Transcript 13_LectureOutline

Lecture Outline
Chapter 13
Physics, 4th Edition
James S. Walker
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Chapter 13
Oscillations about
Equilibrium
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Units of Chapter 13
• Periodic Motion
• Simple Harmonic Motion
• Connections between Uniform Circular
Motion and Simple Harmonic Motion
• The Period of a Mass on a Spring
• Energy Conservation in Oscillatory
Motion
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Units of Chapter 13
• The Pendulum
• Damped Oscillations
• Driven Oscillations and Resonance
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13-1 Periodic Motion
Period: time required for one cycle of periodic
motion
Frequency: number of oscillations per unit
time
This unit is
called the Hertz:
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13-2 Simple Harmonic Motion
A spring exerts a restoring force that is
proportional to the displacement from
equilibrium:
A mass attached to a spring undergoes simple harmonic motion about x = 0
(a) The mass is at its maximum positive value of x. Its velocity is zero, and the force on it points to the left with
maximum magnitude. (b) The mass is at the equilibrium position of the spring. Here the speed has its maximum has its
maximum value, and the force exerted by the spring is zero. (c) The mass is at its maximum displacement in the
negative x direction. The velocity is zero is zero here, and the force points to the right with maximum magnitude. (d)
The mass is at the equilibrium position of the spring with zero force acting on it and maximum speed. (e) The mass has
completed one cycle of its oscillation about x = 0.
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13-2 Simple Harmonic Motion
A mass on a spring has a displacement as a
function of time that is a sine or cosine curve:
Here, A is called
the amplitude of
the motion.
Displaying position versus time for simple harmonic motion
As an air-track cart oscillates about its equilibrium position, a pen attached to it
traces its motion onto a moving sheet of paper. This produces a “strip chart”,
showing that the cart’s motion has the shape of a sine or a cosine.
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13-2 Simple Harmonic Motion
If we call the period of the motion T – this is the
time to complete one full cycle – we can write
the position as a function of time:
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13-2 Simple Harmonic Motion
An air-track cart attached to a spring completes one oscillation every 2.4
s. At t=0, the cart is released from rest at a distance of 0.10 m from its
equilibrium position. What is the position of the cart at (a) 0.30 s, (b) 0.6 s (
c) 2.7 s
X = (0.10 m) cos [( 2π/2.4 s) (0.30 s)] = (0.10 m) cos (π /4) = 7.1 cm
X = (0.10 m) cos [( 2π/2.4 s) (0.60 s)] = (0.10 m) cos (π /2) = 0
X = (0.10 m) cos [( 2π/2.4 s) (2.7 s)] = (0.10 m) cos (9π /4) = 7.1 cm
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13-3 Connections between Uniform Circular
Motion and Simple Harmonic Motion
An object in simple
harmonic motion has the
same motion as one
component of an object
in uniform circular
motion:
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13-3 Connections between Uniform Circular
Motion and Simple Harmonic Motion
Here, the object in circular motion has an
angular speed of
where T is the period of motion of the
object in simple harmonic motion.
Θ=ωt
That is the angular position increases linearly with time
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13-3 Connections between Uniform Circular
Motion and Simple Harmonic Motion
The position as a function of time:
The angular frequency, ω:
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13-3 Connections between Uniform Circular
Motion and Simple Harmonic Motion
The velocity as a function of time:
And the acceleration:
Si unit: m/s2
Both of these are found by taking
components of the circular motion quantities.
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13-3 Connections between Uniform Circular
Motion and Simple Harmonic Motion
As in example 13-1, an air-track cart attached to a spring completes
one oscillation every 2.4 s. At t =0 the cart is released from rest with
the spring stretched 0.10 m from its equilibrium position. What are the
velocity and acceleration of the cart at (a) 0.30 s and (b) 0.60 s?
ω = 2π/T = 2 π/ (2.4 s) = 2.6 rad /s
V = - (0.10 m) (2.6 rad /s) sin [ ( 2π / 2.4 s) (0.30 s)]
V = - (26 cm/s) sin (π /4) = - 18 cm/s
a = - (0.10 m) (2.6 rad /s)2 cos [ ( 2π / 2.4 s) (0.30 s)]
= - (68 cm / s2) cos (π /4) = - 48 cm /s2
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13-4 The Period of a Mass on a Spring
Since the force on a mass on a spring is
proportional to the displacement, and also to
the acceleration, we find that
.
Substituting the time dependencies of a and x
gives
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13-4 The Period of a Mass on a Spring
Therefore, the period is
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13-4 The Period of a Mass on a Spring
On December 29,1997, a United Airlines flight from Tokyo to Honolulu was
hit with severe turbulence 31 minutes after takeoff. Data from the
airplane’s “black box” indicated the 747 moved up and down with an
amplitude of 30.0 m and a maximum acceleration of 1.8g. Treating the upand –down motion of the plane as simple harmonic, find (a) the time
required for one complete oscillation and (b) the plane’s maximum vertical
speed. ( c) what amplitude of motion would result in a maximum
acceleration of 0.50g, everything remaining the same?
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13-4 The Period of a Mass on a Spring
A 0.120 kg mass attached to a spring oscillates with an amplitude of
0.0750 m and a maximum speed of 0.524 m/s. Find (a) the force constant
and (b) the period of motion.
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13-4 The Period of a Mass on a Spring
Now, when a mass m is attached to
a vertical spring, it causes the
spring to stretch. In fact, the
vertical spring is in equilibrium
when it exerts an upward force
equal to the weight of the mass.
That is, the spring stretches by an
amount yo given by
kyo = mg
yo = mg / k.
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13-4 The Period of a Mass on a Spring
A 0.260 kg mass is attached to a vertical spring. When the mass is put into motion,
its period is 1.12 s. (a) How much does the mass stretch the spring when it is at
rest in its equilibrium position? (b) Suppose this experiment is repeated on a
planet where the acceleration due to gravity is twice what it is on Earth. By what
multiplicative factors do the period and equilibrium stretch change?
T = 2 π (m / k)1/2
K = 4π2m / T2 = 4 π2 (0.260 kg) / 1.12 s)2 = 8.18 kg / s2
K = 8.18 N/m
Kyo = mg
yo = mg / k = (0.26 kg)(9.81 m/s2) / 8.18 N/m
yo = 0.312 m
(b) T = 2 π (m / k)1/2
g - 2g
yo = mg / k - 2yo
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13-5 Energy Conservation in Oscillatory
Motion
In an ideal system with no nonconservative
forces, the total mechanical energy is
conserved. For a mass on a spring:
Since we know the position and velocity as
functions of time, we can find the maximum
kinetic and potential energies:
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13-5 Energy Conservation in Oscillatory
Motion
As a function of time,
So the total energy is constant; as the
kinetic energy increases, the potential
energy decreases, and vice versa.
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13-5 Energy Conservation in Oscillatory
Motion
This diagram shows how the energy
transforms from potential to kinetic and
back, while the total energy remains the
same.
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13-5 Energy Conservation in Oscillatory
Motion
A 0.98 kg block slides on a frictionless, horizontal surface with a
speed of 1.32 m/s. The block encounters an unstretched spring with a
force constant of 245 N /m, as shown in the sketch. (a) How far is the
spring compressed before the block comes to rest? (b) How long is
the block in contact with the spring before it comes to rest?
1/2mvo = 1/2KA2
A = Vo (m/k)1/2 = (1.32 m /s )(0.98 kg) (245 N / m)1/2
A = 0.0835 m
Calculate the period of one oscillation
T = 2π (m/k)1/2 = 2π (0.98 kg/ 245 N/m)1/2 = 0.397 s
In moving from the equilibrium position of the spring to
maximum compression, the mass has undergone
one-quarter of a cycle; thus the time is T/4.
t = 1/4T = ¼ (0.397 s) = 0.0993 s
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13-6 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.
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13-6 The Pendulum
Looking at the forces
on the pendulum bob,
we see that the
restoring force is
proportional to sin θ,
whereas the restoring
force for a spring is
proportional to the
displacement (which
is θ in this case).
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13-6 The Pendulum
However, for small angles, sin θ and θ are
approximately equal.
Period of a Pendulum (small amplitude)
T = 2 π (L / g) ½
SI unit : s
Relationship between sin θ and θ.
For small angles measured in radians,
sin θ is approximately equal to θ. Thus,
when considering small oscillations of a
pendulum, we can replace sin θ with θ.
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13-6 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:
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13-6 The Pendulum
A pendulum is constructed from a string 0.627 m long attached to a mass of 0.250 kg. When set
in motion, the pendulum completes one oscillation every 1.59 s. If the pendulum is held at rest
and the string is cut, how long will it take for the mass to fall through a distance of 1.0 m?
T
g = 4 π2 L / T2
g = 4 π2 (0.627 m) / ( 1.59 s)2 = 9.79 m/s2
y = 1/2gt2 or t = (2y / g ) ½
t = (2(1.0 m) / 9.79 m / s2)1/2 = 0.452 s
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13-6 The Pendulum
A physical pendulum is a
solid mass that oscillates
around its center of mass,
but cannot be modeled as a
point mass suspended by a
massless string. Examples:
Examples of physical pendula
In each case, an object of definite size
and shape oscillates about a given pivot
point. The period of oscillation depends
in detail on the location of the pivot point
as well as on the distance l from it to the
center of mass, CM.
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13-6 The Pendulum
In this case, it can be shown that the period
depends on the moment of inertia:
Substituting the moment of inertia of a point
mass a distance l from the axis of rotation
gives, as expected,
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13-7 Damped Oscillations
In most physical situations, there is a
nonconservative force of some sort, 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:
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13-7 Damped Oscillations
This exponential decrease is shown in the
figure:
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13-7 Damped Oscillations
The previous image shows a system that is
underdamped – it goes through multiple
oscillations before coming to rest. A critically
damped system is one that relaxes back to the
equilibrium position without oscillating and in
minimum time; an overdamped system will
also not oscillate but is damped so heavily
that it takes longer to reach equilibrium.
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13-8 Driven Oscillations and Resonance
An oscillation can be driven by an oscillating
driving force; the frequency of the driving force
may or may not be the same as the natural
frequency of the system.
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13-8 Driven Oscillations and Resonance
If the driving frequency
is close to the natural
frequency, the
amplitude can become
quite large, especially
if the damping is small.
This is called
resonance.
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Summary of Chapter 13
• 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.
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Summary of Chapter 13
• The amplitude is the maximum displacement
from equilibrium.
• Position as a function of time:
• Velocity as a function of time:
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Summary of Chapter 13
• Acceleration as a function of time:
• Period of a mass on a spring:
• Total energy in simple harmonic motion:
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Summary of Chapter 13
• Potential energy as a function of time:
• Kinetic energy as a function of time:
• A simple pendulum with small amplitude
exhibits simple harmonic motion
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Summary of Chapter 13
• Period of a simple pendulum:
• Period of a physical pendulum:
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Summary of Chapter 13
• 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
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Summary of Chapter 13
• 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
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