Transcript Oscillatory Motion

Physics 111: Mechanics
Lecture 14
Dale Gary
NJIT Physics Department
Life after Phys 111

The course material of Phys 111 has given you a taste of a wide range
of topics which are available to you as a student. Prerequisite is Phys
121 or Phys 121H.

For those of you who have an interest in gravitation/astronomy, I
suggest the following electives:



For those of you interested in the biological or BME/medical aspects, I
suggest the following electives:


Phys 320, 321 – Astronomy and Astrophysics I and II
Phys 322 – Observational Astronomy
Phys 350 – Biophysics I, Phys 451 - Biophysics II
For those of your interested in light, optics, and photonics, I suggest the
following elective which Federici will be teaching this fall and Fall 2014:

OPSE 301 – Introduction to Optical Science and Engineering
Oscillatory Motion
Periodic motion
 Spring-mass system
 Differential equation of
motion
 Simple Harmonic Motion
(SHM)
 Energy of SHM
 Pendulum
 Torsional Pendulum

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Periodic Motion

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Periodic motion is a motion that regularly returns to a given
position after a fixed time interval.
A particular type of periodic motion is “simple harmonic
motion,” which arises when the force acting on an object is
proportional to the position of the object about some
equilibrium position.
The motion of an object
connected to a spring is a
good example.
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Recall Hooke’s Law

Hooke’s Law states Fs = -kx

Fs is the restoring force.





It is always directed toward the equilibrium position.
Therefore, it is always opposite the displacement from
equilibrium.
k is the force (spring) constant.
x is the displacement.
What is the restoring force for a surface water
wave?
Restoring Force and the
Spring Mass System



In a, the block is displaced to the right of x = 0.
 The position is positive.
 The restoring force is directed to
the left (negative).
In b, the block is at the equilibrium position.
 x = 0
 The spring is neither stretched nor
compressed.
 The force is 0.
In c, the block is displaced to the left of x = 0.
 The position is negative.
 The restoring force is directed to
the right (positive).
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Differential Equation of Motion
Using F = ma for the spring, we have ma  -kx
 But recall that acceleration is the second derivative of
the position:
d 2x

a

So this simple force equation is an example of a
differential equation,
d 2x
m 2  -kx
dt

dt 2
d 2x
k
or
- x
2
dt
m
An object moves in simple harmonic motion whenever its
acceleration is proportional to its position and has the
opposite sign to the displacement from equilibrium.
Acceleration
Note that the acceleration is NOT constant, unlike our
earlier kinematic equations.
 If the block is released from some position x = A, then
the initial acceleration is – kA/m, but as it passes
through 0 the acceleration falls to zero.
 It only continues past its equilibrium point because it
now has momentum (and kinetic energy) that carries it
on past x = 0.
 The block continues to x = – A, where its acceleration
then becomes +kA/m.

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Analysis Model, Simple Harmonic Motion


d 2x
k
What are the units of k/m, in a 
- x
2
dt
m
?
They are 1/s2, which we can regard as a frequency-squared, so let’s
write it as
k
2 
m

Then the equation becomes

A typical way to solve such a differential equation is to simply search
for a function that satisfies the requirement, in this case, that its
second derivative yields the negative of itself! The sine and cosine
functions meet these requirements.
a  - 2 x
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SHM Graphical Representation

A solution to the differential
equation is
x(t )  A cos(t  f )

A, , f are all constants:
A = amplitude (maximum position
in either positive or negative x direction,
 = angular frequency,
k
m
f = phase constant, or initial phase angle. T
A and f are determined by initial conditions.
Remember, the period
and frequency are:
2 
1  

f  

 
T 2 
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Motion Equations for SHM
x(t )  A cos(t  f )
dx
 - A sin(t  f )
dt
d 2x
a (t )  2  - 2 A cos(t  f )
dt
v(t ) 
The velocity is 90o out of phase with
the displacement and the
acceleration is 180o out of phase with
the displacement.
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SHM Example 1

Initial conditions at t = 0 are


x (0)= A
v (0) = 0
This means f = 0
 The acceleration reaches extremes
of  2A at  A.
 The velocity reaches extremes of
 A at x = 0.

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SHM Example 2
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Initial conditions at t = 0 are




x (0)= 0
v (0) = vi
This means f = -  / 2
The graph is shifted one-quarter
cycle to the right compared to the
graph of x (0) = A.
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Consider the Energy of SHM Oscillator


The spring force is a conservative force, so in a frictionless
system the energy is constant
Kinetic energy, as usual, is
K  12 mv2  12 m 2 A2 sin 2 t  f 

The spring potential energy, as usual, is
U  12 kx2  12 kA2 cos2 t  f 

Then the total energy is just
E  K  U  12 kA2 (a constant)
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Transfer of Energy of SHM
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The total energy is contant at all times, and is E  12 kA
(proportional to the square of the amplitude)
Energy is continuously being transferred between potential energy
stored in the spring, and the kinetic energy of the block.
2
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Simple Pendulum

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The forces acting on the bob are the
tension and the weight.
T is the force exerted by the string
mg is the gravitational force
The tangential component of the
gravitational force is the restoring force.
Recall that the tangential acceleration is
d 2
at  r  L  L 2
dt

This gives another differential equation
d 2
g
g

m
sin


m
 (for small  )
2
dt
L
L
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Frequency of Simple Pendulum

The equation for  is the same form as for the spring,
with solution
 (t )   max cos( t  f )
where now the angular frequency is

g
L

2
L
 2
 so the period is T =


g

Summary: the period and frequency of a simple
pendulum depend only on the length of the string and
the acceleration due to gravity. The period is
independent of mass.
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Torsional Pendulum
Assume a rigid object is suspended
from a wire attached at its top to a
fixed support.
 The twisted wire exerts a restoring
torque on the object that is
proportional to its angular position.
 The restoring torque is t  -k 
 k is the torsion constant of the
support wire.
 Newton’s Second Law gives

d 2
t  I  -k  I 2
dt
d 2
k


dt 2
I
Section 15.5