Physics 201: Lecture 29, Pg 1 Lecture 29 Goals Goals Describe oscillatory motion in a simple...

Physics 201: Lecture 29, Pg 1

Lecture 29

• GoalsGoals Describe oscillatory motion in a simple pendulum Describe oscillatory motion with torques Introduce damping in SHM Discuss resonance


Final Exam Details

Sunday, May 13th 10:05am-12:05pm in 125 Ag Hall & quiet room

Format: Closed book Up to 4 8½x1 sheets, hand written only Approximately 50% from Chapters 13-15 and 50% 1-12 Bring a calculator

Special needs/ conflicts:

All requests for alternative test arrangements should be made by Thursday May10th (except for medical emergency)


Mechanical Energy of the Spring-Mass System

Kinetic energy is always

K = ½ mv2 = ½ m(A)2 sin2(t+)

Potential energy of a spring is,

U = ½ k x2 = ½ k A2 cos2(t + )

And 2 = k / m or k = m 2

K + U = constant

x(t) = A cos( t + )

v(t) = -A sin( t + )

a(t) = -2A cos( t + )


SHM is a close as Lake Mendota…

So can you estimate the characteristic frequency for a bobbing in the water?

If you have equilibrium and there is a linear restoring force, then yes with = (k / m)½

F mgB

y0

mgFF By 0 equlibriumAt

gAymgF wB 0


SHM is a close as Lake Mendota…

Deeper than y0 means

and a net force of

A linear restoring force with k = wAg

and boat mass m = Ay0 w so = (g / y0)½

Lighter boats bob more quickly than heavy ones

(if the same size)

FB mg

y

AygF wB gyyAF w )( 0Net


The shaker cart You stand inside a small cart attached to a heavy-duty

spring, the spring is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you.

At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground.

What effect does jettisoning the sandbag at the equilibrium position have on the amplitude of your oscillation?

A. It increases the amplitude.B. It decreases the amplitude.C. It has no effect on the amplitude.

Hint: At equilibrium, both the cart and the bag are moving at their maximum speed.


The shaker cart

Instead of dropping the sandbag as you pass through equilibrium, you decide to drop the sandbag when the cart is at its maximum distance from equilibrium.

What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation?

A. It increases the amplitude.

B. It decreases the amplitude.

C. It has no effect on the amplitude.

Hint: At maximum displacement there is no kinetic energy.


The shaker cart

What effect does jettisoning the sandbag at the cart’s maximum displacement from equilibrium have on the maximum speed of the cart?

A. It increases the maximum speed.

B. It decreases the maximum speed.

C. It has no effect on the maximum speed.

Hint: At maximum displacement there is no kinetic energy.


The Pendulum (using torque)

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.

z = Iz = -mg sin() L

z ≈ mL2z ≈ -mg L

L (d2 /dt2) = -g

compare to max = -kx

d2 /dt2 = (-g/L)

with (t)= 0 cos( t + )

and =(g/L)½

z

L

mg

y

x

T

L sin


The 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.

If small then sin()

0° tan 0.00 = sin 0.00 = 0.00

5° tan 0.09 = sin 0.09 = 0.09

10° tan 0.17 = sin 0.17 = 0.17

15° tan 0.26 = 0.27

sin 0.26 = 0.26

L

m

mg

z

y

x

T


Exercise Simple 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 recalling that = (g/L)½ (A) T1 = T2

(B) T1 > T2

(C) T1 < T2


A 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.

z = I = -| r x F | = (L/2) mg sin()

Irod at end = mL2/3

- mL2/3 L/2 mg -1/3 L d2/dt2 = ½ g

Lmg

z

xCM

T

L

g

2

3


Torsion 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.

Torque is proportional to the angular displacement: = - where is the torsion constant

= (/I)½

I

wire


Exercise Period All of the following torsional pendulum bobs have

the same mass and radius with = (/I)½

Which pendulum rotates the slowest (i.e. has the

longest period) if the wires are identical?

RRRR

(A) (B) (C) (D)


What about Friction?A velocity dependent drag force (A model)

2

2

dtxd

mdtdx

bkx

We can guess at a new solution.

)(cos expA )( )(2

ttxm

bt

With,

22

2

22

mb

mb

mk

o

02

2

xm

k

dt

dx

m

b

dt

xd

and now 02 ≡ k / m

Note


What about Friction?

A damped exponential

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

A)(cos

2expA )( )( tt

mbtx mbo 2/if


Variations in the damping

Small damping time constant (m/b)

Low friction coefficient, b << 2m

Moderate damping time constant (m/b)

Moderate friction coefficient (b < 2m)


Damped Simple Harmonic Motion

A downward shift in the angular frequency There are three mathematically distinct regimes

22 )2/( mbo

mbo 2/ mbo 2/

underdamped critically damped overdamped

mbo 2/


Driven SHM with Resistance Apply a sinusoidal force, F0 cos (t), and now consider what A and b do,

2220

2

0

)()(

/

mb

mFA

b/m small

b/m middling

b large

tm

Fx

m

k

dt

dx

m

b

dt

xd cos 2

2

Not Zero!!!

stea

dy s

tate

am

plitu

de


For Thursday

Review for final!

Physics 201: Lecture 29, Pg 1 Lecture 29 Goals Goals Describe oscillatory motion in a simple...

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