Chapter 15

Oscillatory MotionSHM

Dr. Armen Kocharian

Periodic MotionPeriodic motion is motion of an object that regularly repeats

The object returns to a given position after a fixed time interval

A special kind of periodic motion occurs in mechanical systems when the force 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 simple harmonic motion

Motion of a Spring-Mass System

A block of mass m is attached to a spring, the block is free to move on a frictionless horizontal surfaceWhen the spring is neither stretched nor compressed, the block is at the equilibrium position

x = 0

Hooke’s Law

Hooke’s Law states Fs = - kxFs is the restoring force

It is always directed toward the equilibrium positionTherefore, it is always opposite the displacement from equilibrium

k is the force (spring) constantx is the displacement

More About Restoring ForceThe block is displaced to the right of x = 0

The position is positive

The restoring force is directed to the left

More About Restoring Force, 2The block is at the equilibrium position

x = 0

The spring is neither stretched nor compressedThe force is 0

More About Restoring Force, 3The block is displaced to the left of x = 0

The position is negative

The restoring force is directed to the right

Acceleration

The force described by Hooke’s Law is the net force in Newton’s Second Law

Hooke Newton

x

x

F Fkx ma

ka xm

=− =

= −

Acceleration, cont.The acceleration is proportional to the displacement of the blockThe direction of the acceleration is opposite the direction of the displacement from equilibriumAn object moves with simple harmonic motion whenever its acceleration is proportional to its position and is oppositely directed to the displacement from equilibrium

Acceleration, finalThe acceleration is not constant

Therefore, the kinematic equations cannot be appliedIf the block is released from some position x = A, then the initial acceleration is –kA/mWhen the block passes through the equilibrium position, a = 0The block continues to x = -A where its acceleration is +kA/m

Motion of the BlockThe block continues to oscillate between –A and +A

These are turning points of the motionThe force is conservativeIn the absence of friction, the motion will continue forever

Real systems are generally subject to friction, so they do not actually oscillate forever

Orientation of the Spring

When the block is hung from a vertical spring, its weight will cause the spring to stretchIf the resting position of the spring is defined as x = 0, the same analysis as was done with the horizontal spring will apply to the vertical spring-mass system

Simple Harmonic Motion –Mathematical Representation

Model the block as a particleChoose x as the axis along which the oscillation occursAcceleration

We let

Then a = -ω2x

2

2d x ka xdt m

= = −

2 km

ω =

Simple Harmonic Motion –Mathematical Representation, 2

A function that satisfies the equation is needed

Need a function x(t) whose second derivative is the same as the original function with a negative sign and multiplied by ω2

The sine and cosine functions meet these requirements

Simple Harmonic Motion –Graphical Representation

A solution is x(t) = A cos (ωt + φ)A, ω, φ are all constantsA cosine curve can be used to give physical significance to these constants

Simple Harmonic Motion –Definitions

A is the amplitude of the motionThis is the maximum position of the particle in either the positive or negative direction

ω is called the angular frequencyUnits are rad/s

φ is the phase constant or the initial phase angle

Simple Harmonic Motion, contA and φ are determined uniquely by the position and velocity of the particle at t = 0If the particle is at x = A at t = 0, then φ = 0The phase of the motion is the quantity (ωt+ φ)x (t) is periodic and its value is the same each time ωt increases by 2π radians

An Experiment To Show SHMThis is an experimental apparatus for demonstrating simple harmonic motionThe pen attached to the oscillating object traces out a sinusoidal on the moving chart paperThis verifies the cosine curve previously determined

Period

The period, T, is the time interval required for the particle to go through one full cycle of its motion

The values of x and v for the particle at time t equal the values of x and v at t + T

2T πω

=

FrequencyThe inverse of the period is called the frequencyThe frequency represents the number of oscillations that the particle undergoes per unit time interval

Units are cycles per second = hertz (Hz)

1ƒ2Tωπ

= =

Summary Equations – Period and Frequency

The frequency and period equations can be rewritten to solve for ω

The period and frequency can also be expressed as:

22 ƒTπω π= =

12 ƒ2

m kTk m

ππ

= =

Period and Frequency, contThe frequency and the period depend only on the mass of the particle and the force constant of the spring They do not depend on the parameters of motionThe frequency is larger for a stiffer spring (large values of k) and decreases with increasing mass of the particle

Motion Equations for Simple Harmonic Motion

Remember, simple harmonic motion is not uniformly accelerated motion

22

2

( ) cos ( )

sin ( t )

cos( t )

x t A tdxv Adtd xa Adt

ω φ

ω ω φ

ω ω φ

= +

= = − +

= = − +

Maximum Values of v and a

Because the sine and cosine functions oscillate between ±1, we can easily find the maximum values of velocity and acceleration for an object in SHM

max

2max

kv A Amka A Am

ω

ω

= =

= =

GraphsThe graphs show:

(a) displacement as a function of time(b) velocity as a function of time(c ) acceleration as a function of time

The velocity is 90o

out of phase with the displacement and the acceleration is 180o

out of phase with the displacement

SHM Example 1

Initial conditions at t= 0 are

x (0)= Av (0) = 0

This means φ = 0The acceleration reaches extremes of ±ω2AThe velocity reaches extremes of ± ωA

SHM Example 2

Initial conditions att = 0 are

x (0)=0v (0) = vi

This means φ = − π/2The graph is shifted one-quarter cycle to the right compared to the graph of x (0) = A

Energy of the SHM OscillatorAssume a spring-mass system is moving on a frictionless surfaceThis tells us the total energy is constantThe kinetic energy can be found by

K = ½ mv 2 = ½ mω2 A2 sin2 (ωt + φ)

The elastic potential energy can be found byU = ½ kx 2 = ½ kA2 cos2 (ωt + φ)

The total energy is K + U = ½ kA 2

Energy of the SHM Oscillator, cont

The total mechanical energy is constant The total mechanical energy is proportional to the square of the amplitudeEnergy is continuously being transferred between potential energy stored in the spring and the kinetic energy of the block

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

Energy of the SHM Oscillator, cont

( )2 2

2 2 2

kv A xm

A xω

= ± −

= ± −

Energy in SHM, summary

Molecular Model of SHMIf the atoms in the molecule do not move too far, the force between them can be modeled as if there were springs between the atomsThe potential energy acts similar to that of the SHM oscillator

SHM and Circular MotionThis is an overhead view of a device that shows the relationship between SHM and circular motionAs the ball rotates with constant angular velocity, its shadow moves back and forth in simple harmonic motion

SHM and Circular Motion, 2The circle is called a reference circleLine OP makes an angle φ with the x axis at t = 0Take P at t = 0 as the reference position

SHM and Circular Motion, 3The particle moves along the circle with constant angular velocity ωOP makes an angle θ with the x axisAt some time, the angle between OPand the x axis will be θ = ωt + φ

SHM and Circular Motion, 4

The points P and Q always have the same x coordinatex (t) = A cos (ωt + φ)This shows that point Q moves with simple harmonic motion along the x axisPoint Q moves between the limits ±A

SHM and Circular Motion, 5The x component of the velocity of Pequals the velocity of QThese velocities are

v = -ωA sin (ωt + φ)

SHM and Circular Motion, 6The acceleration of point P on the reference circle is directed radially inwardP ’s acceleration is a = ω2AThe x component is –ω2 A cos (ωt + φ)This is also the acceleration of point Qalong the x axis

SHM and Circular Motion, Summary

Simple Harmonic Motion along a straight line can be represented by the projection of uniform circular motion along the diameter of a reference circleUniform circular motion can be considered a combination of two simple harmonic motions

One along the x-axisThe other along the y-axisThe two differ in phase by 90o

Simple PendulumA simple pendulum also exhibits periodic motionThe motion occurs in the vertical plane and is driven by gravitational forceThe motion is very close to that of the SHM oscillator

If the angle is <10o

Simple Pendulum, 2The forces acting on the bob are T and mg

T is the force exerted on the bob by the stringmg is the gravitational force

The tangential component of the gravitational force is a restoring force

Simple Pendulum, 3In the tangential direction,

The length, L, of the pendulum is constant, and for small values of θ

This confirms the form of the motion is SHM

2

2sintd sF mg mdt

θ= − =

2

2 sind g gdt L L

θ θ θ= − = −

Simple Pendulum, 4The function θ can be written asθ = θmax cos (ωt + φ)The angular frequency is

The period is

gL

ω =

2 2 LTg

π πω

= =

Simple Pendulum, SummaryThe period and frequency of a simple pendulum depend only on the length of the string and the acceleration due to gravityThe period is independent of the massAll simple pendula that are of equal length and are at the same location oscillate with the same period

Physical Pendulum

If a hanging object oscillates about a fixed axis that does not pass through the center of mass and the object cannot be approximated as a particle, the system is called a physical pendulum

It cannot be treated as a simple pendulum

Physical Pendulum, 2The gravitational force provides a torque about an axis through OThe magnitude of the torque is mgd sin θI is the moment of inertia about the axis through O

Physical Pendulum, 3

From Newton’s Second Law,

The gravitational force produces a restoring forceAssuming θ is small, this becomes

2

2sin dmgd Idt

θθ− =

22

2d mgddt I

θ θ ω θ⎛ ⎞= − = −⎜ ⎟⎝ ⎠

Physical Pendulum,4

This equation is in the form of an object in simple harmonic motionThe angular frequency is

The period is

mgdI

ω =

2 2 ITmgd

π πω

= =

Physical Pendulum, 5A physical pendulum can be used to measure the moment of inertia of a flat rigid object

If you know d, you can find I by measuring the period

If I = md then the physical pendulum is the same as a simple pendulum

The mass is all concentrated at the center of mass

Torsional Pendulum

Assume a rigid object is suspended from a wire attached at its top to a fixed supportThe twisted wire exerts a restoring torque on the object that is proportional to its angular position

Torsional Pendulum, 2The restoring torque is τ = −κθ

κ is the torsion constantof the support wire

Newton’s Second Law gives

2

2

2

2

dIdt

ddt I

θτ κθ

θ κ θ

= − =

= −

Torsional Period, 3The torque equation produces a motion equation for simple harmonic motionThe angular frequency is

The period is

No small-angle restriction is necessaryAssumes the elastic limit of the wire is not exceeded

Iκω =

2 IT πκ

=

Damped Oscillations

In many real systems, nonconservative forces are present

This is no longer an ideal system (the type we have dealt with so far)Friction is a common nonconservative force

In this case, the mechanical energy of the system diminishes in time, the motion is said to be damped

Damped Oscillations, contA graph for a damped oscillationThe amplitude decreases with timeThe blue dashed lines represent the envelope of the motion

Damped Oscillation, ExampleOne example of damped motion occurs when an object is attached to a spring and submerged in a viscous liquidThe retarding force can be expressed as R = - b vwhere b is a constant

b is called the damping coefficient

Damping Oscillation, Example Part 2

The restoring force is – kxFrom Newton’s Second LawΣFx = -k x – bvx = maxWhen the retarding force is small

compared to the maximum restoring force we can determine the expression for x

This occurs when b is small

Damping Oscillation, Example, Part 3

The position can be described by

The angular frequency will be

2 cos( )b tmx Ae tω φ

−= +

2

2k bm m

ω ⎛ ⎞= − ⎜ ⎟⎝ ⎠

Damping Oscillation, Example Summary

When the retarding force is small, the oscillatory character of the motion is preserved, but the amplitude decreases exponentially with timeThe motion ultimately ceasesAnother form for the angular frequency

where ω0 is the angular frequency in the absence of the retarding force

220 2

bm

ω ω ⎛ ⎞= − ⎜ ⎟⎝ ⎠

Types of Damping

is also called the natural frequency of the system

If Rmax = bvmax < kA, the system is said to be underdampedWhen b reaches a critical value bc such that bc / 2 m = ω0 , the system will not oscillate

The system is said to be critically dampedIf Rmax = bvmax > kA and b/2m > ω0, the system is said to be overdamped

0km

ω =

Types of Damping, contGraphs of position versus time for

(a) an underdamped oscillator(b) a critically damped oscillator (c) an overdamped oscillator

For critically damped and overdamped there is no angular frequency

Forced Oscillations

It is possible to compensate for the loss of energy in a damped system by applying an external forceThe amplitude of the motion remains constant if the energy input per cycle exactly equals the decrease in mechanical energy in each cycle that results from resistive forces

Forced Oscillations, 2

After a driving force on an initially stationary object begins to act, the amplitude of the oscillation will increaseAfter a sufficiently long period of time, Edriving = Elost to internal

Then a steady-state condition is reached The oscillations will proceed with constant amplitude

Forced Oscillations, 3

The amplitude of a driven oscillation is

ω0 is the natural frequency of the undamped oscillator

( )

0

222 2

0

FmA

bmωω ω

=⎛ ⎞− + ⎜ ⎟⎝ ⎠

Resonance

When the frequency of the driving force is near the natural frequency (ω ≈ ω0) an increase in amplitude occursThis dramatic increase in the amplitude is called resonanceThe natural frequency ω0 is also called the resonance frequency of the system

Resonance, cont

At resonance, the applied force is in phase with the velocity and the power transferred to the oscillator is a maximum

The applied force and v are both proportional to sin (ωt + φ)The power delivered is F . v

This is a maximum when F and v are in phase

Resonance, FinalResonance (maximum peak) occurs when driving frequency equals the natural frequencyThe amplitude increases with decreased dampingThe curve broadens as the damping increasesThe shape of the resonance curve depends on b