PHYS 1443 – Section 003 Lecture #19

1. PHYS 1443 – Section 003Lecture #19 Monday, Nov. 20, 2002 Dr. Jaehoon Yu • Energy of the Simple Harmonic Oscillator • Simple Pendulum • Other Types of Pendulum • Damped Oscillation Today’s homework is homework #19 due 12:00pm, Wednesday, Nov. 27!! PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

2. Announcements • Evaluation today • We have classes next week, both Monday and Wednesday • Remember the Term Exam on Monday, Dec. 9 in the class PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

3. Generalized expression of a simple harmonic motion Angular Frequency Phase constant Amplitude Phase Let’s think about the meaning of this equation of motion What happens when t=0 and f=0? An oscillation is fully characterized by its: What is f if x is not A at t=0? • Amplitude • Period or frequency • Phase constant A/-A What are the maximum/minimum possible values of x? Equation of Simple Harmonic Motion The solution for the 2nd order differential equation PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

4. What is the time for full cycle of oscillation? Since after a full cycle the position must be the same One of the properties of an oscillatory motion The period What is the unit? How many full cycles of oscillation does this undergo per unit time? Frequency 1/s=Hz Let’s now think about the object’s speed and acceleration. Speed at any given time Max speed Max acceleration Acceleration at any given time What do we learn about acceleration? Acceleration is reverse direction to displacement Acceleration and speed are p/2 off phase: When v is maximum, a is at its minimum More on Equation of Simple Harmonic Motion PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

5. Since KE/PE E=KE+PE=kA2/2 -A A Energy of the Simple Harmonic Oscillator How do you think the mechanical energy of the harmonic oscillator look without friction? Kinetic energy of a harmonic oscillator is The elastic potential energy stored in the spring Therefore the total mechanical energy of the harmonic oscillator is Total mechanical energy of a simple harmonic oscillator is a constant of a motion and is proportional to the square of the amplitude Maximum KE is when PE=0 One can obtain speed PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu x

6. Example 13.4 A 0.500kg cube connected to a light spring for which the force constant is 20.0 N/m oscillates on a horizontal, frictionless track. a) Calculate the total energy of the system and the maximum speed of the cube if the amplitude of the motion is 3.00 cm. From the problem statement, A and k are The total energy of the cube is Maximum speed occurs when kinetic energy is the same as the total energy b) What is the velocity of the cube when the displacement is 2.00 cm. velocity at any given displacement is c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. Potential energy, PE Kinetic energy, KE PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

7. L q T m mg s The Pendulum A simple pendulum also performs periodic motion. The net force exerted on the bob is Since the arc length, s, is results Again became a second degree differential equation, satisfying conditions for simple harmonic motion If q is very small, sinq~q giving angular frequency The period only depends on the length of the string and the gravitational acceleration The period for this motion is PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

8. Example 13.5 Christian Huygens (1629-1695), the greatest clock maker in history, suggested that an international unit of length could be defined as the length of a simple pendulum having a period of exactly 1s. How much shorter would out length unit be had this suggestion been followed? Since the period of a simple pendulum motion is The length of the pendulum in terms of T is Thus the length of the pendulum when T=1s is Therefore the difference in length with respect to the current definition of 1m is PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

9. O d q CM dsinq mg Physical Pendulum Physical pendulum is an object that oscillates about a fixed axis which does not go through the object’s center of mass. Consider a rigid body pivoted at a point O that is a distance d from the CM. The magnitude of the net torque provided by the gravity is Then Therefore, one can rewrite Thus, the angular frequency w is By measuring the period of physical pendulum, one can measure moment of inertia. And the period for this motion is Does this work for simple pendulum? PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

10. O Pivot L CM Mg Example 13.6 A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical plane. Find the period of oscillation if the amplitude of the motion is small. Moment of inertia of a uniform rod, rotating about the axis at one end is The distance d from the pivot to the CM is L/2, therefore the period of this physical pendulum is Calculate the period of a meter stick that is pivot about one end and is oscillating in a vertical plane. Since L=1m, the period is So the frequency is PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

11. O qmax P Torsional Pendulum When a rigid body is suspended by a wire to a fixed support at the top and the body is twisted through some small angle q, the twisted wire can exert a restoring torque on the body that is proportional to the angular displacement. The torque acting on the body due to the wire is k is the torsion constant of the wire Applying the Newton’s second law of rotational motion Then, again the equation becomes Thus, the angular frequency w is This result works as long as the elastic limit of the wire is not exceeded And the period for this motion is PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

12. y y y y v w ax P P P P A A A a y A q q q f O O O O x x x x x vx Q Q Q t=t t=0 q=wt+f Simple Harmonic and Uniform Circular Motions Uniform circular motion can be understood as a superposition of two simple harmonic motions in x and y axis. When the particle rotates at a uniform angular speed w, x and y coordinate position become Since the linear velocity in a uniform circular motion is Aw, the velocity components are Since the radial acceleration in a uniform circular motion is v2/A=w2A, the components are PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

13. Example 13.7 A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to the right. A) Determine the x coordinate as a function of time. Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the angular frequency is 8.00rad/s. Therefore the equation of motion in x direction is Since x=2.00, when t=0 However, since the particle was moving to the right f=-48.2o, Find the x components of the particle’s velocity and acceleration at any time t. Using the displcement Likewise, from velocity PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

14. Damped Oscillation More realistic oscillation where an oscillating object loses its mechanical energy in time by a retarding force such as friction or air resistance. Let’s consider a system whose retarding force is air resistance R=-bv (b is called damping coefficient) and restoration force is -kx The solution for the above 2nd order differential equation is Damping Term The angular frequency w for this motion is This equation of motion tells us that when the retarding force is much smaller than restoration force, the system oscillates but the amplitude decreases, and ultimately, the oscillation stops. We express the angular frequency as Where the natural frequency w0 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu

15. More on Damped Oscillation The motion is called Underdamped when the magnitude of the maximum retarding force Rmax = bvmax <kA How do you think the damping motion would change as retarding force changes? As the retarding force becomes larger, the amplitude reduces more rapidly, eventually stopping at its equilibrium position Under what condition this system does not oscillate? The system is Critically damped Once released from non-equilibrium position, the object would return to its equilibrium position and stops. What do you think happen? If the retarding force is larger than restoration force The system is Overdamped Once released from non-equilibrium position, the object would return to its equilibrium position and stops, but a lot slower than before PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu