SHM Simple Harmonic Motion

The oscillation of an object between 2 points for an indefinite period of time with NO LOSS of MECHANICAL ENERGY!

The net external force is a linear restoring force!

** In the real world there is friction which "dampens" the oscillation.

** A forced oscillation has an external driving force applied to balance the energy loss.

bring back to some equilibriurm

Periodic Motion

Circular Motion

*Reminder

Springs!

K ­> spring constant ­> describes the "stretchiness" of the spring

Fspring = ­Kx <­ Hooke's Law

Any FApplied = +Kx, opposes restoring force

equilibrium

­ here means the force is opposite the displacement

+ here means force is in the same direction as displacement

Springs!

equilibrium

x = distance away from equilibrium

F = Restoring force, towards equilibrium

d t v t

a t

In a tire pressure gauge, the air in the tire pushes against a plunger attached to a spring when the gauge is pressed against the tire valve.

Suppose the spring constant of the spring is K = 320 N/m and the bar indicator of the gauge reads 2.0 cm when the gauge is pressed against the tire valve.

What force does the air in the tire apply to the spring?

F

x

slope = K

Area = WORK = Energy!

Work = F x = Area under the curve!

FS = ­Kx

y = mx + b

Work = ΔKE = ­ΔPE When Mechanical Energy is conserved!

Derive Potential Energy of a spring! (consider no final PE, spring at equilibrium)

Also called ELASTIC PE

F

x

slope = K

An object of mass m = 0.2 kg ,attached to a spring, sits on a horizontal frictionless surface. The spring has a spring constant of K= 545 N/m. It is stretched to x = 4.5 cm and then released from rest.

Determine the final translational speed of the object when the final displacement of the spring is:

a) +2.25 cm

b) 0 cm

x = 0 cm x = 4.5 cm

x = 0 cm x = 4.5 cm

Pendulums

θ Restoring Force

always towards

equilibrium

The Period

Springs Pendulums

*F = ­Kx *F = mg

θ Let's measure g in this room

How to go about it?

Find the length of a pendulum that oscillates at 5 Hz.

Then if 5 kg object is attached to the end of the pendulum, and raised through an angle of 25o from equilibrium,

What is the maximum speed of that object?

When ΣFv = 0 and ΣFh = 0 the object is in EQUILIBRIUM!

Translational Equilibrium

Now Look: It'still in Translational Equilibrium

BUT WHAT HAPPENS NOW?

ORQUE!rotating about an axis Each FORCE is

acting through a DISTANCE from the axis of rotation!

Axis of rotation: point about which an object rotates

Lever arm: the shortest distance from rotation axis to a line of action of a force. Will be from line of action to the axis of rotation.

Torque: a VECTOR quantity whose magnitude is the magnitude of the (the lever arm) X (Force)

(moment arm)

by convention rotating ccw is +

rotating cw is ­

Definitions

(pivot, fulcrum)

Cross Product (VECTOR Product)

Multiplying two vectors together with the result being a vector

Unlike the DOT or Scalar product for work in which the result is a scalar

τorque = r X Flever arm force applied

τorque = r X F = |r||F|sinθ

angle between r & F

The result of the cross product is that only the parts of the force and lever arm that are perpendicular to one another

contribute to the τorque

Ways to find τorque:

τ = |r||F|sinθ

τ = r F part of the lever arm to the force

τ = r F part of the force to the lever arm

torque is a vector so it has a magnitude and a direction

force = 3.0N

"pivot point"axis ofrotation

= lever arm X force

= ?

L = 4.0 m

L2

force = 3.0N

= lever arm X force

= ?

L = 4.0 m

L2

= 3.0N (2.0m) = 6.0Nm

force = 3.0N

= lever arm X force

= ?

L = 4.0 m

L2

force = 3.0N

"pivot point"axis ofrotation

= lever arm X force

= ?

L = 4.0 m

L2

600

force = 3.0N

= lever arm X force

= ?

L = 4.0 m

L2

600lever a

rm

300

= 3.0N (2.0m) sin60 0 = 5.2Nm

Static Equilibrium* Translational Equilibrium

The linear acceleration of the center of mass of the object

must be zero.

F = 0

*Rotational Equilibrium

The angular acceleration about any axis must be zero.

τ = 0Problem solving Strategy

1. Draw a Free­Body­Diagram

2. Resolve all forces into components, choosing a convenient coordinate system.

3. Choose a convenient axis for calculating the net torque; choose an origin that will simplify calculations.

4. Write out equation for each condition of equilibrium ­­> then solve equations!

80N p

2m

10m

F2 = ?

The board weighs 32N !

F2 = ?

What is F2 ? System must be in both translational and rotational equilibrium!

Hint: ΣFx=0 ΣFy=0 Σ =0

80N

Np

Fg=32N F2=?

2m 3m8m