# Oscillations

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```					Oscillations
Consider a spring-mass system
undergoing oscillations or vibrations
on a frictionless horizontal surface.
-x                           +x

Fnet = 0 N

X= 0 (equilibrium position)
When the displacement is positive the
restoring force is negative.
+x

Fnet

X= 0
When the displacement is negative the
restoring force is positive.
-x

Fnet

X= 0
Amplitude =     X

The amplitude is equal to the
maximum displacement from the
equilibrium position.
The period (T) is the
length of time it
requires to complete
one oscillation
The frequency (ƒ)is
equal to the number of
oscillations in one
second or how often
an oscillation occurs.
T = 1/ƒ
The unit for period is the second.

1 / s = 1 hertz = 1 Hz
Which represents one oscillation
per second.
Any repetitive motion
such as a mass on a
spring or a pendulum
can be described as
harmonic motion
Simple Harmonic
Motion (SHM)

SHM occurs if the restoring
force is directly proportional
to the displacement.
Therefore in SHM the
restoring force:
F = -kx
Free-body diagram

k x

x
mg
m
Gently place a mass on the spring
so that it can hang at rest
Apply a light
force to extend
the spring
x
k (x+ x)
x

mg
x               New
equilibrium
position
x
k (x+ x)

mg
The force acting on the mass is

F = -k (x+ x) + mg   = -k (x+ mg ) + mg
k

therefore F = -k x
So a good rule of practice … for vertical
springs: find the new equilibrium point,
measure the displacements from there …
and ignore gravity.
Period of a Pendulum
Another example of Simple
Harmonic Motion
Period as a function of Length
Using radian measure x = L  which represents the arc
path of the pendulum or its actual displacement.

x=L                    
Frestore = - mg (sin 
)
Frestore = - kx

x
L mg (sin  )
x=0
Period of a Pendulum vs Length

25

20
Period (s)

15

Period is directly proportional to the
10
________ Root of the Length.

5

0
0   20   40       60         80         100      120       140   160
Lenght (cm)
140

120

100

1/2
80             y = k (x)
Y

60

2
40
y = k (x)                 y = k (x)
20

0
0   2           4         6         8       10   12   14
X
Period as a function of Square Root of the Length

25
Period of the Pendulum

20

Period as a function of Square
Root of the Length
15
^T

10

5

^ L
0
0   2              4            6           8           10          12   14
Square Root of the Length
Equating Acceleration (ax) a
Pendulum to the Centripetal
Acceleration

Ac = 4π2r    Then T2 = 4π2r
T2               Ac
SHM (reference circle)
using Newton’s 2nd Law
Frestore = - Kx = ma

a = -K x
m
which infers that any object in SHM will
have an acceleration that is directly
proportional to its displacement.
If an object is moving in a vertical
circle with constant speed v0 and
radius R. Consider a light source
shining from above that casts a
v0

R

Shadow moves back and forth with properties
equal to the horizontal component of the
circularly moving object.
Can we show that acceleration of the
displacement.

Recall from unit circle that:
At the ends          x = R cos 
v0   vhorizontal = 0
ac = v02 cos
R
R

ac = v02 R(cos) = v02 (x)
R2           R2
+x             Since v and R are constant,
ac is proportional to x
Again since v and R are constant, ac is
proportional to x. So if the acceleration of
the shadow is directly proportional to its
displacement, this motion is SHM.
Recall from unit circle that:
At the ends          x = R cos 
v0   vhorizontal = 0
ac = v02 cos
R
R

ac = v02 R(cos) = v02 (x)
R2           R2
+x
If we reverse our argument, we
will find that for every system
undergoing SHM, you can
visualize a reference circle where
the uniform circular motion
corresponds to the real system…
Correspondence of circle to system
Reference circle             Real System
• Radius (R)                 • Amplitude (A)
• Constant tangential        • Maximum speed (v0 )
speed
• Period of revolution (T)   • Period of oscillation (T)

2πA = v0T             > T = 2πA
v0
Using a spring-mass system, we can
compare the energy stored in the
spring at full extension to the
kinetic energy as it passes through
the equilibrium point
ES = EK
½ kSA2 = ½ m v02      A =    m
v0     kS
A =   m
v0    kS

Substituting into our equation
for the systems period (T):
The period
T = 2πA = 2π m            depends on the
v0              kS   system
properties
To determine the period of a pendulum
we can replace kS with      mg/L

Spring-mass
system
T = 2π m
kS
The period of a
Pendulum      T = 2π               L             pendulum is
independent of
g             mass