421: Oscillations
1
PH421: Homework 30%;
Laboratory reports 35%;
Final 35%. All lab reports will be submitted in class.
Mon
9
Tue
10
~ November 2015 ~
Wed
11
Civic Holiday
Thu
12
Lab & Discussion:
the LCR circuit
Fri
13
Upload data
-HW1 due
16
17
18
-HW2 (1,2) due
19
20
-HW2 due
Formal LCR Lab
Report Due
23
Upload data
24
Fourier Methods &
Impulse Lab
31
1
25
2
-demo pendulum lab
Research in the Physics; HW3(1,2) due
intro to senior thesis
-Formal Fourier
Impulse Lab Due
26
Civic Holiday
27
Civic Holiday
3
4
Review Session I
HW3 due
2
Goal #1: Intro to Formal
Technical Writing
• Two “formal” lab reports (35%) are required. Good
technical writing is very similar to writing an essay
with sub-heading. We want to hear a convincing
story, not a shopping list of everything you did.
• Check out course web-site
3
Goal #2: Intro to Research in Physics
• We introduce the senior thesis research/writing
requirement
• If you’re thinking about about grad school, med
school, etc. and have not started/planned out research
opportunities you are already well behind the
competition. Start now!
- due Feb. Department SURE Science scholarship
- due (very soon) external competitions, REUs, etc.
4
Are oscillations ubiquitous
or are they merely a paradigm?
Superposition of 5
brain neuron activity
REPRESENTING SIMPLE HARMONIC
MOTION
simple
not simple
6
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/shm.gif
Simple Harmonic Motion
y(t)
Watch as time evolves
7
x(t) = A cos (w 0t + f )
phase
angle
amplitude
position
1
T=
=
w0 f
angular freq
(cyclic) freq
x0
-0.25
-A
2p
1
A
f
t=w0
period
0.25
0.75
1.25
1.75
-1
time
determined by initial conditions
determined by physical system
8
1
0.5
Position (cm)
x(t) = A cos (w 0t + f )
0
-0.5
x
-1
0
t
1
2
0
t
1
2
0
t
1
2
6
4
2
Velocity (cm/s)
0
x(t) = Aw 0 cos (w 0t + f + p2 )v
-2
-4
-6
40
20
Acceleration (cm/s2)
0
x(t) = Aw cos (w 0t + f + p )a
2
0
-20
-40
time (s)
9
These representations of the position of a simple harmonic
oscillator as a function of time are all equivalent - there are 2
arbitrary constants in each. Note that A, f, Bp and Bq are
REAL; C and D are COMPLEX.
x(t) is real-valued variable in all cases.
A:
x(t) = Acos(w0t + f )
B:
x(t) = B p cos w 0 t + Bq sin w 0 t
C:
x(t) = C exp(iw 0 t ) + C * exp(-iw 0 t )
D:
x(t) = Re[ D exp(iw 0 t )]
Engrave these on your soul - and know how to derive the
relationships among A & f; Bp & Bq; C; and D .
10
Example: initial conditions
k
m
x(t) = Acos(w0t + f )
k
x(t) = B p cos w 0 t + Bq sin w 0 t
m
x
m = 0.01 kg; k = 36 Nm-1. At t = 0, m is displaced 50mm to
the right and is moving to the right at 1.7 ms-1.
Express the motion in
form A
form B
11
x(t) = Acos(w0t + f )
(
)
x(t) = 57.5 cos éë 60s ùû t - 0.516 mm
-1
x(t) = B p cos w 0 t + Bq sin w 0 t
(
)
(
x(t) = 50mm cos éë 60s -1 ùû t + 28.3mmsin éë 60s -1 ùû t
Bp = Acos f
Bq = -Asin f
A = Bp + Bq
2
tan f = -
)
2
Bq
Bp
12
Using complex numbers: initial conditions. Same
example as before, but now use the "C" and "D" forms
k
k
m
x(t) = C exp(iw 0 t ) + C * exp(-iw 0 t )
x(t) = Re[ D exp(iw 0 t )]
m
x
m = 0.01 kg; k = 36 Nm-1. At t = 0, m is displaced 50mm to
the right and is moving to the right at 1.7 ms-1.
Express the motion in
form C
form D
13

x(t) = C exp(iw 0 t ) + C * exp(-iw 0 t )
x(t )  (25  14.16i)e
i 60s 1t
 (25 14.16i)e
x(t) = Re[ D exp(iw 0 t )]

x(t )  Re 50  28.3i e
i 60s 1t
i 60s 1t
mm
mm
Acos f = Bp = 2Re[C ] = Re[ D]
Asin f = -Bq = 2Im[C] = Im[ D]
D = 2C = A
Im[ D] Im[C ]
tan f =
=
Re[ D] Re[C14]
Clicker Questions
15
A particle executes simple harmonic motion.
When the velocity of the particle is a maximum which one of
the following gives the correct values of potential energy and
acceleration of the particle.
(a)potential energy is maximum and acceleration is maximum.
(b)potential energy is maximum and acceleration is zero.
(c)potential energy is minimum and acceleration is maximum.
(d)potential energy is minimum and acceleration is zero.
16
A particle executes simple harmonic motion.
When the velocity of the particle is a maximum which one of
the following gives the correct values of potential energy and
acceleration of the particle.
(a)potential energy is maximum and acceleration is maximum.
(b)potential energy is maximum and acceleration is zero.
(c)potential energy is minimum and acceleration is maximum.
(d)potential energy is minimum and acceleration is zero.
Answer (d). When velocity is maximum displacement is zero so
potential energy and acceleration are both zero.
17
A mass vibrates on the end of the spring. The mass is replaced
with another mass and the frequency of oscillation doubles.
The mass was changed by a factor of
(a) 1/4
(b) ½
(c) 2
(d) 4
18
A mass vibrates on the end of the spring. The mass is replaced
with another mass and the frequency of oscillation doubles.
The mass was changed by a factor of
(a) 1/4 (b) 1/2 (c) 2 (d) 4
Answer (a). Since the frequency has increased the mass must
have decreased. Frequency is inversely proportional to the
square root of mass, so to double frequency the mass must
change by a factor of 1/4.
19
A mass vibrates on the end of the spring. The mass is replaced
with another mass and the frequency of oscillation doubles. The
maximum acceleration of the mass:
(a)
(b)
(c)
(d)
remains the same.
is halved.
is doubled.
is quadrupled.
20
A mass vibrates on the end of the spring. The mass is replaced
with another mass and the frequency of oscillation doubles. The
maximum acceleration of the mass:
(a)
(b)
(c)
(d)
remains the same.
is halved.
is doubled.
is quadrupled.
Answer (d). Acceleration is proportional to frequency squared. If
frequency is doubled than acceleration is quadrupled.
21
A particle oscillates on the end of a spring and its position as a
function of time is shown below.
At the moment when the mass is at the point P it has
(a) positive velocity and positive acceleration
(b) positive velocity and negative acceleration
(c) negative velocity and negative acceleration
(d) negative velocity and positive acceleration
22
A particle oscillates on the end of a spring and its position as a
function of time is shown below.
At the moment when the mass is at the point P it has
(a) positive velocity and positive acceleration
(b) positive velocity and negative acceleration
(c) negative velocity and negative acceleration
(d) negative velocity and positive acceleration
Answer (b). The slope is positive so velocity is positive. Since the
slope is getting smaller with time the acceleration is negative. 23
Optional Review of Complex Numbers
24
i = -1
Complex numbers
z = a + ib
Re ( z ) = a üï
ý real numbers
Im ( z ) = b ïþ
z = ze
if
Imag
|z|
b
f
a
z = a +b
2
2
Real
Argand diagram
b
tan f =
a
25
Euler’s relation
exp(if) = cos f + isin f
exp(iw 0 t ) = cos(w 0 t ) + i sin(w 0 t )
26
Consistency argument
z = z e if
z = a + ib
If these represent the same thing, then the assumed Euler
relationship says:
a + ib = z cos f + i z sin f
Equate real parts:
Equate imaginary parts:
z = a +b
2
2
a = z cos f
b = z sin f
b
tan f =
a
27
x(t) = Re[ Ae e
if iw 0 t
] = Re[ Ae
i (w 0 t+f )
]
Imag
t = T0/4
p
+f
2
PHASOR
t = 0, T0, 2T0
f
Real
w0t + f
t=t
28
Adding complex numbers is easy in rectangular form
z = a + ib
w = c + id
Imag
z + w = [ a + c] + i [b + d ]
b
a
c
Real
d
29
Multiplication and division of complex numbers is easy in polar form
z = ze
if
iq
w= we
Imag
i [f + q ]
zw = z w e
|z|
qf
q
|w|
f
Real
30
Another important idea is the COMPLEX CONJUGATE of a complex
number. To form the c.c., change i -> -i
z = a + ib
z* = a - ib
Imag
z + z* = [ a + a ] + i [ b - b ] = 2a
b
z = ze
if
z* = z e
if
zz* = z e z e
-if
-if
= z
|z|
a
f
Real
2
The product of a complex number
and its complex conjugate is REAL.
We say “zz* equals mod z squared”
31
And finally, rationalizing complex numbers, or: what to do when
there's an i in the denominator?
a + ib
z=
c + id
a + ib c - id
z=
´
c + id c - id
z=
ac + bd + i ( bc - ad )
c +d
ac + bd ( bc - ad )
= 2
+i 2
2
c +d
c + d2
2
Re ( z )
2
Im ( z )
32