25
Physics 232 Lecture 01 1 Periodic Motion or Oscillations

Lecture 01

Embed Size (px)

DESCRIPTION

g

Citation preview

  • Physics 232 Lecture 01 1

    Periodic Motion or

    Oscillations

  • Physics 232 Lecture 01 2

    Periodic Motion Periodic Motion is motion that repeats about a point of stable equilibrium

    A necessary requirement for periodic motion is a Restoring Force

    Stable Equilibrium

    Unstable Equilibrium

  • Physics 232 Lecture 01 3

    Characteristics

    Amplitude - A The maximum displacement from the equilibrium position

    Period - T The time to complete one cycle of motion, peak to peak or valley to valley

    Frequency - f The number of cycles per unit time

    Tf 1=

    A

    A

    T

  • Physics 232 Lecture 01 4

    Simple Harmonic Motion Consider a mass m attached to a horizontal spring having a spring constant k with the spring unstretched

    We now pull the mass to the right and then let the mass go

    What is the subsequent motion of the mass?

    The restoring force is given by Hookes Law xkmaF ==

  • Physics 232 Lecture 01 5

    Simple Harmonic Motion This is a second order differential equation

    xkdt

    xdm =22

    Since this is a second order differential equation, there are two constants of integration and the general solution is

    ( ) += tAx cos

    where A is the maximum displacement, is a phase angle, and is the angular velocity and is given by

    mk

    =

    A and are determined from the initial, boundary, conditions

  • Physics 232 Lecture 01 6

    Phase Angle

  • Physics 232 Lecture 01 7

    Simple Harmonic Motion The period of the motion is related to by

    kmT

    22 ==

    Note that the period of the motion is independent of the displacement!

    The velocity of the particle is given by

    ( ) +== tAdtdxv sin

    with the maximum velocity being A

  • Physics 232 Lecture 01 8

    Simple Harmonic Motion We have so far that

    ( ) += tAx cos and ( ) +== tAdtdxv sin

    The constants A and are determined from the initial conditions, that is what are x and v at a specified time or some other suitable initial conditions The maximum displacement occurs when the velocity is zero, which occurs at the extreme of motion (two locations: x = A)

    The maximum velocity occurs when the displacement from the equilibrium position is zero (two values: v = )

  • Physics 232 Lecture 01 9

    SHM Energy Considerations We assume that the system is isolated and frictionless

    With the force being given by xkF =It can be shown that there is a potential energy in the system that is 2

    21 xkU =

    The total energy is then given by

    2221

    21 xkvm

    UKEETotal

    +=

    +=

  • Physics 232 Lecture 01 10

    SHM Energy Considerations If we substitute for x and v we find that

    221 AkETotal =

    The total energy is constant

    The energy shifts back and forth between the kinetic energy and the potential energy, but with the sum of the two being constant

  • Physics 232 Lecture 01 11

    SHM Energy Considerations The relationship between kinetic energy, potential energy, displacement, velocity and acceleration can be seen in the following diagram

  • Physics 232 Lecture 01 12

    Choice of Oscillatory Function Note that we used the cosine function in our development of Simple Harmonic Motion

    We could have also used the sine function for our description of SHM

    The only difference is in the phase angle

    ( ) += tAx cos

    ( )'sin += tAx

  • Physics 232 Lecture 01 13

    Simple Harmonic Motion Simple Harmonic Motion can be used to describe motion in many situations under appropriate approximations

    Any potential energy function that under appropriate circumstances can be approximated by a parabolic function will exhibit SHM

  • Physics 232 Lecture 01 14

    Simple Pendulum Consider a mass m suspended from a massless, unstretchable string of length L The forces acting on the mass are as shown The restoring force is the one perpendicular to the string

    sinmgFrestoring =But this is a nonlinear function of However for small angles sinWe then have that

    mgFrestoring =

  • Physics 232 Lecture 01 15

    Simple Pendulum From before we have that mgFrestoring =

    We also have that

    Lmgk =

    this is then becomes the equation of SHM

    Lx=

    This then yields that xL

    mgFrestoring =

    If we then let

    The frequency of oscillation is given by Lgf

    21

    =

    which is independent of the mass attached to the string

  • Physics 232 Lecture 01 16

    Damped Motion Real life situations have dissipative forces

    The dissipative force is often related to the velocity that is in the motion

    The fact that there is a dissipative force, leads to motion that is damped, that is the amplitude decreases with time

    bvFdiss =

    The total force is then the sum of the restoring force and this dissipative force

    vbxkFnet =

    The minus sign indicates that this force is in the opposite direction to the velocity

  • Physics 232 Lecture 01 17

    Damped Motion This net force leads to a slightly more complicated second order differential equation

    022

    =++ kxdtdxb

    dtxdm

    The exact solution to this equation depends upon the damping constant

    There are three possible solutions

    Underdamped: kmb 2

  • Physics 232 Lecture 01 18

    Damped Motion - Underdamped kmb 2

    The general solution for this situation is given by ( ) ( )tttmb eAeAex 22 212/ +=

    This solution looks like

    with mk

    mb

    =4

    2

    2

  • Physics 232 Lecture 01 21

    Damped Motion The most efficient damping, as far getting back to zero amplitude, is the critically damped case

    Note that the overdamped case may yield some interesting behavior depending on the relative values of the parameters

  • Physics 232 Lecture 01 22

    Forced Oscillation It is also possible to drive a system such as an oscillator with an external force that is also time varying

    The general differential equation is of the form

    tFbvxkdt

    xdm dcosmax22

    =++

    where the term on the right hand side is the driving force

    This solution to this equation involves two functions, a complementary function and a particular solution

  • Physics 232 Lecture 01 23

    Forced Oscillation The solution is given by

    ( )( )

    += t

    bmk

    Ftx ddd

    cos)(2222

    max

    with ( )

    = 2

    1

    /2/2tan

    d

    dmk

    mb

    represents the phase difference between the driving force and the resulting motion

    There can be a delay between the action of the driving force and the response of the system

  • Physics 232 Lecture 01 24

    Resonance

    ( ) 2222max

    dd bmk

    FA +

    =

    The term in front of the cosine function represents the amplitude

    If we vary the angular velocity of the driving force, we find that the system has its maximum amplitude when

    mk

    d

    This phenomenon of the amplitude peaking at a driving frequency that is near the natural frequency of the system is known as resonance

  • Physics 232 Lecture 01 25

    Resonance The strength of the amplitude depends upon the magnitude of the damping coefficient

    The smaller the value of b the more pronounced the peak

    The larger the damping the peak becomes broader, less sharp, and shifts to lower frequencies

    If mkb 2> the peak disappears completely

    Periodic MotionorOscillationsPeriodic MotionCharacteristicsSimple Harmonic MotionSimple Harmonic MotionPhase AngleSimple Harmonic MotionSimple Harmonic MotionSHM Energy ConsiderationsSHM Energy ConsiderationsSHM Energy ConsiderationsChoice of Oscillatory FunctionSimple Harmonic MotionSimple PendulumSimple PendulumDamped MotionDamped MotionDamped Motion - UnderdampedDamped Motion- Critically DampedDamped Motion - OverdampedDamped MotionForced OscillationForced OscillationResonanceResonance