Periodic Motion and Theory of Oscillations
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Periodic Motion and Theory of Oscillations m 0 X x F Restoring force F x = -kx is a linear functi displacement x from equilibrium position x= Harmonic oscillator: ma x = - k a x m k x dt x d 2 2 2 2 , 0 ) sin( ) ( 2 1 ) cos( ) ( t A dt dx t v f T t A t x Initial conditions at t=0: sin cos 0 0 A v A x 0 0 2 2 0 2 0 tan , x v v x A Simple harmonic motion: Position, velocity, and acceleration are periodic, sinusoidal functions of time. Oscillator equation:
description
Periodic Motion and Theory of Oscillations. a x. Harmonic oscillator: ma x = - kx. Restoring force F x = -kx is a linear function of displacement x from equilibrium position x=0. m. 0. Oscillator equation:. X. Initial conditions at t=0:. Simple harmonic motion: - PowerPoint PPT Presentation
Transcript of Periodic Motion and Theory of Oscillations
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
mkx
dtxd
222
2
,0
)sin()(
21)cos()(
tAdtdxtv
fTtAtx
Initial conditions at t=0:
sincos
0
0
AvAx
0
02
202
0 tan,xvvxA
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
21)(cos)(sin
221
21
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 14.69, 14.77)
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
121
21 mvkymgymvEUUKE 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
21
21
mgkv
kmg
kmvyA
kmv
kmg
kmgymvkymgy
y2=y0+A
(d) kmT
mk
22,
Applications of the Theory of Harmonic OscillationsOscillations of Balance Wheel in a Mechanical Watch
ITft
IdtdI zzz
22,)cos(
0,, 2
2Newton’s 2nd law for rotation yields
Exam Example 31: SHM of a thin-rim balance wheel (problems 14.41-14.43)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
60
120
0 2rR
rRUU
Displacement fromequilibrium x = r – R0
Restoring force
7
0
13
00
07
013
0
0
0 111212Rx
Rx
RU
rR
rR
RU
drdUFr
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
Hzmkf 11106.5
21
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
Imgd
dtd
,022
2
mgdIT 2
Simple pendulum: I = md2
gdT 2
Example: Find length d for the period to be T=1s.
cmmssmdgTd 2525.0)14.3(4
)1(/8.94 2
22
2
2
Exam Example 32: Physical Pendulum (problem 14.88, 14.53)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 ,
dcm
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 Imgd
dtd 2,02
2
2
X
y
0
(d) Object’s moment of inertia 2,
2232
322 2
1 TLg
ImgdmLII
θ
gM
Damped Oscillations
Springs in the automobile’s suspension system: oscillation with ω0
The shock absorber: damping γ
Damped OscillationsFrictional force f = - b vxdissipates mechanical energy. Newton’s 2nd law: max = -kx - bvx
Differential equation of the damped harmonic oscillator:
mb
mk
xdtdx
dtxd
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)1/2Damping power: 2
xx bvfvdtdE
20
22,121 ,21 tt eCeCx
Fourier analysis:
titi
ti
eAeAtx
iiAex21
21
2202,1
20
2
)(
02
Forced Oscillations and Resonance
mtFx
dtdx
dtxd )(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)(
mFA d
