Page 1Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 1Physics 207, Lecture 19, Nov. 5Goals:Goals:••Chapter 14Chapter 14 Understand and use energy conservation in oscillatory systems. Understand the basic ideas of damping and resonance.••Chapter 15Chapter 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••AssignmentAssignment HW8, Due Wednesday, Nov. 12th Monday: Read all of Chapter 15.Physics 207: Lecture 19, Pg 2SHM So Far The most general solution is x(t) = A cos(ωt + φ)where A = amplitude ω = (angular) frequency = 2π f = 2π/Tφ = phase constant For SHM without friction, The frequency does not depend on the amplitude ! This is true of all simple harmonic motion! The oscillation occurs around the equilibrium point where the force is zero! Energy is a constant, it transfers between potential and kineticmk=ωVelocity: v(t) = -ωA sin(ωt + φ)Acceleration: a(t) = -ω2A cos(ωt + φ)Simple Pendulum:Lg=ωPage 2Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 3The 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 decreasePhysics 207: Lecture 19, Pg 4The 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 theelastic potential energy at maximum displacementand the kinetic energy at equilibrium must remain constant.Page 3Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 5The 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.Physics 207: Lecture 19, Pg 6What about Vertical Springs? For a vertical spring, if y is 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 j kmF= -kyy = 0Uky=122kydydUF −=−=22dtydmWhich has solution y(t) = A cos( ωt + φ)ω=kmwherePage 4Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 7Exercise 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 negativeacceleration ? Remember: velocity is slope and acceleration is the curvaturety(t)(a)(b)(c)y(t) = A cos( ωt + φ)v(t) = -A ω sin( ωt + φ)a(t) = -A ω2cos( ωt + φ)Physics 207: Lecture 19, Pg 8Example A mass m = 2 kg on a spring oscillates 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 ω ? What is the spring constant k ?General relationships E = K + U = constant, ω = (k/m)½So at maximum speed U=0 and ½ mv2= E = ½ kA2thus k = mv2/A2= 2 x (2) 2/(0.1)2 = 800 N/m, ω = 20 rad / sec kxmPage 5Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 9Home Exercise Initial Conditions A mass hanging from a vertical spring is lifted a distance dabove 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)?kmy0d (A) v(t) = - vmaxsin( ωωωωt ) a(t) = -amax cos( ωωωωt ) (B) v(t) = vmaxsin( ωt ) a(t) = amax cos( ωt )(C) v(t) = vmaxcos( ωt ) a(t) = -amax cos(ωt )(both vmaxand amaxare positive numbers)t = 0Physics 207: Lecture 19, Pg 10Exercise Simple Harmonic Motion You are sitting 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 ω = (g / L)½(A) T1 = T2(B) T1 > T2(C) T1 < T2T1T2Page 6Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 11Energy 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.-A A0xUUKEPhysics 207: Lecture 19, Pg 12SHM 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 H2molecule lookssomething like this:-A A0xUUKEUxPage 7Physics 207 – Lecture 19Physics 207: Lecture 19, Pg 13See: http://hansmalab.physics.ucsb.eduSHM and quadratic potentials Curvature reflects the spring constantor modulus (i.e.,
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