Practice Questions Flashcards
general transverse wave
y(x,t) = Acos{kx±ωt+φ}
where φ is the offset
k =
2 pi / λ
positive propogation
to the left with a negative sign
i.e. kx - ωt
negative propogation
to the right with a positive sign
i.e. kx + ωt
the offset, φ, can be determined via
the boundary conditions
phase change is
ωt
dy/dt =
v
dv/dt =
a
displacement corresponds to which component
the REAL component
max values of displacement when
cos{kx±ωt} = 1
{kx±ωt} = 2 pi n
Show y(x,t) is a solution to the wave equation
- take the derivatives of y(x,t)
- substitute the derivatives into the wave equation
- solve to find a known relation
Energy density =
E/δx
δx =
∫ dx
progressive transverse wave
y(x,t) = y(x-ct)
where u = x-ct
dy/dt for a progressive transverse wave =
dy/du du/dt
dy/dx for a progressive transverse wave =
dy/du du/dx
units of impedance Z
kgs^-1
to convert from g to kg
x10^-3
linear density
rho, p
Transmitted energy coefficient T =
Et/Ei = C^2/A^2 Z2/Z1 = 4Z1Z2/(Z1+Z2)^2
mass derivation thingy
- Draw a Diagram
- Evaluate yi + yr = yt at boundary conditions (A+B=C)
- F = ma
- Fnet = T dy(t)/dx - T (dy(i)/dx + dy(r)/dx)
- substitute to F = ma
- evaluate derivatives at Boundary conditions
- multiply by -i
- C = A + B
- kT = Zω
for progressive waves to not be reflected
use impedance matching
impedance matching
Z2 = √(Z1Z3)
and
L = λ/4
Reflected Energy Coefficient R =
Er/Ei = B^2/A^2 Z1/Z1 = (Z1-Z3/Z1+Z3)^2
if at t = 0 the string is in equilibrium then the Epot =
0
ms =
ρL
calculating the energy residing in a mode
- Standing wave y(x,t)
- find if A or B is zero for the boundary conditions
- simply original y(x,t)
- take the derivative and evaluate for boundary conditions
- make use of orthogonality of sin function
- integrate LHS and RHS separately
- Solve for An or Bn
- Substitute into En
E is proportional to
the velocity squared
plucked string diagram for L/2
see notes
released from rest at time t=0
displaced
- find the derivative of y(n) then apply boundary conditions
- rewrite y(n) based on 1. then again apply boundary conditions
Pressure
P = nRT/V
number of moles n
n = M/m0
where m0 is the molecular weight
density
p = M/V
speed in pistons
c = √(γP/p)
P is pressure
and p is density
longitudinal waves in a solid bar
wave equation
normal wave equation but replace c^2 for Y/p
where Y is youngs modulus and p is density
standing waves where one end is fixed and the other free
Ψ(x,t) = Asin(kx+θ)cos(ωt+φ)
at the free end of a thin bar
we have balanced forces
T = YA ∂Ψ/∂x = 0
k value for a string where one is fixed and one is free to move
k(n) =((2n-1)π )/ (2l)
where l/L is length
Elastic energy
E = 1/2 k Δl^2
where k is the spring constant
spring constant k
k = F/Δl
Youngs modulus Y
Y = (F/A) / (l/Δl)
A membrane is a
2D wave equation
general solution to the 2D wave equation
z = 2iAexp{i(ωt-k(x)x)} sin(nπy/b)
2iAexp{i(ωt-k(x)x) representing a wave travelling the positive x-direction at velocity c = ω/k(x)
sin(nπy/b) shape of propagation in y direction
cut off frequency requires kx
to be real
for kx to be real
kx > 0
group velocity
v(g) = ∂ω/∂k
how long does it take a wave burst to travel
t = x/v(g)
phase velocity
v(p) = ω/k
and v(p) = c^2/v(g)
if a membrane is clamped round all edges
z = 0
and hence
k(x) = nπ/a and k(y) = nπ/b
dispersion equation
ω = 2 √(K/m) sin(ka/2)
where K the spring constant = YA/a
for max sin(ka/2) = 1
amu units
measured in gmol^-1 amu
spring constant units
Nm^-1
the wavefunction for a slit of finite width is
Ψ(x) = Aasinc(k(x)a/2)
Intensity can be found via
|Ψ(x)|^2
The diffraction pattern from the infinitely thin triple slit apparatus
Ψ(x) = Ao{1+2cos(k(x)d)}
Infinitely narrow double-slit problem , double slit
Ψ(x) = Ao cos(k(x)a/2)
Fraunhofer diffraction pattern of a rectangular aperture diagram
see notes
Fraunhofer diffraction intensity of a rectangular aperture
I(x) = Io sinc^2(k(x)a/2) sinc^2(k(y)b/2)
for an airy disc
θ = 1.22λ/D
the distance from the centre of the fringes
y = fθ
convolution of a wave function
Ψ = Ψ1 x Ψ2
simply multiply
two narrow slits or two pinholes
Ψ = Ao cos(k(y)d/2)
one end free and one end fixed boundary conditions
Ψ(x) = Ae^(ikx) + Be^(-ikx)
Ψ(BC) usually Ψ(0) = A + B
A = - B
F = ma ->
T dy/dx |join = m d^2y/dt^2 | join
2D wave equation
∂^z/∂x^2 + ∂^z/∂y^2 = 1/c^2 ∂^2z/∂t^2
solving the 2D wave equation to find the general solution
start from 2D wave equation
z = X(x)Y(y)T(t)
z = A{sinkx x } {sinky y} e^iωt
{coskx x} {cosky y}
evaluate at boundary conditions
z = 0 at x = 0, y = 0 -> cosines are zero
z = 0 at x=a, y =b -> kx = nxπ/a and ky = nyπ/b
leads to normal modes of the form see crib sheet for ω
with the general solution ( again see crib sheet for z = )
what happens if the long axis of the rectangular aperture is made shorter
the maximum in the kx axis will spread out
what happens if the rectangular aperture is replaced with an elliptical aperture of the same length and width
more elliptical fringes with similar spacings. The pattern will tend towards the airy disc and ring pattern, stretched out, with the intensity at the zero points increased compared to the rectangular.
the diffraction pattern produced by plane illumination of a circular aperture is
an airy disc
Rayleigh criterion
for two point sources to be resolved, the bright peak in the image from one source should be no closer than the first minimum in the diffraction pattern arising from the other sources
how does the airy disc of the diffraction pattern give rise to the Rayleigh Criterion
due to closely spaced concentric rings from the pattern needing to be resolved as hard to determine the separation
A two-dimensional membrane diagram for y = 0 and y = b
k(v) = k(x) i(v hat) + k(y) j(v hat)
a non dispersive medium
does not depend on ω so v(g) = ω/k
a dispersive medium
does depend on ω so v(g) ≠ ω/k => v(p) = ∂ω/∂k
solutions to the 1D wave equation may be obtained by
separation of variables
y = X(x)T(t)
∂^2T/∂t^2 X(x) = c^2T(t) ∂^2X/∂x^2
X(x) = X(0) sin(kx+φ)
T(t) = T(0) sin(ωt+θ)
y = X(x)T(t)
evaluate at BC
gives the normal modes of the string
proportions of the total stored energy
E(n) ∝ ω(n)^2 A^2
if the string is free at x = L, the end must be
parallel to the axis i.e. ∂y/∂x = 0
how can the Fourier transform technique be used to find the angular distribution of the intensity in a Fraunhofer diffraction pattern
for a particular wavelength, the field amplitude, expressed in terms of its spatial frequency components, is the Fourier Transform of the aperture function
A’(k(x),k(y)) = ( ∫∫ A) (x,y) exp{-i[k(x)x+k(y)y] dxdy
where kx and ky are the spatial frequencies
A(x,y) describes the transmission through the aperture
the aperture is opaque other than for
-a/2 ≤ y ≤ a/2
and
-b/2 ≤ x ≤ b/2
derive
I(k(x),k(y)) = I(0) sinc^2 (k(x)b/2) sinc^2 (k(y)a/2)
A’(k(x),k(y))
= (b/2 ∫ -b/2) exp{-ik(x)x}dx (a/2 ∫ -a/2) exp{-ik(y)y}dy
solve for first integral = bsinc(k(x)b/2)
similar result for second integral
So A’(k(x),k(y)) = basinc(k(x)b/2)sinc(k(y)a/2)
squared gives intensity
for k(y) = 0 sketch the variation of intensity of k(x)
see notes
Convolution theorem
Fourier transform (FT) of the convolution of two functions equals the product of the individual Fourier transformations of the functions
an array of identical apertures is equivalent to
the convolution of a single aperture with an array of delta functions located at the centres of the apertures in the array
deriving the 1D wave equation
draw a piece of string
-Tcosθ(1) + Tcosθ(2) = 0
F(net) = T(∂y/∂x|x(0)+δx/2 - ∂y/∂x|x(0)-δx/2)
F = ma and m = pδx
divide by δx
LHS = T ∂^2y/∂x^2
waves on a string are
non-dispersive
a loaded string of one or more point masses is
dispersive
“matching”
part of the wave is reflected and the part is transmitted at the join
the relative reflected amplitude will be determined by the impendence of the two strings where energy will be conserved.
a correctly chosen matching piece ensures
that there is no reflected amplitude back into the first string, ensuring that the incident energy propagates into the other string
With the aid of a diagram what boundary conditions must be satisfied at the location of a ring of mass, m
for small θ, Tsinθ = Ttanθ
but tanθ = ∂y/∂x
so upwards force on m = - Tsinθ = -T∂y/∂x |(x(mass)
F(net) = -T[∂y(i)/∂x + ∂y(r)/∂x]
if there is no dissipation
reflected energy = incident energy
ring of mass assumes
a steady state motion, so the total mass motion energy is fixed
conservation of the energy of the moving mass is not relevant for calculating reflected energy
warming of the spring material
reflected amplitude will be reduced and energy reflection will be <1, corresponding to energy lost, i.e. dissipated in the spring
force acting on the nth mass
F(n) = K(δ(a_n+1)-δ(a_n) - K(δ(a_n)-δ(a_n-1)
where the first term is the net extension of spring to the right of an
and the second term is the net extension of spring to the left
Equation of motion for linear chain of atoms
F(n) = mδ(d dot) (an) = m d^2δ(an)/dt^2
= -ω^2 δ(an)
candidate solution for linear chain of atoms
δ(an) = α exp{i(ωt-ka(n))}
sketch the form of ω-k relationship
at k = π/a adjacent masses move in anti-phase ; there is no wave propogation
see notes for diagram
acoustic mode
lower frequency ‘acoustic’ branch corresponds to wave propagation through the chain
optical mode
higher frequency ‘optical’ branch has adjacent masses in the unit cell moving in anti-phase
impedence is
the ratio of transverse force to transverse velocity
-T dy/dx and dy/dx = -1/c dy/dt
sketch of the velocity profile
see notes
Fresnel-Kirchoff diagram
see notes
i -
is the phase offset
exp{ikr}/r
is the amplitude of a
dΣ
the vector element, pointing inwards, is the element of the surface of integration
r -
is the distance from the area element to the point of interest, P
n’(v)
is the unit vector in the direction of propagation of the illuminating plane
n(v)
is in the direction from the aperture to the point P
obliquity factor
(n+n’/2) . dΣ = dΣ
How the Fresnel-Kirchoff improves the Huygens-wavelet approach
The Fresnel-Kirchhoff treatment correctly includes wavelength dependence.
It also provides the correct prediction of phase.
The obliquity factor removes what would otherwise be a backward propagating wave from the spherical Huygens wavelets.
when applied to the diffraction of light the Fresnel-Kirchhoff result is not complete in its description , why?
The Fresnel-Kirchhoff treatment calculates the diffraction of a scalar field.
Light is a vector quantity.
The principal omission is that
the scalar treatment cannot include consideration of polarization effects.
fraunhoffer diffraction is observing
in the far field where the diffracted waves are essentially parallel plane waves.
The focusing lens converges these to form a more local image in the focal plane of the lens - where the screen should be placed
the shape of a transverse wave as it propogates is
unchanged
dispersion
when the speed of wave propagation in the medium is not independent of frequency. There is a non-linear w-k relationship
When transmitting information through a medium the maximum throughput will be
when the individual bits of information take a minimum amount of transmission time. But a signal that is short in time occupies an inversely-related large frequency range.
In propagating through a dispersive medium, the frequency components of the individual bits of information will
travel at different speeds and so an initially short duration of the signal will smear out to occupy a longer time interval.
In addition, the peak amplitude will decrease as the signal propagates.
Both limit the transmission speed possible.
for double square use method of
subtraction
double square = [large square - small square]FT
= FT[large square] - FT[small square]
k(v) is the
propagating vector
k(x) and k(y) are the
x- and y- directed components of the propagation vector
k(v) = k(x) i(v hat) + k(y) j(v hat)
spatial frequencies
k(x) = kθ(x) and k(y) = kθ(y)
θ(y) and θ(x) are
the angle subtended at the origin of the optic axis
to include an offset in a Fourier transform
exp{offset}
how does an offset affect the diffraction pattern
the diffraction pattern is unchanged
the pattern is shifted by a spatial frequency offset
experimental set up for fraunhofer diffraction
two converging lenses plane waves entering the first with a slit after the 1st lens and a screen after the 2nd lens
Fixed end means
the direction and displacement are inverted i.e.
-Acos(kx+wt)
how to find the one cycle
integrate over lambda rather than L
double slit can be described as
a delta function
if f < f(min)
no steady-state propagation in the x-direction is possible and only standing waves in the y-direction persist.
phase velocity is
the velocity of those parts of the wave that have equal phase
group velocity is
the velocity with which the overall envelope of the wave propagates through space
v(max) =
√(k/m) A
particle velocity is
the simple harmonic velocity of the oscillator about its equilibrium position
for no boundary ,
Z(1) = Z(2)
B/A = 0
if the boundary is fixed,
B/A = -1
If the boundary is free,
B/A = 1
n = 1 mode is the
fundamental mode or the 1st harmonic.
you only get perfect nodes if
the incident and reflected waves have identical amplitudes, i.e. Incident wave energy = reflected wave energy
adding mass to the period equation
reduces the k values for odd harmonics -> reducing the frequenices.
Even harmonics are not affected.
in the generalised form of the wave equation
(Γ dy/dt and qy) allow for damping factors
the term exp(-Γt/2) provides the decay of the wave envelope due to dissipation; the term exp(±kc’t-kx) provides the oscillatory part of the wave solution.
v(g) = c
no dissipation
v(g) < c
normal dissipation
v(g) > c
anomalous dissipation
small spatial features correspond to
large spatial frequencies and so a big diffraction pattern
