Oscillations and Simple Harmonic Motion:

Mechanics C

Oscillatory Motion

Oscillatory Motion is repetitive back and forth motion about an equilibrium position

Oscillatory Motion is periodic.Swinging motion and vibrations are forms of

Oscillatory Motion.

Objects that undergo Oscillatory Motion are called Oscillators.

Simple Harmonic Motion

The time to complete one full cycle of

oscillation is a Period.

T 1f

f 1T

The amount of oscillations per second is called frequency and is measured in Hertz.

What is the oscillation period for the broadcast of a 100MHz FM radio station?

Heinrich Hertz produced the first artificial radio waves back

in 1887!

T 1f

11108Hz

110 8s 10ns

Simple Harmonic Motion

The most basic of all types of oscillation is depicted on

the bottom sinusoidal graph. Motion that follows

this pattern is called simple harmonic motion or SHM.

Simple Harmonic Motion

An objects maximum displacement from its equilibrium position is

called the Amplitude (A) of the motion.

What shape will a velocity-time graph have for SHM?

Everywhere the slope (first derivative) of the position graph is zero, the velocity

graph crosses through zero.

2cos tx t AT

We need a position function

to describe the motion above.

Mathematical Models of SHM

2cos tx t AT

cos 2x t A ft

cosx t A t

1Tf

2T

x(t) to symbolize position as a function of

time

A=xmax=xmin

When t=T, cos(2π)=cos(0)

x(t)=A

Mathematical Models of SHM

sinv t A t

cosx t A t

d x tv t

dt

In this context we will call omega Angular

Frequency

What is the physical meaning of the product (Aω)?

maxv AThe maximum speed of an oscillation!

Recall: Hooke’s LawHere is what we want to do: DERIVE AN EXPRESSION THAT DEFINES THE DISPLACEMENT FROM EQUILIBRIUM OF THE SPRING IN TERMS OF TIME.

0)(2

2

2

xmk

dtxd

dtxdmkx

dtxdamakx

maFkxF Netspring

WHAT DOES THIS MEAN? THE SECOND DERIVATIVE OF A FUNCTIONTHAT IS ADDED TO A CONSTANT TIMES ITSELF IS EQUAL TO ZERO.What kind of function will ALWAYS do this?

Example:

An airtrack glider is attached to a spring, pulled 20cm to the right, and

released at t=0s. It makes 15 oscillations in 10 seconds.

What is the period of oscillation?15

10sec11.5oscilationsf HzT

1 1 0.671.5

T sf Hz

Example:

An airtrack glider is attached to a spring, pulled 20cm to the right, and

released at t=0s. It makes 15 oscillations in 10 seconds.

What is the object’s maximum speed?

max2Av AT

max

0.2 21.88 /

0.67m

v m ss

Example:

An airtrack glider is attached to a spring, pulled 20cm to the right, and

released at t=0s. It makes 15 oscillations in 10 seconds.

What are the position and velocity at t=0.8s?

cos 0.2 cos 0.8 0.0625x t A t m s m

sin 0.2 sin 0.8 1.79 /v t A t m s m s

Example:

A mass oscillating in SHM starts at x=A and has period T. At what time, as

a fraction of T, does the object first pass through 0.5A?

2cos

( ) 0.5

tx t AT

x t A

20.5 cos tA AT

1cos 0.52T t

2 3T t

6

Tt

Model of SHM

When collecting and modeling data of SHM your mathematical model had a value as shown below:

x(t) Acos t

x(t) Acos t C What if your clock didn’t start at x=A or x=-A?

This value represents our initial conditions. We call it the phase angle:

x(t) Acos t

SHM and Circular Motion

Uniform circular motion projected onto one dimension is simple harmonic motion.

SHM and Circular Motion

x(t) Acos

ddt

t

x(t) Acos t

Start with the x-component of position of the particle in UCM

End with the same result as the spring in SHM!

Notice it started at angle zero

Initial conditions:

t 0

We will not always start our clocks at one amplitude.

x(t) Acos t 0

vx (t) Asin t 0

vx (t) vmax sin t 0

The Phase Constant:

t 0

Phi is called the phase of the oscillation

Phi naught is called the phase constant or phase shift. This

value specifies the initial conditions.

Different values of the phase constant correspond to different starting points on the circle and thus to

different initial conditions

Phase Shifts:

An object on a spring oscillates with a period of 0.8s and an amplitude of 10cm. At t=0s, it is 5cm to the left of

equilibrium and moving to the left. What are its position and direction of motion at t=2s?

x(t) Acos t 0

x0 5cm Acos 0 Initial conditions:

0 cos 1 x0

A

cos 1 5cm

10cm

120

23 rads

From the period we get:

2T

2

0.8s7.85rad /s

An object on a spring oscillates with a period of 0.8s and an amplitude of 10cm. At t=0s, it is 5cm to the left of

equilibrium and moving to the left. What are its position and direction of motion at t=2s?

x(t) Acos t 0

7.85rad /s

0 23 rads

A 0.1mt 2s

x(t) 0.1cos 7.85 2 23

x(t) 0.05m

We have modeled SHM mathematically. Now comes the physics.

Total mechanical energy is conserved for our SHM example of a spring with

constant k, mass m, and on a frictionless surface.

E K U 12

mv2 12

kx2

The particle has all potential energy at x=A and x=–A, and the particle has purely kinetic energy at x=0.

At turning points:

E U 12

kA2

At x=0:

E k 12

mvmax2

From conservation:

12

kA2 12

mvmax2

Maximum speed as related to amplitude:

vmax km

A

From energy considerations:

From kinematics:

Combine these:

vmax km

A

vmax A

km

f 1

2km

T 2 mk

a 500g block on a spring is pulled a distance of 20cm and released. The subsequent oscillations are measured to

have a period of 0.8s. at what position or positions is the block’s speed 1.0m/s?

The motion is SHM and energy is conserved.

12

mv2 12

kx2 12

kA2

kx2 kA2 mv2

x A2 mk

v2

x A2 v2

2

2T

2

0.8s7.85rad /s

x 0.15m

Dynamics of SHM

Acceleration is at a maximum when the particle is at maximum and minimum displacement from x=0.

ax dvx (t)

dt

d Asin t dt

2Acos t

Dynamics of SHM

Acceleration is proportional to the

negative of the displacement.

ax 2Acos t

ax 2x

x Acos t

Dynamics of SHM

As we found with energy considerations:

ax 2x

F max kx

max kx

ax km

x

According to Newton’s 2nd Law:

ax d2xdt 2

Acceleration is not constant:

d2xdt 2

km

x

This is the equation of motion for a mass on a spring. It is of a general

form called a second order differential equation.

2nd-Order Differential Equations:

Unlike algebraic equations, their solutions are not numbers, but functions.

In SHM we are only interested in one form so we can use our solution for many objects undergoing SHM.

Solutions to these diff. eqns. are unique (there is only one). One common method of solving is guessing the

solution that the equation should have…

d2xdt 2

km

xFrom

evidence, we expect

the solution:

x Acos t 0

2nd-Order Differential Equations:

Let’s put this possible solution into our equation and see if we guessed right!

d2xdt 2

km

x

IT WORKS. Sinusoidal oscillation of SHM is a result of Newton’s laws!

x Acos t 0

d2xdt 2 2Acos t

dxdt

Asin t

2Acos t km

Acos t

2 km

Vertical springs oscillate differently than horizontal springs because there are 2 forces acting.

The equilibrium position gets shifted downward

What about vertical oscillations of a spring-mass system??

Fnet kL mg 0Hanging at rest:

kL mg

L mk

g

this is the equilibrium position of the system.

Now we let the system oscillate. At maximum:

But:

Fnet k L y mg

Fnet kL mg ky

kL mg 0So:

Fnet ky

Everything that we have learned about horizontal oscillations is equally valid for

vertical oscillations!

You need to show how to derive the Period of a Pendulum equationT = 2∏√l/g

The Pendulum

Fnet t mgsin ma t

d2sdt 2 gsin

Equation of motion for a pendulum

s L

Small Angle Approximation:

d2sdt 2 gsin

When θ is about 0.1rad or less, h and

s are about the same.

sin

cos 1

tan sin 1

d2sdt 2 g

sL

Fnet tm d2s

dt 2 mgsL

The Pendulum

Equation of motion for a pendulum

d2sdt 2

gsL

gL

(t) max cos t 0

x(t) Acos t 0

A Pendulum Clock

What length pendulum will have a period of exactly 1s?

gL

T 2 Lg

g T2

2

L

L 9.8m/s2 1s2

2

0.248m

Conditions for SHM

Notice that all objects that we look at are described

the same mathematically.

Any system with a linear restoring force will undergo simple

harmonic motion around the equilibrium position.

A Physical Pendulum

d2dt 2

mglI

I mgd mglsin

when there is mass in the

entire pendulum, not just the bob.

Small Angle Approx.

mgl

I

Damped Oscillations

All real oscillators are damped oscillators. these are any that slow

down and eventually stop.a model of drag force for

slow objects:

Fdrag bv

b is the damping constant (sort of like a coefficient of friction).

Damped Oscillations

F Fs Fdrag kx bv ma

kx b dxdt

m d2xdt 2 0

Another 2nd-order diff eq.

Solution to 2nd-order diff eq:

x(t) Ae bt / 2m cos t 0

km

b2

4m2

02

b2

4m2

Damped Oscillations

x(t) Ae bt / 2m cos t 0

A slowly changing line that provides a border to

a rapid oscillation is called the envelope of

the oscillations.

http://www.youtube.com/watch?v=IqK2r5bPFTM&feature=related

Driven Oscillations

Not all oscillating objects are disturbed from rest then allowed to move undisturbed.

Some objects may be subjected to a periodic external force.

DrivenOscillations

All objects have a natural frequency at which they tend to vibrate when disturbed.

Objects may be exposed to a periodic force with a particular driving frequency.

If the driven frequency matches

the natural frequency of an

object, RESONANCE occurs

SpringBest example of simple harmonic oscillator.

T = 2m/k

m

Simple PendulumActs as simple harmonic oscillator only when angle of swing is small.

T = 2L/g

Conical PendulumNot really a simple harmonic oscillator, but equation is similar to simple pendulum.

T = 2L(cos )/g

The small angle approximation for a simple pendulum

mg

mgcos

mgsinglT

TLgLg

ifLgmLLmg

IFr

pendulum

2

2,

0)(

sin,sin)()(sin

sin2

A simple pendulum is one where a mass is located at the end of string. The string’s length represents the radius of a circle and has negligible mass.

Once again, using our sine function model we can derive using circular motion equations the formula for the period of a pendulum.

If the angle is small, the “radian” value for theta and the sine of the theta in degrees will be equal.

Torsional PendulumTwists back and forth through equilibrium position.

T = 2I/

Physical PendulumAnything that doesn’t fall into any of the other categories of pendulums.

T = 2I/

= Mgd

K1 + U1 = K2 + U2◦K = 1/2mv2

◦U = mgh1/2mv1

2+mgh1 =1/2mv22+ mgh2

v12 + 2gh1 = v2

2+ 2gh2

Energy Conservation in Pendulums

h

The Physical Pendulum

mgdIT

TImgdImgd

ifImgd

LdIdmg

IFr

pendulumphysical

2

2,

0)(

sin,2,sin

sin

A physical pendulum is an oscillating body that rotates according to the location of its center of mass rather than a simple pendulum where all the mass is located at the end of a light string.

It is important to understand that “d” is the lever arm distance or the distance from the COM position to the point of rotation. It is also the same “d” in the Parallel Axes theorem.

ExampleA spring is hanging from the ceiling. You know that

if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the:

kkkxFs )3)((330Spring Constant

Angular frequency 5

1102

mk

mk

110 N/m

4.7 rad/s

ExampleA spring is hanging from the ceiling. You know

that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the:

Amplitude

Frequency and Period

Stated in the question as 1.5 m

7.4222

7.42

22

T

f

Tf

0.75 Hz

1.34 s

ExampleA spring is hanging from the ceiling. You know

that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the:

2

22

)5.1)(110(21

21

21

U

kAkxU sTotal Energy

Maximum velocity )7.4)(5.1(Av

123.75 J

7.05 m/s

Position of mass at maximum velocity

Maximum acceleration of the mass

Position of mass at maximum acceleration

At the equilibrium position

)5.1()7.4( 22Aa 33.135 m/s/s

At maximum amplitude, 1.5 m

A spring is hanging from the ceiling. You know that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the:

a) Find the length of a simple seconds pendulum.

b) What assumption have you made in this calculation?

A “seconds pendulum” beats seconds; that is, it takes 1 s for half a cycle.

Problem #2

A thin uniform rod of mass 0.112 kg and length 0.096 m is suspended by a wire through its center and perpendicular to its length. The wire is twisted and the rod set to oscillating. The period is found to be 2.14 s. When a flat body in the shape of an equilateral triangle is suspended similarly through its center of mass, the period is 5.83. Find the rotational inertia of the triangle.

Problem #3

A uniform disk is pivoted at its rim. Find the period for small oscillations and the length of the equivalent simple pendulum.

THE

END