simple harmonic motion. oscillations mass on a spring particle moving in uniform circular motion me...

Simple Harmonic Motion

Oscillations

Mass on a spring

Particle moving in uniform circular motion

Me walking down lecture hall steps

Oscillation or vibration – the motion of an object that regularly repeats itself, back and forth, over the same path.

That motion is periodic, meaning it has a certain frequency and period.

Springs and Simple Harmonic Motion

€

rF S = −k

r x = m

r a

r a = −

k

mr x

€

ETotal =1

2mv 2 +

1

2kx 2

Eat turning point =1

2kA2

1

2kA2 =

1

2mv 2 +

1

2kx 2

mv 2 = kA2 − kx 2

v = ±k

mA2 − x 2

( )

Recall the spring:

Circles and Simple Harmonic MotionA fundamental parallel can be drawn between SHM (mass on a spring), and uniform circular motion.

€

x = Acos θ( )

θ = θ 0 +ωt

x = Acos ωt +θ 0( )

What is ω for a spring?

Circles and Simple Harmonic Motion

€

1

2kA2 =

1

2mvmax

2

vmax = Ak

m

T =2πA

vmax

= 2πm

k

f =1

T=

1

2π

k

m

ω = 2πf =k

m

A fundamental parallel can be drawn between SHM (mass on a spring), and uniform circular motion.

Position Conceptualized

€

x(t) = Acos(ωt) = Acos2πt

T

⎛

⎝ ⎜

⎞

⎠ ⎟

€

x(t) = Asin(ωt) = Asin2πt

T

⎛

⎝ ⎜

⎞

⎠ ⎟

Depending upon the initial conditions…

t = 0

t = 0

T

Velocity

€

v =k

mA2 − x 2

( )

=k

mA2 − Acos ωt +θ 0( )( )

2 ⎛ ⎝ ⎜ ⎞

⎠ ⎟

= Ak

m1 − cos ωt +θ 0( )

2

= Ak

msin ωt +θ 0( )

2

= −Ak

msin ωt +θ 0( )

Acceleration

€

a =F

m

=1

m−kx( )

= −k

mA

⎛

⎝ ⎜

⎞

⎠ ⎟cos ωt +θ 0( )

All together now…

€

x(t) = Acos ωt +θ 0( )

v(t) = −Ak

msin ωt +θ 0( )

a(t) = −k

mA

⎛

⎝ ⎜

⎞

⎠ ⎟cos ωt +θ 0( )

€

xmax = A

vmax = Aω

amax = Aω 2

ExampleAs mass-spring system oscillates with an amplitude of 3.0 cm. If the spring constant is 270 N/m and object has a mass of 0.50 kg, determine the period, the maximum speed and maximum acceleration of the mass.

€

T = 2π mk = 2π 0.5

270 = 0.270 s

€

vmax = Aω = A km = 0.03 270

0.5 = 0.697 ms

€

amax = Aω 2 = Ak

m= 0.03

270

0.5=16.2 m

s2

3.0 cm

The Pendulum

€

FRestoring = −mgsin θ( )

≈ −mgs

l

⎛

⎝ ⎜

⎞

⎠ ⎟

= −mg

l

⎛

⎝ ⎜

⎞

⎠ ⎟s

= −keff x

€

keff ≡mg

lx ≡ s

The Pendulum

€

keff ≡mg

lx ≡ s

€

ω =k

m=

mgl( )

m=

g

l

T =2π

ω= 2π

l

g

ExampleA simple pendulum has a length of 52.3 cm and makes 83.9 complete oscillations in 2.00 min after being pulled to the side by 0.25m.

Find the period of the pendulum, the acceleration due to gravity at the location of the pendulum, and the maximum speed of the pendulum.

€

T =time

oscillation=

120

83.9=1.43 s

€

Tpend = 2π lg

g =2π

T

⎛

⎝ ⎜

⎞

⎠ ⎟2

l

=2π

1.43

⎛

⎝ ⎜

⎞

⎠ ⎟2

0.523

=10.1 ms2

€

vmax = A gl

= 0.25 10.10.523

=1.10 msec

Grandfather ClockYour grandfather clock is running slowly. Every 24 hours, you find that it has lost 2 minutes. How will you fix this problem?

€

Tdesired = 2.0 s

Tactual =time for n oscillations

n oscillations

n = 24 hr( ) 60 minhr( ) − 2 min( ) 30 osc

min( )

= 43140

Tactual =24 * 3600

43140= 2.00278 s

€

Tpendulum = 2πl

g

l =T

2π

⎛

⎝ ⎜

⎞

⎠ ⎟2

g

lactual =2.00278

2π

⎛

⎝ ⎜

⎞

⎠ ⎟2

9.8( ) = .996 m

ldesired =2.00

2π

⎛

⎝ ⎜

⎞

⎠ ⎟2

9.8( ) = .993 m

Shorten the pendulum by 3 mm.