REPRESENTING SIMPLE HARMONIC
MOTION
simple
not simple
0
http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/shm.gif
Simple Harmonic Motion
y(t)
Watch as time evolves
1
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
2
determined by physical system
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)
3
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 .
4
Example: initial conditions
k
m
k
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 x(t) = Acos
(w0t + f )
form B
x(t) = B p cos w 0 t + Bq sin w 0 t
5
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
6
)
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.
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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.
8
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
9
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.
10
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.
11
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.
12
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
13
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
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slope is getting smaller with time the acceleration is negative.
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
15
Euler’s relation
exp(if) = cos f + isin f
exp(iw 0 t ) = cos(w 0 t ) + i sin(w 0 t )
16
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
17
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
18
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
19
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
20
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”
21
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 )
22
Using complex numbers: initial conditions. Same
example as before, but now use the "C" and "D" forms
k
m
k
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
x(t) = C exp(iw 0 t ) + C * exp(-iw 0 t )
form D
x(t) = Re[ D exp(iw 0 t )]
23

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 =
= 24
Re[ D] Re[C ]