Periodic Motion and Theory of Oscillations

m

0 XxFRestoring force Fx = -kx is a linear function ofdisplacement x from equilibrium position x=0.

Harmonic oscillator: max = - kxax

m

kx

dt

xd 22

2

2

,0

)sin()(

21)cos()(

tAdt

dxtv

fTtAtx

Initial conditions at t=0:

sin

cos

0

0

Av

Ax

0

02

202

0 tan,x

vvxA

Simple harmonic motion:Position, velocity, andacceleration are periodic,sinusoidal functions of time.

Oscillator equation:

Energy in Simple Harmonic MotionTotal mechanical energy E=K+U in harmonic oscillations is conserved:

constkAttk

mkAkxmvE

222

2222

2

1)(cos)(sin

22

1

2

1

Example: Non-adiabatic perturbation of mass(a) M → M + m at x=0 results in a change of velocity due to momentum conservation: Mvi=(M+m)vf, vf= Mvi/(M+m), hence, Ef= MEi/(M+m), Af= Ai[M/(M+m)]1/2, Tf = Ti [(M+m)/M]1/2

(b) M → M + m at x=A (v=0) does not changevelocity, energy, and amplitude;only the period is changed again due to an increase of the total mass Tf = Ti [(M+m)/M]1/2

Exam Example 30: A Ball Oscillating on a Vertical Spring (problems 14.38, 14.83)

y

0

y0

y1=y0-A

v0

v1=0

Data: m, v0 , k Find: (a) equilibrium position y0;(b) velocity vy when the ball is at y0;(c)amplitude of oscillations A; (d) angular frequency ω and period T of oscillations.

Unstrained→

Equilibrium

Lowest position

Solution: Fy = - ky

(a) Equation of equilibrium: Fy – mg = 0, -ky0 = mg , y0 = - mg/k (b) Conservation of total mechanical energy

20

200

20 2

1

2

1

2

1mvkymgymvEUUKE yelasticgrav

kmgvmkygyvv y /)/2( 22000

20

(c) At the extreme positions y1,2 = y0 ± A velocity is zero and

2

20

202

0

20

2

2,120

22,12,1 1

2

1

2

1

mg

kv

k

mg

k

mvyA

k

mv

k

mg

k

mgymvkymgy

y2=y0+A

(d) k

mT

m

k 2

2,

Applications of the Theory of Harmonic Oscillations

Oscillations of Balance Wheel in a Mechanical Watch

ITft

Idt

dI zzz

22,)cos(

0,,2

2

Newton’s 2nd law for rotation yields

Exam Example 31: SHM of a thin-rim balance wheel (problems 14.41,14.97)

Data: mass m, radius R , period T

R

Questions: a) Derive oscillator equation for a small angular displacement θ from equilibrium position starting from Newton’s 2nd law for rotation. (See above.)b) Find the moment of inertia of the balance wheel about its shaft. ( I=mR2 )c) Find the torsion constant of the coil spring.

2)/2(/2 TRmIT

(mass m)

Vibrations of Molecules due to van der Waals Interaction

F

6

0

12

00 2

r

R

r

RUU

Displacement fromequilibrium x = r – R0

Restoring force

7

0

13

00

0

7

0

13

0

0

0 111212

R

x

R

x

R

U

r

R

r

R

R

U

dr

dUFr

Approximation of small-amplitude oscillations:

|x| << R0 , (1+ x/R0)-n ≈ 1 – nx/R0,

Fr = - kx , k = 72U0/R02

m m

Example: molecule Ar2 , m = 6.63·10-26kg,U0=1.68·10-21 J, R0= 3.82·10-10 m

Hzm

kf 11106.5

2

1

2

Potential well formolecular oscillations

Simple and Physical PendulumsNewton’s 2nd law for rotation of physical pendulum:

Iαz = τz , τz = - mg d sinθ ≈ - mgd θ

I

mgd

dt

d

,022

2

mgd

IT 2

Simple pendulum: I = md2

g

dT 2

Example: Find length d for the period to be T=1s.

cmmssm

dgT

d 2525.0)14.3(4

)1(/8.9

4 2

22

2

2

Exam Example 32: Physical Pendulum (problem 14.99, 14.54)Data: Two identical, thin rods, each of mass m and length L,are joined at right angle to form an L-shaped object. Thisobject is balanced on top of a sharp edge and oscillates.

Find: (a) moment of inertia for each of rods;(b) equilibrium position of the object’s center of mass;(c) derive harmonic oscillator equation for smalldeflection angle starting from Newton’s 2nd law for rotation;(d) angular frequency and period of oscillations.

Solution: (a) dm = m dx/L ,

d

cm

2

0

21 )3/1()/( mLdxxLmI

L

(b) geometry and definition xcm=(m1x1+m2x2)/(m1+m2)→ ycm= d= 2-3/2 L, xcm=0

m m

(c) Iαz = τz , τz = - 2mg d sinθ ≈ - 2mgd θ I

mgd

dt

d 2,02

2

2

X

y

0

(d) Object’s moment of inertia 2

,22

32

3

22 2

1 TL

g

I

mgdmLII

θ

gM

Damped Oscillations

Springs in the automobile’s suspension system: oscillation with ω0

The shock absorber: damping γ

Damped OscillationsFrictional force f = - b vx

dissipates mechanical energy. Newton’s 2nd law:

max = -kx - bvxDifferential equation of the damped harmonic oscillator:

m

b

m

k

xdt

dx

dt

xd

2,

02

20

202

2

220,)cos( tAex tGeneral solution:

underdamped (γ < γcr)(instability if γ<0)

overdamped (γ > γcr)

Critical damping γcr = ω0 , bcr=2(km)1/2

Damping power: 2xx bvfv

dt

dE

20

22,121 ,21 tt eCeCx

Fourier analysis:

titi

ti

eAeAtx

iiAex

2121

2202,1

20

2

)(

02

Forced Oscillations and Resonance

m

tFx

dt

dx

dt

xd )(2 2

02

2

Amplitude of a steady-state oscillationsunder a sinusoidal driving force F = Fmax cos(ωdt)

222220

max

4)( ddm

FA

At resonance, ωd ≈ ω, driving force does positive work all the timeWnc = Ef – Ei >0, and even weak force greatly increases amplitude of oscillations.

vF

Example: laser ( ← → )→

Parametric resonance is another typeof resonance phenomenon, e.g. L(t).

Forced oscillator equation:

(self-excited oscillation of atoms and field)

0

max0 2)(

m

FA d