Ch. 13 Oscillations About Equilibrium. Periodic Motion Motion that repeats itself over a fixed and...

Ch. 13 Oscillations About Equilibrium

Periodic Motion

• Motion that repeats itself over a fixed and reproducible period of time.• The revolution of a planet about its

sun is an example of periodic motion. The highly reproducible period (T) of a planet is also called its year.

•Mechanical devices on earth can be designed to have periodic motion. These devices are useful timers. They are called oscillators.

Simple Harmonic Motiona form of periodic motion of a particle,

in which the acceleration is always directed towards some equilibrium point and is proportional to the displacement from this point. Abbreviation SHM

Simple Harmonic Oscillators

A device that undergoes SHM.

Most common examples springs and pendulums.

Simple Harmonic Motion You attach a weight to a spring, stretch the spring

past its equilibrium point and release it. The weight bobs up and down with a reproducible period, T.

• Plot position vs. time to get a graph that resembles a sine or cosine function. The graph is "sinusoidal", so the motion is referred to as simple harmonic motion.

• Springs and pendulums undergo simple harmonic motion and are referred to as simple harmonic oscillators.

Amplitude

• Maximum displacement from equilibrium. • Related to energy. (Potential Energy)

Period (T) • Length of time required for one

oscillation. Units: seconds (s)

Can be found by # of oscillations divided by the time for the oscillations.

Frequency • How fast the oscillator is

oscillating. • f = 1/T • Unit: Hz or s-1

Sample Problem

• Determine the amplitude, period, and frequency of an oscillating spring using Lab Qwest and the motion sensors. See how this varies with the force constant of the spring and the mass attached to the spring.

Springs

Springs are a common type of simple harmonic oscillator.

Our springs are "ideal springs", which means • They are massless. • They are both compressible and extensible. They will follow Hooke's Law. • F = -kx

Ex: Calculate the period of a 200-g mass attached to an ideal spring with a force constant of 1000 N/m.

k = 1000 N/m, m = 200 g= 0.2 kgT = ?T = 2√(m/k)T = 2√(0.2/1000)T = 0.089 s

Ex: A 300-g mass attached to a spring undergoes simple harmonic motion with a frequency of 25 Hz. What is the force constant of the spring?

m = 0.3 kg, f = 25 Hzk = ? SOLVE FOR k: T = 2√(m/k)k = (42m)/T2 solve for T = 1/f = 1/25T = 0.04 s k = 42(0.3)/(.04)2

k = 7402.2 N/m

Ex: An 80-g mass attached to a spring hung vertically causes it to stretch 30 cm from its unstretched position. If the mass is set into oscillation on the end of the spring, what will be the period?

Ex: An 80-g mass attached to a spring hung vertically causes it to stretch 30 cm from its unstretched position. If the mass is set into oscillation on the end of the spring, what will be the period?

m = 0.08 kg, x = 0.03 mT = ?T = 2√(m/k) need to find kF = Fs – Fg = ma, not accelerating

Kx = mg k = mg/x = 0.08(10)/(0.3)

k = 2.67 N/mT = 2√(m/k)T = 2√(0.08/2.67)T = 1.088 s

Sample Problem You wish to double the force

constant of a spring. You • A. Double its length by connecting

it to another one just like it. • B. Cut it in half.• C. Add twice as much mass.• D. Take half of the mass off.

Sample Problem You wish to double the force

constant of a spring. You • A. Double its length by connecting

it to another one just like it. • B. Cut it in half.• C. Add twice as much mass.• D. Take half of the mass off.

Conservation of EnergySprings and pendulums obey conservation of

energy. • The equilibrium position has high kinetic

energy and low potential energy. • The positions of maximum displacement have

high potential energy and low kinetic energy. • Total energy of the oscillating system is

constant.

Sample problem. A spring of force constant k = 200 N/m is attached to

a 700-g mass oscillating between x = 1.2 and x = 2.4 meters. Where is the mass moving fastest, and how fast is it moving at that location?

Going to use Conservation of Energy.

U1 + K1 = U2 + K2

Midpoint or equilibrium position of the oscillation?It is 1.8 m, so what is the actual displacement during

the oscillation? x = 0.6 m, from either side of the equilibrium position

½ kx2 = ½ mv2

v = √(kx2/m) v = √(200(0.6)2/0.7) v = 10.1 m/s

Sample problem. A spring of force constant k = 200 N/m is attached to

a 700-g mass oscillating between x = 1.2 and x = 2.4 meters. What is the speed of the mass when it is at the 1.5 meter point?

Going to use Conservation of Energy again, you may pick your second point

4

U1 + K1 = U2 + K2 = U3 + K3 = U4 + K4 v2 = 10.1 m/sxmax (1 or 4) = 0.6 mv1 or 4 = 0 m/sx3 = 0.3 mx2 = 0 m

½ kx12 + ½mv1

2 = ½ kx32 + ½mv3

2 v3 = √[2(½ kx1

2 - ½ kx32) /m]

v3 = √[2(½(200)(0.6)2 - ½(200)(0.3)2) /0.7]v3 = 8.78 m/s

Sample problem. A 2.0-kg mass attached to a spring oscillates with an

amplitude of 12.0 cm and a frequency of 3.0 Hz. What is its total energy?

Pendulums The pendulum can be thought of as a simple

harmonic oscillator. The displacement needs to be small for it to work properly.

Sample problem Predict the period of a pendulum consisting of a 500

gram mass attached to a 2.5-m long string.m = 0.5 kg, l = 2.5 m, g = 10 m/s2

T = ?T = 2√(l/g) T = 2√(2.5/10)T = 3.14 s

Sample problem Suppose you notice that a 5-kg weight tied to a string

swings back and forth 5 times in 20 seconds. How long is the string?

m = 5 kg, t = 20 s, # of oscillations = 5l = ?T = 2√(l/g) solve for l and we need to find T.T = 20/5T = 4 s l = (T2g)/(42) l = [42(10)]/(42) l = 4.05 m

Sample problem The period of a pendulum is observed to be T.

Suppose you want to make the period 2T. What do you do to the pendulum?

T = 2√(l/g)T = 2√(4l/g) T = √(4) [2√(l/g)]T = 2 [2√(l/g)] = 2T

Conservation of EnergyPendulums also obey conservation of energy. • The equilibrium position has high kinetic

energy and low potential energy. • The positions of maximum displacement have

high potential energy and low kinetic energy. • Total energy of the oscillating system is

constant.