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.

Physics chapter 11 3

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.

Physics chapter 11 4

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)

Physics chapter 11 5

Simple Harmonic Motion

= 2f =2T

f =1

T

a = - k

mx

T = 2m

k

mass & springxa 2

Physics chapter 11 6

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?

Physics chapter 11 7

Example

Write an equation for the force constant:

F kx

k Fx

mgx

Physics chapter 11 8

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

Physics chapter 11 9

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?

Physics chapter 11 10

Example 2

Physics chapter 11 11

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.

Physics chapter 11 12

Example 2

f 1

2gA

1

29.8m / s2

0.06m

= 2.0 Hz

amax 2A 42f 2A g

Physics chapter 11 13

Simple Pendulum

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

2 =g

L

T = 2L

g

Physics chapter 11 14

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

Physics chapter 11 15

Energy in SHM

2 21 1

2 2E mv kx

2max

1

2E mv

21

2E kA

Physics chapter 11 16

Simple Harmonic Motion

x Acost

Simple Harmonic Motion

-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

Physics chapter 11 17

Motion Equations

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

Physics chapter 11 18

Motion Equations

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

a = – 2xamax = 2A at x = ±A

Physics chapter 11 19

x vs. t

x = Acoswt

Simple Harmonic Motion

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8

time (s)

x

Physics chapter 11 20

x and v vs. t

Simple Harmonic Motion

-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.

Physics chapter 11 21

x, v, and a vs. t

Simple Harmonic Motion

-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

Physics chapter 11 22

More equations

km

m

kf

2

1

2

k

m

fT 2

1

Physics chapter 11 23

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.

Physics chapter 11 24

Example

•

x = ?

vmax = ?

amax = ?

Physics chapter 11 25

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

Physics chapter 11 26

Example

v = – Asintv = – 1.6sin8 t m/s

vmax = A= 2πfA

vmax = 1.6 m/s

Physics chapter 11 27

Example

a = – 2Acosta = – 12.82cos8 t m/s2

amax = 2A

amax = 12.82 m/s2