SPRING-MASS OSCILLATORS

AP Physics

Unit 8

Recall Hooke’s Law

• Applied force (Fapplied) stretches or compresses spring from its natural length

• Restoring force (Fr) acts to return spring to lowest energy state

kxF

FF

r

appliedr

An energy approach to SHM

• Stretched/compressed spring stores elastic potential energy: Us = ½kx2

• When released, mass oscillates about its equilibrium position as PE KE etc– Amplitude of oscillation is xmax

– At x=0, Us = 0 so K is maximized

– At x=A, Us is maximized, so K=0

Us, max = ½kA2

Example #1

• A 2.0 kg block is attached to an ideal spring with a force constant of 500 N/m. The spring is stretched 8.0 cm and released.

• When the block is 4.0 cm from equilibrium– what is the total

energy of the system?– what is the velocity of

the block?

JE

kAE

KUE

o

o

ooo

6.1

)08.0)(500(2102

1 22

By energy conservation, E=1.6J at every spring position

smv

Jv

Emvkx

/1.1

6.1)0.2(21)04.0(5002

121

21

22

22

An energy approach to SHM

Since energy is conserved and– at x=A, Us is maximized and K=0

– at x=0, Us = 0 so K is maximized

it follows that2max

2

21

21 mvkA

thereforem

kAv max

SHM and the Reference Circle

• Motion of the shadow cast by a particle moving in a vertical circle mimics SHM– Amplitude corresponds to the radius of the circle– Period of the oscillation corresponds to the period

of the UCM

T

A

T

Rv

22 or

v

AT

2

sincem

kAv max

km

vk

mvT

2

2

Example #2

A 2.0kg block is attached to a spring with a force constant of 300 N/m.

Determine the period and frequency of the oscillations.

kmT 2

sT 51.03000.22

HzT

f 9.11

Vertical Spring-Mass Oscillators

• As it turns out, the behavior is the same regardless of the orientation, i.e. gravity does not affect the period or frequency of the oscillations.

• Sounds improbable? Let’s see why it is not…

Vertical SMOs

• Consider a spring with constant k on which a mass m is hung, stretching the spring some distance x

• The spring is in equilibrium:

Fapplied= Fr or kx=mg

• If the spring is further displaced by some amount A, the restoring force increases to

k(x+A) while the weight remains mg

Vertical SMOs

The net force on the block is nowF= k(x+A)- mg

But since kx=mg, the force on the block is F= kA. This is Hooke’s Law!

Instead of oscillating about the natural length of the spring as happens with a horizontal SMO, oscillations of a vertical SMO are about the point at which the hanging mass is in equilibrium!

Example #3

• A 1.5 kg block is attached to the end of a vertical spring with a constant of 300 N/m. After the block comes to rest, it is stretched an additional 2.0 cm and released.– What is the frequency of

the oscillation?– What are the maximum &

minimum amounts of stretch in the spring?

Hzf

mkf

3.25.1300

2

12

1

cmmx

x

mgkx

9.4049.0

)8.9)(5.1(300

Since A=2.0 cm, the spring is stretched a maximum of 6.9 cm and a minimum of 2.9 cm