Physics 121 Newtonian Mechanics Instructor Karine Chesnel April, 7, 2009

Chapter 15 Oscillatory motion Spring motion Harmonic equation and solutions Frequence and period Energy of harmonic oscillator Comparison between harmonic motion and circular motion

Ch.15 Oscillatory motion4/7/09 Back to Spring motion An object attached to a spring, experiences a force that tends to bring it back to position at rest L0L0 x 0 F x F A B Compressed Extended The expression of the spring force is Applying the Newton’s second law Equation of the motion

x Harmonic motion Can also be written Equation of the motion is A general solution to this harmonic equation is Harmonic equation Amplitude Frequency Phase constant (phase at t=0) The values of the constants are given by the initial conditions At t = 0 Ch.15 Oscillatory motion4/7/09

Harmonic motion Let’s check that this solution verifies the harmonic equation A solution for the motion is We find that We indeed find an harmonic equation or Ch.15 Oscillatory motion4/7/09

Frequency and period A solution for the motion is with The period T of the function corresponds to a full cycle The phase variation is 2  T Thus The function x(t) is sinusoidal with time x(t) t Ch.15 Oscillatory motion4/7/09

Frequency and period A solution for the motion is The period T is T  is also called the angular frequency The frequency is with x(t) t Ch.15 Oscillatory motion4/7/09

Quiz # 35 In a pinball game, a ball of mass m= 250g is launched by compressing a spring of stiffness constant k= 7 N/m Assuming the board is horizontal and frictionless, how long it will take between the spring is released and the ball separates from the spring? A0.29 s B0.59 s C1.18 s D1.34 s The time required for the system (spring + pinball) to go from one end to the other end is one half of the oscillation period Ch.15 Oscillatory motion4/7/09

Position, velocity and acceleration Position, velocity and acceleration are all sinusoidal functions T x(t) t Position Velocity Acceleration t V(t) Phase quadrature t a(t) Opposite phase Ch.15 Oscillatory motion4/7/09

Initial conditions with The two constants : - amplitude A - phase  can be determined using the initial conditions, or the conditions at a certain time. For example if we know what is: - the position x(0) at t=0 - the velocity V(0) at t=0 We have the relationship We can calculate the values of constant A and  Ch.15 Oscillatory motion4/7/09

Quiz # 36 An oscillator made from an object of mass 350g, attached to a spring of stiffness constant k=8 N/m At some specific time, when the position of the oscillator is 8cm from the center, it is measured that the linear velocity is 1.35 m/s. What is the maximum amplitude of this oscillator ? A 8.2 cm B12 cm C29 cm D5.4 cm The maximum amplitude of the oscillator is The angular frequency is Thus the maximum amplitude is Ch.15 Oscillatory motion4/7/09

Energy The kinetic energy of the system is The potential energy is The total mechanical energy is Knowing that Finally Ch.15 Oscillatory motion4/7/09

Energy 0t U K 0x U K Ch.15 Oscillatory motion4/7/09

Energy 0x 1 L0L0 x 0 F F x Max compression: U max, K=0 Center U=0, K max Max extension U max, K=0 Ch.15 Oscillatory motion4/7/09

Harmonic motion and circular motion x y  x The position along the axis x is If the motion is uniform circular Angular speed Angle at t=0 R Ch.15 Oscillatory motion4/7/09

Example Motion of the wheels in old steam train The pressure coming form the steam is converted into mechanical energy to push the pistons. The motion of the pistons back and forward is converted into circular motions of the wheels V  Ch.15 Oscillatory motion4/7/09

The pendulum (punctual)  L m 0 Newton’s second law Projected along the tangential direction Remaining in a small angular range Harmonic equation t According to kinematics Ch.15 Oscillatory motion4/7/09

The pendulum (punctual)  L m 0 The equation for motion is The solution for the angle position is With The angular speed does not depend on the mass m The period of the oscillation is Ch.15 Oscillatory motion4/7/09

Quiz # 37 A pendulum clock made of a metal ball of mass 800g attached an adjustable string of length L= 15cm, has a period of 0.77s. To what length should the string be adjusted in order for this pendulum to have a period of 1second? A12 cm B18 cm C25 cm D33 cm The period of the pendulum is given by The distance L is given by = 24.8 cm Ch.15 Oscillatory motion4/7/09

Next Class Homework assignment No homework April 9 7pm Problems 15:1-5 April 9 9am