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Periodic Motion and Theory of Oscillations m 0 X Restoring force F x = -kx is a linear function of displacement x from equilibrium position x=0. Harmonic.

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Presentation on theme: "Periodic Motion and Theory of Oscillations m 0 X Restoring force F x = -kx is a linear function of displacement x from equilibrium position x=0. Harmonic."— Presentation transcript:

1 Periodic Motion and Theory of Oscillations m 0 X Restoring force F x = -kx is a linear function of displacement x from equilibrium position x=0. Harmonic oscillator: ma x = - kx axax Initial conditions at t=0: Simple harmonic motion: Position, velocity, and acceleration are periodic, sinusoidal functions of time. Oscillator equation:

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: Mv i =(M+m)v f, v f = Mv i /(M+m), hence, E f = ME i /(M+m), A f = A i [M/(M+m)] 1/2, T f = T i [(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 T f = T i [(M+m)/M]1/2

3 Exam Example 30: A Ball Oscillating on a Vertical Spring (problems 14.38, 14.83) y 0 y0y0 y 1 =y 0 -A v0v0 v1=0v1=0 Data: m, v 0, k Find: (a) equilibrium position y 0 ; (b) velocity v y when the ball is at y 0 ; (c)amplitude of oscillations A; (d) angular frequency ω and period T of oscillations. Unstrained→ Equilibrium Lowest position Solution: F y = - ky (a) Equation of equilibrium: F y – mg = 0, -ky 0 = mg, y 0 = - mg/k (b) Conservation of total mechanical energy (c) At the extreme positions y 1,2 = y 0 ± A velocity is zero and y 2 =y 0 + A (d)

4 Applications of the Theory of Harmonic Oscillations Oscillations of Balance Wheel in a Mechanical Watch Newton’s 2 nd 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 2 nd law for rotation. (See above.) b) Find the moment of inertia of the balance wheel about its shaft. ( I=mR 2 ) c) Find the torsion constant of the coil spring. (mass m)

5 Vibrations of Molecules due to van der Waals Interaction Displacement from equilibrium x = r – R 0 Restoring force Approximation of small-amplitude oscillations: |x| << R 0, (1+ x/R 0 ) -n ≈ 1 – nx/R 0, F r = - kx, k = 72U 0 /R 0 2 m m Example: molecule Ar 2, m = 6.63·10 -26 kg, U 0 =1.68·10 -21 J, R 0 = 3.82·10 -10 m Potential well for molecular oscillations

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

7 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. This object 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 small deflection angle starting from Newton’s 2 nd law for rotation; (d) angular frequency and period of oscillations. Solution : (a) dm = m dx/L, d cm (b) geometry and definition x cm =(m 1 x 1 +m 2 x 2 )/(m 1 +m 2 )→ y cm = d= 2 -3/2 L, x cm =0 mm (c) Iα z = τ z, τ z = - 2mg d sinθ ≈ - 2mgd θ X y 0 (d) Object’s moment of inertia θ

8 Damped Oscillations Springs in the automobile’s suspension system: oscillation with ω 0 The shock absorber: damping γ

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

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


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