Periodic Motion

Periodic Motion

Chapter 11: Sections 1-3, 5

Physics chapter 11 2

Oscillation

A complete fluctuation of the value of a quantity above and below some median value.


Simple Harmonic MotionEquilibrium position - Central position where

forces are balanced.Amplitude (A) - the maximum displacement

from equilibrium.(m)Period (T) - Time for one complete

oscillation.(s)Frequency (f)- Number of oscillations that

occur in a given unit of time. (cyc/s or Hz)Angular frequency (w) - the frequency given

in rad/s.


Simple Harmonic Motion

The displacement vector and the vector for the restoring force are always in opposite directions.

Hooke's LawF = kxThe force a spring exerts is proportional to

the distance that the spring is stretched or compressed.

k = F⁄x (N/m)



= 2f =2T

f =1

T

a = - k

mx

T = 2m

k

mass & springxa 2


Example

When a ball is suspended from a spring, the spring stretches by 70 mm.

If the ball oscillates up and down, what is its period?

What is its frequency?


Example

Write an equation for the force constant:

F kx

k Fx

mgx


Example

We don't know the mass of the ball.Write the equation for the period of

oscillation:T 2

mk

1k

xmg

T 2mxmg

2xg

20.07m

9.8m / s2 0.53s

f 1T1.89 Hz


Example 2

A piston undergoes simple harmonic motion in a vertical direction with an amplitude of 6 cm.

A coin is placed on top of the piston. What is the lowest frequency at which the

coin will be left behind by the piston on its downstroke?


Example 2


Example 2

The coin will leave the piston when the downward acceleration of the latter exceeds the acceleration of gravity g.

The maximum downward acceleration of piston occurs at the highest point of its motion.


Example 2

f 1

2gA

1

29.8m / s2

0.06m

= 2.0 Hz

amax 2A 42f 2A g


Simple Pendulum

All of the mass may be considered to be located at the end or bob.

2 =g

L

T = 2L

g


Example

A certain simple pendulum has a period on the earth of 0.500 s. What is its period on the surface of the moon, where g = 1.67 m/s2?

1.21s


Energy in SHM

2 21 1

2 2E mv kx

2max

1

2E mv

21

2E kA



x Acost


-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8

time (s)

po

siti

on

(m

)

x


Motion Equations

x = 0.5cos2πt (in standard units)Amplitude = 0.5 mAngular Frequency = 2π rad/sf = /2πT = 1/f


Motion Equations

x = Acostv = – Asinta = – 2Acost vmax = A= 2πfA where x = 0.

a = – 2xamax = 2A at x = ±A


x vs. t

x = Acoswt


-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8

time (s)

x


x and v vs. t


-2-1.5

-1-0.5

00.5

11.5

2

0 1 2 3 4 5 6 7 8

time (s)

x

v

v = – wAsinwtvmax = wA= 2πfA where x = 0.


x, v, and a vs. t


-3

-2

-1

0

1

2

3

0 1 2 3 4 5 6 7 8

time (s)

x

v

a

a = – w2Acoswta = w2xamax = w2A at x = ±A


More equations

km

m

kf

2

1

2

k

m

fT 2

1


ExampleAs shown on the next slide, a rotating

wheel drives a piston by means of a long connecting rod pivoted at both ends.

The wheel's radius is 20 cm and it is turning at 4.0 rev/s.

A) Write an equation for the displacement of the piston as a function of time.

B) Find the maximum speed and acceleration of the piston.


Example

•

x = ?

vmax = ?

amax = ?


Example

x = Acos t

A = 20 cm = 0.2 m = 4 rev/s x 2 rad/1 rev = 8 rad/s

x = 20cos8 t cmx = 0.2cos8 t m


Example

v = – Asintv = – 1.6sin8 t m/s

vmax = A= 2πfA

vmax = 1.6 m/s


Example

a = – 2Acosta = – 12.82cos8 t m/s2

amax = 2A

amax = 12.82 m/s2

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Periodic Motion