oscillations lecture: ap mechanics

8/12/2019 Oscillations lecture: AP Mechanics

1/9

Given a potential energy graph, oscillations will occur between turning points

determined by

txUE

E

tx

Even the most asymmetric well can be approximated by a parabola if the systemstays close to the bottom.


2/9

2

2

1kxU

Simple Harmonic Motion k and x need not refer tospring parameters, but the

spring-mass system is an

important example

kx

dx

dUF

2

2

dt

xdmmakx


3/9

02

2

xm

k

dt

xdStandard form of harmonic oscillator equation

Solution: tAtx cos mk2

andA determined by initial conditions

Exercise: A spring-mass with k = 25 N/m and m = 5 kg is pulled out 2 m and released.

Find its position at any later time.

amplitude phase angle

phase

angular frequency

2cos0 Ax 0sin0 Av

tAtv sin

2

0

A

ttx 5cos2


4/9

The motion will have completed one full cycle when the phase has

increased by 2. This time interval is called the periodof the oscillation.

2

tTt

2T

Energy Considerations

How can the total energy be constant when both K and U depend on time?

22

2

1

2

1kxmvE

22 cos2

1sin2

1 tAktAmE

ttkAE 222 cossin2

1 2

2

1kAE


5/9

Vertical Springs

When a spring mass-system

vibrates vertically, an extra

force, gravity is involved.How does this affect the

motion?

k

mx

m

kx

kxmgnetF

gxm

k

dt

xd

2

2

? Rewrite: 02

2

k

mgx

m

k

dt

xd

Define new displacement:kmgxx Note that:

2

2

2

2

dtxd

dtxd

Then we have: 02

2

xm

k

dt

xdGravity does not change the period.

All effects of gravity can be ignored if we choose to measure X from the newequilibrium point.

mg


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Example: A 1 kg mass hangs vertically from a spring with k = 20 N/m. A .1 kg dart

moving at 20 m/s is fired into the mass and sticks into it. Find the period and

amplitude of the resulting harmonic motion.

kg1

fPP 0

V1.1201.

smV 82.1

The new equilibrium point is

mx

f55.

20

101.1

0

The period just depends on k and the mass oscillating:

s

k

mT 47.1

20

1.122

Use energy conservation with x referred to new equilibrium point:

fEE 0

222 202

105.20

2

182.11.1

2

1A

mA 43.

The old equilibrium point is

mxi 50.

20

1010


7/9

Other Oscillators

Torsion Pendulum

As wire twists, a restoring

torque is exerted.

k

I

Fixed axis rotation

I

2

2

dt

dIk

0

2

2

I

k

dt

d

k

IT 2 tt mcos


8/9

Simple Pendulum

T

mg

L

ILmg sin

2mLI

0sin2

2

L

g

dt

d? Small angle approximation:

sin

02

2

L

g

dt

d

g

LT 2


9/9

Compound Pendulum

H

CM IHmg sin

0sin2

2

I

mgH

dt

d

Small angle approximation:

sin0

2

2

I

mgH

dt

d

mgH

IT 2

About pivot

oscillations lecture: ap mechanics

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