Physics 111: Lecture 24 Today’s Agenda

Physics 111: Lecture 24, Pg 1

Physics 111: Lecture 24

Today’s Agenda

Introduction to Simple Harmonic Motion Horizontal spring & mass

The meaning of all these sines and cosines Vertical spring & mass The energy approach The simple pendulum The rod pendulum


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

HorizontalSpring


SHM Dynamics

At any given instant we know that F = ma must be true.

But in this case F = -kx and ma =

So: -kx = ma =

k

x

m

F = -kx a

m d xdt

2

2

d xdt

km

x2

2 a differential equation for x(t)!

m d xdt

2

2


SHM Dynamics...

d xdt

km

x2

2

d xdt

x2

22

km

Try the solution x = A cos(t)

tsinAdtdx

xtcosAdt

xd 222

2

This works, so it must be a solution!

define

Where is the angularfrequency of motion


SHM Dynamics...

y = R cos = R cos (t)

But wait a minute...what does angular frequency have to do with moving back & forth in a straight line ??

Movie (shm)

x

y

-1

1

0

1 12 2

3 3

4 45 5

6 62

Shadow


SHM Solution

We just showed that (which came from F = ma)

has the solution x = A cos(t) .

This is not a unique solution, though. x = A sin(t) is also a solution.

The most general solution is a linear combination of these two solutions!

x = B sin(t)+ C cos(t)

d xdt

x2

22

tsinCtcosBdtdx

xtcosCtsinBdt

xd 2222

2

ok


Derivation:

x = A cos(t + ) is equivalent to x = B sin(t)+ C cos(t)

x = A cos(t + )

= A cos(t) cos - A sin(t) sin

where C = A cos() and B = A sin()

It works!

= C cos(t) + B sin(t)

We want to use the most general solution:

So we can use x = A cos(t + ) as the most general solution!


SHM Solution...

Drawing of A cos(t ) A = amplitude of oscillation

T = 2/

A

A


SHM Solution...

Drawing of A cos(t + )


SHM Solution...

Drawing of A cos(t - /2)

A

= /2

= A sin(t)!


Lecture 24, Act 1Simple Harmonic Motion

If you added the two sinusoidal waves shown in the top plot, what would the result look like?

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-0.20

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0.40

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-1.00

-0.50

0.00

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1.000 10

0

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(a)

-2.00

-1.50

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0.00

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2.00

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(c)-0.6

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0

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Lecture 24, Act 1 Solution

Recall your trig identities:

2

BA2

BA2BA coscoscoscos

tt coscos

22a cos

2b

bta cosSo

Where

The sum of two or more sines or cosines having the same frequency is just another sine or cosine with the same frequency.

The answer is (b).Prove this with Excel


What about Vertical Springs?

We already know that for a vertical spring if y is measured from

the equilibrium position

The force of the spring is the negative derivative of this function:

So this will be just like the horizontal case:

-ky = ma =

j

k

m F = -ky

y = 0

U ky12

2

kydydUF

m d ydt

2

2

Which has solution y = A cos(t + )

km

where

VerticalSpring


SHM So Far

The most general solution is x = A cos(t + ) where A = amplitude

= frequency = phase

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!

km


The 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

SimplePendulum


Aside: sin and cos for small

A Taylor expansion of sin and cos about = 0 gives:

...!5!3

sin53

...!4!2

1cos42

and

So for << 1, sin 1cos and


The Simple Pendulum...

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

d = Lsin L for small

so = -mg L

But = I I = mL2 L

dm

mg

z

mgL mL ddt

22

2

ddt

2

22 g

Lwhere

Differential equation for simple harmonic motion!

= 0 cos(t + )


Lecture 24, Act 2Simple Harmonic Motion

You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1.

Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2.

Which of the following is true:

(a) T1 = T2

(b) T1 > T2

(c) T1 < T2



We have shown that for a simple pendulum gL

T Lg

2Since T 2

If we make a pendulum shorter, it oscillates faster (smaller period)



L1

L2

Standing up raises the CM of the swing, making it shorter!

T1 T2

Since L1 > L2 we see that T1 > T2 .


The 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 torque about the rotation (z) axis is

= -mgd = -mg(L/2)sin -mg(L/2) for small

In this case

So = I becomes

Ldmg

z

L/2

xCM

I 13

2mL

2

22

dtdmL

31

2Lmg

ddt

2

22 3

2gL

where

d I


Lecture 24, Act 3Period

(a) (b) (c)

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

LR

LS

RS L32L RS L

23L RS LL

PhysicalPendulum


LR

LS

S = P if RS L32L

Lecture 24, Act 3Solution

SS L

gR

R L2g3


Recap of today’s lecture

Introduction to Simple Harmonic Motion (Text: 14-1)Horizontal spring & mass

The meaning of all these sines and cosines Vertical spring & mass (Text: 14-3) The energy approach (Text: 14-

2) The simple pendulum (Text: 14-3) The rod pendulum

Look at textbook problems Chapter 14: # 1, 13, 33, 55, 93

Physics 111: Lecture 24 Today’s Agenda

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