Physics 116 ELECTROMAGNETISM AND OSCILLATORY MOTION Lecture 4 - - PowerPoint PPT Presentation

• R. J. Wilkes

Email: ph116@u.washington.edu

Physics 116

ELECTROMAGNETISM AND OSCILLATORY MOTION Lecture 4

Pendulum motion Damped SHM

Oct 4, 2011

• Homework set 1 due Thursday by 5 pm
• Suggestion: Check the New York Times Tuesdays =

science page day

• Clickers: Registration web page is open

https://catalyst.uw.edu/webq/survey/wilkes/142773

All registered 116 students got an email yesterday with this link (to your @u address) (if you are not registered for 116, you cannot access the page) if you did not get an email about this, and are registered for the course, see me “Survey questions” capture the info we need to connect your clicker ID to your name

Bring your clicker every day from now on

• Typo in formula yesterday:

Announcements

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• Now that we know a(t), we can find period T without calculus:
• We defined
• So
• This tells us that T increases for larger m, or smaller k

Period for SHM

F = ma → − kx = ma → − kx = −m Aω 2 cos ωt

( )

{ }

So − Aω 2 cos ωt

( )= −

k m ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Acos ωt

( )

ω 2 = k m ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ → ω = k m

ω = 2π f = 2π 1 T ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ T = 2π 1 ω ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ → T = 2π m k

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Also: this meant “newton-meters”, not a minus sign!

Yesterday:

Today

Lecture Schedule

(up to exam 1)

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SHM: The story so far…

2 2 2 2 2 2

2 1 ), ( sin 2 1 ), ( cos 2 1 ) cos( ) ( ) sin( ) ( ) cos( ) ( 2 1 2 2 kA K U E t kA K t kA U t A t a t A t v t A t x k m T m k T f ma kx ma F = + = = = − = − = = = → = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = = − → = ω ω ω ω ω ω ω π ω π π ω

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• A pendulum oscillates about its rest position

IF its max displacement is small, this motion is SHM

• SHM occurs only when restoring force is F = -kx
• For small angles (you decide what’s small!),

So F is proportional to θ (for larger θ, nonlinear so not exactly SHM)

Pendulum motion as example of SHM

http://paws.kettering.edu/~drussell/Demos/Pendulum/Pendula.html

θ θ θ θ mg mg ≈ → ≈ sin sin

Initial height (max angle) of bob determines E U=mgh (h=initial height relative to rest position) E swaps between K and U as with spring-mass system: K=mgh when bob is at h=0 “Galileo’s pendulum” demo illustrates this…

If m of string ~ 0 this is a simple pendulum If we have to worry about m of the string or other support, it’s a physical pendulum

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• Similar triangles for string and vertical, and for mg components
• So for small displacements, θ = s / L

Examples / applications

θ

s How long must a simple pendulum be to have period T = 2 s? (so it ticks every 1 second) g L T k m T L mg k s L mg mg Frestoring π π θ 2 2 = → = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − =

( ) ( )

( )(

) ( )

m s s m gT L g L T g L T 994 . 2 2 / 8 . 9 2 2 2

2 2 2 2 2 2 2

= = = → = = π π π π

Minus sign means: Opposite displacement s 4-Oct-11 Physics 116 - Au11 7

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• If mass of the bob’s support (string, rod, etc) is not negligible, we

have to take into account its moment of inertia

– The support itself is an extended object rotating about its pivot

• So if it has moment of inertia I about its pivot point

where l = distance of CM of support from P

Physical pendulum (review 114!)

θ

So rod has smaller T than simple pendulum of same L Notice: m doesn’t matter!

2

2 l l m I g T π = g L mL mL g L T π π 2 2

2 2

= =

For a simple pendulum, bob = a point mass ( I = m l 2), and l = L, so If the pendulum is a solid rod, ( I = m L2 / 3), and CM is located at its center ( l = L / 2) g L L m mL g L m I g T 3 2 2 2 3 2 2 2

2 2 2

π π π = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = l l

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• Physical pendulum: Hoop on a peg

– recall: parallel axis theorem

I = I (about some parallel axis) + mR2 Where R= distance between axes

• Solid disk hung on a peg

Examples / applications ( ) ( )

g D T g R mR mR g R m I g T R mR mR mR I π π π π 2 : Notice 2 2 2 2 2 2

2 2 2 2 2 center about 2

= = = = = = + = l l l So period is same as simple pendulum with L=diameter of hoop (but true only for this particular example!)

R R

( )

g R mR mR g R m I g T mR mR mR I 2 3 2 2 / 3 2 2 2 3 2 1 : Now

2 2 2 2 2 center about 2

π π π = = = = + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = l l

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• Physical pendulum: ballistic pendulum

– Method for measuring projectile speed – Shoot bullet into wood block suspended by strings 1.45 kg wood block, L=0.745 m, 0.0095 kg bullet Block rises 0.124 m Strategy: use energy considerations to find initial speed of block Use momentum conservation (review114: inelastic collisions) to find v0

Examples / applications

Here, v = speed of bullet+block just after collision

block block block bullet

U K K K const K U E = → = → = + =

Always! At time of collision ( initial ) (at max height)

s m m s m gh v v M m K gh M m U / 56 . 1 124 . / 8 . 9 ( 2 2 ) ( 2 1 ) (

2 2

= = = + = = + = v0 m/s 240 kg 0.0095 m/s 1.56 ) kg 1.45 kg 0.0095 ( ) ( ) ( = + = + = + = m v M m v v M m mv = Conservation of momentum in the collision (about 860 km/hr)

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• Spring-mass vs pendulum

– Mass m stretches spring (constant k) by distance L when attached – What is period of mass-spring system?

Examples / applications So spring-mass system has same period as a simple pendulum with L = displacement of mass when attached to spring

g L k m T g L k m mg kL mg kx F π π 2 2

spring

= = = → = → = =

L

m

k

m

L

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• If friction is significant, total energy is not constant

E = U + K = ½ kA2 diminishes with time: oscillation gets damped out Common case: friction force is proportional to speed (example: air resistance) For this case, amplitude A decreases exponentially with time – If damping is too great, we never get even the first half cycle – If damping is just right (critical damping) we get just the 1st half cycle – Critical damping is a desirable feature in many applications

• For example, car springs: shock absorbers (= dampers) are set to

provide critical damping for a comfortable ride

Damped oscillations

v b Fdamping − =

) 2 / exp( ) ( ) ( m bt t A t A − =

The exponential function imposes an envelope on the oscillations

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Damped oscillations

• 1.5
• 1
• 0.5

0.5 1 1.5 0.1 0.2 0.3 0.4 0.5

t / T = fraction of period displacement

exp(-bt/2m)

• scillator
• 1.5
• 1
• 0.5

0.5 1 1.5 0.5 1 1.5

t / T = fraction of period displacement

exp(-bt/2m)

• scillator
• 1.5
• 1
• 0.5

0.5 1 1.5 0.1 0.2 0.3 0.4 0.5

t / T = fraction of period displacement

exp(-bt/2m)

• scillator
• Underdamped (oscillates

several cycles)

• Overdamped (quits

immediately)

• Critically damped

(smooth return to zero)

Exponential “envelope” Oscillator motion 4-Oct-11 Physics 116 - Au11 13