Physics 207: Lecture 20, Pg 1 Lecture 20 Goals: Chapter 14 Chapter 14  Compare and contrast different systems with SHM.  Understand energy conservation (transfer) in SHM.  Understand the basic ideas of damping and resonance. Chapter 15 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 Assignment  HW9, Due Wednesday, Apr. 7 th  Tuesday (after break!!!): Read all of Chapter 15

Physics 207: Lecture 20, Pg 2 SHM So Far The most general solution is x = A cos(  t +  ) where A = amplitude  = (angular) frequency  = phase constant l For SHM without friction,  The frequency does not depend on the amplitude !  This is true of all simple harmonic motion! l The oscillation occurs around the equilibrium point where the force is zero! l Energy is a constant, it transfers between potential and kinetic

Physics 207: Lecture 20, Pg 3 The “Simple” Pendulum l A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.  F y = ma y = T – mg cos(  ) = ma c = m v T 2 /L  F x = ma x = -mg sin(  ) where x = L tan  If  small then x  L  and sin(  )   0 ° tan 0.00 = sin 0.00 = ° tan 0.09 = sin 0.09 = ° tan 0.17 = sin 0.17 = ° tan 0.26 = 0.27 sin 0.26 = 0.26  L m mg z y x T

Physics 207: Lecture 20, Pg 4 The “Simple” Pendulum l A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.  F y = ma y = T – mg cos(  ) = ma c = m v T 2 /L  F x = ma x = -mg sin(  ) where x = L tan  If  small then x  L  and sin(  )   dx/dt = L d  /dt a x = d 2 x/dt 2 = L d 2  /dt 2 so a x = -g  = L d 2  / dt 2  L d 2  / dt 2 + g  = 0 and  =   cos(  t +  ) or  =   sin(  t +  ) with  = (g/L) ½  L m mg z y x T

Physics 207: Lecture 20, Pg 5 SHM So Far The most general solution is x(t) = A cos(  t +  ) where A = amplitude  = (angular) frequency = 2  f = 2  /T  = phase constant Velocity: v(t) = -  A sin(  t +  ) Acceleration: a(t) = -  2 A cos(  t +  ) Simple Pendulum: Hooke’s Law Spring: Spring constant Inertia

Physics 207: Lecture 20, Pg 6 SHM So Far For SHM without friction l The frequency does not depend on the amplitude ! l The oscillation occurs around the equilibrium point where the force is zero! l Mechanical Energy is constant, it transfers between potential and kinetic energies.

Physics 207: Lecture 20, Pg 7 The shaker cart l 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. l At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground. l What effect does jettisoning the sandbag at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude.

Physics 207: Lecture 20, Pg 8 The shaker cart l 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. l What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude.

Physics 207: Lecture 20, Pg 9 The shaker cart l What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the maximum speed of the cart? It increases the maximum speed. It decreases the maximum speed. It has no effect on the maximum speed.

Physics 207: Lecture 20, Pg 10 What about Vertical Springs? l For a vertical spring, if y is measured from the equilibrium position l Recall: force of the spring is the negative derivative of this function: l This will be just like the horizontal case: -ky = ma = j k m F= -ky y = 0 Which has solution y(t) = A cos(  t +  ) where

Physics 207: Lecture 20, Pg 11 Exercise Simple Harmonic Motion l 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 t y(t) (a) (b) (c) y(t) = A cos(  t +  ) v(t) = -A  sin(  t +  ) a(t) = -A   cos(  t +  )

Physics 207: Lecture 20, Pg 12 Home Exercise l A mass m = 2 kg on a spring oscillates (no friction) with amplitude A = 10 cm. At t = 0 its speed is at a maximum, and is v=+2 m/s  What is the angular frequency of oscillation  ?  What is the spring constant k ? General relationships E = K + U = constant,  = (k/m) ½ So at maximum speed U=0 and ½ mv 2 = E = ½ kA 2 thus k = mv 2 /A 2 = 2 x (2) 2 /(0.1) 2 = 800 N/m,  = 20 rad / sec k x m

Physics 207: Lecture 20, Pg 13 Home Exercise Initial Conditions l A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. l Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction)? k m y 0 d (A) v(t) = - v max sin(  t ) a(t) = -a max cos(  t ) (B) v(t) = v max sin(  t ) a(t) = a max cos(  t ) (C) v(t) = v max cos(  t ) a(t) = -a max cos(  t ) (both v max and a max are positive numbers) t = 0

Physics 207: Lecture 20, Pg 14 Home Exercise Initial Conditions l A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction): k m y 0 d (A) v(t) = - v max sin(  t ) a(t) = -a max cos(  t ) (B) v(t) = v max sin(  t ) a(t) = a max cos(  t ) (C) v(t) = v max cos(  t ) a(t) = -a max cos(  t ) (both v max and a max are positive numbers) t = 0

Physics 207: Lecture 20, Pg 15 The “Torsional” Pendulum l A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements.   z = I  z = -mg sin(  ) L   z ≈ mL 2  z ≈ -mg  L L (d 2  /dt 2 ) = -g  d 2  / dt 2 = (-g/L)  with  (t)=  0 sin  t or  0 cos  t and  =(g/L) ½ or if a true horizontal torsional pendulum I  z = -  with  =(  /I) ½  L m mg z y x T

Physics 207: Lecture 20, Pg 16 Exercise Simple Harmonic Motion l You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T 1. l Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T 2. Which of the following is true recalling that  = (g / L) ½ (A) T 1 = T 2 (B) T 1 > T 2 (C) T 1 < T 2 T1T1 T2T2

Physics 207: Lecture 20, Pg 17 Energy in SHM l For both the spring and the pendulum, we can derive the SHM solution using energy conservation. l The total energy (K + U) of a system undergoing SMH will always be constant! l This is not surprising since there are only conservative forces present, hence energy is conserved. -AA0 x U U K E

Physics 207: Lecture 20, Pg 18 SHM and quadratic potentials l SHM will occur whenever the potential is quadratic. l For small oscillations this will be true: l For example, the potential between H atoms in an H 2 molecule looks something like this: -AA0 x U U K E U x

Physics 207: Lecture 20, Pg 19 See: SHM and quadratic potentials l Curvature reflects the spring constant or modulus (i.e., stress vs. strain or force vs. displacement) Measuring modular proteins with an AFM U x

Physics 207: Lecture 20, Pg 20 What about Friction? A velocity dependent drag force (A model) We can guess at a new solution. With, and now  0 2 ≡ k / m Note

Physics 207: Lecture 20, Pg 21 What about Friction? A damped exponential if

Physics 207: Lecture 20, Pg 22 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)

Physics 207: Lecture 20, Pg 23 Damped Simple Harmonic Motion l A downward shift in the angular frequency l There are three mathematically distinct regimes underdamped critically damped overdamped

Physics 207: Lecture 20, Pg 24 Driven SHM with Resistance Apply a sinusoidal force, F 0 cos (  t), and now consider what A and b do,   b/m small b/m middling b large   Not Zero!!! steady state amplitude

Physics 207: Lecture 20, Pg 25 Resonance-based DNA detection with nanoparticle probes Change the mass of the cantilever & change the resonant frequency Su et al., APL 82: 3562 (2003)

Physics 207: Lecture 20, Pg 26 Exercise Resonant Motion l Consider the following set of pendulums all attached to the same string D A B C If I start bob D swinging which of the others will have the largest swing amplitude ? (A)(B)(C)

Physics 207: Lecture 20, Pg 27 Chapter 15, Fluids l 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

Physics 207: Lecture 20, Pg 28 Fluids (Ch. 15) l 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 l 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.

Physics 207: Lecture 20, Pg 29 Fluids l An intrinsic parameter of a fluid  Density units : kg/m 3 = g/cm 3  (water) = x 10 3 kg/m 3 = g/cm 3  (ice) = x 10 3 kg/m 3 = g/cm 3  (air) = 1.29 kg/m 3 = 1.29 x g/cm 3  (Hg) = 13.6 x10 3 kg/m 3 = 13.6 g/cm 3

Physics 207: Lecture 20, Pg 30 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.  Force (a vector) in a fluid can be expressed in terms of pressure (a scalar) as: l Another parameter: Pressure

Physics 207: Lecture 20, Pg 31 What is the SI unit of pressure? A. Pascal B. Atmosphere C. Bernoulli D. Young E. p.s.i. Units : 1 N/m 2 = 1 Pa (Pascal) 1 bar = 10 5 Pa 1 mbar = 10 2 Pa 1 torr = Pa 1 atm = x10 5 Pa = 1013 mbar = 760 Torr = 14.7 lb/ in 2 (=PSI)

Physics 207: Lecture 20, Pg 32 F If the pressures were different, fluid would flow in the tube! Pressure vs. Depth l For a uniform 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? F Imagine a tube that would connect two regions at the same depth.

Physics 207: Lecture 20, Pg 33 Lecture 20 Assignment Assignment  HW9, Due Wednesday, Apr. 7 th  Tuesday: Read all of Chapter 15