Adv physics
Download
1 / 14

Adv Physics - PowerPoint PPT Presentation


  • 185 Views
  • Uploaded on

Adv Physics. Chapter 13 Sections 1 & 2. Simple Harmonic Motion. Any motion that repeats itself over the same path in a fixed time due to a restoring force ex – simple pendulum (gravity) mass on spring (force of spring) - restoring force is proportional to displacement.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Adv Physics' - primo


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Adv physics

Adv Physics

Chapter 13

Sections 1 & 2


Simple harmonic motion
Simple Harmonic Motion

  • Any motion that repeats itself over the same path in a fixed time due to a restoring force

    ex – simple pendulum (gravity)

    mass on spring (force of spring)

    - restoring force is proportional to displacement


Hooke s law
Hooke’s Law

  • For small displacements from equilibrium, a spring applies a force to a mass that is proportional to its displacement from equilibrium and opposite in direction


Hooke s law cont d
Hooke’s Law cont’d.

F = -kx where F = force

k = spring constant

(-indicates stiffness of

spring)

x = displacement

Negative in equation since force is always opposite the direction of displacement


Sample problem
Sample Problem

A 76 N crate is attached to a spring (k = 450 N/m). How much displacement is caused by the weight of this crate?


Sample problem1
Sample Problem

A spring of k = 1962 N/m loses its elasticity if stretched more than 50 cm. What is the mass of the heaviest object the spring can support without being damaged?


Sample problem2
Sample Problem

If a mass of 0.55 kg attached to a vertical spring stretches the spring 2 cm from its original equilibrium position, what is the spring constant?


Simple harmonic oscillators
Simple Harmonic Oscillators

  • All SHOs cycle through acceleration, velocity, force and energy

  • Pendulum

    at release pt.(max amplitude) – max F

    - max PE, min KE

    -max a

    -min v (v=0)


Simple harmonic oscillators1
Simple Harmonic Oscillators

bottom of swing - min F (F=0)

(amp = 0) - min PE, max KE

- min a (a=0)

- max v

max swing - max F

(max amp) - max PE, min KE

- max a

- min v (v=0)


Terms to know
Terms to Know

  • Period – time it takes to complete one cycle of motion

    T = time/#cycles

    [T] = sec

  • Frequency – number of cycles per second

    f = #cycles/time

    [f] = 1/sec = Hertz, Hz


Periods
Periods

  • Period of simple pendulum

    ____

    T = 2π √ (l/g)

  • Period depends on length and accel due to g

  • Period does not depend on mass or how far you pull it back (if Θ < 15 degrees)


Periods1
Periods

  • Period of a mass-spring oscillator

    _____

    T = 2π √ (m/k)

  • Period depends on stiffness of spring and mass on spring

  • Period does not depend on how far you pull it back


Mass spring oscillators
Mass Spring Oscillators

Us = ½ kx2

KE = ½ mv2

Total energy = ½ kA2

___________

Since E = Us + KE v=√ (k/m) (A2 – x2)

____

Max v when x = 0 so vmax=√ (k/m)A


Sample problem3
Sample Problem

A mass spring system is in SHM in the horizontal direction. If the mass is 0.25 kg, the spring constant is 12 N/m, and the amplitude is 15 cm, (a) what would be the maximum speed of the mass, and (b) where would this occur? (c) What would be the speed at a half amplitude position?


ad