3. Simple Harmonic Motion
Those of you who have had a course in either number theory or abstract algebra will find this
material familiar. On the other hand, this may be the first time those of you from an applied
mathematics, physics, or engineering background will have seen these ideas.
3.1.
Ways to express simple harmonic motion on R
A simple harmonic motion, or sinusoid is a function f (t) that satisfies the differential equation
of simple harmonic motion:
f 00 (t) + λ2 f (t) = 0
(3.1)
For fixed λ, the set of all solutions of this is a two-parameter family of functions, which is to
say that it takes two numbers to describe any particular one of the simple harmonic motions of
that frequency. These numbers can be supplied in any of three ways:
As cosine and sine:
As amplitude and phase:
In complex form:
f (t) = a cos λt + b sin λt
f (t) = R cos(λt − ϕ)
f (t) = ceiλt + c∗ e−iλt
(the numbers are a and b)
(the numbers are R and ϕ)
(the numbers are c and c∗ )
All of these are different but equivalent forms for expressing the same thing – so there must be
equations linking these forms.
The relationship between cosine/sine form and amplitude/phase form:
(
a = R cos ϕ
b = R sin ϕ
(3.2)
This looks just like the formulas for conversion between rectangular and polar coordinates. As in
that case, the other direction goes as follows:
p
R = a2 + b2
(3.3)
b
b
π
π
ϕ = tan−1
or tan−1
+ π (or or − ), depending on quadrant
(3.4)
a
a
2
2
ϕ may take on any of infinitely many possible values which are equivalent modulo 2π. (In doing
these formulas we have tacitly assumed that f (t) takes on only real values.)
The relationship between cosine/sine form and complex form:
In all cases:
(
a = c + c∗
b = i(c − c∗ )


c = a − ib
2
a + ib

c∗ =
2
In the case that f (t) is real-valued:
c∗ = c
12
(3.5)
(3.6)
(3.7)
(
a = 2 Re c
b = −2 Im c
(3.8)
The relationship between amplitude/phase form and complex form:
Assume that f (t) is real-valued:
(
R = 2|c| = 2|c∗ |
ϕ = − arg c = arg c∗
(3.9)
(that is, if c = reiθ , then R = 2r and ϕ = −θ.)
One thing we notice is that for real-valued functions, there is an intimate relationship between
the amplitude/phase form and the complex form. If we choose one of the complex coefficients (c or
c∗ – and if f is real, they are complex conjugates of each other), then its modulus (absolute value)
is (apart from a factor of 2) the amplitude and its argument is (apart from a possible sign change)
the phase. We will make little further direct use of the amplitude/phase form, since the complex
form serves many of the same purposes.
The rest of this section is a sort of glossary of terms associated with simple harmonic motion.
Just for the sake of the argument, assume that the t in f (t) is time and that t is measured in
seconds. Under that assumption, this table the units of measurement for each of these concepts.
NAME
FORMULA
UNITS
(sec−1 )
Amplitude:
λ = 2πν
2π
1
T =
=
λ
ν
1
λ
ν= =
T
2π
p
p
a2 + b2 = 2|c|2 + 2|c∗ |2
(the units of f (t))
Phase:
can mean either ϕ or arg c
(radians; i.e., dimensionless)
Angular frequency:
Period:
Frequency:
(sec)
cycles
= Hz
sec
Note that the amplitude equals 2|c| = R if f is real.
If we choose to write all our expressions based on the frequency ν rather than the angular
frequency λ – and we will make that very choice throughout the semester – then the differential
equation is f 00 (t) + 4π 2 ν 2 f (t) = 0 and the equation of motion is
f (t) = a cos 2πνt + b sin 2πνt = R cos(2πνt − ϕ) = ce2πiνt + c∗ e−2πiνt
3.2.
(3.10)
The Hartley, or “cas” form of simple harmonic motion
This section introduces a fourth way to express the very same simple harmonic motion. The
purposes of this section may not be entirely clear. What is being developed here is the groundwork
for a variation of the Fourier transform called the Hartley transform, which has some computational
advantages over the Fourier transform because of its use of real rather than complex arithmetic.
Recall that eiθ = cos θ + i sin θ. If you look at a pre-calculus book that treats complex numbers
and DeMoivre’s Theorem, you will find this alternative notation: cis θ = cos θ + i sin θ. By vague
analogy to this, we introduce the following:
13
Definition:
cas θ = cos θ + sin θ
(3.11)
√
From this, cas λt is a simple harmonic motion of amplitude 2 whose phase is offset by 45◦ one
π
way or the other (that is, by ± ) from either cos λt or sin λt. If we then write cas(−λt), we have
2
another simple harmonic motion, offset in phase by 90◦ from cas λt, and suitable for use as a basis
element. Start with the same f (t) that we had in (3.1) - (3.10) and write it as follows:
f (t) = d cas(λt) + decas(−λt)
(3.12)
The derivation of the formulas connecting this form to the three previous ones is left to the reader;
in any case, here are those formulas:




d = a + b
2
(3.13)


a
−
b

de =
2



a = d + de
(3.14)


b = d − de


∗


d = (1 + i)c + (1 − i)c
2
(3.15)

∗

(1
−
i)c
+
(1
+
i)c

de =
2


e


c = (1 − i)d + (1 + i)d
2
(3.16)

 ∗ (1 + i)d + (1 − i)de

c =
2




d = R cas ϕ
2
(3.17)


R
cas(−ϕ)

de =
2




e2
R = 2 d2 + (d)
(3.18)


π
ϕ = angle(d, −d)
e +

4
The function named “angle” in the second equation of (3.18) is intended to stand for the whole
process of converting rectangular to polar coordinates; if we were to use this function consistently,
we would be able to replace the mess that is equation (3.4) with the simple statement that ϕ =
angle(a, b).
14
3.3.
Periodic simple harmonic motion: SHM on TP
Since the Greek letter “ν” is sometimes hard to read, as is also the Roman letter “t”, we will
now start using “x” instead of “t” and “s” instead of “ν”. What would it take for a function f (x)
to be a simple harmonic motion that is also periodic of period P ? P would have to be an integer
multiple of the period of the motion. [Saying a function is periodic of period P does not preclude
it from being periodic with a shorter period as well.] If the frequency of f is s, then the period is
1/s, and we must have (1/s) | P. This means that P = n/s for some integer n. This in turn forces
s = n/P. Lets repeat this calculation. Assume that f (x) = e2πisx for some real number s and that
f is periodic of period P. Then we must have for all real x that
e2πisx = e2πis(x+P ) = e2πisx+2πisP = e2πisx e2πisP
e2πisP = 1
from which
(3.19)
which in turn happens only if sP is an integer (positive, negative, or zero). Thus sP = n and
s = nP, n ∈ Z
(3.20)
Whether we permit negative frequencies depends on the style in which we choose to express
our simple harmonic motions. If we work in cosine/sine or amplitude/phase form, we can restrict
our attention to nonnegative frequencies, but if we choose complex or cas form, we must also use
negative frequencies. Whichever mode we choose, (and we will choose the complex form the vast
bulk of the time), equation (3.20) gives us the only frequencies that result in simple harmonic
motions periodic of period P.
We claim that any finite linear combination of functions periodic of period P is itself a function
periodic of period P. (If you don’t believe this claim, verify it yourself.) We are most interested in
linear combinations of simple harmonic motions periodic of period P. Within the class of functions
periodic of period P, we define a trigonometric polynomial of degree N to be a function S(x) of
one of the following forms:
(Real form)
N a[0] X
2πkx
2πnx
S(x) =
a[n] cos
+
+ b[n] sin
2
P
P
(3.21)
k=1
(Complex form)
S(x) =
N
X
c[n]e2πikx/P
(3.22)
k=−N
We note that a trigonometric polynomial of degree N has one constant term and terms that
represent non-constant simple harmonic motions of N different frequencies. Each such frequency
requires two terms, each with its own coefficient (a[k] and b[k] or c[k] and c[−k]). Overall, a
trigonometric polynomial of degree N has 2N + 1 terms. (More formally, the set of all possible
trigonometric polynomials of degree no greater than N is a (2N + 1)-dimensional vector space.)
If we take a limit in (3.21) or (3.22) as the degree N tends to ∞, we get a trigonometric series.
If we pick the coefficients in a certain way, it is a Fourier series – one of the main topics of this
course.
3.4.
Simple harmonic sequences: SHM on Z
A simple harmonic motion is inherently a continuous function, but we can take any continuous
function whatsoever and sample it. That is, we can lay down a sequence of regularly spaced points
15
and evaluate the function at those points, keep those values as a sequence, and throw away all of
the other information. Assume we have a simple harmonic motion f (x) = e2πisx . Now sample this
1
at zero and at every point to the right or left of zero at intervals of exactly , where q is some
q
positive number. If we wish to sample it sparsely, we let q be small; if we want a fine and detailed
sampling, we let q be large. That is, we pick the points x[n] to be
x[n] =
n
q
(3.23)
Having done this, we define a sampled sequence g[n] by letting g[n] = (x[n]). But then,
g[n] = e2πisn/q
(3.24)
Such a g[n] is a simple harmonic motion on Z. We would like to find out what the set of all
possible frequencies for such motions is. At first glance, it appears that we can use any real number
s as the frequency (with the usual proviso that s and −s stand for the same frequency and we
need both to completely describe the motion). But there is a catch. To illustrate the catch, lets
consider a particular
case
(that is, we are sampling exactly at the integers). If
in which
q= 1
2 x
2 9x
we look at cos
and cos
as functions of a continuous real variable x, we see
π 8
π 8
two different simple harmonic motions: one with frequency 1/8 and the other one with frequency
9/8, which is nine times as fast. In Figure 3.1, we have plotted both of these curves on
the
2 n
interval [0, 8]. Now sample both of these at every integer. That is, let f [n] = cos
and
π 8
2 9n
let g[n] = cos
. The only information we now have about these two sequences is that
π 8
shown on the by the filled dots on the graph of Figure 3.1. If those dots are all the information
that we have, then we can’t tell any difference between the two graphs. The frequency 9/8 has the
frequency 1/8 as an alias.
Figure 3.1
Let’s explore this phenomenon a little more. Let’s take simple harmonic motions in exponential
form, sampled at intervals of 1/q, of frequencies s and r. When are the following two functions
16
identical?
?
f [n] = e2πisn/q ≡ g[n] = e2πirn/q
This happens when
f [n]
= 1 for all n, which is to say that for all integers n,
g[n]
e2πisn/q−2πirn/q = 1
n
e2πi(s−r)n/q = e2πi(sr)/q = 1
(3.25)
The only way for (3.25) to be true for all integers n is for e2πi(s−r)/q to be 1, which happens
only when (s − r)/q is an integer. Another way to say this is that the phenomenon happens if and
only if q | (s − r). To repeat,
f [n] = e2πisn/q is the same sequence as g[n] = e2πirn/q iff s ≡ r
(mod q)
(3.26)
1
9
≡ (mod 1).
8
8
This means that we can let s ∈ [0, q) and we will have the set of all possible frequencies of simple
harmonic motion on Z (sampled every 1/q) . If we were to go beyond this interval, we would just
be giving new names to functions that we already have. More succinctly: the set of all frequencies
of simple harmonic motion on Z (sampled every 1/q) is the set Tq .
The example illustrated in Figure 3.1 works because
3.5.
Periodic discrete simple harmonic motion: SHM on PN
Suppose we have a function f (x) which happens to be a simple harmonic motion of frequency
s. Now let us suppose two more things. The first that it happens to be periodic of period P. From
k
section 3.3, we learn that the only possible values of the frequencies s are s =
for some integer
P
k. The second is that rather than considering it to be a function on a continuous domain, we will
1
sample its values at a sequence of x’s at a spacing of to turn it into a sequence. We suppose that
q
1
there are an integer number of sample spacings in each overall period P ; this requires that qP
q
be an integer. Let N = qP. Our sequence is g[n] and
n
= e2πisn/q = e2πi(k/P )(n/p) = e2πikn/N
(3.27)
g[n] = f
q
Another way to say this is to once again write ζ = e2πi/N as the first N ’th root of unity, in
which case our simple harmonic motion is g[n] = ζ kn . At first glance, we can let k be any integer,
but the aliasing phenomenon discovered in section 3.4 applies here just as well. To say that the
k
frequency of this simple harmonic motion is s =
only makes sense if you equate all frequencies
P
k
j
s that are equivalent mod q. But if s =
and r = , then to say that s ≡ r (mod q) is to
P
P
j−k
say that (s − r) = mq for some integer n, which implies that
= mq, which implies that
P
j − k = mqP = mN. That is, the simple harmonic motion whose frequency is indexed by the
integer j is the same sequence as that motion whose frequency is indexed by the integer k if and
only if k ≡ j (mod N ). This says that the set of integers k which can index simple harmonic
motions on PN is itself exactly PN .
17
3.6.
Summary
For functions f (x) on R :
A typical simple harmonic motion is e2πisx , where x ∈ R
(3.28)
For functions f (x) on TP (that is, functions on R periodic of period P ):
A typical simple harmonic motion is e2piikx/P , where k ∈ Z
(3.29)
For functions (sequences) f [n] on Z :
A typical simple harmonic motion is e2πisn/q, where s ∈ Tq
(3.30)
For functions f [n] on PN (that is, sequences periodic of period N ):
A typical simple harmonic motion is e2πikn/N, where k ∈ PN
3.7.
(3.31)
Exercises
Derivations:
3.1. Derive the identities (3.2), (3.3), and (3.4).
3.2. Derive the identities (3.5), (3.6), (3.7), and (3.8).
Problems:
3.3. Consider functions f [n] on PN with N = 8.
(a) Let f [n] = e2πin/8 . Find the values of this function for n = 0, 1, 2, . . . , 7. Plot two graphs
(preferably on two sets of axes, one above the other), with one graph showing the real
part of this function and the other graph showing the imaginary part.
(b) Plot the same sort of graph for g[n] = e2πi(7n)/8 .
1
.
8
(d) What frequency of simple harmonic motion is g[n]? Can you answer this question in
more than one way? [Hint: are there any other names for 7?]
(c) Explain why f [n] is a simple harmonic motion of frequency
(e) Let c1 and c2 be arbitrary complex scalars. Is c1 e2πin/8 + c2 e2πi(7n)/8 a simple harmonic
motion? Why or why not?
18