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PHY 231 1 PHYSICS 231 Lecture 34: Oscillations & Waves Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom Period T 6 3 Frequency f 1/6 1/3 ½ (m/k) 6/(2) 3/(2) 2/(2 (2)/6 (2)/3 (2 k m f T 2 2 1

PHY 231 1 PHYSICS 231 Lecture 34: Oscillations & Waves Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom Period T 6 3 2 Frequency

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PHY 2311

PHYSICS 231Lecture 34: Oscillations & Waves

Remco ZegersQuestion hours: Thursday 12:00-13:00 & 17:15-

18:15Helproom

Period T 6 3 2Frequency f 1/6 1/3 ½(m/k) 6/(2) 3/(2) 2/(2) (2)/6 (2)/3 (2)/2

k

m

fT

221

PHY 2312

Harmonic oscillations vs circular motion

t=0 t=1 t=2

t=3 t=4

v0=r=A

v0

=t=t

A

v0

vx

PHY 2313

time (s)

A

-A

-kA/m

kA/m

velocity v

a

x

A(k/m)

-A(k/m)

xharmonic(t)=Acos(t)

vharmonic(t)=-Asin(t)

aharmonic(t)=-2Acos(t)

=2f=2/T=(k/m)

PHY 2314

Another simple harmonic oscillation: the pendulum

Restoring force: F=-mgsinThe force pushes the mass mback to the central position.

sin if is small (<150) radians!!!

F=-mg also =s/Lso: F=-(mg/L)s

PHY 2315

pendulum vs spring

parameter spring pendulum

restoring force F

F=-kx F=-(mg/L)s

period T T=2(m/k)T=2(L/g)*

frequency f f=(k/m)/(2)

f=(g/L)/(2)

angular frequency

=(k/m) =(g/L)

*

g

L

Lmg

mT 2

/2

PHY 2316

example: a pendulum clock

The machinery in a pendulum clock is keptin motion by the swinging pendulum.Does the clock run faster, at the same speed,or slower if:a) The mass is hung higherb) The mass is replaced by a heavier massc) The clock is brought to the moond) The clock is put in an upward accelerating

elevator?

L m moon elevator

faster

same

slower g

LT 2

PHY 2317

example: the height of the lecture room

demo

22

2

25.04

2

TgT

L

g

LT

PHY 2318

damped oscillations

In real life, almost all oscillations eventually stop due to frictional forces. The oscillation is damped. We can alsodamp the oscillation on purpose.

PHY 2319

Types of damping

No dampingsine curve

Under dampingsine curve with decreasingamplitude

Critical dampingOnly one oscillations

Over dampingNever goes through zero

PHY 23110

Waves

The wave carries the disturbance, but not the water

Each point makes a simple harmonic vertical oscillation

position x

position y

PHY 23111

Types of waves

Transversal: movement is perpendicular to the wave motion

wave

oscillation

Longitudinal: movement is in the direction of the wave motion

oscillation

PHY 23112

A single pulse

velocity v

time to time t1

x0 x1

v=(x1-x0)/(t1-t0)

PHY 23113

describing a traveling wave

While the wave has traveled onewavelength, each point on the ropehas made one period of oscillation.

v=x/t=/T= f

: wavelengthdistance betweentwo maxima.

PHY 23114

example2m

A traveling wave is seento have a horizontal distanceof 2m between a maximumand the nearest minimum andvertical height of 2m. If itmoves with 1m/s, what is its:a) amplitudeb) periodc) frequency

2m

a) amplitude: difference between maximum (or minimum) and the equilibrium position in the vertical direction (transversal!) A=2m/2=1m

b) v=1m/s, =2*2m=4m T=/v=4/1=4sc) f=1/T=0.25 Hz

PHY 23115

sea waves

An anchored fishing boat is going up and down with thewaves. It reaches a maximum height every 5 secondsand a person on the boat sees that while reaching a maximum, the previous waves has moves about 40 m awayfrom the boat. What is the speed of the traveling waves?

Period: 5 seconds (time between reaching two maxima)Wavelength: 40 m

v= /T=40/5=8 m/s

PHY 23116

Speed of waves on a string

LM

Fv

/

F tension in the string mass of the string per unit length (meter)

example: violin

L M

screwtension T

v= /T= f=(F/)

so f=(1/)(F/) for fixed wavelength the frequency willgo up (higher tone) if the tension is increased.

PHY 23117

example

A wave is traveling through thewire with v=24 m/s when thesuspended mass M is 3.0 kg.a) What is the mass per unit length?b) What is v if M=2.0 kg?

a) Tension F=mg=3*9.8=29.4 N v=(F/) so =F/v2=0.05 kg/m b) v=(F/)=(2*9.8/0.05)=19.8 m/s

PHY 23118

bonus ;-)

The block P carries out a simple harmonic motion with f=1.5HzBlock B rests on it and the surface has a coefficient ofstatic friction s=0.60. For what amplitude of the motion doesblock B slip?

The block starts to slip if Ffriction<Fmovement

sn-maP=0smg=maP so sg=aP

ap= -2Acos(t) so maximally 2A=2fAsg=2fA A= sg/2f=0.62 m