Oscillations about equilibrium
(Synonym:Vibration)
What means Oscillation?
Oscillation is the periodic variation, typically in time, of some measure
as seen, for example, in a swinging pendulum.
Many things oscillate/vibrate: Periodic motion
(a motion that repeats itself over and over)
Pulse
Oscillations are the origin of the
sensation of musical tone
……….. in Aerospace: Orbits
Electrical/Computer:
LRC resonance in circuits
Physics: Atomic Vibrations, String Theory, Electromagnetic Waves
Why does something vibrate/oscillate?
Whenever the system is displaced from equilibrium, a restoring force pulls it back,
but it overshoots the equilibrium position.
……more examples
….heart beat, breathing, sleeping, taking shower, eating, chewing, blinking, drinking
….motion of planets, stars, motion of electrons, atoms
….wind (Tacoma Bridge)
….vocal cords, ear drums
Parameter used to describe vibrations
Period T Time taken to complete one
cycle of the vibration. Units: s
Frequency f = 1/T
Number of vibration cycles per
second. Units: 1/s (Hz, Hertz)
Amplitude A Maximum
displacement from equilibrium
position
One cycle is time take for a pendulum:
center – right – center – left –
center
Pendulum
T
Special cases of periodic motion
Simple harmonic motion (SHM)
occurs when the restoring force
(the force directed toward a stable
equilibrium point) is proportional to
the displacement from equilibrium.
For instance when the restoring
force is F = - k x.(Hook’s Law)
Simple harmonic motion
The displacement from equilibrium can be describes as a
cosinusoidal function
A
Period T
Equation of motion
for SHM
∂2 x
m 2 = −kx
∂t
∂2 x k
+ x=0
2
m
∂t
∂2 x
2
+
ω
x=0
2
∂t
Simple harmonic motion is the projection of circular
motion on the x-axis
ω
θ
Angular velocity
is NOT necessarily
the same as
Angular frequency
Displacement, Velocity and Acceleration of SHM
x(t ) = A cos(ωt + φ )
dx(t )
= − Aω sin(ωt + φ )
v(t ) =
dt
dv(t )
a(t ) =
= − Aω 2 cos(ωt + φ ) = −ω 2 x(t )
dt
A is the amplitude of the
motion, the maximum
displacement from
equilibrium, Aω=vmax, and
Aω2 =amax.
Mass-Spring Java applet
Mass-Spring-System
A Spring always pushes or pulls mass back towards equilibrium
position. The time period can be calculated from Hooke´s Law:
F = −kx F = ma ⇒ ma = −kx
m[− Aω 2 cos(ωt )] = −k[ A cos(ωt )]
ω2 =
k
m
or T = 2π
m
k
Independent from amplitude!
(Application: measure the mass of astronauts in
space) Heavier mass – slower oscillations
Stiffer spring (greater k) – rapid oscillations
The period of oscillation is T =
1 2π
=
ω
f
Mass Spring-System in vertical setup
When a mass-spring system is oriented
vertically, it will exhibit SHM with the same
period and frequency as a horizontally
placed system.
Same formulae as for the horizontal setup
but the system oscillates around a new
equilibrium position y0.
y0 = mg/k
Energy Conservation in Oscillatory Motion
Potential Energy of simple harmonic
motion
Energy: E = U + K
U: Potential Energy
K: Kinetic Energy
Spring
Epot = U = ½ k x2
Ekin = K = ½ m v2
Turning points:
E = Umax + 0
(Displacement and U at maximum)
Minimum:
E = Kmax + 0
(Velocity and K at maximum)
Total energy of system
E = U + K = ½ k (A cos(ωt))2 + ½ m (A ω sin(ωt))2
= ½ k A2 cos2(ωt) + ½ m A2 ω2 sin2(ωt)
E = ½ k A2 cos2(ωt) + ½ m A2 k/m sin2(ωt) = ½ k A2 (cos2(ωt)+
sin2(ωt))
And therefore:
E = ½ k A2
Exercise 1: The period of oscillation of an object in an ideal
mass-spring system is 0.50 sec and the amplitude is 5.0 cm.
What is the speed at the equilibrium point?
1 2 1 2 1 2 1 2
At equilibrium x=0: E = K + U = mv + kx = kA = mv
2
2
2
2
Since E=constant, at equilibrium (x = 0) the KE must be
a maximum. Here v = vmax = Aω.
The amplitude A is given, but ω is not.
2π
2π
=
= 12.6 rads/sec
ω=
T
0.50 s
and v = Aω = (5.0 cm )(12.6 rads/sec) = 62.8 cm/sec
Exercise 2: The diaphragm of a speaker has a mass of 50.0 g
and responds to a signal of 2.0 kHz by moving back and forth
with an amplitude of 1.8×10-4 m at that frequency.
(a) What is the maximum force acting on the diaphragm?
(
)
2
2
(
)
F
=
F
=
ma
=
m
A
ω
=
mA
2
π
f
=
4
π
mAf
∑
max
max
2
2
The value is Fmax=1400 N.
(b) What is the mechanical energy of the diaphragm?
Since mechanical energy is conserved, E = KEmax = Umax.
U max =
KEmax
1 2
kA
2
1 2
= mvmax
2
The value of k is unknown so use KEmax.
KEmax =
1 2
1
1
2
2
mvmax = m( Aω ) = mA2 (2πf )
2
2
2
The value is KEmax= 0.13 J.
Exercise 3: The displacement of an object in SHM is given
by:
y (t ) = (8.00 cm )sin[(1.57 rads/sec)t ]
What is the frequency of the oscillations?
Comparing to y(t)= A sinωt gives A = 8.00 cm and ω = 1.57 rads/sec.
The frequency is:
ω 1.57 rads/sec
f =
2π
=
= 0.250 Hz
2π
Other quantities can also be determined:
2π
2π
=
= 4.00 sec
The period of the motion is T =
ω 1.57 rads/sec
xmax = A = 8.00 cm
vmax = Aω = (8.00 cm )(1.57 rads/sec) = 12.6 cm/sec
amax = Aω 2 = (8.00 cm )(1.57 rads/sec) = 19.7 cm/sec2
2
A torsional pendulum is an oscillator for which
the restoring force is torsion. For example,
suspending a bar from a thin wire and winding it
by an angle θ, a torsional torque
is produced, where is a characteristic property of the
wire, known as the torsional constant. Therefore, the
equation of motion is
where I is the moment of inertia. But this is just a
simple harmonic oscillator with equation of
motion
where
is the initial angle,
is the angular frequency, and
is the phase constant.
Pendulum
A mass, called a bob, suspended from a fixed point so that it can
swing in an arc determined by its momentum and the force of gravity.
The length of a pendulum is the distance from the point of
suspension to the center of gravity of the bob. Chance observation of
a swinging church lamp led Galileo to find that a pendulum made
every swing in the same time, independent of the size of the arc. He
used this discovery in measuring time in his astronomical studies. His
experiments showed that the longer the pendulum, the longer is the
time of its swing.
If we assume the angle θ is small, for then we can approximate sin θ
with θ (expressed in radian measure). (As an example, if θ = 5.00° =
0.0873 rad, then sin θ = 0.0872, a difference of only about 0.1%.)
With that approximation and some rearranging, we then have
∂ 2θ mgL
+
θ =0
2
∂t
I
Physical Pendulum,
Small amplitude
UPendulum
For smaller displacements, the movement of
= mgh = mgL (1-cosθ) an ideal pendulum can be described
mathematically as simple harmonic motion
(like the mass-spring), as the change in
potential energy at the bottom of a circular arc
is nearly proportional to the square of the
displacement. Real pendulums do not have
infinitesimal displacements, so their behaviour
is actually of a non-linear kind.
The Physical Pendulum
A "physical" pendulum has
extended size and is a
generalization of the bob pendulum.
An example would be a bar rotating
around a fixed axle. A simple
pendulum can be treated as a
special case of a physical pendulum
with moment of inertia I. ( I = ∑ miri2)
Period of a physical pendulum
(Note: l is now the length from the
suspension point to the center of
mass CM instead of L)
Example:
Simple Pendulum: I = mL2
Leg: I = 1/3 mL2
Simple Pendulum
All the mass of a simple
pendulum is concentrated in the
mass m of the particle-like bob,
which is at radius L from the
pivot point. Thus, we can
substitute I = mL2 for the
rotational inertia of the
pendulum.
L
T = 2π
g
for small amplitudes!!
Exercise 4: A clock has a pendulum that performs one full
swing every 1.0 sec. The object at the end of the string
weights 10.0 N. What is the length of the pendulum?
L
T = 2π
g
Solving for L:
(
)
9.8 m/s 2 (1.0 s )
gT 2
L=
=
= 0.25 m
2
2
4π
4π
2
Pivot
Length: L
Mass: M
θ
ICM= 1/12 ML2
CM
Parallel-Axis Theorem
IPivot= 1/12 ML2 + M (½ L)2 = 1/3 ML2
mg
The period is
1
ML2
I
2L
3
T = 2π
= 2π
= 2π
1
gMl
3
g
gM ( L)
2
Ring
Disc
r
r
CM
IPivot= Mr2 + Mr2 = 2Mr2
CM
IPivot= ½ Mr2 + Mr2 = 3/2 Mr2
I
2 Mr 2
T = 2π
= 2π
gMl
gMr
I
= 2π
T = 2π
gMl
2r
T = 2π
g
3r
T = 2π
2g
3
Mr 2
2
gMr
http://lectureonline.cl.msu.edu/%7Emmp/applist/damped/d.htm
http://physics.usask.ca/~pywell/p121/Images/tacoma.avi
http://www.walter-fendt.de/ph14e/resonance.htm