OSCILLATIONS: I

1. An oscillation is any self-repeating motion. This motion is characterized by:
a) The period , which is the time for completing one full cycle.
b) The frequency 1/ ,which is the number of
T
f= T cycles per second. (Another
frequently used symbol is ).
c) The amplitude A, which is the maximum displacement from equilibrium (or the
size of the oscillation).
2. Why does a syste
ν
m oscillate? It does so because a force is always directed towards
a central equilibrium position. In other words, the force always acts to return the
object to its equilibrium position. So th
restore
e object will oscillate around the equilibrium
position. The restoring force depends on the displacement , where
is the distance away from the equilibrium point, the negative si
F = −kΔx Δx
gn shows that the
force acts towards the equilibrium point, and is a constant that gives the strength of
the restoring force.
k
3. Let us understand these matters in the context of a spring tied
to a mass that can move freely over a frictionless surface. The
force ( ) (or - because the extension will be
ca
F x = −kx kΔx Δx
2
lled for short). In the first diagram is positive, is
negative in the second, and zero in the middle one. The energy
stored in the spring, ( ) 1 , is positive in the first and
2
th
x x x
U x = kx
2
ird diagrams and zero in the middle one. Now we will use Newton's second law to
derive a differential euation that describes the motion of the mass: From ( )
and it follows that
F x kx
ma F m d
= −
=
2
2 2
2 2 , or 0 where . This is
the equation of motion of a simple harmonic oscillator (SHO) and is seen widely in
many different branches of physics. Although we have derived it
x kx dx x k
dt dt m
= − +ω = ω ≡
for the case of a
mass and spring, it occurs again and again. The only difference is that , which is
called the oscillator frequency, is defined differently depending on the situation.
4. In
ω
order to solve the SHO equation, we shall first learn how to differentiate some
elementary trignometric functions. So let us first learn how to calculate d cos
dt
starting from the basic defini
ωt
tion of a derivative:
Compressed
Stretched
( ) ( ) ( )
( ) ( )
cos and cos . Take the difference:
( ) cos cos
sin sin( / 2)
Start : x t t x t t t t
x t t x t t t t
t t t
ω ω
ω ω
ω ω ω
= +Δ = + Δ
+ Δ − = + Δ −
= − Δ + Δ
sin as becomes very small.
cos sin .
(Here you should know that sin for small , easily proved by drawing trian
t t t
d t t
dt
ω ω
ω ω ω
θ θ θ
≈ − Δ Δ
∴ =−
≈ gles.)
You should also derive and remember a second important result:
sin cos .
(Here you should know that cos 1 for small .)
5. What
d t t
dt
ω ω ω
θ θ
=
≈
( )
( )
2
2
2
2
2
2
happens if you differentiate twice?
sin cos sin
cos sin cos .
So twice differentiating either sin
d t d t t
dt dt
d t d t t
dt dt
t
ω ω ω ω ω
ω ω ω ω ω
ω
= =−
= − = −
2
2
2
or cos gives the same function back!
6. Having done all the work above, now you can easily see that any function of the
form ( ) cos sin satisfies . But what do , a, b represe
t
x t a t b t d x x
dt
ω
= ω + ω =−ω ω
( )
nt ?
a) The significance of becomes clear if your replace by 2 in either sin
or cos . You can see that cos 2 cos 2 cos . That is, the
function merely repeats it
t t t
t t t t
π
ω ω
ω
π
ω ω ω π ω
ω
+
⎛⎜ + ⎞⎟= + =
⎝ ⎠
self after a time 2 / . So 2 / is really the period of the
motion , 2 2 . The frequency of the oscillator is the number of
complete vibrations per unit time: 1 1 so 2
2
T T m
k
k
T m
π ω π ω
π
π ν
ω
ν ω πν
π
= =
= = =
[ ] 1
2 .
Sometimes is also called the angular frequency. Note that dim , from
it is clear that the unit of is radian/second.
b) To understand what and mean let us note that fro
k
T m
T
a b
π
ω ω
ω
−
= =
=
m ( ) cos sin it
follows that (0) and that d ( ) sin cos (at 0). Thus,
dt
is the initial position, and is the initial velocity divided by .
x t a t b t
x a xt a t b t b t
a b
ω ω
ω ω ω ω ω
ω
= +
= =− + = =
44
Mg
P
θ C
θ
c) To understand what and mean let us note that from ( ) cos sin it
follows that (0) and that d ( ) sin cos (at 0). Thus,
dt
is the initial position, and
a b xt a t b t
x a xt a t b t b t
a b
ω ω
ω ω ω ω ω
= +
= =− + = =
( )
is the initial velocity divided by .
d) The solution can also be written as: ( ) cos . Since cos and sin never
beome bigger than 1, or less than -1, it follows that . For obviou
m
m m
x t x t
x x x
ω
= ω +φ
− ≤ ≤+ s reason
is called the amplitude of the motion. The frequency of the simple harmonic motion
is independent of the amplitude of the motion.
e) The quantity is called the phase of t
m x
θ =ωt +φ he motion. The constant is called the
A different value of just means that the origin of time has been
chosen differently.
7. . Put 0
phase constant.
φ
φ
Energy of simple harmonic motion φ =
2
for convenience, and so imagine a mass
whose position oscillates like cos . Let us first calculate the potential energy:
1 1
2 2
m
m
x x t
U kx kx
= ω
= = 2 2
2
2 22 2 2 2
cos .
Now calculate the kinetic energy:
1 v 1 1 sin 1 sin
2 2 2 2
The sum of potential + kinetic is:
m m
t
K m m dx m x t kx t
dt
ω
= = ⎛⎜ ⎞⎟ = ω ω = ω
⎝ ⎠
( )
2 2 2 2
2 2 2 2
1 cos 1 sin
2 2
1 cos sin 1 .
2 2
Note that this is independent of time and energy goes from kinetic to potential,
m m
m m
E K U kx t kx t
kx t t kx
ω ω
ω ω
= + = +
= + =
( 2 2)
then
back to kinetic etc.
8. From the above, you can see that v . From this it is clear that
the speed is maximum at 0 and that the speed is zero at .
9. Putting two springs i
m
m
dx k x x
dt m
x xx
= =± −
= =±
1 2
1 2
1 2
n parallel makes it harder to stretch them, and . In
series they are easier to stretch, and . So a mass will oscillate faster in
the first case as compared to the
eff
eff
k k k
k k k
k k
= +
⎛ ⎞
=⎜⎝ + ⎟⎠
second.