Simple Harmonic Motion

Things that vibrate

§ 13.1–13.3

Hooke’s Law

• Force Law: F = –kx

• F = force exerted by the spring

• k = spring constant (characteristic of the particular spring)

• x = distance spring is displaced from equilibrium

Conditions of Motion

• Newton’s second law: F = ma

• Force F depends on position by Hooke’s law: F = –kx

–kx = ma

• tells how motion changes at each position

• second-order ordinary differential equation

–kx = md2xdt2

Group Work

a. Plot the mass’s position vs. time.

b. Select at least four events on the plot. At each event, draw arrows to indicate the mass’s velocity and acceleration.

A Hooke’s law spring is fixed at one end and has a mass attached to the other end. The spring is initially at its neutral position. The mass is pulled, stretching the spring. The mass is then released.

CPS Question

The net force on a Hooke’s law object is

A. zero at the top and bottom

B. maximum at the top and bottom

C. minimum but not zero at the top and bottom

CPS Question

The acceleration of a Hooke’s law object is

A. zero at the top and bottom

B. maximum at the top and bottom

C. minimum but not zero at the top and bottom

CPS Question

The velocity of a Hooke’s law object is

A. maximum at the equilibrium position

B. maximum at the top and bottom

C. maximum midway between equilibrium and top or bottom

Simulation Spreadsheet

• Describe how position changes with time.• What is the velocity when position is at an extreme? At

0?• What is the force when velocity is at an extreme? At 0?• Where is max/min U?• Where is max/min K?

From class web site, right-click on the “Numerical simulation” link.

Enter parameters to produce 1–2 cycles.

Simulation Spreadsheet

• What happens to the period when k changes?• What happens to the period when m changes?

From class web site, right-click on the “Numerical simulation” link.

Enter parameters to produce 1–2 cycles.

CPS Question

All other things being equal, if the mass of a Hooke’s law oscillator increases, its period

A. decreases.

B. does not change.

C. increases.

CPS Question

All other things being equal, if the stiffness k of a Hooke’s law spring increases, its period

A. decreases.

B. does not change.

C. increases.

Group Work

• First MasteringPhysics problem, parts A-E.

Uniform Circular Motion

• Centripetal force F = mv2/r inwards

• Constant magnitude F0; direction depends on position

F0

F0

F0

F0

F0

F0

• Force in y-direction is proportional to –y

Uniform Circular Motion

• Angle changes at a steady rate.

• Projection on y-axis has Hooke’s law force.

• So, projection on y-axis must have Hooke’s law motion too!

• What is the projection of an angle on the y-axis?

Group Work

Verify that

if x(t) = A cos (t + ),

where• A and are any real constants

–kx = md2xdt2

• =   k/m

Equation of Motion

x(t) = A cos (t + ), • Amplitude A • Angular frequency • Initial phase angle

Period and Frequency

• Period T– time of one cycle (units: s)

• Frequency f– cycles per unit time (units: 1/s = Hz)– f = 1/T

• Angular frequency – radians per unit time (units: 1/s = Hz)– = 2f– =   k/m

Another form

x(t) = A cos (t + ) = C cos (t) + S sin (t)

where

C = A cos () and S = –A sin ()

Group Work

• Fourth MasteringPhysics problem (Exercise 13.4).

Group Work

• Second MasteringPhysics problem (“Harmonic Oscillator Acceleration”).

Group Work

• Fifth MasteringPhysics problem (Exercise 13.10), parts A–C.

Initial Conditions

• Given m, k, x0, and v0, how does one find the equations of motion?– m and k give .

– x0, v0, and give A.

– x0/A and v0/(A) give .

• Now you can do MasteringPhysics Exercise 13.26.

• Can practice/explore with “SHM parameters” spreadsheet

CPS Question

The potential energy of an oscillating mass is greatest

A. at its extreme positions.

B. at its equilibrium (middle) position.

C. between the middle and an extreme position.

CPS Question

The kinetic energy of an oscillating mass is greatest

A. at its extreme positions.

B. at its equilibrium (middle) position.

C. between the middle and an extreme position.

Constant Mechanical Energy

E = 1/2 kx2 + 1/2 mv2

Energy

• Potential energy of a stretched spring :

PE = kx212

• Conservation of energy:

(This of course ignores the sullen reality of energy dispersal by friction and drag.)

PE + KE = constant = kA212

where A is the oscillation amplitude.

Energy

y

time

KE = 0

KE = 0

KE = max PE = 0

PE = max

PE = max

Group Work

• First MasteringPhysics problem, parts F and G

Effect of Gravity

• Less than you might expect:

• Changes equilibrium position x = 0

• Does not change k

Spring + Gravity

position

forc

e

0

0

spring alone

gravity

Spring + Gravity

position

forc

e

0

0

spring alone

gravity

0

spring + gravity

Spring + Gravity

position

net

forc

e

0

0

different equilibrium position

same k