Chapter 13: Oscillatory MotionsSimple harmonic motion

Spring and Hooke’s law• When a mass hanging from a spring and in equilibrium, the Newton’s 2nd law says:

This means the force due to the spring is equal to the force by gravity and opposite in direction when the spring is stretched.

gsgsy FFFFmaF 0

• Hooke’s law states that increasing the weight by equal amounts increases the stretch of the spring by equal amount. Therefore, the force due to the spring must be proportional to the stretch of the spring.

constant. spring theis where kxkFs

x is deviation from the spring w/o weight

This is also true when the spring shrinks

y

Simple harmonic motion

Simple harmonic motion (SHM)

• Let’s study a motion of the mass m. When the mass is attached to the spring, the spring stretches by x0. Then lift the mass by A and release it.

maFFmaF gsy

mamgxxk )( 0• Since in equilibrium ,0 0 mgkxFF gs

d

and )/(2

2

dt

x

dt

dx

dt

d

dt

dvaxmkamakx

xxm

k

dt

xd 22

2

y y y

Equation for SHM

• The initial stretch is x0-x and from Hooke’s law:

Simple harmonic motion

Simple harmonic motion (SHM) (cont’d)

xxm

k

dt

xd 22

2

• Solution:

)cos()( tAtxA

xxAx 0

0 arccos ,cos)0( As

AfTf :amplitude ,/1: period ,2/ :Frequency

xxatAdt

tdvta

xAxvtAdt

tdxtv

22

22

)(),cos()(

)(

)(),sin()(

)(

Hz s

velocity

acceleration

(rad/s)frequency angular :m

k

phase constant

Simple harmonic motion

Simple harmonic motion (SHM) (cont’d)

• Solution:

)cos()( tAtx

• What is SHM/SHO?

A simple harmonic motion is the motionof an oscillating system which satisfiesthe following condition:

1. Motion is about an equilibrium position at which point no net force acts on the

system.2. The restoring force is proportional to and oppositely directed to the displacement.3. Motion is periodic.

t=0t=-

Acos

f=

By Dr. Dan Russell, Kettering University

Simple harmonic motion Connection between SHM and circular motion

• For an object in circular motion, the angular velocity is defined as,

tdt

d

• The tangential velocity is related to the angular velocity : rv

• The centripetal acceleration is also related to the angular velocity:

222 )( r

r

r

r

va

• The position, velocity and acceleration of the object as a function of time are:

)()cos()(cos

)sin()(sin

)cos()(cos

222 txtrtara

trtvrv

trtxrx

phasor a called is Vector r

SHM!

Simple harmonic motion

Displacement, velocity and acceleration in SHM

• Displacement

)cos()( tAtx

• Velocity

)sin()(

)( tAdt

tdxtv

)cos()(

)( 2 tAdt

tdvta

• Acceleration

0

Note: 222 /)0()0( vxA

Energy in SHM

Energy conservation

UKE Energy conservation in a SHM

dx

dUkxFs

No friction22

2

1 ;

2

1kxUmvK

const. 2

1

2

1 22 kxmvE

EkxmvkA 222

2

1

2

1

2

1

22 xAm

kv

BTW:

2

0 2

1 kxdxFUU

x

s Ch.7

2

Energy in SHM

Energy conservation in a SHM (cont’d)

const. 2

1

2

1 22 kxmvE

ene

rgy

ene

rgy

distance from equilibrium pointTime

E

kinetic energy

potential energy

Applications of SHM Simple pendulum

• The forces on the mass at the end are gravity and the tension. The tension, however, exerts no torque about the top of the string.

angle smallfor sinsin 2

gg

mmgI

gfT

gf

g

21

,2

1

2 ,

pendulum simple a offrequency Angular

g

dt

d

dt

d

2

2

mg

Physical pendulum

• A simple pendulum has all its mass concentrated at a point and oscillates due to gravitational torques. Objects that do not have their mass concentrated at a point also oscillate due to gravitational torques.

sinsinI

mgrImgrI

I

mgr

pendulum physical a offrequency Angular

Applications of SHM

Angular SHM

• An angular version of SHM is called torsion oscillation and shown on the right.• A disk suspended by a wire experiences a restoring torsion when rotated by a small angle :

Idt

dI

kxF

2

2

c.f.

m

k

I c.f.

:SHMangular an offrequency Angular

Applications of SHM

: torsion constant

Oscillation with friction• In real world dissipative forces such as friction between a block and a table exist. Such a dissipative force will decrease the amplitude of an oscillation – damped oscillation.

Damped oscillations

The friction reduces the mechanical energy of the system as timepasses, and the motion is said to be damped.

A simple example of damped oscillation

Damped oscillations

• Consider a simple harmonic oscillation with a frictional damping force that is directly proportional to the velocity of the oscillating object.

mabvkxF

2

2

dt

xdm

dt

dxbkx

If the damping force is relatively small, the motion is described by:

2

2

)2/(

4'

where)'cos()(

m

b

m

k

tAetx tmb

A simple example of damped oscillation

Damped oscillations (cont’d)

2

2

)2/(

4'

where)'cos()(

m

b

m

k

tAetx tmb

tmbAe )2/(

By Dr. Dan Russell, Kettering University

An example of resonantly driven damped harmonic oscillator

PushPush

Wait 1 Wait 1 periodperiod

Forced oscillations and resonance Driving force

Driving force (cont’d)

Forced oscillations and resonance

• The additional force that pushed by the person in the animation on the previous page is called a driving force.

• When a periodically varying driving force with angular frequency d is applied to a damped harmonic oscillator, the resulting motion is called a forced oscillation.

)cos( : force driving dt d=

d=

d=By Dr. Dan Russell, Kettering University

Forced oscillation and resonance

Forced oscillations and resonance

Fixed Moving/driving force

Damped

Damped SHM Forced damped SHM

tFtF dcos)( max

2

2

4'

m

b

m

k

natural frequency

Forced oscillation and resonance (cont’d)

Forced oscillations and resonance

2222

max

)( dd bmk

FA

Amplitude for a forced damped oscillation:

mk

mk

d

d

/near

maximum a hasA ,When 2

resonance:

natural frequency

The fact that there isan amplitude peak atdriving frequencies closeto the natural frequencyof the system is calledresonance

angular freq. of driving force

A

Problem 1

Exercises

h=0.40 m

m=0.20 kg

M=2.2 kg

k=400 N/m

a) The speed of the pan and the steak immediately after the collision (total inelastic collision):

m/s 6.24.2

2.2)m 40.0)(m/s 80.9(2 2

Mm

Mvv if

Initial speed of the meat just before thecollision: ghvi 2Final speed of the meat-pan just after thecollision:

Problem 1(cont’d)

Exercises

h=0.40 m

m=0.20 kg

M=2.2 kg

k=400 N/m

b) The amplitude of the subsequent motion:

When the steak hits the pan, the pan isMg/k above the new equilibrium position.

)/( , / where

]/[ 2222222

MmkkMgx

vxAxAv

f

ffff

So the amplitude is:

m. 21.0)(

2 22

Mmk

ghM

k

MgA

c) The period:

s. 49.0)(

2

k

MmT

Problem 2

Exercises

Each:M/2, R

k

stretched by xand then released

cylindersrolls w/oslipping

Ra

MRIIfR

frictionfkxfMa

cmcm

2)2/1(,

):(

.)2/3(/

22

xxM

k

RIM

kxa

cm

.2

32

2

k

MT

f

kx

Problem 3

Exercises

L LTwo identical, thin rods, each with mass mand length L, are joined at right angles toform an L-shaped object. This object isbalanced on top of a sharp edge. If the L-shaped object is deflected slightly, it oscillates.Find the frequency of the oscillation.

Solution:

The moment of inertia about the pivot:22 )3/2()3/1(2 mLmL

The center of gravity is located when balanced at a distance

)22/(Ld below the pivot.L L

)22/(Ld

Think the L-shaped object as a physicalpendulum and is represented by the centerof gravity. The period T is:

mgd

IT 2

Problem 4

Exercises

F1=-k1x1 F2=-k2x2

F1=-k1x

F2=-k2x

Find the effective spring constant.

FxkFxkF 222111

21, xxxxkF eff

2121 k

F

k

Fxx

k

Fx

eff

21

21

kk

kkkeff

xkFxkFxkF

FFF

eff 2211

21

,,

21 kkkeff