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Home , Oscillation

Physics 106 Lecture 11 Oscillations – I SJ 7th Ed.: Chap 15.1 to 3, 15.5

• Oscillating systems – Characterization – Examples: they are everywhere – Key concepts • Simple : a paradigm – Representation for displacement, velocity, acceleration – Oscillator equation • SHM example: oscillator (Hooke’s law) • Energy relations, exam ple • SHM example: torsion • SHM example: simple pendulum

Definition of an Oscillator : A system executing periodic, repetitive behavior • System state (t) = state(t+T) = …= state(t+NT) • T = period = to complete one complete cycle • State can mean: position / velocity, electric and magnetic fields, others…

Example: simple G G Hooke' s Law restoring : F = - kx

1 2 1 2 E = constant = K + U = mv + kx mech el 2 2

Oscillation range limited by: Kmax = Umax Spring oscillator is the pattern for many systems

Can represent it also by circular motion e.g., x-component of a mass on a rotating stick)

1 Oscillating systems are everywhere in nature: MECHANICAL: • pendula (swinging objects on cables) • auto suspensions • music/ (e.g,string instruments) • guitar, piano, drums, bells, horns, reeds… • rotatingg, machines, vehicle wheels, ,,, orbits, structures, y pendulum spring y(t) x(t) θ(t) m Simple v (t) v(t) L ω(t) y systems θ θ ay(t) a(t) α(t) x m x(t) vx(t) ax(t) vibrating string uniform circular motion

Electrical examples: • , radios driven by oscillator circuits • atomic and molecular energy states • light, x-rays, IR, radio, other EM • antennas, transmitters • the stable shapes of solid objects

Simple Harmonic Oscillators applied to solids

• Simple harmonic oscillators are good models of a wide variety of physical phenomena • Molecular example – If the atoms in the molecule do not move too far, the between them can be modeled as if there were springs between the atoms – The acts similar to that of the SHM oscillator

simple oscillator realistic potential

2 Water Waves: Harmonic oscillations of the level at a point

disturbance y1(t) = y10 cos(2πft) The phase angle v Φ sets the

y2(t) = y20 cos(2πft + φ) v

Some concepts for oscillations

restoring A force causes the system to return to some equilibrium force: state periodically and repeat the motion Resonant oscillation period, determined by physics of the natural system alone. Disturb system to start, then let it go. : ElExamples: pend dllkilitiulum clock, violin string

undamped Idealized case, no energy lost, motion persists forever oscillations: Example: orbit of electrons in atoms and molecules

damped Oscillation dies away due to loss of energy, converted to heat oscillations: or another form. Example: a swing eventually stops simple Undamped natural oscillation with F = - kx (Hooke’s Law); harmonic i.e . restoring force is proportional to the displacement away oscillation: from the equilibrium state

forced External periodic force drives the system motion at it’s own oscillations: frequency/period, may not be the resonant frequency

3 Characterizing oscillations

Amplitude xmax • maximum displacement away from equilibrium (disturbance) displacement

Period T • time for motion to repeat

Frequency f = 1/T • # of cycles per time unit • [Frequency] = Hertz (Hz) = t-1

Angular frequency: time • ω = 2πf (units: rad/s) • ωT = 2π (radians) • T = 2π / ω

“Phase”: where in the cycle the system is “Phase constant” or “phase angle”: pick what amplitude part of the cycle is at t = 0

Amplitude and phase constant specify the initial conditions for an oscillator

Frequency and amplitude of an oscillator

‰ 11.1. The figures show plots of the amplitude of two harmonic oscillators versus time. When object B is compared to object A, which of the following correctly describe the and amplitude?

A) B has larger frequency and larger amplitude than A B) B has larger frequency and smaller amplitude than A C) B has larger period and larger amplitude than A D) B has smaller frequency and smaller amplitude than A E) B has larger period and larger amplitude than A

ω = 2πf = 2π/T x(t) = xm cos(ωt + ϕ)

4 Hooke’s Law (Spring)Oscillator F(t) = − kx(t) F F Restoring force – use Second Law d2x(t) F(t) = m a(t) = m = − k x(t) -xm +xm dt2 Force equation No losses 2 d x(t) k Prototype for oscillator k ∴ = − x(t) equations describes periodic ω = m dt2 m motion with resonant frequency:

Equations of the form above have sine (or cosine) solutions:

x(t) = xm cos(ωt + ϕ) Anggqyular frequency ω = 2πf = 2π/T Initial condition If x(t=0) = xm then phase constant φ = 0, +/- 2π, etc examples: If x(t=0) = 0 then phase constant φ = π/2, 3π/2, etc

stiff spring / small mass Æ high frequency, short period big mass / weak spring Æ low frequency, long period

Example: auto suspension trades off spring stiffness against un-sprung

Representation of … …as the x-component of a vector rotating at constant speed

r = xm= x(t=0) displacement phase angle phase constant at time t (radians), equals zero for sketch r ωτ x(t) = xm cos(ωt + ϕ) x(t) amplitude angular frequency y(t) would be represented as sin(ωt+φ)

-xm +xm d(cos[f(t)]) df(t) velocity at time t Note : = − sin[f(t)] dt dt dx(t) vx (t) ≡ = −ωxm sin(ωt + ϕ) T dt sketch shows φ =0 maximum v ≡ −ωx since clock starts velocity m m when ppghase angle = 0 If φ = π/2, clock starts acceleration at time t at ωt = - π/2 dv (t) a (t) ≡ x = −ω2x cos(ωt + ϕ) x dt m 2 d2x(t) solutions of maximum a ≡ −ω x 2 equations like acceleration m m ≡ −ω x(t) dt2 this oscillate:

5 Motion Equations for Simple Harmonic Motion xt( ) = A cos (ω t+ φ ) dx A = x vA==−ωωφsin( t + ) max dt d 2 x aA==−ω 2 cos(ωφ t + ) dt 2 • Simple harmonic motion is one-dimensional - directions can be denoted by + or - sign • Simple harmonic motion is not uniformly accelerated motion • The sine and cosine functions oscillate between ±1. The maximum values of velocity and acceleration for an object in SHM are: k vAA==ω max m k aAA==ω 2 max m

Position and velocity of an oscillator

‰ 11.2. The figure shows the displacement of a harmonic oscillator versus time. When the motion has progressed to point A on the graph, which of the following correctly describe the position and velocity?

A) The position and velocity are both positive B) The position and velocity are both negative C) The position is negative, the velocity is zero D) The position is positive, the velocity is negative E) The position is negative, the velocity is positive

x(t) = xm cos(ωt + ϕ) dx(t) v (t) ≡ = −ωx sin(ωt + ϕ) x dt m

6 Example: Spring Oscillator in natural, undamped oscillation

Let k = 65 N/m, xm = 0.11 m at t = 0, m = 0.68 kg a) Find ω, f, T x(t) = xm cos(ωt + ϕ) ω = k/m = 65 / 0.68 = 9.78 rad / s vx (t) = −ωxm sin(ωt + ϕ) f = ω / 2π = 1.56 Hz T = 1/f = 0.64 s = 640 ms 2 ax (t) = −ω xm cos(ωt + ϕ) b) Find the amplitude o f the oscillati o ns

xm = 0.11 m c) Find the maximum speed and when it is reached

vm = amplitude of v(t) = ωxm = 9.78 × 0.11 = 1.1 m/s at t = T/4, 3T/4, etc d) Find the maximum acceleration and when it is reached 2 2 2 am = amplitude of a(t) = ω xm = (9.78) × 0.11 = 11 m/s at t = 0, T/2, same as x(t) e) Find phase constant

Match initial conditions at t = 0 : x(t) = xm = xm cos(ϕ) ⇒ cos(ϕ) = 1 ∴ ϕ = 0, + / - 2π, etc f) The formula for the motion is: x(t) = 0.11× cos(9.78t + 0)

Energy applied to simple harmonic oscillators

E = K + U = 1 mv2 + 1 kx2 constant if no mech el 2 x 2 or forces other than the spring

Recall: F = −kx (Hookes Law restoring force)

xf W = Fdx = - U = work done in stretching spring from 0 to x ∫0 el f xf 1 ∴ U = (−)(−)k xdx = kx2 el ∫0 2 f 2 Oscillator x(t) = xm cos(ωt + ϕ) where: ω = k/m solutions: vx (t) = −ωxm sin(ωt + ϕ) vx, max = ωxmax Energy: E = 1 m(−ωx )2 sin2 (ωt + φ) + 1 kx2 cos2 (ωt + φ) mech 2 max 2 max = 1 kx2 sin2 (ωt + φ) + cos2 (ωt + φ) 2 max [] this always = 1 Result: 1 2 1 2 E = kx = mv for no losses (damping) mech 2 max 2 x,max

7 Energy in SHM summary K xt() =+ A cos (ω t φ ) dx vA==−ωωφsin( t + ) dt U dx2 aA==−ω 2 cos(ω t + φ) dt 2

Simple Pendulum SHM, small amplitude oscillation of angular coordinate • Mass m swinging at the end of a massless string • Angular displacement θ(t) oscillates: − θm ≤ θ ≤ +θm • provides restoring torque: τ = − m g L sin(θ) 2 • Apply angular d θ 2 τ = Ια = Ι (force equation) Ι ≡ mL Second Law: dt2 • AlApply sma ll θ θ3 θ5 sin(θ) = θ − + − ..... ≈ θ (radians) approximation: 3! 5! mgL no ∴ α(t) ≈ − θ(t) torque mL2 2 • Another oscillator equation: torque d θ(t) g d2x(t) = − θ(t) acceleration and displacement ≡ −ω2x(t) 2 dt2 dt L have same time dependence phase constant • Solution displacement θ(t) = θ cos (ωt + φ) (radians) oscillates: at t m amplitude natural angular frequency g 2π L natural frequency ω ≡ period T = = 2π no mass dependence! L ω g

• Amplitude θm and phase constant φ specify initial state at t = 0… ... If at rest with θ = θm at t = 0, then φ = 0 • alternate: use linear s(t) ≡ L θ(t) atang(t) ≡ α(t)L = − (g/L) s(t) displacement (arc length):

8 Torsion Pendulum SHM involving angular coordinate

• A restoring torque acts when τ = −κθ (κ positive) disk is twisted from rest: • Device oscillates − θ ≤ θ ≤ +θ with amplitude: m m • Apply Second Law angular): d2θ τ = Iα = I (force equation) 2 I dt d2θ(t) κ • Oscillator equation form: = − θ(t) acceleration and displacement dt2 I have same time dependence

• Solution oscillates, as did linear oscillator: phase constant angular displacement (radians) at time t θ(t) = θm cos (ωt + φ) amplitude natural frequency Natural κ frequency ω ≡ No amplitude Note: θ(t) = displacement Ι dependence! ωt+φ = “phase” 2π Ι Does depend on φ = “phase constant” Period T = = 2π ω κ mass distribution I

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