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Example: Prove that the angular of a vertical with a spring constant k and a hanging m is still given by k ω = m

Physics 106 Lecture 12 – II SJ 7th Ed.: Chap 15.4, Read only 15.6 & 15.7

• Recap: SHM using phasors (uniform circular ) • Phys ica l pendul um exampl e • Damped harmonic oscillations • Forced oscillations and . • Resonance examples and discussion – music – structural and mechanical engineering – • Sample problems • Oscillations summary chart

1 Goals

Oscillations in the presence of

Oscillations in the presence of oscillating external

Damped Oscillations

• Non-conservative may be present – is a common nonconservative force – No longer an ideal system (such as those dealt with so far) • The mechanical of the system diminishes in , motion is said to be damped • The motion of the system can be decaying oscillations if the damping is “weak”.

• If damping is “strong”, motion may die away without oscillating. • Still no driving force, once system has been started

2 Add Damping: Emech not constant, oscillations not simple • Spring oscillator as before, but with dissipative force

Fdamp neglect such as the system in the figure, with vane moving in fluid.

Fdamp viscous drag force, proportional to

Fdamp = −bv

• Previous force equation gets a new, damping force term d2x(t) dx(t) Fnet = m = − kx()(t) − b dt2 dt

new term d2x(t) b dx(t) k + = − x(t) dt2 m dt m

Solution for Damped oscillator equation

new term d2x(t) b dx(t) k + = − x(t) dt2 m dt m

bt − 2 Solution: 2m k b x(t) = xme cos(ω't + φ) ω'≡ − modified m 2 oscillations exponentially altered ω’ can be real 4m decaying envelope frequency or imaginary k ω = : natural frequency 0 m

22 ωω'(/2)≡−0 bm

• Recover undamped solution for b Æ 0

3 Damped physical systems can be of three types

bt − 2 Solution: 2m k b x(t) = xme cos(ω't + φ) ω'≡ − damped m 4m2 oscillations

UdUnder damped : small bk< 2 m bk2 < , for which ω is positive. 4mm2

Critically damped: bkm= 2 2 b k 2 ≈ ≡ ω0 for which ω'≈ 0 4m2 m 2 Overdamped: b k 2 > ≡ ω0 for which ω' is imaginary 4m2 m Math Review: cos(ix )==+ cosh( x ) ( exx e− ) / 2 sin(ix )==− sinh( x ) ( exx e− ) / 2 cos(ix+ y )=− cos( ix )cos( y ) sin( ix )sin( y )

Types of Damping, cont.

a) an underdamped oscillator b) a critically damped oscillator c) an overdamped oscillator

For critically damped and overdamped oscillators there is no periodic motion and the angular frequency ω has a different meaning

4 b2 k << ≡ ω2 Weakly damped oscillator : 4m2 m 0

bt − kb2 − 2m ω ' ≡≈ω0 xt()= xem cos(ω0 t+ϕ ) m 4m2

bt - 2m Xm = xm()(t) ≈ xme slow decay of amplitude envelope

≈ cos(ω0t + φ) small fractional change in amplitude during one complete cycle

b2 k << ≡ ω2 Weakly damped oscillator : 4m2 m 0 bt kb2 − ω ' ≡≈− ω0 xt()= xe2m cos(ω t+ϕ ) m 4m2 m 0 bt - Amplitude : X = A 2m m xm(t) ≈ xme

small fractional slow decay change in amplitude of amplitude envelope during one complete cycle

≈ cos(ω0t + φ)

Velocity with weak damping: find derivative bt − d 2m maximum velocity v(t) = x(t) ≈ vme sin(ω't + φ) dt vm = − ω0xm

altered exponentially frequency ~ ω decaying envelope 0

5 Mechanical energy decays exponentially in an “weakly damped” oscillator (small b)

E = K(t) + U(t) = 1 mv2(t) + 1 kx2(t) mech 2 2

bt − 2m xt()=+ xem cos(ω0 t ϕ ) Velocity with weak damping: find derivative bt − d 2m maximum velocity v(t) = x(t) ≈ vme sin(ω't + φ) dt vm = − ω0xm

altered exponentially frequency ~ ω decaying envelope 0

bt − ⎛⎞b 2m xem ⎜⎟−+cos(ω0 tϕ ) term is negligible, because b is small.. ⎝⎠2m

Mechanical energy decays exponentially in an “weakly damped” oscillator (small b)

E = K(t) + U(t) = 1 mv2(t) + 1 kx2(t) mech 2 2 Substitute previous solutions: bt bt − − 2m 2m x(t) = xme cos(ω't + φ) v(t) ≈ −ω0xme sin(ω't + φ)

E = 1 mω2x2 e−bt / msin2(ω't + φ) mech 2 0 m + 1 kx2 e−bt / mcos2(ω't + φ) 2 m As always: cos2(x) + sin2(x) = 1

2 k Also: ω ≡ 0 m ∴ E (t) = 1 kx2 e−bt / m mech 2 m

exponential decay at twice Initial mechanical energy the of amplitude decay

6 Damped physical systems can be of three types

bt − 2 Solution: 2m k b x(t) = xme cos(ω't + φ) ω'≡ − damped m 2 exponentially altered ω’ can be real 4m oscillations decaying envelope frequency or imaginary

2 Underdamped: b k 2 << ≡ ω0 for which ω'≈ ω0 4m2 m ƒ The restoring force is large compared to the damping force. ƒ The system oscillates with decaying amplitude 2 Critically damped: b k 2 ≈ ≡ ω0 for which ω'≈ 0 4m2 m ƒ The restoring force and damping force are comparable in effect. ƒ The system can not oscillate; the amplitude dies away exponentially 2 Overdamped: b k 2 > ≡ ω0 for which ω' is imaginary 4m2 m ƒ The damping force is much stronger than the restoring force. ƒ The amplitude dies away as a modified exponential ƒ Note: Cos( ix ) = Cosh( x )

Forced (Driven) Oscillations and Resonance ƒ An external driving force starts oscillations in a stationary system ƒ The amplitude remains constant (or grows) if the energy input per cycle exactly equals (or exceeds) the energy loss from damping

ƒ Eventually, Edriving = Elost and a steady-state condition is reached ƒ Oscillations then continue with constant amplitude

ƒ Oscillations are at the driving frequency ωD

FD(t) = F0 cos(ωDt + φ')

FD(t) Oscillating driving force applied to a dampe d osc illat or

7 Equation for Forced (Driven) Oscillations k ω = natural frequency ω = 0 0 m ωD = driving frequency of external force External driving force function:

FD(t) = F0 cos(ωDt + φ') dx() t d2 x () t FFt==( ) -b - k x(t) m net D dt dt 2 FD(t)

Solution for Forced (Driven) Oscillations dx() t d2 x () t FFt==( ) -b - k x(t) m net D dt dt 2

FD(t) = F0 cos(ωDt + φ')

Solution (steady state solution):

x(t) = Acos(ωDt + φ)

F0 /m where A = 2 2 2 bωD 2 (ωD − ω0 ) + ( ) m FD(t)

The system always oscillat es at the driving frequency ωD in steady-state The amplitude A depends on how k ω ω ω = close D is to natural frequency 0 0 m “resonance”

8 Amplitude of the driven oscillations: F /m A = 0 bω (ω2 − ω2 )2 + ( D )2 D 0 m ƒ The largest amplitude oscillations occur at or resonance

near RESONANCE (ωD ~

ω0) As damping becomes weaker Æ resonance sharpens & amplitude at resonance increases.

Resonance

ƒ At resonance, the applied force is in phase with the velocity and the Fov transferred to the oscillator is a maximum. ƒ The amplit ud e of resonant oscillations can become enormous when the damping is weak, storing enormous amounts of energy

Applications: • buildings driven by earthquakes • bridges under wind load • all kinds of radio devices, microwave • other numerous applications

9 Forced resonant torsional oscillations due to wind - Tacoma Narrows Bridge

Roadway collapse - Tacoma Narrows Bridge

10 Twisting bridge at resonance frequency

Breaking glass with voice

Another Breaking glass with voice

Yet another breaking glass with voice

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