1 / 10

Periodic Motion and Theory of Oscillations

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:

fayre
Télécharger la présentation

Periodic Motion and Theory of Oscillations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Periodic Motion and Theory of Oscillations ax Harmonic oscillator: max = - kx Restoring force Fx = -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: Position, velocity, and acceleration are periodic, sinusoidal functions of time.

  2. Energy in Simple Harmonic Motion Total mechanical energy E=K+U in harmonic oscillations is conserved: 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 change velocity, energy, and amplitude; only the period is changed again due to an increase of the total mass Tf = Ti [(M+m)/M]1/2

  3. Exam Example 30: A Ball Oscillating on a Vertical Spring(problems 14.38, 14.83) y Data: m, v0 , k y2=y0+A • Find:(a) equilibrium position y0; • (b) velocity vy when the ball is at y0; • amplitude of oscillations A; • (d) angular frequency ω and • period T of oscillations. Unstrained→ 0 v0 y0 Equilibrium Solution: Fy = - ky • Equation of equilibrium: • Fy – mg = 0, -ky0 = mg , y0 = - mg/k • (b) Conservation of total mechanical energy Lowest position y1=y0-A v1=0 (c) At the extreme positions y1,2 = y0 ± A velocity is zero and (d)

  4. Applications of the Theory of Harmonic Oscillations Oscillations of Balance Wheel in a Mechanical Watch (mass m) Newton’s 2nd law for rotation yields R Exam Example 31: SHM of a thin-rim balance wheel(problems 14.41,14.97) Data: mass m, radius R , period T 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.

  5. Vibrations of Molecules due to van der Waals Interaction Potential well for molecular oscillations m m Displacement from equilibrium x = r – R0 Restoring force Approximation of small-amplitude oscillations: |x| << R0 , (1+ x/R0)-n ≈ 1 – nx/R0, Fr = - kx , k = 72U0/R02 Example: molecule Ar2 , m = 6.63·10-26kg, U0=1.68·10-21 J, R0= 3.82·10-10 m

  6. Simple and Physical Pendulums Newton’s 2nd law for rotation of physical pendulum: Iαz = τz , τz = - mg d sinθ ≈ - mgd θ Simple pendulum: I = md2 Example: Find length d for the period to be T=1s.

  7. Exam Example 32: Physical Pendulum (problem 14.99, 14.54) 0 X Data:Two identical, thin rods, each of mass m and length L, are joined at right angle to form an L-shaped object. This object is balanced on top of a sharp edge and oscillates. m m θ d 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 small deflection angle starting from Newton’s 2nd law for rotation; (d) angular frequency and period of oscillations. cm y Solution: (a) dm = m dx/L , (b) geometry and definition xcm=(m1x1+m2x2)/(m1+m2)→ ycm= d= 2-3/2 L, xcm=0 (c) Iαz = τz , τz = - 2mg d sinθ ≈ - 2mgd θ (d) Object’s moment of inertia

  8. Damped Oscillations Springs in the automobile’s suspension system: oscillation with ω0 The shock absorber: damping γ

  9. Damped Oscillations Frictional force f = - b vx dissipates mechanical energy. Newton’s 2nd law: max = -kx - bvx Differential equation of the damped harmonic oscillator: Fourier analysis: General solution: underdamped (γ < γcr) (instability if γ<0) overdamped (γ > γcr) Critical damping γcr = ω0 , bcr=2(km)1/2 Damping power:

  10. Forced Oscillations and Resonance Forced oscillator equation: Amplitude of a steady-state oscillations under a sinusoidal driving force F = Fmax cos(ωdt) At resonance, ωd ≈ ω, driving force does positive work all the time Wnc = Ef – Ei >0, and even weak force greatly increases amplitude of oscillations. Example: laser ( ←→ )→ (self-excited oscillation of atoms and field) Parametric resonance is another type of resonance phenomenon, e.g. L(t).

More Related