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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

- 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.

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

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 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 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

- 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 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

- 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

- 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)

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

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 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?

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