Transcript File - Mr MAC`s Physics

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 Motion
a 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 kg
T=?
T = 2p√(m/k)
T = 2p√(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 Hz
k = ? SOLVE FOR k: T = 2p√(m/k)
k = (4p2m)/T2
solve for T = 1/f = 1/25
T = 0.04 s  k = 4p2(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 m
T=?
T = 2p√(m/k) need to find k
SF = Fs – Fg = ma, not accelerating
Kx = mg  k = mg/x = 0.08(10)/(0.3)
k = 2.67 N/m
T = 2p√(m/k)
T = 2p√(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 Energy
Springs 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
½ kx12 + ½mv12 = ½ kx32 + ½mv32
v3 = √[2(½ kx12 - ½ kx32) /m]
v2 = 10.1 m/s
xmax (1 or 4) = 0.6 m
v1 or 4 = 0 m/s
x3 = 0.3 m
x2 = 0 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 = 2p√(l/g)
T = 2p√(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 = 5
l=?
T = 2p√(l/g) solve for l and we need to find T.
T = 20/5
T=4s
l = (T2g)/(4p2)
l = [42(10)]/(4p2)
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 = 2p√(l/g)
T = 2p√(4l/g)
T = √(4) [2p√(l/g)]
T = 2 [2p√(l/g)] = 2T
Conservation of Energy
Pendulums 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.