Simple Harmonic Motion

Swings and Springs

Periodic motion

Periodic motion Harmonic motion

Motion that repeats itself

Frequency

Frequency = f

= number of oscillations in 1

second

Unit = 1/s = hz

Period = T

= time for 1 oscillation/cycle

Unit = s

Tf

fT

1

1

Simple Harmonic Motion (SHM)

Simple cosine wave describes SHM

)cos()( tAtx

Mathematical description

A = amplitude = center to max height or trough ω = angular frequency φ = phase angle = where (in radians) the wave started

)cos()( tAtx

fT

12

2f

x(t), v(t), a(t)

Position is given by the cosine wave. Differentiate for velocity and acceleration. Note: a = -ω2x (characteristic of SHM)

)cos(

)sin(

)cos()(

2

tAdt

dva

tAdt

dxv

tAtx

Identify SHM

constant) (positive

constant) (positive

xa

constant) (positive

constant) (positive

xF

Motion obeys these relationships:

Forces look like this:

Springs

Find the period of a spring:

xm

ka

kxF

s

s

k

m

fT 2

1

m

k

m

kf

2

1

2

SHM problem

A block of mass 680 g is fastened to a spring of spring constant 65 N/m. The block is pulled 11 cm from its equilibrium position at x=0 on a frictionless surface and released from rest at t=0.

a) What force does the spring exert on the block just before the block is released?

NmmNkxF 2.7)11.0)(/65(

… SHM problem

b) What are the angular frequency, the frequency, and the period of the resulting oscillation?

k = 65 N/m m = 680 g x = 11 cm

rad/s 78.9

kg 68.0

N/m 65

m

k

hz 56.12

rad/s 8.9

2

f

s 64.0hz 56.1

11

fT

… SHM problem

c) What is the amplitude of oscillation?

k = 65 N/m m = 680 g x = 11 cm

= 9.8 rad/s f = 1.6 hz T = 0.64 s

xmax = 11 cm

… SHM problem

d) What is the block’s maximum speed?

k = 65 N/m m = 680 g x = 11 cm

= 9.8 rad/s f = 1.6 hz T = 0.64 s

)sin(

)cos()(

tAdt

dxv

tAtx

m/s 1.1)m 11.0)(rad/s 8.9()( maxmax xv

Hint:

… SHM problem.

e) What is the magnitude of the block’s maximum acceleration?

k = 65 N/m m = 680 g x = 11 cm

= 9.8 rad/s f = 1.6 hz T = 0.64 s

)cos(

)cos()(

2

2

2

tAdt

xda

tAtx

22

max

2

max m/s 11)m 11.0()rad/s 8.9()( xa

Hint:

Physical Pendulum

Most pendulums in the real world are not even approximately “simple”.

Period of a pendulum

Simple pendulum:

Physical pendulum:

I = rotational inertia (moment of inertia)

about the pivot point

h = distance from pivot to center of mass

m = mass of physical pendulum

g

LT 2

mgh

IT 2

Swings (pendulums)

Derive period of a pendulum from torque

g

lT 2

Superposition of waves

Waves add mathematically:

Wave 1:

Wave 2:

)cos()( 1111 tAtx

)cos()( 2222 tAtx

)cos()cos()()( 22211121 tAtAtxtx

Superposition of Waves

Fourier transform: Play with waves of different frequencies: http://teacher2.smithtown.k12.ny.us/winters/Si

mulations.htm

Original time domain wave

Frequency domain wave