Physics 1501: Lecture 26, Pg 1 Physics 1501: Lecture 26 Today’s Agenda l Homework #9 (due Friday Nov. 4) l Midterm 2: Nov. 16 l Katzenstein Lecture: Nobel

Physics 1501: Lecture 26, Pg 1

Physics 1501: Lecture 26Physics 1501: Lecture 26TodayToday’’s Agendas Agenda

Homework #9 (due Friday Nov. 4) Midterm 2: Nov. 16

Katzenstein Lecture: Nobel Laureate Gerhard t’HooftFriday at 4:00 in P-36 …

TopicsSimple Harmonic Motion – masses on springs Pendulum Energy of the SHO


New topic (Ch. 13) New topic (Ch. 13) Simple Harmonic Motion (SHM)Simple Harmonic Motion (SHM)

We know that if we stretch a spring with a mass on the end and let it go the mass will oscillate back and forth (if there is no friction).

This oscillation is called Simple Harmonic Motion,and is actually very easy to understand...

km

km

km


SHM So FarSHM So Far

We showed that (which came from F=ma)

has the most general solution x = Acos(t + )

where A = amplitude

= frequency

= phase constant

For a mass on a spring

The frequency does not depend on the amplitude !!!We will see that this is true of all simple harmonic

motion ! The oscillation occurs around the equilibrium point where

the force is zero!


The Simple PendulumThe Simple Pendulum

A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small

displacements.

L

m

mg

z


The Simple Pendulum...The Simple Pendulum...

Recall that the torque due to gravity about the rotation (z) axis is = -mgd.

d = Lsin Lfor small

so = -mg L

L

dm

mg

z

where

Differential equation for simple harmonic motion !

= 0 cos(t + )

But = II=mL2


The Rod PendulumThe Rod Pendulum

A pendulum is made by suspending a thin rod of length L and mass M at one end. Find the frequency of oscillation for small displacements.

Lmg

z

xCM


The Rod Pendulum...The Rod Pendulum...

The torque about the rotation (z) axis is

= -mgd = -mg{L/2}sin-mg{L/2}for small

In this case

Ldmg

z

L/2

xCM

where

d I

So = Ibecomes


Lecture 26, Lecture 26, Act 1Act 1PeriodPeriod

(a) (b) (c)

What length do we make the simple pendulum so that it has the same period as the rod pendulum?

LR

LS


Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, that we know where the CM is located and what the moment of inertia I about the axis is.

The torque about the rotation (z) axis for small is (sin ≈ )

= - Mgd ≈ - MgR

General Physical PendulumGeneral Physical Pendulum

d

Mg

z-axis

R

xCM

where

= 0 cos(t + )


Lecture 26, Lecture 26, Act 2Act 2Physical PendulumPhysical Pendulum

A pendulum is made by hanging a thin hoola-hoop of diameter D on a small nail. What is the angular frequency of oscillation of the hoop

for small displacements ? (ICM = mR2 for a hoop)

(a)

(b)

(c)

D

pivot (nail)


Torsion PendulumTorsion Pendulum

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


Torsion Pendulum...Torsion Pendulum...

Since = -k = Ibecomes

I

wire

Similar to “mass on spring”, except

I has taken the place of m (no surprise)

where


Lecture 26, Lecture 26, Act 3Act 3PeriodPeriod

All of the following pendulum bobs have the same mass. Which pendulum rotates the fastest, i.e. has the smallest period? (The wires are identical)

RRRR

A) B) C) D)


Spring-mass system

Pendula

General physical pendulum

» Simple pendulumTorsion pendulum

Simple Harmonic OscillatorSimple Harmonic Oscillator

dMg

z-axis

R

xCM

= 0 cos(t + )

k

x

m

FF = -kx aa

I

wire

x(t) = Acos(t + )

where


Energy of the Spring-Mass SystemEnergy of the Spring-Mass System

Add to get E = K + U

1/2 m (A)2sin2(t + ) + 1/2 k (Acos(t + ))2

Remember that

U~cos2K~sin2

E = 1/2 kA2

so, E = 1/2 kA2 sin2(t + ) + 1/2 kA2 cos2(t + )

= 1/2 kA2 [ sin2(t + ) + cos2(t + )]

= 1/2 kA2

ActiveFigure


Energy in SHMEnergy in SHM

For both the spring and the pendulum, we can derive the SHM solution using energy conservation.

The total energy (K + U) of a system undergoing SMH will always be constant!

This is not surprising since there are only conservative forces present, hence energy is conserved.

-A A0 s

U

U

KE


SHM and quadratic potentialsSHM and quadratic potentials SHM will occur whenever the potential is quadratic. Generally, this will not be the case: For example, the potential between

H atoms in an H2 molecule lookssomething like this:

-A A0 x

U

U

KEU

x


SHM and quadratic potentials...SHM and quadratic potentials...However, if we do a Taylor expansion of this function about the minimum, we find that for smalldisplacements, the potential IS quadratic:

U

x

U(x) = U(x0 ) + U(x0 ) (x- x0 )

+ U (x0 ) (x- x0 )2+....

U(x) = 0 (since x0 is minimum of potential)

x0

U

x Define x = x - x0 and U(x0 ) = 0

Then U(x) = U (x0 ) x 2


SHM and quadratic potentials...SHM and quadratic potentials...

U

x

x0

U

x

U(x) = U (x0) x 2

Let k = U (x0)

Then:

U(x) = k x 2

SHM potential !!


What about Friction?What about Friction? Friction causes the oscillations to get

smaller over time This is known as DAMPING. As a model, we assume that the force due

to friction is proportional to the velocity.


What about Friction?What about Friction?

We can guess at a new solution.

With,



What does this function look like?(You saw it in lab, it really works)



There is a cuter way to write this function if you remember that exp(ix) = cos x + i sin x .


Damped Simple Harmonic MotionDamped Simple Harmonic Motion

Frequency is now a complex number! What gives?Real part is the new (reduced) angular frequencyImaginary part is exponential decay constant

underdamped critically damped overdamped

ActiveFigure


Driven SHM with ResistanceDriven SHM with Resistance

To replace the energy lost to friction, we can drive the motion with a periodic force. (Examples soon).

Adding this to our equation from last time gives,

F = F0 cos(t)


Driven SHM with ResistanceDriven SHM with Resistance

So we have the equation,

As before we use the same general form of solution,

Now we plug this into the above equation, do the derivatives, and we find that the solution works as long as,


Driven SHM with ResistanceDriven SHM with Resistance So this is what we need to think about, I.e. the amplitude

of the oscillating motion,

Note, that A gets bigger if Fo does, and gets smaller if b or m gets bigger. No surprise there.

Then at least one of the terms in the denominator vanishes and the amplitude gets real big. This is known as resonance.

Something more surprising happens if you drive the pendulum at exactly the frequency it wants to go,


Driven SHM with ResistanceDriven SHM with Resistance Now, consider what b does,

b small

b middling

b large


Dramatic example of resonanceDramatic example of resonance In 1940, turbulent winds set up a torsional vibration in the

Tacoma Narrow Bridge


Dramatic example of resonanceDramatic example of resonance

when it reached the natural frequency


Dramatic example of resonanceDramatic example of resonance

it collapsed !

Other example: London Millenium Bridge


Lecture 26, Lecture 26, Act 4Act 4Resonant MotionResonant Motion

Consider the following set of pendula all attached to the same string

D

A

B

CIf I start bob D swinging which of the others will have the largest swing amplitude ?

(A) (B) (C)

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Physics 1501: Lecture 26, Pg 1 Physics 1501: Lecture 26 Today’s Agenda l Homework #9 (due Friday Nov. 4) l Midterm 2: Nov. 16 l Katzenstein Lecture: Nobel