. What about Vertical Springs? • For a vertical spring, ifyis measured from the equilibrium position • Recall: force of the spring is the negative derivative of this function: • This will be just like the horizontal case:-ky = ma = j k y = 0 F= -ky m Which has solution y(t) = A cos( t + ) where

9. y(t) (a) (c) t (b) Exercise Simple Harmonic Motion • A mass oscillates up & down on a spring. It’s position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration ? Remember: velocity is slope and acceleration is the curvature y(t) = A cos( t + ) v(t) = -A  sin( t + ) a(t) = -A 2 cos( t + )

10. T1 T2 Exercise Simple Harmonic Motion • You aresitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1. • Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2. Which of the following is true recalling that w = (g / L)½ (A)T1 = T2 (B)T1 > T2 (C) T1 < T2

11. U K E U x -A 0 A Energy in SHM • For both the spring and the pendulum, we can derive the SHM solution using energy conservation. • The total energy (K + U) of a system undergoing SMH will always be constant! • This is not surprising since there are only conservative forces present, hence energy is conserved.

12. U K E U x -A 0 A SHM and quadratic potentials • SHM will occur whenever the potential is quadratic. • For small oscillations this will be true: • For example, the potential betweenH atoms in an H2 molecule lookssomething like this: U x

13. U x SHM and quadratic potentials • Curvature reflects the spring constant or modulus (i.e., stress vs. strain or force vs. displacement) Measuring modular proteins with an AFM See: http://hansmalab.physics.ucsb.edu

14. What about Friction?A velocity dependent drag force (A model) We can guess at a new solution. and now w02 ≡k / m Note With,

15. What about Friction? A damped exponential if

16. Variations in the damping Small damping time constant (m/b) Low friction coefficient, b << 2m Moderate damping time constant (m/b) Moderate friction coefficient (b < 2m)

17. Damped Simple Harmonic Motion • A downward shift in the angular frequency • There are three mathematically distinct regimes underdamped overdamped critically damped

18. Driven SHM with Resistance • Apply a sinusoidal force, F0 cos (wt), and now consider what A andb do, Not Zero!!! b/m small steady state amplitude b/m middling b large w  w  w0

19. Resonance-based DNA detection with nanoparticle probes Change the massof the cantilever & change the resonant frequency Su et al., APL 82: 3562 (2003)

20. Stick - Slip Friction • How can a constant motion produce resonant vibrations? • Examples: • Strings, e.g. violin • Singing / Whistling • Tacoma Narrows Bridge • …

21. Dramatic example of resonance • In 1940, a steady wind set up a torsional vibration in the Tacoma Narrows Bridge 

22. A short clip • In 1940, a steady wind sets up a torsional vibration in the Tacoma Narrows Bridge 

23. Dramatic example of resonance • Eventually it collapsed 

24. Exercise Resonant Motion • Consider the following set of pendulums all attached to the same string B D C A If I start bob D swinging which of the others will gain the most mechanical energy (assuming virtually no friction) ? (A) (B) (C)

25. Chapter 15, Fluids • This is an actual photo of an iceberg, taken by a rig manager for Global Marine Drilling in St. Johns, Newfoundland. The water was calm and the sun was almost directly overhead so that the diver

26. Fluids (Ch. 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.

27. Fluids • An intrinsic parameter of a fluid • Density units : kg/m3 = 10-3 g/cm3 r(water) = 1.000 x 103 kg/m3 = 1.000 g/cm3 r(ice) = 0.917 x 103 kg/m3 = 0.917 g/cm3 r(air) = 1.29 kg/m3 = 1.29 x 10-3 g/cm3 r(Hg) = 13.6 x103 kg/m3 = 13.6 g/cm3

28. n A Fluids • Another parameter: Pressure • Any force exerted by a fluid is perpendicular to a surface of contact, and is proportional to the area of that surface. • Force (avector) in a fluid can be expressed in terms of pressure (a scalar) as:

29. What is the SI unit of pressure? • Pascal • Atmosphere • Bernoulli • Young • p.s.i. Units : 1 N/m2 = 1 Pa (Pascal) 1 bar = 105 Pa 1 mbar = 102 Pa 1 torr = 133.3 Pa 1 atm = 1.013 x105 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in2 (=PSI)

30. 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! • p(y) = p0 - y g r • Gauge pressure (subtract p0) F2 = F1+ m g = F1+ rVg F2 /A = F1/A + rVg/A p2 = p1 - rg y

31. Lecture 20 • Assignment • HW8, Due Wednesday, Apr. 8th • Tuesday: Read all of Chapter 15