Oscillating Systems Flashcards
complex number basics
a=rcos theta
b=rsin theta
r = sqrt(a^2+b^2)
theta = arctan(b/a)
derivation of euler’s identity
a+ib
rcostheta+irsintheta
sub in taylor expansion of costheta and sintheta
terms cancel
left with re^itheta
e^ipi=cospi+isinpi=-1=i^2
if i represents a rotation by 90 degrees into the imaginary plane, complex numbers can be represented using
polar coordinates
(r,theta)
phase
how much something leads/lags the main motion
in complex numbers, phase is angle theta to the
imaginary plane
if z=A0e^iwt, the angle is changing with
time
therefore vector is rotating
rotation when wt=2pi
2pi
T=
2pi/w
features of SHM
motion confined within +/-A
T between 2 successive occasions where x and dx/dt repeat
has relative phase phi
sinusoidal variations
small displacements
restoring force directly prop. to displacement from eqm
Hooke’s law
F=-kx
large angles
period gets longer and is a function of initial starting angle
no longer SHM
Fourier’s Theorem
any function that repeats regularly can be built up from a set of sinusoidal functions of appropriate periods and amplitudes
superposition
fourier series: more terms=
more square the wave
can better approximate system with more terms
expression for mass on a spring
F=ma
F=-kx
equate and sub in a=d2x/dt2
solving DE:
m(d2x/dt2)=-kx
guess a sinusoidal solution and differentiate twice
x=Asin(wt+phi 0)
v=…
a=-w^2x
use w^2=k/m
d2x/dt2=-k/m x so solution
what turns cosine to sine
phase of pi/2
what variables completely define SHM?
A,w,phi0
derivation of phi0 = arctan(-v0/wx0)
take x=Acos(wt+phi0) when t=0
dx/dt=v0=…
v0/w=…
v0/x0w= -Asin phi0 / A cosphi0= -tanphi0
derivation of A=sqrt(x0^2+v0^2/w^2)
square x0 and v0
v0^2/w^2=…
x0^2+v0^2/w^2
to solve equation of motion of harmonic oscillator, need to find function for which
the double derivative leads back to original function
x in complex form
Z=Ae^i(wt+phi0)
SHM may be described by the projection of
a particle in uniform circular motion
velocity on argand diagram
perpendicular to position vecotr
angle of wt+phi0+pi/2
acceleration on argand diagram
anti-parallel to position (phase 180)
angle wt+phi0+pi
total energy of system
kinetic + potential
have x=Acos(wt+phi0), v=-wAsin(wt+phi0)
potential energy, V
V = - integral Fdx = integral kx dx
=1/2 kx^2
sub in x
potential energy V for a linear force F is
quadratic
for small displacements approximate taylor expansion as linear
have x=Acos(wt+phi0), v=-wAsin(wt+phi0)
kinetic energy, T
=1/2 m (dx/dt)^2
sub in w^2=k/m
cancel terms
total energy
E=T+v
sub in previous expressions
sin^2+cos^2=1
E=1/2kA^2
why is total energy constant throughout motion
only depends on A and k
time dependence PE and KE oscillate out of phase with each other by
180 degrees
potential energy maximal if
displacement maximal
kinetic energy maximal if
velocity is at its peak/trough
idealised system for pendulum
massless string
extenstionleess string
no air resistance
no friction at pivot
restoring force for a displacement of s=l phi
F=-mg sin phi
how to get DE:
d2phi/dt2 + g/l sin phi =0
equate restoring force and f=ma
primary bonds
ionic
covalent
metallic
secondary bonds
van der waals
ionic
exchanging of e- forming ion
covalent
sharing of e-
metallic
sea of e- shared by all atoms
VDW
irregularities in e- distribution creates dipoles that attract one another
what 2 forces determine the potential of an ionic molecule
coulomb attraction between + and - ions
quantum mechanical effect within Pauli exclusion principle
quantum mechanical effect within Pauli exclusion principle
as two ions go towards each other, electron clouds overlap, e- move into states of higher energy
seen as repulsive force, potential parameterised as B/r^9
Vr=
Fcoulomb + B/r^9
ionic potential
must be a position where distance R between ions where sum of attractive and repulsive forces is zero
must be minimum
diff to find turning point
(around tp approximate as quadratic)
how to find force constant k
second derivative of potential
find dv/dr where R=r
get an expression for B
find d2v/dr2 and sub in B
k is only dependent on
bond length R
how could bond length be measured
if vibration frequency measured
would allow a measurement of how far atoms are from each other
reduced mass
μ = m1m2/m1+m2
general method for verifying something is a solution to a DE
take solution and differentiate twice and try to simplify to required form
as long as system is linear, the result of two or more harmonic vibrations is
the sum of individual vibrations
superposition angle
alpha 2 - alpha 1
result of rotation by (wt+alpha1)
if rotation of frequencies is incommensurable eg root2
result will not be periodic
combination of frequencies close to each other
effect called beats
resulting vibration has frequency that is average of the two initial frequencies known as
beat frequency/ fast oscillation
resulting amplitude varies periodically with time and is known as
modulating envelope/ slow oscillation
modulating envelope depends on
difference between both frequencies
how to show modulating envelope depends on difference between frequencies
assume two vibrations have equal amplitude
set w1t=a+b, w2t=a-b
set x=x1+x2
use trig identities on formula sheet with
a=w1t-b
b=a-w2t
sub into 2cosacosb
get expression: x=(modulation)(beat)
superposition in two dimensions
if w1 and w2 not commensurable
movement of point limited to rectanglel of 2A1 by 2A2
entire rectangle ‘filled’
superposition in two dimensions
w1 and w2 are commensurable
have periodic 2D orbit known as Lissajous figure
lissajous figures - straight line
identical frequency, no relative phase
y=A2/A1 x
lissajous figure - cirlce
x=A1coswt
y=A2cos(wt+pi/2)
square x and y and add gets cirlce with radius A
lissajous figure for identical frequency but arbitrary amplitude and phase
ellipse
homogeneous eqn
RHS=0
general solution of a non-homogeneous eqn
sum of general solution of related homogeneous eqn and particular solution of non-homogeneous
SHM with damping DE
assume forces linear and equate
md2x/dt2=-kx-bv
rearrange and divide by m
b/m=2 gamma, k/m=w^2 (sub in)
use integrating factor z(t)=e^lambda t
diff integrating factor and sub in for a,v,x
diide by e^lambda t to get characteristic eqn
find lambda by taking roots of quadratic
if gamma^2 < w0^2
root -ve and therefore imaginary
if gamma^2=w0^2
root 0 so toor vanishes, resulting in one special solution
if gamma2>w0^2
root +ve so real
dissipative term
represents resistive force which is velocity dependent
(expect term prop. to dx.dt)
damping characterised by
gamma which has same dimensions as frequency
light damping
root -ve and imaginary
total energy for light damping
E=1/2kA^2 e^-2gammat
decay constant of 2 gamma
quality factor
Q=w/2 gamma
describes how damping factor acts on system
as gamma approaches 0, very little damping
w^2=
w0^2-gamma^2
beat frequency
difference between 2 individual frequencies
think of as one wave laps the other
beat period
point where both waves line up
heavy damping
system experiences no oscillations
once excited, slowly moves back to eqm position
critical damping possible solution
z(t)=te^-gamma t
what type of damping takes longer to reach eqbm
light
although heavy damping is more aggressive in initial steepness…
it takes longer than critical damping to reach eqbm
(depending on initial conditions, may pass eqbm point once before settling)
what type of damping reaches eqbm first
critical
useful applications in mechanics
resonances
amplitude of an oscillation can become very large even if periodic driving force is small, if the driving frequency w is close to the natural frequency w0
example of resonance
tapping on wine glass till it smashes
pushing a swing: need to push in phase to keep going higher
resonance catastrophe
amplitudes beyond what physical system can allow
destroys system
equation of motion for a damped oscillator with a periodic external force
m d2x/dt2 = -kx - bdx/dt +Focoswt
equation of motion for a damped oscillator with a periodic external force in exponential form
d2z/dt2 + 2gamma dz/dt + wo^2z = F0e^iwt
after some time, solution to the homogeneous case will have been
damped away
just leaves special solution corresponding to external force
transition period
addition of a damped oscillator at the natural frequency to the forced oscillation at the driving frequency
steady state
transient effects die away and see only the effects of the driving force (solutions of inhomogeneous equation)
resonance curve
frequency dependence of A and delta
resonance frequency
max amplitude can be obtained
how to find resonance frequency
differentiating to find turning point
w «_space;wres
system oscillates in phase with A=F/mw0^2 approx = Fo/k
w=wres
system oscillates with A much larger than that of driving force but pi/2 out of phase
w»wres
system is pi out of phase and oscillates in opposite direction to driving force with oscillations having small A
for small damping coefficient, phase change
prominent, strong peak and fast phase change around the natural frequency
for large damping coefficient, phase change
resonance frequency moves closer to zero and resonance peak vanishes also resulting in slow phase change
higher quality factor means higher A at wres and
sharper resonance curve
mass on string for w approaching 0
direction of motion pi/2 out of phase with direction of force applied
mass on string for w approaching infinity
pi out of phase
largest amplitude when
w=w0
power absorbed by driven oscillator
P=dW/dt = Fdx/dt = Fv
for undamped oscillator, no dissipative effects so
mean power absorbed P bar =0 (steady state solution for x(t))
power only depends on
Q
eg: high Q glass absorbs more power than low
system with Q=50 will absrob 50x as much power as Q=1
good at absorbing power =
good at oscillating = low damping coefficient
Tacoma bridge collapse
during storm, wind blowing against the bridge caused it it start twisting
wind provided external periodic force that matched natural frequency of the bridge
caused amplitude of torsion to become so great the bridge collapsed
