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Home , Oscillation, Power (physics), Simple harmonic motion

Chapter 15

Oscillations and and Waves

• Simple • in SHM • Some Oscillating Systems • Damped Oscillations • Driven Oscillations •

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 2

Simple harmonic motion (SHM) occurs when the restoring (the force directed toward a stable equilibrium point ) is proportional to the from equilibrium.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 3 Characteristics of SHM

• Repetitive motion through a central equilibrium point. • Symmetry of maximum displacement. • Period of each cycle is constant. • Force causing the motion is directed toward the equilibrium point (minus sign). • F directly proportional to the displacement from equilibrium. = - ω2 x Displacement

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 4 A Simple (SHO)

Frictionless surface

The restoring force is F = −kx.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 5 Two Springs with Different

Frictionless surface

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 6 SHO Period is Independent of the

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 7 The Period and the Angular

2π T = . The period of is ω

where ω is the of k the oscillations, k is the ω = m constant and m is the of the block.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 8 Simple Harmonic Motion

At the equilibrium point x = 0 so, a = 0 also. When the stretch is a maximum, a will be a maximum too.

The at the end points will be zero, and it is a maximum at the equilibrium point.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 9 Representing Simple Harmonic Motion

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 10 Representing Simple Harmonic Motion

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 11 Representing Simple Harmonic Motion

Position - x max = A

Velocity - v max = ωA

2 Acceleration - a max = ω A

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 12 A simple harmonic oscillator can be described mathematically by:

x() t = Acos ωt dx v() t = = -A ωsin ωt dt dv a() t = = -A ω2cos ωt dt where A is the amplitude of the motion, the maximum Or by: displacement from equilibrium, A ω = v , and x() t = Asin ωt max Aω2 = a dx max. vt=() =A ωcos ωt dt dv a() t = = -A ω2 sin ωt dt

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 13 - Circular Functions

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 14 Projection of

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 15 Circular Motion is the superposition of two linear SHO that are 90 0 out of phase with each other

y t A = sin(ω )

x t A = cos(ω )

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 16 Shifting Trig Functions

sin  The minus sign means that the x = A  ω t - φ  cos  phase is shifted to the right. sin t  A plus sign indicated the phase x=A{} 2 π - ϕ  cos T  is shifted to the left

π  x = Asinω t - 2  π π x= A() sin ωt cos2 - sin 2 cos ωt () x= A sin ωt (0) - (1)cos ωt x = -Acos ωt

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 17 Shifting Trig Functions

π  sinω t-  =0 Shifted Trig Functions 2  1.50

sin( ωt) π 1.00 ω δ ωt- =0 sin( t- ) 2 0.50 π ωt = 0.00 2 -3.00 -2.00 -1.00 0.00 1.00 2.00 3.00 -0.50 π 1 1 T t= ; = -1.00 2ω ω 2π π T T -1.50 t = = 2 2 π 4

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 18 Energy

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 19 Equation of Motion & Energy

Assuming the table is frictionless:

F = - kx = ma ∑ x x

k 2 Classic form for SHM a() t=- xt=-()() ω x t x m

12 1 2 Also, EKt=+=()() Ut mvt() + kxt() 2 2

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 20 Spring

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 21 Spring Total Energy

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 22 Approximating Simple Harmonic Motion

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 23 Approximating Simple Harmonic Motion

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 24 Potential and

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 25 The period of oscillation of an object in an ideal mass-spring system is 0.50 sec and the amplitude is 5.0 cm. What is the at the equilibrium point?

At equilibrium x = 0:

1 1 1 E = K +U = mv 2 + kx 2 = mv 2 2 2 2

Since E = constant, at equilibrium (x = 0) the

KE must be a maximum. Here v = v max = A ω.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 26 Example continued:

The amplitude A is given, but ω is not.

2π 2π ω = = =12 .6 rads/sec T 0.50 s

and v = Aω = (5.0 cm 12 )(.6 rads/sec = 62 .8 )cm/sec

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 27 The diaphragm of a speaker has a mass of 50.0 g and responds to a of 2.0 kHz by moving back and forth with an amplitude of 1.8 ×10 −4 m at that frequency.

(a) What is the maximum force acting on the diaphragm?

2 2 2 2 ∑ F = Fmax = ma max = m()Aω = mA ()2πf = 4π mAf

The value is F max =1400 N.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 28 Example continued:

(b) What is the of the diaphragm?

Since mechanical energy is conserved, E = K max = U max.

1 The value of k is unknown so use K . U = kA 2 max max 2 1 1 1 1 2 2 2 2 2 K = mv Kmax = mv max = m()Aω = mA ()2πf max 2 max 2 2 2

The value is K max = 0.13 J.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 29 Example: The displacement of an object in SHM is given by:

y()()()t = 00.8 cm sin []57.1 rads/sec t

What is the frequency of the oscillations?

Comparing to y(t) = A sin ωt gives A = 8.00 cm and ω = 1.57 rads/sec. The frequency is:

ω 57.1 rads/sec f = = = .0 250 Hz 2π 2π

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 30 Example continued: Other quantities can also be determined:

2π 2π The period of the motion is T = = = 00.4 sec ω 57.1 rads/sec

xmax = A = 8.00 cm

vmax = Aω = (8.00 cm 1.57 )( rads/sec =12 .6 cm/sec ) 2 2 2 amax = Aω = (8.00 cm 1.57 )( rads/sec =19 .7 ) cm/sec

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 31 What About ?

When a mass-spring system is oriented vertically, it will exhibit SHM with the same period and frequency as a horizontally placed system.

The effect of gravity is canceled out.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 32 Why We Ignore Gravity with Vertical Springs

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 33 The Simple

A simple pendulum is constructed by attaching a mass to a thin rod or a string. We will also assume that the amplitude of the oscillations is small.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 34 The Simple Pendulum

The pendulum is best described using polar coordinates. The origin is at the pivot point. The coordinates are (r, φ). The r-coordinate points from the origin along the rod. The φ- coordinate is perpendicular to the rod and is positive in the counter clockwise direction.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 35 F= − mgsin φ = ma φ Apply ’s 2 nd ∑ φ Law to the pendulum v2 bob. ∑ Fr = T − mgcos φ = m r

If we assume that φ <<1 rad, then sin φ ≈ φ and cos φ ≈1, the angular frequency of oscillations is then:

∑ Fφ =− mgsin α φ = maφ = mL −mgsin αφ = mL g φ α = − (g / L )sin ω = L φα = − (g / L ) 2π L The period of oscillations is T = = 2π ω g

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 36 Example: A has a pendulum that performs one full swing every 1.0 sec. The object at the end of the string weighs 10.0 N. What is the length of the pendulum?

L T = 2π g

gT 2 ()9.8 m/s 2 ()1.0 s 2 L = = = 0.25 m Solving for L: 4π 2 4π 2

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 37 The gravitational potential energy of a pendulum is U = mgy. Taking y = 0 at the lowest point of the swing, show that y = L(1-cos θ).

θ Lcos θ L L

y = L 1( − cos θ ) y=0

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 38 The Physical Pendulum

A physical pendulum is any rigid object that is free to oscillate about some fixed axis. The period of oscillation of a physical pendulum is not necessarily the same as that of a simple pendulum.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 39 The Physical Pendulum

I T = 2 π MgD

I is the of about

the given axis. The I cm from the table will need to be modified using the parallel axis theorem.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 40 Compound Pendulum I T = 2 π MgD

I = I rod + I disk

M = m rod + M disk

D = from the axis to the center of mass of the rod and disk.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 41 Damped Oscillations

When dissipative such as are not negligible, the amplitude of oscillations will decrease with time. The oscillations are damped.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 42 Damped Oscillations Equations dx2 dx m +b +m ω x = 0 dt2 dt 0 -t  x(t) = A exp cos(ω' t + δ) 0 2τ 

2 ' b  k m ω = ω0 1 -   ω0 = τ = ; bc = 2m ω 0 2m ω0  m b

For b > b c the system is overdamped. For b = b c the system is critically damped. The object doesn’t oscillate and returns to its equilibrium posion very rapidly.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 43 Damped Oscillations

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 44 Graphical representations of damped oscillations:

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 45 Damped Oscillations

•Overdamped : The system returns to equilibrium without oscillating. Larger values of the the return to equilibrium slower.

•Critically damped : The system returns to equilibrium as quickly as possible without oscillating. This is often desired for the damping of systems such as doors. •Underdamped : The system oscillates (with a slightly different frequency than the undamped case) with the amplitude gradually decreasing to zero.

Source: Damping @ Wikipedia

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 46 Damped Oscillations

The larger the damping the more difficult it is to assign a frequency to the oscillation.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 47 Damped Oscillations

E α A 2

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 48 Forced Oscillations

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 49 Forced Oscillations and Resonance

A force can be applied periodically to a damped oscillator (a forced oscillation).

When the force is applied at the natural frequency of the system, the amplitude of the oscillations will be a maximum. This condition is called resonance .

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 50 Forced Oscillations Equations dx2 dx m +b +m ω x = F cos ωt dt2 dt 0 0

ma friction spring applied force

x = Acos(ω δ t - )

F bω A = 0 tan δ = 2 2 22 22 m( ω2- ω 2 ) m ( ω0 - ω ) + b ω 0

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 51 Energy and Resonance d2 x dx m +b +m ω x = F cos ωt dt2 dt 0 0

At resonance v and F o are in phase dx v= =- ωAsin( ωt - δ) x dt π v = - ωAsin( ωt - ) = + ω Acos t x 2 -t  Energy α A2 = A 2 exp 0 τ  ω m Q = ω τ = 0 -t  0 E =1 m ω2A 2 = E exp b 2 0 τ  1 2 2 E0 =2 m ω A 0 ; τ = m b

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 52 Transfer ω Q = 0 ∆ω

The in the system, represented by “b” keeps the amplitude from going to infinity.

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 53 Tacoma Narrows Bridge

Nov. 7, 1940

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 54 Tacoma Narrows Bridge

Nov. 7, 1940

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 55 Tacoma Narrows Bridge

The first Tacoma Narrows Bridge opened to traffic on July 1, 1940. It collapsed four months later on November 7, 1940, at 11:00 AM (Pacific time) due to a physical phenomenon known as aeroelastic flutter caused by a 67 kilometres per hour (42 mph) wind. The bridge collapse had lasting effects on science and engineering. In many undergraduate physics texts the event is presented as an example of elementary forced resonance with the wind providing an external periodic frequency that matched the natural structural frequency (even though the real cause of the bridge's failure was aeroelastic flutter[1]). Its failure also boosted research in the field of bridge / aeroelastics which have themselves influenced the designs of all the world's great long-span bridges built since 1940. - Wikipedia http://www.youtube.com/watch?v=3mclp9QmCGs

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 56

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 57 End of Chapter Problems

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 58 Chap 14 - #92

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 59 Extra Slides

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 60 The Full Equation

∂2y 1 ∂ 2 y - = 0 ∂x2 v 2 ∂ t 2

y(t) = Asin(kx - ωt) x t   y(t) = Asin 2 π - 2π 2π    k = ; ω = 2 πf = λ T   λ T Traveling with the wave the phase is constant

x t dx dt - =Constant - = 0 λ T λ T dx λ = = λ f = v Wave velocity dt T

MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 61