Simple Harmonic Motion
Many mechanical systems in nature move repeatedly back and forth around a central position. In this
course we will call this behavior - simple harmonic motion (SHM). We use the example of a mass on a
spring oscillating back and forth to illustrate the equations of SHM.
A mass-spring is pulled down (about - 0.15 m) from its resting, equilibrium position (0 m) and let go – it
then oscillates freely up and down as indicated in the graph below:
amplitude - xo
period - T
Some definitions:
displacement: ( x ) measured distance from the equilibrium position at any time.
amplitude: ( xo ) maximum displacement from the equilibrium position (here about .15 m).
T = time period: time of one complete oscillation (one up/down cycle – here about 1.57 seconds).
f = frequency: the number of oscillations in one second (here less than one – about .64 Hz (units: 1/s or
Hertz (Hz))). Frequency is the inverse of the time period: f = 1 T
angular frequency: w = 2p T or w = 2p f : the frequency in radians (here about 4.0 rad/s ( = 2p 1.57 s).
To find the displacement position as a function of time (t) – we can write the general equation of the
graph above:
x ( t ) fi x = xo cos w t
(where:
xo = amplitude, w = angular frequency, sign should actually be negative (-cos, started down))
From previous experience we know the velocity should be the slope of the displacement vs. time graph –
mathematically it is the derivative of the displacement function:
v (t ) =
(where:
d ( x ( t ))
dt
=
d ( xo cos w t )
= - w xo sin w t fi
dt
vo = max velocity amplitude)
v = - vo sin w t
( vo = w xo )
In the graph below – it can be seen that the displacement and velocity graphs are out-of-phase (not in
synch). From our equations – this makes sense – if displacement is a cos function, velocity should be
the displacement’s –sin function.
displacement
velocity
In the same way we find the acceleration:
a (t ) =
dv ( t ) d ( -w xo sin w t )
=
= - w 2 xo cos w t fi
dt
dt
( a(t) = -w
2
x(t)
)
a = -w 2 x is virtually the definition of SHM:
the acceleration is a) proportional but b) in the opposite direction from the displacement from the
equilibrium position. They are related by w 2 . This can be seen below.
acceleration
displacement
Since forces and accelerations go together ( F = ma ), the restorative spring force shows a similar
relationship known as Hooke’s Law: F = -kx (F is proportional but oppose direction to x, where k is
known as the spring constant (units: N/m)). The spring constant can be found as the slope of the F vs.
displacement graph below (about 10 N/m).
(
)
If F = ma = - kx = m -w 2 x , we can see that: a =
-k
x and that w =
m
k 2p
=
m
T
m
k
What is the Work done to stretch the spring? – what would be the elastic potential energy stored in the
spring as it is stretched away down from equilibrium? Since the work = Ep = Fd (or in this case Fx), it
should equal the area under the curve of out Force vs. displacement graph: and since it is a triangle - it
would be 1 2 (base)(height), thus:
From the second relationship we can easily find the time period (T) for the spring: T = 2p
1
1
1
1
(height)(base) = (F)(x) = (kx)(x) fi E p = kx 2
2
2
2
2
gives the equation for elastic potential energy of the spring = Ep
Data from graph above shows:
1
1
1
(height)(base) = (F)(x) = (1.6N )(.16m) = .128J
2
2
2
1
1
E p = kx 2 =
10 N m (.16m)2 = .128J
2
2
(
)
If there is no loss in energy as the spring oscillates back and forth the total energy = Etot will always be
equal to the original potential energy = Ep at the original maximum displacement = xo. Total energy =
Etot will also be equal to the kinetic energy = Ek at the maximum velocity = vo. Total energy will always
equal the sum of the kinetic and potential energies at any time: Etot =Ek + Ep.
1
1
k
vo = ± w xo2 or v = ± w xo2 - x 2
Thus: Etot = mvo2 = kxo2 fi vo2 = xo2 fi
2
2
m
Sub this v into Ek:
2
1
1
1
Ek = mv 2 = m w xo2 - x 2
fi Ek = w 2 xo2 - x 2 anytime
2
2
2
1
and Etot = Ekmax = w 2 xo2 for Ekmax using max amplitude xo
2
(
)
(
)
IB equations given:
w = 2p T
other important relationships:
w = 2p f
f =1 T
w=
k 2p
=
m
T
a=
(
-k
x = -w 2 x
m
F = ma = - kx = m -w 2 x
x = xo sin w t and x = xo cos w t
v = vo cos w t and v = - vo sin w t
v = ± w xo2 - x 2
(
)
1
Ek = w 2 xo2 - x 2
2
1
Etot = Ekmax = w 2 xo2
2
Ep =
1 2
kx
2
)
T = 2p
m
k