1

Oscillations

Readings: Chapter 14

2

Oscillation: Periodic Motion

T – period of motion

1f

T - frequency

3

Oscillation: Periodic Motion: Simple Harmonic Motion

Simple Harmonic Motion – sinusoidal oscillation

2( ) cos cos 2x t A t A ftT

or 2( ) sin sin 2x t A t A ftT

The most general expression for sinusoidal motion

0 02( ) cos cos 2x t A t A ftT

0 - phase constant A - amplitude

0( ) cos 2 90 sin 2x t A ft A ft If then

00 90

4

Oscillation: Periodic Motion: Simple Harmonic Motion

Example: Uniform circular motion

( ) cos cosx t A A t

x- component:

- angular frequency

5

2( ) cos cos 2x t A t A ftT

6

Oscillation: Periodic Motion: Simple Harmonic Motion

0 0 02( ) cos cos 2 cosx t A t A ft A tT

0 - phase constant

A - amplitude T – period of motion (units – s)

1f

T - frequency (units – hertz – Hz =1/s)

- angular frequency (units – rad/s)

12 2fT

A

A

7

0 0 02( ) cos cos 2 cosx t A t A ft A tT

0 - phase constant

Phase constant specifies the initial position of the oscillator

8

0( ) cosx t A t

Simple Harmonic Motion: Position, Velocity, and Acceleration

0

0 max 0

cos( )( )

sin sin

d tdx tv t A

dt dtA t v t

maxv A

0

2 20

sin( )( )

cos ( )

d tdv ta t A

dt dtA t x t

9

0( ) cosx t A t

Simple Harmonic Motion: object oscillating on a spring

2 20( ) cos ( )a t A t x t

Newton’s second law:

2ma F m x kx km

F kx Hooke’s law:

10

0( ) cosx t A t

Simple Harmonic Motion: Conservation of energy

Kinetic energy:

0( ) sinv t A t

2

2 2 20sin

2 2mv m

K A t

Potential energy:

2

2 20cos

2 2kx k

U A t

Total energy:

km

2 2 2 2 20 0

2 2 2 20 0

sin cos2 2

sin cos2 2

m kE K U A t A t

k kA t t A

11

2

2 2 2 2 20 0sin sin

2 2 2mv m k

K A t A t

2

2 20cos

2 2kx k

U A t km

12

0( ) cosx t A t

Example: Find the relation between kinetic and potential energy at x=A/3.

Then

2 22 12 2 9 18 9kx k A kA

U E

Then the kinetic energy at this point is

2

2kE K U A

The total energy is the sum of kinetic and potential energy. At x=A the kinetic energy is 0.

At x=A/3 the potential energy is

2 2 24 82 18 9 9k k k

K E U A A A E

2

2

11884

9

kxUK kA

13

If force is proportional to displacement (it is not necessary the spring system)

then

0( ) cosx t A t

Simple Harmonic Motion:

Solution of this equation:

ma F kx

km

F kx

2

2

d xa

dt2

2

d xm kxdt

14

Oscillations about equilibrium position

( ) ( )netF k L y w ky k L w ky

Equilibrium:

0( ) cosy t A t

k L w w

Lk

Net force:

Oscillations about equilibrium position

15

The pendulum: small-angle approximation

0w g

k ml l

, sinnet t tF w w

then

If y is the arc length then

yl

, sinnet t ty

F w wl

If y<<l then

, 0sinnet t ty y

F w w w k yl l

The net force is the sum of two forces: tension and gravitational force.

Tangential component of the net force is

0( ) cosy t A t

0k mg gm ml l

16

If there is a friction then there will be energy loss

2

2k

E A

The energy is determined by the amplitude of oscillations, so the energy loss means that the amplitude is decreasing