plus pendulums revisited. simple harmonic motion oscillatory motion (repeated motion) only shm if...

Plus Pendulums Revisited

Simple Harmonic MotionOscillatory Motion (repeated motion)Only SHM if

Oscillations are about an equilibrium positionThe restoring force is proportional to the object’s

displacement from equilibriumExamples include:

Tacoma Narrows Bridge Collapse! http://www.youtube.com/watch?v=ASd0t3n8Bnc

Springs with masses on the endPendulums with VERY small angles (< 10 degrees)

For most pendulum motion we use energy or circular motion.

Consider a mass oscillating horizontally across a frictionless surface:For a, b, and c :Label Fe, KE, EPE, v, a, and x. Numbers are not required, except for zeroes. You may use max and min to describe quantities.

Horizontal Oscillating System (cont.)(b) This is defined as the equilibrium position,

where ∆x=0, and as such has no net force. Fe=0According to Newton’s Second Law, net force

is proportional to acceleration. If net force here is zero, then a=0

When you stretch a spring, you give it elastic PE. When you let it go, the EPE is converted to KE. At equilibrium the spring has max KE

KE=1/2mv2, so if KE is maximum at equilibrium, then I also have max v (velocity)

Horizontal Oscillating System (cont.)(a) and (c) are almost identical in terms of these

quantities. At (a) the spring is stretched to its maximum while still allowing it to oscillate. This is called the amplitude (A). So, ∆x=+A

At (c) the spring is compressed the maximum amount while still allowing for oscillation. This is the same amplitude as (a). So, ∆x=-A

Hooke’s Law says Fe=-k∆x. If I apply what I know about amplitude at either end, I get

(a) max Fe, pointing to the left(b) max Fe, pointing to the right

Horizontal Oscillating System (cont.)(a) and (c) According to Newton’s Second Law,

net force is proportional to acceleration. If net force here is at a maximum, then I have max a

(a) and (c) At either end, while oscillating the mass has to stop to return to equilibrium. At both ends v=0

(a) and (c) At either end, if v=0, then I have no kinetic energy. Elastic potential energy is EPE=1/2k∆x2 , so if my spring is at maximum compression or stretched to a maximum, I have max EPE

Frequency (f) vs Period (T)Number of oscillations Time for one

oscillation per one secondUnit : Hertz (Hz) Unit : Seconds

(s)

These quantities are inverses, so…

time

oscf

osc

timeT

Tf

1

fT

1

How to predict periods for oscillators:Pendulum:

SHM (springs):

g

lT 2

k

mT 2

A 67 kg boy swings from a vine with a period of oscillation on 2.34 seconds. What is the length of the vine?

(Not PAP)A mass is allowed to oscillate on the end of a spring of constant 140 N/m. If the mass oscillates with a period of 1.26 seconds, what is the value of the mass?

(PAP) An 125N is hung on the end of a spring, stretching it 23 cm. The mass is then replaced with an unknown mass and allowed to oscillate. If the mass oscillates with a period of 1.26 seconds, what is the value of the mass?

Do the following practice problems:1. When a 55 g mass is attached to a spring, it

makes 37 vibrations in 10.0 seconds. What is the system’s frequency? What is its period?

2. An unknown load oscillates from the end of a spring, of constant 55 N/m, with a period of 6.58 seconds. What is the mass of the load?

3. A mass oscillates horizontally at the end of a spring of constant 85N/m. If the system has an amplitude of 11 cm, what is the maximum kinetic energy? What is the maximum potential energy?