Chapter 14: Simple Harmonic Motion Evaluations and Lab exam are on Monday , Dec 3 during class time in SIMS 209. Lab exam will be on pre-labs, demos, and theory in the pre-labs Final Exam: SIMS 209 Section of 9:30am: exam Monday, Dec 10 at 11:30pm Section of 11:00am: exam Thursday, Dec 6 at 8:00am

11-5 Angular Momentum and Torque for a Rigid Object Example 11-8: Atwood’s machine. An Atwood machine consists of two masses, mA and mB, which are connected by an inelastic cord of negligible mass that passes over a pulley. If the pulley has radius R0 and moment of inertia I about its axle, determine the acceleration of the masses mA and mB, and compare to the situation where the moment of inertia of the pulley is ignored. Solution: First find the angular momentum of the system, and then apply the torque law. The torque is dL/dt, and a is dv/dt; taking the derivative of L and solving for a gives the solution in the text.

11-5 Angular Momentum and Torque for a Rigid Object For a rigid object, we can show that its angular momentum when rotating around a particular axis is given by: No matter where we choose the point O, L= Iω, this is valid when the rotation axis is along the symmetry axis Figure 11-15. Calculating Lω = Lz = ΣLiz. Note that Li is perpendicular to ri and Ri is perpendicular to the z axis, so the three angles marked φ are equal.

14.1 What is a Spring? Anything that can be modeled by Hooke’s Law. Images: http://boomeria.org/chemlectures/bonding/modes.jpg http://www.fitzz.com/images/products/zenzu-ball-chair-03.jpg http://t3.gstatic.com/images?q=tbn:ANd9GcTLaT30czBN2P2_sfuwuY4x8aFwh4GBGs9dCkubdLAFZCv1cSn

14-1 Oscillations of a Spring If an object vibrates or oscillates back and forth over the same path, each cycle taking the same amount of time, the motion is called periodic. The mass and spring system is a useful model for a periodic system. Figure 14-1. Caption: A mass oscillating at the end of a uniform spring.

14-1 Oscillations of a Spring We assume that the surface is frictionless. There is a point where the spring is neither stretched nor compressed: this is the equilibrium position. We measure displacement from that point (x = 0 on the previous figure). The force exerted by the spring depends on the displacement:

14-1 Oscillations of a Spring The minus sign on the force indicates that it is a restoring force—it is directed to restore the mass to its equilibrium position. k is the spring constant. The force is not constant, so the acceleration is not constant either.

14-1 Oscillations of a Spring Displacement is measured from the equilibrium point. Amplitude is the maximum displacement. A cycle is a full to-and-fro motion. Period is the time required to complete one cycle. Frequency is the number of cycles completed per second. Figure 14-2. Caption: Force on, and velocity of, a mass at different positions of its oscillation cycle on a frictionless surface.

14-1 Spring-restoring force X=0 x=-A x=A Restoring Force = -kx amax occurs when x=A x=A X=0 x=-A F x Restoring Force At x=-A X=0 x=-A x=A F m x At x=A

14-1 Oscillations of a Spring If the spring is hung vertically, the only change is in the equilibrium position, which is at the point where the spring force equals the gravitational force. Figure 14-3. Caption: (a) Free spring, hung vertically. (b) Mass m attached to spring in new equilibrium position, which occurs when ΣF = 0 = mg – kx0. Has the same Period and frequency as if it were horizontal

spring-sine wave Equilibrium x= v= a= max x= v= a= -A max x= v= a= A max x= v= a= -A max x= v= a= A max

The time interval for one repetition of the cycle in a simple harmonic motion is called the A) frequency B) period C) amplitude D) phase Question 12

A mass on a spring is oscillating back and forth on a frictionless surface. Where in the motion is the velocity the largest A) at x = +/-A B) somewhere between x=0 and x= +/-A C) at x=0 D) we need more info Question x=A X=0 x=-A 13

14-1 Oscillations of a Spring Example 14-1: Car springs. When a family of four with a total mass of 200 kg step into their 1200-kg car, the car’s springs compress 3.0 cm. (a) What is the spring constant of the car’s springs, assuming they act as a single spring? (b) How far will the car lower if loaded with 300 kg rather than 200 kg? Figure 14-4. Caption: Photo of a car’s spring. (Also visible is the shock absorber, in blue—see Section 14–7.) Solution: a. k = F/x = 6.5 x 104 N/m. b. x = F/k = 4.5 cm

Problem 2 2. (I) An elastic cord is 65 cm long when a weight of 75 N hangs from it but is 85 cm long when a weight of 180 N hangs from it. What is the “spring” constant k of this elastic cord?

14-2 Simple Harmonic Motion Any vibrating system where the restoring force is proportional to the negative of the displacement is in simple harmonic motion (SHM), and is often called a simple harmonic oscillator (SHO). Substituting F = -kx into Newton’s second law gives the equation of motion: with solutions of the form: Ф is the phase

14-2 Simple Harmonic Motion Substituting, we verify that this solution does indeed satisfy the equation of motion, with: Figure 14-5. Caption: Sinusoidal nature of SHM as a function of time. In this case, x = A cos (2πt/T). The constants A and φ will be determined by initial conditions; A is the amplitude, and φ gives the phase of the motion at t = 0.

14-2 Simple Harmonic Motion The velocity can be found by differentiating the displacement: These figures illustrate the effect of Ф: Figure 14-6. Caption: Special case of SHM where the mass m starts, at t = 0, at the equilibrium position x = 0 and has initial velocity toward positive values of x (v > 0 at t = 0). Figure 14-7. Caption: A plot of x = A cos (ωt +φ) when φ < 0.

14-2 Simple Harmonic Motion Because then

Problem 3 3.(I) The springs of a 1500-kg car compress 5.0 mm when its 68-kg driver gets into the driver’s seat. If the car goes over a bump, what will be the frequency of oscillations? Ignore damping.

Problem 4 (I) (a) What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then released from rest, and whose period is 0.66 s? (b) What will be its displacement after 1.8 s?

14-2 Simple Harmonic Motion The velocity and acceleration for simple harmonic motion can be found by differentiating the displacement: Figure 14-8. Caption: Displacement, x, velocity, dx/dt, and acceleration, d2x/dt2, of a simple harmonic oscillator when φ = 0.