SIMPLE HARMONIC MOTION:
NEWTON’S LAW
simple
not simple
PRIOR READING:
Main 1.1, 2.1
Taylor 5.1, 5.2
0
http://www.myoops.org/twocw/mit/NR/rdonlyres/Physics/8-012Fall-2005/7CCE46AC-405D-4652-A724-64F831E70388/0/chp_physi_pndulm.jpg
The simple pendulum
r; r = L
q
T
Energy approach
q
0
0
dq '
ò t ' = qò
m
mg
t
t=
2
E - mgLcm (1- cosq '))
(
I
q
òq
0
dq '
2mgLcm
1- cosq max - (1- cosq '))
(
2
mLcm
t=
q
òq
0
dq '
2g
cosq '- cosq max )
(
Lcm
1
Lcm = L
q
T
The simple pendulum
Energy approach
t=
m
q
dq '
òq
2g
cosq '- cosq max )
(
Lcm
0
mg
g
w0 º
L
angular frequency
w t = arcsin
q
q max
t=
q
òq
0
q0
- arcsin
q max
f
dq '
g 2
q max- q '2
L
(
q
q'
=
arcsin
w0
q max q
1
)
0
Þ q ( t ) = q max sin (w t + f )
2
r; r = L
q
T
The simple pendulum
m
t = Iq
t = r ´ åF = ?
mg
Newton
Known torque
ˆ
t = Lmgsinq Ä
Iq = mL2q ·ˆ
g
q = - sin q
L
This is NOT a restoring force proportional to displacement
(Hooke’s law motion) in general, but IF we consider small
motion, IT IS! Expand the sin series …
sin q = q -
q3
3!
+
q5
5!
+ ...
3
q
L
T
m
mg
The simple pendulum
in the limit of small angular displacements
g
q = - sin q ®
L
g
g
q = - q or q + q = 0
L
L
k
Compare with x + x = 0
m
What is q(t) such that the above equation is obeyed?
q is a variable that describes position
t is a parameter that describes time
"dot" and "double dot" mean differentiate w.r.t. time
g, L are known constants, determined by the system.
4
REVIEW PENDULUM
q
g
q+ q=0
L
L
T
q (t) = Ce
m
mg
pt
C, p are unknown (for now) constants,
possibly complex
q (t) = p 2Ce pt = p 2q (t)
Substitute:
g
pq+ q=0
L
2
g
p = ±i
= ±iw 0
L
p is now known (but C is not!). Note that w0 is NOT a new
quantity! It is just a rewriting of old ones - partly
5
shorthand, but also "w" means "frequency" to physicists!
REVIEW PENDULUM
q
L
TWO possibilities …. general
solution is the sum of the two and it
must be real (all angles are real).
T
m
iw 0t
q (t) = Ce
+ C 'e
-iw 0t
mg
If we force C' = C* (complex conjugate of C), then
x(t) is real, and there are only 2 constants, Re[C],
and Im[C]. A second order DEQ can determine only
2 arbitrary constants.
q (t) = Ceiw t + C *e-iw t
0
0
Simple harmonic motion
6
REVIEW PENDULUM
q
L
T
m
q (t) = Ceiw t + C *e-iw t
0
Re[C], Im[C] chosen to fit initial
conditions. Example:
q(0) = 0 rad and dqdt(0) = 0.2 rad/sec
mg
q (0) = C e
iw 0 0
0
+C*e
-iw 0 0
1
q (0)  iw0C eiw 0  iw0C * e  iw 0
0
1
1
0
1
0 = C + C* = 2 Re[C]
0.2rad / s  iw0  C  C *  iw0 2i Im[C ]
Þ Re[C] = 0
0.2
0.1
 Im[C ] 

2w0
w0
C  0i
0.1
w0
cartesian

0.1
w0
i

e 2 ; C*  ?
polar
7
REVIEW PENDULUM
q
L
T
m
mg
q (t )  Ceiw t  C *eiw t ; C 
0
q (t) =
q (t) =
q (t) =
0.1
w0
0.1
w0
0.1
w0
0
i
p
e e
2
(e
iw 0 t
+
i(w 0 t+p /2 )
0.1
w0
+e
e
-i
0.1
w0
p
2
e
i

e2
-iw 0 t
-i(w 0t+p /2)
)
( 2 cos (w t + p / 2))
0
æ
ö
q (t) =
cos çw 0 t + p / 2 ÷
w0
è
f ø
0.2
A
8
Remember, all these are equivalent forms. All of them have a
known w0=(g/L)1/2, and all have 2 more undetermined
constants that we find … how?
q (t) = A cos (w 0t + f )
q (t) = Bp cos w 0t + Bq sin w 0t
q (t) = C exp ( iw 0t ) + C * exp ( -iw 0 t )
q (t) = Re éë D exp ( iw 0t ) ùû
Do you remember how the A, B, C, D constants are related?
If not, go back and review until it becomes second nature!
9
q
L
T
m
mg
g
w0 =
L
L
T = 2p
g
The simple pendulum
("simple" here means a point mass; your
lab deals with a plane pendulum)
g
q=- q
L
q(t) = qmax cos(w 0t + f )
simple harmonic motion
( potential confusion!! A “simple”
pendulum does not always execute
“simple harmonic motion”; it does so
only in the limit of small amplitude.)
Period does not
depend on qmax, f
10
Free, undamped oscillators – other examples
k
m
L
No friction
k
m
x
I
C
q
1
q=q
LC
mx = -kx
r; r = L
q
Common notation for all
T
m
mg
g
q»- q
L
y +w y = 0
2
0
• The following slides simply repeat the previous
discussion, but now for a mass on a spring, and for a
series LC circuit
12
REVIEW MASS ON IDEAL SPRING
F(x) = mx
k
k
m
F(x) = -kx
m
-kx = mx
x
k
x+ x=0
m
Newton
Particular type of force.
m, k known
Linear, 2nd order
differential equation
What is x(t) such that the above equation is obeyed?
x is a variable that describes position
t is a parameter that describes time
"dot" and "double dot" mean differentiate w.r.t. time
m, k are known constants
13
REVIEW MASS ON IDEAL SPRING
k
k
k
x+ x=0
m
m
x(t) = Ce
m
x
pt
C, p are unknown (for now) constants,
possibly complex
x(t) = p 2Ce pt = p 2 x(t)
Substitute:
k
p x+ x=0
m
2
k
p = ±i
= ±iw 0
m
p is now known. Note that w0 is NOT a new quantity! It is
just a rewriting of old ones - partly shorthand, but also
“w”
14
means “frequency” to physicists!
x(t) = A cos (w 0t + f )
k
k
m
A, f chosen to fit initial conditions:
x(0) = x0 and v(0) = v0
x0 = A cos f
v0 = -w 0 Asin f
m
x
Square and add:
Divide:
x02 +
v02
w
2
0
(
)
= A 2 cos 2 f + sin 2 f = A 2
-v0
= tan f
w 0 x0
15
x(t) = A cos (w 0t + f )
æ k
æ -v0 ö ö
x(t) = x + 2 cos ç
t + arctan ç
÷ø ÷
w0
m
w
x
è
è
0 0 ø
2
0
v02
x(t) = Acos f cos w 0t - Asin f sin w 0t
Aeif iw 0t Ae-if -iw 0t
x(t) =
e +
e
2
2
x(t) = Re éë Aeif eiw 0t ùû
2 arbitrary constants
(A, f because 2nd
order linear
differential equation
16
x(t) = A cos (w 0t + f )
Position:
• A, f are unknown constants - must be determined from initial
conditions
• w0, in principle, is known and is a characteristic of the
physical system
Velocity:
dx
º x(t) = -w 0 Asin (w 0 t + f )
dt
d x
2
º
x(t)
=
w
0 A cos (w 0 t + f )
2
dt
= -w 02 x(t)
2
Acceleration:
This type of pure sinusoidal motion with a single frequency is
called
SIMPLE HARMONIC MOTION
17
THE LC CIRCUIT DIFFERENTIAL EQUATION
q
VC =
C
2
dI
d q
VL = L = L 2 = Lq
dt
dt
L
I
C
q
VL + VC = 0
Kirchoff’s law
(not Newton this time)
q
L q  0
C
1
q=q
LC
k
x x
m
Same differential equation
as the SHO spring!
What is inductance??
d
L
dI
It is how much magnetic flux is created in the
inductor coil by a given current I, in a wire.
What is the voltage change across an inductor?
A voltage change occurs WHEN there is change in
magnetic flux (i.e. some of the energy is ‘converted’ to
a magnetic field)
d  d  dI
dI
VL 

L
dt
dI dt
dt
19
THE LC CIRCUIT
VL + VC = 0
L
I
C
q
Kirchoff’s law
(not Newton this time)
dI
d 2q
VL = L = L 2 = Lq
dt
dt
q
VC =
C
1
q=q
LC