Periodic

 Periodic motion is motion of an object that Week 14: Chapter 15 regularly returns to a given after a fixed time interval  A special kind of periodic motion occurs in Oscillatory Motion mechanical systems when the acting on the object is proportional to the position of the object relative to some equilibrium position  If the force is always directed toward the equilibrium position, the motion is called

Motion of a Spring- System Hooke’s Law

 A block of mass m is  Hooke’s Law states F = - kx attached to a spring, the s block is free to move on a  Fs is the restoring force frictionless horizontal  It is always directed toward the equilibrium position surface  Therefore, it is always opposite the displacement from  Use the active figure to vary equilibrium the initial conditions and observe the resultant  k is the force (spring) constant motion  x is the displacement  When the spring is neither stretched nor compressed, the block is at the equilibrium position  x = 0

More About Restoring Force More About Restoring Force, 2

 The block is displaced  The block is at the to the right of x = 0 equilibrium position  The position is positive  x = 0  The restoring force is  The spring is neither directed to the left stretched nor compressed  The force is 0

1 More About Restoring Force, 3

 The block is displaced  The force described by Hooke’s Law is the to the left of x = 0 net force in Newton’s Second Law  The position is negative

 The restoring force is FFHooke Newton directed to the right kx max k ax x m

Simple Harmonic Motion – Motion of the Block Mathematical Representation

 The block continues to oscillate between –A  Model the block as a particle and +A  The representation will be particle in simple harmonic motion model  These are turning points of the motion  Choose x as the axis along which the oscillation  The force is conservative occurs dx2 k   Acceleration In the absence of friction, the motion will ax2 continue forever dt m 2 k  Real systems are generally subject to friction, so  We let   they do not actually oscillate forever m

 Then a = -2x

Simple Harmonic Motion – Simple Harmonic Motion – Mathematical Representation, 2 Graphical Representation

 A function that satisfies the equation is  A solution is x(t) = A needed cos (t +    Need a function x(t) whose second derivative is A,  are all the same as the original function with a negative constants sign and multiplied by 2  A cosine curve can be  The sine and cosine functions meet these used to give physical requirements significance to these constants

2 Simple Harmonic Motion – Definitions Simple Harmonic Motion, cont

 A is the amplitude of the motion  A and  are determined uniquely by the  This is the maximum position of the particle in position and velocity of the particle at t = 0 either the positive or negative direction  If the particle is at x = A at t = 0, then  = 0   is called the angular frequency  The phase of the motion is the quantity (t +  Units are rad/s )   is the phase constant or the initial phase  x (t) is periodic and its value is the same each angle time t increases by 2 radians

Period Frequency

 The period, T, is the time interval required  The inverse of the period is called the for the particle to go through one full cycle of frequency its motion  The frequency represents the number of  The values of x and v for the particle at time t oscillations that the particle undergoes per equal the values of x and v at t + T unit time interval 1  2 ƒ  T  T 2   Units are cycles per second = hertz (Hz)

Summary Equations – Period An object of mass m is hung from a spring and set into oscillation. The period of the oscillation is measured and and Frequency recorded as T. The object of mass m is removed and replaced with an object of mass 2m. When this object is set into oscillation, what is the period of the motion?  The frequency and period equations can be rewritten to solve for  2 A. 2T 2ƒ T B. 1.4T2T

 The period and frequency can also be C. T expressed as: D. 0.7TT / 2 mk1 T 2ƒ E. T/2 km2

3 Motion Equations for Simple Period and Frequency, cont Harmonic Motion xt()  A cos ( t  )  The frequency and the period depend only on dx the mass of the particle and the force vAsin( t  ) constant of the spring dt 2  They do not depend on the parameters of dx 2 aA cos( t  ) motion dt 2  The frequency is larger for a stiffer spring  Simple harmonic motion is one-dimensional and so (large values of k) and decreases with directions can be denoted by + or - sign increasing mass of the particle  Remember, simple harmonic motion is not uniformly accelerated motion

Maximum Values of v and a Graphs

 The graphs show:  Because the sine and cosine functions  (a) displacement as a oscillate between ±1, we can easily find the function of time maximum values of velocity and acceleration  (b) velocity as a function of time for an object in SHM  (c ) acceleration as a function of time k  The velocity is 90o out of phase with the vAAmax  m displacement and the acceleration is 180o out k aAA 2 of phase with the max m displacement

SHM Example 1 SHM Example 2

 Initial conditions at t = 0  Initial conditions at are t = 0 are  x (0)= A  x (0)=0  v (0) = 0  v (0) = vi  This means  = 0  This means  =  /2  The acceleration  The graph is shifted reaches extremes of ± one-quarter cycle to the 2A at A right compared to the  The velocity reaches graph of x (0) = A extremes of ± A at x = 0

4 Energy of the SHM Oscillator, Energy of the SHM Oscillator cont

 Assume a spring-mass system is moving on a  The total mechanical frictionless surface energy is constant  The total mechanical  This tells us the total energy is constant energy is proportional to the square of the amplitude  The kinetic energy can be found by  2 2 2 2 Energy is continuously  K = ½ mv = ½ m A sin (t + ) being transferred between  The elastic potential energy can be found by potential energy stored in 2 2 2 the spring and the kinetic  U = ½ kx = ½ kA cos (t + ) energy of the block 2  The total energy is E = K + U = ½ kA  Use the active figure to investigate the relationship between the motion and the energy

Energy of the SHM Oscillator, cont Energy in SHM, summary

 As the motion continues, the exchange of energy also continues  Energy can be used to find the velocity

k vAx22  m   22A x 2

Importance of Simple Harmonic Oscillators SHM and Circular Motion

 Simple harmonic oscillators  This is an overhead view of are good models of a wide a device that shows the variety of physical relationship between SHM phenomena and circular motion  Molecular example  As the ball rotates with  If the atoms in the molecule constant angular speed, its do not move too far, the shadow moves back and between them can forth in simple harmonic be modeled as if there were motion springs between the atoms  The potential energy acts similar to that of the SHM oscillator

5 SHM and Circular Motion, 2 SHM and Circular Motion, 3

 The circle is called a  The particle moves reference circle along the circle with constant angular  Line OP makes an velocity  angle  with the x axis  OP makes an angle  at t = 0 with the x axis  Take P at t = 0 as the  At some time, the angle reference position between OP and the x axis will be  t + 

SHM and Circular Motion, 4 SHM and Circular Motion, 5

 The points P and Q always have the same x  The x component of the coordinate velocity of P equals the velocity of Q  x (t) = A cos (t + )  These velocities are  This shows that point Q moves with simple  v = -A sin (t + ) harmonic motion along the x axis  Point Q moves between the limits ±A

SHM and Circular Motion, 6 Simple Pendulum

 The acceleration of point P  A simple pendulum also exhibits periodic motion on the reference circle is directed radially inward  The motion occurs in the vertical plane and is  P ’s acceleration is a = 2A driven by gravitational force  The x component is  –2 A cos (t + ) The motion is very close to that of the SHM  This is also the acceleration oscillator of point Q along the x axis  If the angle is <10o

6 Simple Pendulum, 2 Simple Pendulum, 3

 The forces acting on the  In the tangential direction, bob are the tension and ds2 the weight Fmgmsin   t 2  T is the force exerted on dt the bob by the string  The length, L, of the pendulum is constant, and for   mg is the gravitational small values of  force dg2 g  The tangential component 2 sin   of the gravitational force is dt L L a restoring force  This confirms the form of the motion is SHM

Simple Pendulum, 4 Simple Pendulum, Summary

 The function  can be written as  The period and frequency of a simple  =  cos (t + ) pendulum depend only on the length of the max string and the acceleration due to gravity  The angular frequency is  The period is independent of the mass g  All simple pendula that are of equal length   L and are at the same location oscillate with the  The period is same period 2 L T 2  g

Clicker Question Physical Pendulum A grandfather clock depends on the period of a pendulum to keep correct time. Suppose a grandfather clock is calibrated correctly and then a mischievous  If a hanging object oscillates about a fixed child slides the bob of the pendulum downward on the axis that does not pass through the center of oscillating rod. The grandfather clock will run: mass and the object cannot be approximated as a particle, the system is called a physical A. slow pendulum B. fast  It cannot be treated as a simple pendulum C. correctly D. depending on the weight of the bob.

7 Physical Pendulum, 2 Physical Pendulum, 3

 The gravitational force  From Newton’s Second Law, provides a about an axis through O d 2 mgdsin I 2  The magnitude of the dt torque is  The gravitational force produces a restoring mgd sin  force  I is the about the axis  Assuming  is small, this becomes through O 2 dmgd  2 2   dt I

Physical Pendulum,4 Physical Pendulum, 5

 This equation is in the form of an object in  A physical pendulum can be used to measure simple harmonic motion the moment of inertia of a flat rigid object  If you know d, you can find I by measuring the  The angular frequency is period mgd  If I = md2 then the physical pendulum is the   I same as a simple pendulum  The period is  The mass is all concentrated at the 2 I T 2  mgd

Torsional Pendulum Torsional Pendulum, 2

 Assume a rigid object is suspended from a wire attached at its top to a fixed support  The restoring torque is    The twisted wire exerts a restoring torque on   is the torsion constant of the object that is proportional to its angular the support wire position  Newton’s Second Law gives d 2  I dt 2 d 2   dt2 I

8 Torsional Period, 3 Damped Oscillations

 The torque equation produces a motion equation for  In many real systems, nonconservative simple harmonic motion  forces are present  The angular frequency is   I  This is no longer an ideal system (the type we have dealt with so far) I  Friction is a common nonconservative force  The period is T  2   In this case, the mechanical energy of the  No small-angle restriction is necessary system diminishes in time, the motion is said  Assumes the elastic limit of the wire is not exceeded to be damped

Damped Oscillation, Example Damped Oscillations, Graph

 One example of damped  A graph for a damped motion occurs when an object oscillation is attached to a spring and  The amplitude decreases submerged in a viscous liquid with time  The retarding force can be  The blue dashed lines  expressed as Rvb where b represent the envelope of is a constant the motion  b is called the damping  Use the active figure to vary coefficient the mass and the damping constant and observe the effect on the damped motion

Damping Oscillation, Damping Oscillation, Equations Equations, cont

 The restoring force is – kx  The position can be described by  From Newton’s Second Law  b t  2m Fx = -k x – bvx = max xAe cos( t  )  When the retarding force is small compared  The angular frequency will be to the maximum restoring force we can

determine the expression for x 2  This occurs when b is small kb   mm2

9 Damping Oscillation, Example Summary Types of Damping

 When the retarding force is small, the oscillatory  If the restoring force is such that b/2m <  , character of the motion is preserved, but the o amplitude decreases exponentially with time the system is said to be underdamped  The motion ultimately ceases  When b reaches a critical value bc such that  Another form for the angular frequency bc / 2 m = 0 , the system will not oscillate 2  2 b The system is said to be critically damped 0  2m  If the restoring force is such that bvmax > kA and b/2m >  , the system is said to be  where 0 is the angular frequency in the absence of the o retarding force and is called the natural frequency of the overdamped system

Types of Damping, cont Forced Oscillations

 Graphs of position  It is possible to compensate for the loss of versus time for energy in a damped system by applying an  (a) an underdamped oscillator external force  (b) a critically damped  oscillator The amplitude of the motion remains  (c) an overdamped constant if the energy input per cycle exactly oscillator equals the decrease in mechanical energy in  For critically damped each cycle that results from resistive forces and overdamped there is no angular frequency

Forced Oscillations, 2 Forced Oscillations, 3

 After a driving force on an initially stationary  The amplitude of a driven oscillation is object begins to act, the amplitude of the F oscillation will increase 0 A  m 2  After a sufficiently long period of time, E 2 driving 22 b = E 0  lost to internal m  Then a steady-state condition is reached   is the natural frequency of the undamped  The oscillations will proceed with constant 0 oscillator amplitude

10 Resonance Resonance, cont

 When the frequency of the driving force is  At resonance, the applied force is in phase

near the natural frequency (  ) an with the velocity and the power transferred to increase in amplitude occurs the oscillator is a maximum  This dramatic increase in the amplitude is  The applied force and v are both proportional to called resonance sin (t + )    The power delivered is Fv  The natural frequency  is also called the  This is a maximum when the force and velocity are in resonance frequency of the system phase  The power transferred to the oscillator is a maximum

Resonance, Final

 Resonance (maximum peak) occurs when driving frequency equals the natural frequency  The amplitude increases with decreased damping  The curve broadens as the damping increases  The shape of the resonance curve depends on b

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