Physics 201: Lecture 1

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Transcript Physics 201: Lecture 1

Physics 201: Chapter 14 – Oscillations (cont’d)

 General Physical Pendulum & Other Applications  Damped Oscillations  Resonances 4/30/2020 Physics 201, UW-Madison 1

4/30/2020

Simple Harmonic Motion: Summary

 

k m

Force:

d

2

s

   2

s dt

2

k k m m s 0 0 s

Solution:

θ =

A

cos(

t +

)

 

g L

 

L

Physics 201, UW-Madison 2

Question 1

The amplitude of a system moving with simple harmonic motion is doubled. The total energy will then be 4 times larger 2 times larger the same as it was half as much quarter as much U

at 

x

1  2

kx

2

A

,

v

  1 2

mv

2 0

U

 1 2

kA

2

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Question 2

A glider of mass m is attached to springs on both ends, which are attached to the ends of a frictionless track. The glider moved by 0.2 m to the right, and let go to oscillate. If m = 2 kg, and spring constants are k 1 = 800 N/m and k 2 N/m, the frequency of oscillation (in Hz) is = 500 approximately 6 Hz 2 Hz 4 Hz 8 Hz 10 Hz f

  2  

k

/ 2 

m

 800  500 2 2   4 Hz

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The harmonic oscillator is often a very good approximation for (non harmonic) oscillations with small amplitude:

U

1 2

kx

2

for small x

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General Physical Pendulum

4/30/2020   Suppose we have some arbitrarily shaped solid of mass

M

hung on a fixed axis, and that we know where the CM is located

and

what the moment of inertia

I

= Mg d Mg

R  about the axis is.

The torque about the rotation (

z

) axis for small  is (using sin   ) 

MgR

   I 

d

2 

dt

2

z-axis

R xCM d d

2     2 

dt

2    0 cos  

t where

    

MgR

I

Mg

Physics 201, UW-Madison 6

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Torsional Oscillator

  Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia

I

about this axis is known. The wire acts like a “rotational spring.”  When the object is rotated, the wire is twisted. This produces a torque that opposes the rotation.

 In analogy with a spring, the torque produced is proportional to the displacement: 

= -

k  

I wire

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Torsional Oscillator…

 Since 

= -

k   

= I

  becomes 

k

 

I d

2 

dt

2

wire d

2 

dt

2    2 

where

 

k

I 

I This is similar to the “mass on spring” Except I has taken the place of m (no surprise).

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Damped Oscillations

  In many real systems, nonconservative forces are present  The system is no longer ideal  Friction/drag force are common nonconservative forces In this case, the mechanical energy of the system diminishes in time, the motion is said to be

damped

  The amplitude decreases with time The blue dashed lines on the graph represent the

envelope

motion of the

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Damped Oscillation, Example

    One example of damped motion occurs when an object is attached to a spring and submerged in a viscous liquid The retarding force can be expressed as

F

= b v where

b

is a constant 

b

is called the

damping coefficient

The restoring force is –

kx

From Newton’s Second Law 

F x

= -

k x

bv x

=

ma x or, 4/30/2020 Physics 201, UW-Madison 10

Damped Oscillation, Example

When b is small enough, the solution to this equation is: with (This solution applies when b<2mω 0 .)

Time constant: 4/30/2020 Physics 201, UW-Madison 11

Energy of weakly damped harmonic oscillator:

 The energy of a weakly damped harmonic oscillator decays exponentially in time.

The potential energy is ½kx 2 : 4/30/2020 Time constant is the time for E to drop to 1/e .

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Types of Damping

      0 is also called the

natural frequency

of the system If

F

max =

bv

max <

kA

, the system is said to be

underdamped

When

b

oscillate reaches a critical value

b c

such that

b c

/ 2 m =  0 , the system will not  The system is said to be

critically damped

If

F

max =

bv

max >

kA

and

b

/2

m

>  0 , the system is said to be

overdamped

For critically damped and overdamped there is no angular frequency

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Forced Oscillations

  It is possible to compensate for the loss of energy in a damped system by applying an external sinusoidal force (with angular frequency  ) The amplitude of the motion remains constant if the energy input per cycle exactly equals the decrease in mechanical energy in each cycle that results from resistive forces    After a driving force on an initially stationary object begins to act, the amplitude of the oscillation will increase After a sufficiently long period of time,

E

driving  Then a steady-state condition is reached =

E

lost to internal  The oscillations will proceed with constant amplitude

The amplitude of a driven oscillation is 4/30/2020

 

0 is the natural frequency of the undamped oscillator Physics 201, UW-Madison 14

Resonance

       When    0 an increase in amplitude occurs This dramatic increase in the amplitude is called

resonance

The natural frequency resonance frequency  0 is also called the At resonance, the applied force is in phase with the velocity and the power transferred to the oscillator is a maximum  The applied force and

v

proportional to sin (  

t

+ are both  )  The power delivered is

F .

v

» This is a maximum when

F

in phase and

v

are Resonance (maximum peak) occurs when driving frequency equals the natural frequency The amplitude increases with decreased damping The curve broadens as the damping increases

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Resonance Applications:

4/30/2020 Extended objects have more than one resonance frequency. When plucked, a guitar string transmits its energy to the body of the guitar. The body’s oscillations, coupled to those of the air mass it encloses, produce the resonance patterns shown.

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Question 3

A 810-g block oscillates on the end of a spring whose force constant is k=60 N/m. The mass moves in a fluid which offers a resistive force proportional to its speed – the proportionality constant is b=0.162 N.s/m.

F s

F R

ma m d

2

x

 

kx dt

2  Write the equation of motion and solution.

 Solution:

x

(

t

) 

b dx dt Ae

b

2

m t

cos( 

t

  )  What is the period of motion?

0.730 s

 

k m

  

b

2

m

  2  Write the displacement as a function of time if at

t = 0, x = 0

and at

t = 1

s,

x = 0.120

m.

4/30/2020

Solution:

x

(

t

)  0.182

e

 0.100

t

sin(8.61

t

  )

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