1. Simple Harmonic Motion
mass on a spring
F = m a
Newton’s Law
F = – k x
spring force
m a = – k x
a = – (k / m) x
acceleration is proportional to x
but in opposite direction 

2. d2x / dt2 = – (k / m) x 2 = k / m Differential Equation x = A cos  t
try a solution:
dx / dt = – A  sin  t
d2x / d2t = – A 2 cos  t
= – 2 x
this is a solution provided
2 = k / m

3. A is the amplitude T is the period Plot of Solution
• how far I pull the mass down at the start (t = 0)
• the maximum displacement
T is the period
• the time after which the motion repeats

4.  = 1 / T   = 2  / T = 2   is the frequency
• how many time per second the motion repeats
cos  t = cos  (t + T) = cos ( t +  T) 
 T = 2 
 = 2  / T = 2   
is the angular frequency

5. Initial Conditions x = A cos( t + ) general solution   = k / m 
is fixed by k and m 
A and  are determined by
the initial conditions
(how we start the oscillation)

6. x = A cos ( t + ) xmax = A v = –  A sin ( t + ) vmax =  A
occurs when x = 0
a = – 2 A cos ( t + )
amax = 2 A
occurs when x = xmax

7. = B cos  t + C sin  t x0 = B  v0 =  C
Another form of the general solution
x = A cos( t + ) = A cos  t cos
= A cos  t cos – A sin  t sin
B = A cos 
C = – A sin 
= B cos  t + C sin  t
B and C are set by initial conditions
at t = 0
x0 = B
v0 =  C 
B is the initial displacement
C is the initial velocity / 

8. d2 / dt2 = – (g / l )  Simple Pendulum l  sin    
F = m a = m l  l 
F = – m g sin 
m l  = – m g sin 
m g sin 
m g
 = – (g / l ) sin 
d2 / dt2 = – (g / l ) sin 
not equation of SHO
but let  be small
 sin    
d2 / dt2 = – (g / l ) 
SHO equation

9. d2x / dt2 = – (k / m) x x = A cos  t  = k / m
pendulum motion is approximately SHO
the approximation is better for smaller angles
Compare
d2x / dt2 = – (k / m) x
x = A cos  t
solution
 = k / m
with
d2 / dt2 = – (g / l ) 
 = A cos  t
solution
 = g / l

10. Simple Harmonic Oscillator?
What is the Energy of a
Simple Harmonic Oscillator?
K = ½ m v2
U = ½ k x2
Substitute SHO solution for x, v

11. = k A2 sin2 t K = m v2 = m 2A2 sin2 t U = k x2 = k A2 cos2 t
Energy of Simple Harmonic Oscillation
x = A cos  t
v = – A sin t
K = m v2 = m 2A2 sin2 t
= k A2 sin2 t
U = k x2 = k A2 cos2 t 1 2 –
both K and U oscillate with time

12. K = k A2 sin2 t 1 2 – 1 2 –
U = k A2 cos2 t

13. Total Energy E = K + U = k A2 = k A2 sin2 t + k A2 cos2 t1 2 – 1 2 –
E = K + U 1 2 –
a constant!
TOTAL Energy does NOT change with time
= k A2
K cannot be negative
turning point
turning point K

14. How to Calculate the Force from the Potential Energy
U = –  F • dr = –  (Fx dx + Fy dy + Fz dz)
dU = – Fx dx – Fy dy – Fz dz
Fx = – dU / dx
Fy = – dU / dy
Fz = – dU / dz
the force is minus the
slope of the potential
F > 0
F < 0
F = 0

15. Why do we care about SHO? Because any potential energy curve looks
(spring and pendulum are fun, but not really fundamental physics…)
Because any potential energy curve looks
like a simple harmonic oscillator near its minimum. U x
Close to min,
it looks like SHO
Zoom in

16. Energy Diagram of Diatomic Molecule
e.g. O2
let r be atomic separation r
r0 equilibrium separation U
dU / dr < 0 repulsive
U  0 as r  0 r r0
dU / dr > 0 attractive
equilibrium point

17.  U r r0 near equilibrium point looks like SHO
approx. parabolic potential
atoms displaced from equilibrium
behave like simple harmonic
oscillator as long as displacement
from equilibrium point is not too large
for small excitations we can model
a diatomic molecule as two point masses
connected by a spring 

18. Complex Numbers introduce i =  – 1 real numbers
natural numbers
whole numbers
integers
rational numbers
irrational numbers
real numbers
all algebraic equations cannot be
solved using the real numbers
e.g x2 = – 1
introduce i =  – 1

19. i times a real number is an imaginary number i b e.g. i 5.3
a complex number a + i b is the sum
of a real number and an imaginary number
e.g i 5.3
real part
imaginary part

20. One of the Most Remarkable
Theorems in all of Mathematics
our number system is now complete
any algebraic equation has a
solution that is a complex number

21. a + i b Complex Plane = A cos  + i A sin  tan  = b / a
any complex number can be represented by
a point in an abstract 2 – dimensional space
a + i b
imaginary
axis
= A cos  + i A sin  
real axis
point labeled by
A and 
tan  = b / a
A =  a2 + b2

22. ei  = cos  + i sin  e is the base of the natural logarithm ln
The Euler Relation
One of the Most Remarkable
Relations in all of Mathematics
ei  = cos  + i sin 
e is the base of the natural logarithm ln
e = • • •
e  1 + 
for small 
from this relation and the properties
of exponents we get all of trigonometry

23. d(A et) / dt =  A et d2(A et) / dt2 = 2 A et
Tool for Solving Differential Equations
taking derivatives of an
exponential function is easy
d(A et) / dt =  A et
d2(A et) / dt2 = 2 A et
its as simple as multiplication

24. x = Aei t x = A cos t Pretend Solution is a Complex Number
rather then
then we can substitute it into the differential
equation and easily take the derivatives
of course we have to remember that the
actual physical solution is the real part of
our assumed solution Re(Aei t)

25. d2x / dt2 + (k / m) x = 0 x = Aei t ( – 2 + (k / m)) Aei t = 0
try as a solution:
x = Aei t
( – 2 + (k / m)) Aei t = 0
– 2 + (k / m) = 0
2 = (k / m) is the natural frequency
2 = (k / m)

26.  r = R cos  t A tunnel through the Earth F = – G m M / r2
= – (G m M / R3) r •
m a = – (G m M / R3) r
d2r / dt2 = – (G M / R3) r
SHO equation with  = GM / R3
 r = R cos  t

27. = 84.3 min Same as the time it takes a satellite
T = 2  /  = 2   R3 / G M
= 2   R/ g
= 6.28  6.37 x 106 m / m / s2
= 5.06 x 103 s
= 84.3 min
Same as the time it takes a satellite
to orbit the Earth right at the surface

28. Exit