Download SHM - Red Hook Central Schools

Displacement
What are the similarities and
differences between graphs?
Mass on spring
Both show oscillatory or cyclic
motion
Oscillatory/Periodic Motion
Repetitive Motion with a period and frequency.
•
Motion can be driven internally - mass on spring.
• Externally – pendulum or tides.
Guitar string
EEG Beta Waves
SHM – special case
• What do you notice?
Vocabulary
• Period T: time required for one cycle of
periodic motion (sec).
• Frequency: number of oscillations per unit
time.
unit is Hertz:
What are the period and frequency of the ECG
• T = 1 sec.
• f = 1 hz,
•

•

•
Angular Frequency - w (rad/sec)
w = 2pf.
remember angular speed from circular motion?
w = Dq/t
w = 2 p rad/s
• Equilibrium (0): the spot the mass would come to
rest when not disturbed – Fnet = 0.
• Displacement: (s or x): distance from equilibrium.
• Amplitude (xo) – max displacement from eq.
Representation of Oscillatory Motion
• Observe the motion of a bobbing mass. Where is the:
• Displacement positive?
• Displacement negative?
• Displacement zero?
• velocity positive?
• Where is the velocity negative?
• Where is the velocity zero?
Simple Harmonic Motion – SHM
• Isochronous period.
• Restoring Force directly proportional to displacement.
Double displacement, double force etc.
• Further from equilibrium, more force directed toward it.
• Force and displacement in opposite directions.
Pendulum is not SHM
• Fnet not directly opposite s
• for small displacement angles it approximates
SHM.
Free Body Diagram Mass on
Spring
•
•
•
•
Remember Hooke’s Law?
F = -kx.
F is the restoring force of the spring.
Complete sheet.
Hwk Intro
•
Simple Harmonic Motion
2 Conditions.
• 1. Acceleration/Fnet proportional to
displacement.
• 2. Acceleration/Fnet directed toward
equilibrium.
• Defining Equation for SHM
• a = -w2x
Graphs Of SHM
Acceleration - displacement
• Since F = - kx.
• ma = - kx.
• a = -k x
•
m
• a a–x
• Graph a (Y axis) vs. displacement on
the X
a = − kx
• Negative gradient = when displacement is positive,
acceleration and restoring force are negative
• ∴ a ∝ − x
• experimental evidence shows k = ω2, where ω is the
angular velocity, which is be ω = 2Πf :
• a = − ω2x
Sketching and
interpreting graphs of
SHM
v=0 v=v v=
0
MAX
-2.0
0.0
2.0
x
EXAMPLE: The displacement x vs. time t for a system
undergoing SHM is shown here.
x-black
(+)
( -)
(+)
( -)
(+) t
Sketch in red the velocity vs. time graph.
SOLUTION: At the extremes, v = 0.
At x = 0, v = vMAX. The slope determines sign of vMAX.
v-red
Sketching and interpreting
graphs of simple harmonic
motion
v=0
-2.0
v=
vMAX
0.0
v=
0
x
2.0
EXAMPLE: The displacement x vs. time t for a system
undergoing SHM is shown here.
x-black
t
Sketch in blue the acceleration vs. time graph.
SOLUTION: Since a  -x, a is just a reflection of x.
Note: x is a sine, v is a cosine, and a is a – sine
wave.
v-red
(different
scale)
a-blue
(different
scale)
Plot of displacement, velocity, acceleration,
on same graph.
Phase and Phase Difference
Two waves in Phase
Phase and Phase Difference
Phase shift = p
4. What is the phase shift?
Phase Difference
• The graphs all have the same period, but
• velocity leads displacement by ¼ of a period 90o.
• Acceleration leads velocity by ¼ of a period 90o.
• Displacement is 90o out of phase with velocity and 180o out of
phase with acceleration.
• When the phase difference is 0o or 360o, the systems are “in phase”.
x
-2.0 0.0 2.0
EXAMPLE: The kinetic energy
vs. displacement for a system
undergoing SHM is shown in
the graph. The system consists
of a 0.125-kg mass on a spring.
(a) Determine the maximum
velocity of the mass.
SOLUTION:
When the kinetic energy is maximum, the velocity
is also maximum. Thus 4.0 = (1/ 2)mvMAX2 so that
4.0 = (1/ 2)(.125)vMAX2  vMAX = 8.0 ms-1.
x
-2.0 0.0 2.0
ET
EXAMPLE: The kinetic energy
vs. displacement for a system
undergoing SHM is shown in
EK
the graph. The system consists
of a 0.125-kg mass on a spring.
(b) Sketch EP and determine the
EP
total energy of the system.
SOLUTION:
Since EK + EP = ET = CONST, and since EP = 0 when
EK = EK,MAX, it must be that ET = EK,MAX = 4.0 J.
Thus the EP graph will be the “inverted” EK graph.
x
-2.0 0.0 2.0
EXAMPLE: The kinetic energy
vs. displacement for a system
undergoing SHM is shown in
the graph. The system consists
of a 0.125-kg mass on a spring.
(c) Determine the spring
constant k of the spring.
SOLUTION: Recall EP = (1/2)kx2.
Note that EK = 0 at x = xMAX = 2.0 cm. Thus
EK + EP = ET = CONST  ET = 0 + (1/ 2)kxMAX2 so that
4.0 = (1/ 2)k 0.0202  k = 20000 Nm-1.
-2.0 0.0 2.0x
EXAMPLE: The kinetic energy
vs. displacement for a system
undergoing SHM is shown in
the graph. The system consists
of a 0.125-kg mass on a spring.
(d) Determine the acceleration
of the mass at x = 1.0 cm.
SOLUTION:
From Hooke’s law, F = -kx we get
F = -20000(0.01) = -200 N.
From F = ma we get -200 = 0.125a  a = -1600 ms-2.
-2.0
0.0
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(a) How do you know that the
mass is undergoing SHM?
SOLUTION:
In SHM, a  -x. Since F = ma, then F  -x also.
The graph shows that F  -x. Thus we have SHM.
2.0
x
x
-2.0 0.0 2.0
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(b) Find the spring constant of
the spring.
SOLUTION: Use Hooke’s law: F = -kx.
Pick any F and any x. Use k = -F / x.
Thus k = -(-5.0 N) / 1.0 m = 5.0 Nm-1.
F = -5.0 N
x = 1.0 m
x
-2.0 0.0 2.0
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(c) Find the total energy of the
system.
SOLUTION: Use ET = (1/2)kxMAX2. Then
ET = (1/2)kxMAX2 = (1/2)  5.0  2.02 = 10. J.
x
-2.0 0.0 2.0
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(d) Find the maximum speed of
the mass.
SOLUTION: Use ET = (1/2)mvMAX2.
10. = (1/2)  4.0  vMAX2
vMAX = 2.2 ms-1.
x
-2.0 0.0 2.0
EXAMPLE: A 4.0-kg mass is
placed on a spring’s end and
displaced 2.0 m to the right.
The spring force F vs. its
displacement x from equilibrium
is shown in the graph.
(e) Find the speed of the mass
when its displacement is 1.0 m.
SOLUTION: Use ET = (1/2)mv 2 + (1/2)kx 2. Then
10. = (1/2)(4)v 2 + (1/2)(5)12
v = 1.9 ms-1.
Graphical Treatment
Equations of SHM
Displacement, x, against time
x = xo cos wt
start point at max ampl.
** Set Calculator in Radians.
Displacement against time
x = xosinwt
•
•
•
•
Speed = d/t = 2pR/T but
2p /T = w so
v = wr but r the displacement x so.
v = wx.
Velocity against time v = vocoswt
Starting where?
Midpoint = max
velocity.
Equations of Graphs
• x = xo cos wt
x = xo sin wt
• v = -vo sin wt
v = vo cos wt
• a = -aocos wt
-ao sin wt
• Released from
top.
Released equilibrium.
Ex 2. A mass on a spring is oscillating
with f = 0.2 Hz and xo = 3 cm. What is
the displacement of the mass 10.66 s
after its release from the top?
• x = xo cos wt
xo = 3 cm
 w = 2pf. = 0.4 p Hz =1.26 rad/s.
• t = 10.66 s
• x = 0.03 cos (1.26 x 10.66) = 0.019 m
• You must use radians on calculator.
SHM and Circular Motion
• Use the relationship to derive
equations.
If an object moving with constant speed in a
circular path is observed from a distant point (in
the plane of the motion), it will appear to be
oscillating with SHM.
The shadow of a pendulum bob moves with
s.h.m. when the pendulum itself is either
oscillating (through a small angle) or moving in a
circle with constant speed, as shown in the
diagram.
For any s.h.m. we can find a corresponding circular
motion.
When a circular motion "corresponds to" a given s.h.m.,
i) the radius of the circle is equal to the amplitude
of the s.h.m.
ii) the time period of the circular motion is equal
to the time period of the s.h.m.
Derive Relationship between accl & w for SHM.
From circular motion ac = v2/r and vc = 2pr/T
Oscillating systems have acceleration too.
• But w = 2p/T
•
•
•
•
•
vc = 2pr/T
vc = wr
But ac = v2/r
So ac = (wr)2/r
but r is related to displacement x.
For any displacement:
a = -w²x
ao = -w²xo
The negative sign shows Fnet & accl
direction opposite displacement.
Derivation of accl in Hamper pg 76.
Or use 2nd derivative of displacement.
Ex 3. A pendulum swings with f = 0.5 Hz.
What is the size & direction of the
acceleration when the bob has displacement
of 2 cm right?
• a = -w²x
• w = 2pf
= p
• a = -(p)2 (0.02 m) = -0.197 m/s2. left.
Ex 4: A mass is bobbing on a spring with a
period of 0.20 seconds. What is its angular
acceleration at a point where its
displacement is 1.5 cm?
•
•
•
•
•
w = 2p/T
.w = 31 rad/s
a = -w²x
a = (31rad/s)(1.5 cm) = 1480 cm/s2.
15 m/s2.
To find the velocity of an oscillating mass or
pendulum at any displacement:
When the mass is at equilibrium, x = 0, and
velocity is maximum:
vo = ± wxo.
Derivation on H pg 77.
Ex 5. A pendulum swings with f = 1 Hz and
amplitude 3 cm. At what position will be its
maximum velocity &what is the velocity?
At max velocity vo = wxo.
w = 2pf = 2p(1) = 2p rad/s
vo = (2p rad/s)(0.03)
vo = 0.188 m/s
vo = 0.2 m/s
Hwk Hamper pg 75- 77 Show equations
and work, hand in virtual solar system lab.
Mechanical Universe w/questions
• http://www.learner.org/resources/series4
2.html?pop=yes&pid=565
SHM, Hooke’s Law & k.
For a mass undergoing SHM on a spring,
what is the relation between angular
frequency w, and k the spring constant?
• Use Hooke’s law and make
substitutions to derive a relation in
terms of angular frequency, k, and
mass.
F= - kx.
ma = - k x.
So a = -k x
m
a = -w² x
So w2 = -k/m
How can you tell this graph depicts SHM?
Acceleration directly a displacement, opposite direction.
What is the slope? w2 or k/m
Where is the amplitude?
Energy in SHM
Pendulum is not SHM
• Fnet not directly opposite s
• for small displacement angles it approximates
SHM.
Energy Conservation in Oscillatory Motion
In an ideal system the total mechanical energy
is conserved. A horizontal mass on a spring:
Horizontal mass no PEg.
Determining the max KE & PE:
•
•
•
•
•
•
•
•
What is PE for a stretched spring.
PE = ½ k x2 so PEmax =
PE = ½ k xo
What is the equation for KE?
KE = ½ mv2
Determine an equation for KEmax for SHM.
vo = wxo,
KEmax = ½ m(w2xo 2),
At any point:
• KE = ½ mw2 (xo2 - x2 )
• How could you determine PE from Etot?
• Subtract KE from Etot.
Since the total E will always equal the
max KE (or PE), we can calculate the
number of Joules of total E from the
KE equation:
ET = ½ mw2xo2
Ex 5: A 200-g pendulum bob is oscillating with
Amplitude = 3 cm, and f = 0.5 Hz. How much
KE will it have as it passes through the origin?
• KEmax = ½m w2xo 2,
• xo = 0.03 m
 w = p.
• KE = 8.9 x 10-4J.
6. The amplitude of an oscillating mass on a
vertical spring is 8.0 cm. The spring
constant is 74 N/m. Find the total energy of
the oscillator and the PE and KE at a
displacement = 4.8 cm.
• Et = 0.24 J.
• Ek = Et – Ep = 0.16 J.
Energy Conservation in Oscillatory Motion
The total energy is constant; as the kinetic
energy increases, the potential energy
decreases, and vice versa.
Energy Conservation in Oscillatory Motion
The E transforms from potential to kinetic &
back, the total energy remains the same.
The Pendulum
A simple pendulum consists of a mass m (of
negligible size) suspended by a string or rod of
length L (and negligible mass).
The angle it makes with the vertical varies with
time as a sine or cosine.
The Pendulum
The restoring force is
actually proportional
to sin θ, whereas the
restoring force for a
spring is proportional
to the displacement x.
Units
•
•The Pendulum
• Damped Oscillations
• Driven Oscillations and Resonance
Mass on a Spring
Since the force on a mass on a spring is
proportional to the displacement, and also to
the acceleration,
Make substitutions to find the relationship
between T and k.
Given that for a mass on a spring
Derive an equation that relates the period
of the oscillation to k, and m.
The Period of a Mass on a Spring
Therefore, the period is
How does T change as mass increases? Sketch it!
Mass on Spring
Period of pendulum
The Pendulum
Substituting θ for sin θ allows us to treat the
pendulum in a mathematically identical way to
the mass on a spring.
Therefore, we find that the period of a pendulum
depends only on the length of the string:
The Pendulum
For small angles, sin θ and θ are approximately
equal.
Damped Oscillations
In most physical situations, there is a force,
which will tend to decrease the amplitude of the
oscillation, and which is typically proportional
to the speed:
This causes the amplitude to decrease
exponentially with time:
Damped Oscillations
This exponential decrease is shown in the figure:
The image shows a system that is lightly damped – it goes
through multiple oscillations before coming to rest.
Damped Oscillations
A critically damped system is one that
relaxes back to the equilibrium position
without oscillating through it. An
overdamped system will also not oscillate
but is damped so heavily that it takes
longer to reach equilibrium.
Driven Oscillations and Resonance
An oscillation can be driven by an oscillating
driving force; the f of the driving force may or may
not be the same as the natural f of the system.
Driven Oscillations and Resonance
If the driving f is close
to the natural
frequency, the
amplitude can become
quite large, especially
if the damping is small.
This is called
resonance.
Summary
• Period: time required for a motion to go
through a complete cycle
• Frequency: number of oscillations per unit time
• Angular frequency:
• Simple harmonic motion occurs when the
restoring force is proportional to the
displacement from equilibrium.
Summary
• The amplitude is the maximum displacement
from equilibrium.
• Position as a function of time:
• Velocity as a function of time:
Summary
• Acceleration as a function of time:
• Period of a mass on a spring:
• Total energy in simple harmonic motion:
Summary
• Potential energy as a function of time:
• Kinetic energy as a function of time:
• A simple pendulum with small amplitude
exhibits simple harmonic motion
Summary
• Period of a simple pendulum:
• Period of a physical pendulum:
Summary
• An oscillating system may be driven by an
external force.
• This force may replace energy lost to friction,
or may cause the amplitude to increase greatly
at resonance
• Resonance occurs when the driving frequency
is equal to the natural frequency of the system
Summary
• Oscillations where there is a nonconservative
force are called damped.
• Underdamped: the amplitude decreases
exponentially with time:
• Critically damped: no oscillations; system
relaxes back to equilibrium in minimum time
• Overdamped: also no oscillations, but
slower than critical damping