PHYS 218sec. 517-520

ReviewChap. 13

Periodic Motion

What you have to know

• Periodic motions (oscillations)

• Physical quantities describing periodic motions

• Equation of motion for simple harmonic motion

• Various examples of harmonic motion

• Pendulums

• We skip the parts which have an asterisk (*) in the textbook.

Physical quantities

( )

( )

( )

Amplitude : maximum displacement from equilibrium the maximum value of

Cycle: one complete round trip

Period : the time for one cycle, unit s (sec)

Frequency : the number of cycles in unit time, uni

A x

T

f

®

( )( )

1t Hz s

Angular frequency : 2 fw w p

-=

=

Equilibrium point

Basic relations

1 1 2, , 2f T f

T f T

pw p= = = =

Restoring force

Simple harmonic motion (SHM)

This is the simplest case when the restoring force is directly proportional to the displacement from the equilibrium

2

2

2

2

In this case, the restoring force is in the form of .

Becasue ,

: This is the basic equation for SHM

xF kx

d xF ma m

dt

d x kx

dt m

=-

= =

=-

( )

22

2

Its solution is sinusoidal so sine or cosine function;

cos cos

Therefore, the SHM can be described by

x A A t

d xx

dt

q w

w

= =

=- for SHM

k

mw=

Then it follows that

1 1, 2

2 2

k mf T

m f k

wp

p p= = = =

does not depend on A

Ex 13.2 SHM

A spring: 6 N force causes a displacement of 0.03 m

Attach a 0.5 kg object to the end of the spring, pull it a distance of 0.02 m nad then release it

Spring constant

2

The first condition determines the spring constant

6.0 N200 N/m 200 kg/s

0.03 m

FF kx k

x= Þ = = = =

, f, T of this SHM

2

Since 0.5 kg,

200 kg/s20 rad/s

0.5 kg

20 rad/s 1 13.2 /s 3.2 Hz, 0.31 s

2 2 3.2 Hz

m

k

m

f Tf

w

wp p

=

= = =

= = = = = = =

Displacement in SHM

22

2Equation of motion:

The general solution of this equation is

cos( ) : amplitude, : phase angle

The two constants and can be determined

by initial conditions such as the position

d xx

dt

x A t A

A

w

w ff

f

=-

= +

0 0

and velocity at 0

At 0, , where cos

t

t x x x A f

=

= = =

( )( ) ( )

( )2

2 22

2 2

velocity in SHM

cos sin

acceleration in SHM

cos

Therefore, the velocity oscillates between and

and the acceleration oscillates between and

x

x

dx dv A t A t

dt dt

d x dva A t x

dt dtA A

A A

w f w w f

w w f w

w w

w w

= = + =- +

= = =- + =-

-

- .

( )

0 0 0 0

2 222 2 2 2 2 2 2 0 0

0 0 2 2 2

2 22 2 20 0

0 02 2

0 0

0 0

At 0, and , where cos and sin

cos and sin 1 cos sin

Therefore,

Also

sintan arctan

cos

t x x v v x A v A

x vx A v A

A A

v vA x A x

v vA

x A x

f w f

f w ff fw

w w

w fw ff

f

= = = = =-

\ = = Þ = + = +

= + Þ = +

æ ö- ÷ç ÷= =- Þ = -ç ÷ç ÷çè ø

Displacement in SHM (2)

Energy in SHM

( ) ( )In SHM, cos , sinx A t v A tw f w w f= + =- +

Total energy

( ) ( )

( ) ( ){ }

2 2 2 2 2 2 2

2 2 2 2

2

1 1 1 1sin cos

2 2 2 21

sin cos using 21

2

E mv kx mA t kA t

kkA t t

m

kA

w w f w f

w f w f w

= + = + + +

= + + + ¬ =

= constant It should be so because of energy conservation

max

max

Since contains a sine function and has a cosine function,

when , 0 and

when 0,

x v

x x v

x v v

=± =

= =±2 2 2

2 2

Since the energy relation gives

1 1 1,

2 2 2

we get

kA mv kx

kv A x

m

= +

=± -

Ex 13.4 Velocity, acceleration, and energy in SHM

200 N/m, 0.5 kg and the oscillating mass is released from rest at 0.02 m.k m x= = =

Maximum and minimum velocity

( )

2 2

max

min

As we derived in the previous page,

.

200 N/m0.02 m 0.40 m/s

0.5 kg

0.40 m/s

kv A x

m

kv A

m

kv A

m

=± -

Þ = = ´ =

=- =-

Note that the minimum value is NOT 0!

Maximum acceleration

( )( )

2

2max max

Since ,

200 N/m 0.02 m8.0 m/s

0.50 kg

ka x x

m

k kAa x

m m

w=- =-

= = = =

Ex 13.4 Cont’d

The body has moved halfway to the center from its original position

2 2

2

The position of the body is / 2 0.01 m.

Then,

0.35 m/s

4.0 m/s

x A

kv A x

mk

a xm

= =

=- - =-

=- =-

The body is moving from x=A/2 to x=0. Thus we choose the negative sign

for the velocity.

Ex 13.5 Energy and momentum in SHM

1v

O

M

m

This is a collision problem.

always consider momentum conservation!

To use momentum conservation,

we should first know the velocity of the body.

The velocity can be obtained by energy consideration.

Þ

21 1

2 21 1 1 1 1

1Before the collision, the total energy is

2

1 1At 0, 0, and

2 2

E kA

kx U E kA Mv v A

M

=

= = = = Þ =

Velocity at x=0

Collision

( )1 2 2 1

in -directionBy momentum conservation ,

0M

M v M m v v vM

x

m+ = + Þ =

+

The putty m is moving in y-direction.

Ex 13.5 Cont’d

After the Collision

( ) ( )

2

22

2 2 1 1

2

Now the body ( ) is moving with at 0.

Then the total energy becomes

1 1

2 2

1The total energy can also be written in terms of the amplitude as ,

2th

M m v x

M ME M m v M m v E

M m M m

E kA

+ =

æ ö æ ö÷ ÷ç ç= + = + =÷ ÷ç ç÷ ÷ç çè ø è ø+ +

=

2 22 1 2 1

erefore,

1 1

2 2

The period is

mass2

spring constant

Then the period after the collision is

2

M MkA kA A A

M m M m

T

M mT

k

p

p

æ ö÷ç= Þ =÷ç ÷çè ø+ +

=

+=

Ex 13.5 Cont’d

If the Collision happens when M is at x=A

The velocity of the block is then zero.

There is no motion and the collision just causes the change of

the mass from to .

Since the amplitude does not change, the

total energy of the system does not

M m

change.

The period depends on the mass, so

2

after the collision.

M mT

kp

+=

Vertical SHM

l

lD

l

F k l= D

F mg=

In equilibrium,

k l mgD =

l xD -

l

F k l= D

F mg=

equilibrium oscillation

x

( )

2

Therefore,

SHM

netF k l x mg kx

d x kx

dt m

= D - - =-

=- Þ

Angular SHM

: angular displacementq

2

2

Equation of motion (from Newton's 2nd law)

If the restoring torque is proportional to the angular displacement

Therefore, we get

SHM

The angular frequency is

I

d

dt I

I

t a

t kq

q ka q

kw

=

=-

= =- Þ

=

Simple pendulum

L

q

T

mg

x

2

2

Equation of motion along the direction

sin (cf )

Therefore,

sin SHM

The angular frequency is

x

F mg x L

d x gg g x

dt L

g

L

q q q

q q

w

=- =

=- » - =- Þ

=sinmg q

cosmg q

The Taylor expansion

When is small

sin

q

q q;

Physical pendulum

Any real pendulum

pivot

d

mg

q

You can treat this object as a point-like object which has the mass of the object

at its center-of-mass position.

Then it becomes similar to the case of simple pendulum.

( )( )

2

2

sin

When is small, sin ,

SHM

angular frequency

mg d I

d mgd

dt I

mgd

I

t q a

q q q

qq

w

=- =

»

Þ =- Þ

=2

2

. For simple pendulum,

then we have

cf I md

mgd g

md dw

=

= =

Forced oscillations & resonance

Damped oscillations

These topics will not be summarized here.

But you should read the textbook for these topics.