Chapter 13• Hooke’s Law: F = - kx• Periodic & Simple Harmonic Motion • Springs & Pendula• Waves• Superposition Next Week!

Review Physics 2A:Springs, Pendula & Circular Motion

Elastic SystemsSmall Vibrations

F kx= −

Natural Frequency of Objects& Resonance

All objects have a natural frequency of vibration or oscillation. Bells, tuning forks, bridges, swings and atoms all have a natural frequency that is related to their size, shape and composition. A system being driven at its natural frequency will resonate and produce maximum amplitude and energy.

Coupled OscillatorsMolecules, atoms and particles are modeled as coupled oscillators.

Waves Transmit Energy through coupled oscillators. Forces are transmitted between the oscillators like springs

Coupled oscillators make the medium.

The Oscillation of Nothing

Hooke’s Law

An elastic system displaced from equilibrium oscillates in a simple way about its equilibrium position with

Simple Harmonic Motion.

Hooke’s Law describes the elastic response to an applied force.

Elasticity is the property of an object or material which causesit to be restored to its original shape after distortion.

Ut tensio, sic vis - as the extension, so is the force

Robert Hooke (1635-1703)

•Leading figure in Scientific Revolution•Contemporary and arch enemy of Newton•Hooke’s Law of elasticity•Worked in Physics, Biology, Meteorology, Paleontology•Devised compound microscope•Coined the term “cell”

Hooke’s LawIt takes twice as much force to stretch a spring twice as far.

The linear dependence of displacement upon stretching force:

appliedF kx=

Hooke’s LawStress is directly proportional to strain.

( ) ( )appliedF stress kx strain=

Hooke’s Law

•The applied force displaces the system a distance x.

•The reaction force of the spring is called the “Restoring Force” and it is in the opposite direction to the displacement.

RestoringF kx= −

An Ideal SpringspringF kx= −

An ideal spring obeys Hooke’s Law and exhibits Simple Harmonic Motion.

Spring Constant k: Stiffness

•The larger k, the stiffer the spring•Shorter springs are stiffer springs•k strength is inversely proportional to the number of coils

Spring Question

Each spring is identical with the same spring constant, k.Each box is displaced by the same amount and released.Which box, if either, experiences the greater net force?

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Hooke’s Law: F = - k x

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Hooke’s Law: F = - k x

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Hooke’s Law: F = - k x

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Hooke’s Law: F = - k x

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Hooke’s Law: F = - k x

Periodic MotionPosition vs Time: Sinusoidal Motion

1: (# / sec), [ ]Frequency f cycles f Hz= =Τ

Terms:

Displacementof Mass

: / , [ ] secPeriod T time cycle T= =

: [ ]Amplitude A m=

2 : 2 , [ ] /Angular Frequency f rad sπω π ω= = =Τ

SHM and Circular Motion

( ) cos ( )x t A tθ=

http://www.edumedia-sciences.com/en/a270-uniform-circular-motion

( ) sinv t A tω ω= − 2( ) cosa t A tω ω= −

222 , , t t c

R vv R a R a RR

πω α ω= = = = =Τ

Simple Harmonic Motion

( ) cos ( )x t A tθ=

Displacementof Mass

( )2

t t tθπ

ω ==Τ

Simple Harmonic Motion( ) cosx t A tω=

Displacementof Mass

2π

ω =Τ

2a xω= −

Notice:

( ) sinv t A tω ω= −2( ) cosa t A tω ω= −

Newton’s 2nd Law?F ma=

2( )m xω= −2k mω=

km

ω =

kx=−

( ) cosx t A tω=

2( ) cosa t A tω ω= −2a xω= −

angularfrequency

2π

ω =Τ

( ) sinv t A tω ω= −

2 mTk

π=

Simple Harmonic Motionkm

ω =2T πω

=

Does the period depend on the displacement, x?

The period depends only on how stiff the spring is and how much inertia there is.

Finding The Spring ConstantA 2.00 kg block is at rest at the end of a horizontal spring on a frictionless surface. The block is pulled out 20.0 cm and released. The block is measured to have a frequency of 4.00 hertz. a) What is the spring constant?

1 12 2

kfm

ωπ π

= = =Τ

2 24k f mπ=

22 2

24 (4 / ) (2 ) mk s kgm

π= 21263 /N m=

( ) sinv t A tω ω= −2( ) cosa t A tω ω= −

( ) cosx t A tω=

Harmonic Motion Described

k2 , m

mTk

π ω= =

restoringF kx= −

sinv A tω ω= −

2 cosa A tω ω= −

max

@ t2odd

v A

n

ωπω

=

=

2max

@a A

t nω

ω π==

Maximum Velocity and Acceleration

Simple Harmonic MotionConsider a m = 2.00 kg object on a spring in a horizontal frictionless plane. The sinusoidal graph represents displacementfrom equilibrium position as a function of time. A) What is the amplitude of motion?B) What is the period, T? frequency, f ?C) What is the equation representing the displacement?

A) A = . 08 m

B) T = 4s f = 1/T = .25Hz2.08 cos4

m tsπ

= .08 cos2

m tπ=C)

2( ) cosx t A tπ=

Τ

Simple Harmonic MotionConsider a m = 2.00 kg object on a spring in a horizontal plane.The sinusoidal graph represents displacement from equilibrium position as a function of time.

E) What is the acceleration of the object at t = 1s? zero!

sinv A tω ω= −

( ) .08 cos2

x t m tπ=

maxv Aω= − 2A π= −

Τ2.084

msπ

= −

max .13v m= −

D) What is the speed of the object at t = 1s?

Simple Harmonic MotionConsider a m = 2.00 kg object on a spring in a horizontal plane.The sinusoidal graph represents displacement from equilibrium position as a function of time.

F) What is the spring constant?

( ) .08 cos2

x t m tπ=

2 mTk

π=2

2

4 mk π=

Τ

2

2

4 (2 )(4 )

kgs

π=

24.94 /k N m=

Finding Maximum Velocity

( ) sinv t A tω ω= − maxv Aω=kAm

=

A 2.00 kg block is at rest at the end of a horizontal spring on a frictionless surface. The block is pulled out 20.0 cm and released. The block is measured to have a frequency of 4.00 hertz. a) What is the spring constant? b) What is the maximum velocity?

2

max1263 /(.02 )

2N mv mkg

= 0.5 /m s=

21263 /k N m=

SHM QuestionThe position of a simple harmonic oscillator is given by

where t is in seconds.

a) What is the period of this oscillator ?b) What is the maximum velocity of this oscillator?

x t t m ( ) ( . ) cos= ⎛⎝⎜

⎞⎠⎟

0 53π

( ) sinv t A tω ω= −

23π πω= =

Τ( ) cosx t A tω= 6sΤ =a)

b) maxv Aω= 2A π=

Τ

max20.56

v msπ

= 0.52 /m s=

Energy in a Spring

212springKE mv=

212springPE kA=

Mechanical Energy is Conserved!

Energy in a Spring: Quiz QuestionWhat speed will a 25g ball be shot out of a toy gun if the spring (spring constant = 50.0N/m) compressed 0.15m? Ignore friction and the mass of the ball.

Use Energy!

spring ballPE KE=

2 21 12 2

kx mv=

kv xm

=

50.0 / (.15 ) 6.7 /.025

N mv m m skg

= =

max

Avω=

=

Note: this is also

Spring QuestionA 1.0-kg object is suspended from a spring with k = 16 N/m. The mass is pulled 0.25 m downward from its equilibrium position and allowed to oscillate. What is the maximum kinetic energy of the object?

Maximum kinetic energy happens at the equilibrium position.

( ) sinv t A tω ω= − maxv Aω= kAm

=

2max max

12

KE mv=2

12

km Am

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

212

A k= axmPE=0.5J=

2max max

12

KE mv=

0gPE =

Simple PendulumFor small angles, simple pendulums exhibit Simple

Harmonic Motion:

Simple PendulumFor small angles, simple pendulums exhibit Simple

Harmonic Motion:

2 Lg

πΤ =

sinF mg mgθ θ= − ≈ −

/s Lθ =

/F mgs L kx≈ − → −

/ , k mg L x s= =

2 mT kπ=

For angles less than 15 degrees:

Restoring Force:

Moon Clock

At what rate will a pendulum clock run on the moon where gM = 1.6m/s2, in hours?

2EarthE

Lg

πΤ = 2MoonM

Lg

πΤ =

Divide:

2

2

MMoon

Earth

E

LgLg

π

π

Τ=

ΤE

Moon EarthM

gg

Τ = Τ 2.45h=