P3 Waves & Diffraction > reflections > Flashcards

reflections Flashcards

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1
Q

standing waves result when

A

waves travelling in opposite directions interact

such interactions are often the result of waves reflecting from boundaries

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2
Q

how waves behave in situations of reflecting from boundaries is determined by

A

their characteristic impedance

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3
Q

the x=0 end is moved to create a sinusoidal wave propagating along +ve x axis

to do this we need to apply

A

time-varying force

at any given moment the force must be whatever is required to balance the transverse component of the string tension

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4
Q

if the tension in the string is T then the y-component of this is

A

Tsin theta

where theta is small

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5
Q

the counterbalancing force of the y-component of tension has a y-component of

A

F = -Tsin theta
approx = -T tan theta

= - T ∂y/∂x at x=0

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6
Q

relating ∂y/∂x and ∂y/∂t

A

∂y/∂t = -c d∂/∂x

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7
Q

transverse force needed in terms of ∂y/∂x and ∂y/∂t

A

= - T ∂y/∂x at x=0
= T/c ∂y/∂t at x=0

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8
Q

simplifications for transverse force needed expression

A

T/c is constant for given T and p
∂y/∂t at x=0 is the transverse velocity
c=sqrt(T/p)

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9
Q

final expression for transverse force needed using the simplifications

A

F(t) = root(Tp)∂y/∂t at x=0

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10
Q

the factor root(Tp) is defined as

A

the characteristic impedance Z of the stretched string

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11
Q

any medium through which waves propagate will

A

present an impedance to those waves

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12
Q

in lossless media, the impedance is

A

real and will be determined by the two energy storing parameters, inertia and elasticity

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13
Q

in lossy media, a

A

complex term is introduced into the impedance

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14
Q

Z=

A

transverse force applied / resulting transverse velocity

= F/v = root(Tp)

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15
Q

since the transverse velocity is determined by the inertia and the elasticity, the impedance is

A

also governed by these properties

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16
Q

impedance is a measure of

A

how hard it is to move the string up and down

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17
Q

impedance exists because the

A

string is under tension

so there is an opposing transverse component from that tension

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18
Q

very useful expression when dealing with reflections at boundaries

A

F = Zv

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19
Q

the importance of impedance is that if you join strings of non-matched impedances, you get

A

some energy (or power) reflected back to the source

(entirely equivalent to the importance of impedance matching in electrical circuits)

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20
Q

complex numbers can be expressed in terms of

A

real and imaginary parts

OR

in terms of magnitude and phase

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21
Q

euler equation

A

e^ipi +1 =0

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22
Q

i^-1=

A

-i

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23
Q

e^ipi =

A

-1 = i^2

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24
Q

consider two different strings, joined at x=0

if the strings are not considered at the x=0 points, then

A

the tension cannot vary between the two strings

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25
the two strings joined at x=0 have different linear densities so
there are different wave velocities in the two sections hence different impedances
26
an incident wave travelling along the string meets the discontinuity in impedance at
position x=0
27
define our incident wave travelling along string 1 in the +ve x direction as
yi = Aexp(i(wt-k1x)) for simplicity assuming no offset phase at t=0
28
in the general situation when the incident wave hits the boundary, there will be
a reflected wave yr and a transmitted wave yt
29
reflected and transmitted wave represented by
yr = Bexp(i(wt-k1x)) yt = Cexp(i(wt-k2x)) assuming k1 does not = k2 since p1 does not = p2
30
why is angular frequency of the oscillation the same through the two strings
if it wasn't, where would the extra cycles appear/disappear from/to at the junction
31
to explore what happens at the junction boundary, we want to
find relationships linking A, B and C do this by making use of boundary conditions (we have three variables so will need two boundary conditions)
32
two boundary conditions needed
1. geometrical boundary conditions 2. dynamical boundary conditions
33
1. geometrical boundary condition
the displacement at x=0 must be continuous (there can be no discontinuity of displacement), so at all time yi(x=0) + yr(x=0) = yt(x=0) A+B=C
34
2. dynamical boundary condition: the transverse force at x=0, T∂y/∂x must be
continuous (continuous slope) otherwise there would be a non-zero net force acting on an infinitely small string element which would result in a non-physical infinite acceleration
35
2. dynamical boundary condition: at the boundary, the net transverse force due to the components of the tensions in the two strings is
Fnet = T2sin[theta2] - T1sin[theta1]
36
2. dynamical boundary condition: assuming small amplitude derivations of the strings, we can use
small angle approx to make the same approximations as before Fnet = T2tan[theta2] - T1tan[theta1] Fnet=T2 ∂yt/∂x - T1 ∂(yi+yr)/∂x
37
2. dynamical boundary condition: for the general case we are considering here, where there can be NO
net transverse force at a massless boundary, Fnet=0 so T2 ∂yt/∂x = T1 ∂(yi+yr)/∂x
38
2. dynamical boundary condition: also have the requirement that T1=T2=T so
so ∂yt/∂x = ∂(yi+yr)/∂x
39
in general, what needs to be continuous
product of tension and string gradient T∂y/∂x
40
if the tensions in the two strings are the same then
it is the string gradient ∂y/∂x that is continuous
41
the only way we can constrain a join where there is no mass is if
the join is constrained to slide on a rod
42
to apply the dynamical boundary condition, we need to
differentiate ∂yi/∂x ∂yr/∂x ∂yt/∂x
43
have set the junction at x=0 so ∂yi/∂x+∂yr/∂x=
∂yt/∂x simplify and then sub in geometric boundary condition A+B=C
44
why do we multiply the simplified dynamical boundary condition by T
to bring in characteristic impedance then sub in k=w/c simplifies to B/A = (Z1-Z2)/(Z1+Z2)
45
does the dynamical boundary condition make sense?
1. would expect no reflection if no boundary (z1=z2) so sub in and B/A=0 GOOD 2. expect inversion if boundary fixed (z2=inf) so B/A=-1 GOOD 3. expect no inversion if boundary free (z2=0) sub in and B/A=1 GOOD
46
ratios of A,B and C are entirely depending on
the ratios of impedances independent of w and hold for waves of all frequencies
47
what happens to previous analysis if there is now a mass at the boundary of the two strings - geometric
geometrical bc unaffected since displacement at x=0 must still be continuous
48
what happens to previous analysis if there is now a mass at the boundary of the two strings - dynamical
possibility of a non-zero force that will cause mass to accelerate so instead of continuous gradient, possibility of kinks at the join
49
mass at boundary - new dynamical bc
obtained from F=ma net transverse force must equal the product of the mass and its acceleration Fnet= T2∂yt/∂x - t1∂(yi+yr)/∂x = M ∂2yt/∂t2 then same as before with derivatives
50
result for dynamical bc for mass between the two strings
B/A=[Z1-Z2-iMw]/[Z1+Z2+iMw]
51
is result for dynamical bc for mass between the two strings physically reasonable
if mass was huge - B/A=-1 so like hitting a fixed end - GOOD if mass really small - recover first result - GOOD
52
since Z1=root(Tp1), the power arriving at the boundary with the incident wave is
Pi=1/2A^2w^2Z1
53
energy conservation in situation of reflected and transmitted waves
start with 'after' ie Pr+Pt and recover 1/2A^2w^2Z1=Pi hence energy is conserved
54
since power is energy per unit time, the ratios of reflected to incident power and transmitted to incident power are
the same as the equivalent energy ratios and E prop to A^2 Er/Ei = B^2Z1/A^2Z1 = (B/A)^2 = (Z1-Z2/Z1+Z2)^2
55
the transmitted energy coefficient is given by
Et/Ei = C^2Z2/A^2Z1 = (C/A)^2 Z2/Z1 = 4Z1Z2/(Z1+Z2)^2
56
if Z1=Z2 then
no energy is reflected and the impedances are said to be matched
57
impedance matching is a very important practical problem as
long distance cables carrying energy must be accurately matched at all joints to avoid wasteful reflection of energy
58
the way to match two media with different impedances is to
insert a third element with a carefully chosen impedance and length
59
set up of three strings together
boundary between strings 1 and 2 at x=0 and boundary between strings 2 and 3 at x=l assume neither join is constrained so that the tension will be equal in each string (but Zs will be different due to different ps)
60
three strings joined together - 5 travelling waves to consider
1. incident wave 2. reflected wave from 1st boundary 3. transmitted wave from 1st boundary 4. reflected wave from 2nd boundary 5. transmitted wave the originates at 2nd boundary
61
boundary conditions for the three string joined together
4 in total - a geometrical and dynamical at each join
62
three strings joined together - what is the goal?
derive an expression for A3/A1 this will allow us to show that with suitable choice for middle section, we can achieve no reflected energy back "perfect matching"
63
to get a relationship for A3/A1 we need to
eliminate variables B1,A2 and B2
64
determining A3/A1 condition 1: geometric condition at x=0
displacement must be continuous A1+B1=A2+B2
65
determining A3/A1 condition 2: dynamical condition at x=0
applying same dynamical arguments as before Z1(A1-B1) = Z2(A2-B2)
66
determining A3/A1 condition 3: geometric condition at x=l
A2exp(-ik2l) + B2exp(eik2l) = A3
67
condition 3: geometric condition at x=l why is there no exp(-ik2x) term on the RHS
to avoid writing any explicit phase factors along with A3, we define A3 to be complex only interested in the energy flow so can do this as concerned with squared magnitude of A3 makes maths easier
68
determining A3/A1 condition 4: dynamical condition at x=l
Z2(A2exp(-ik2l) - B2exp(ik2l)) = Z3A3
69
determining A3/A1 condition 4: dynamical condition at x=l STEPS
begin Z1(A1-B1) = Z2(A2-B2) get A1 in terms of A2 and B2 define Z1/Z2 = r12 to eliminate B2 define Z2/Z3=r23 to eliminate A2 define r13=r12r23 and use properties of exp set central length l=lambda2/4and simplify
70
the condition r12=r23 corresponds to choosing
Z1/Z2 = Z2/Z3 Z2=sqrt (Z1Z3)
71
we achieve no backreflection of energy by
1. choosing the matching string to be a quarter wavelength long and 2. having an impedance that is the harmonic mean of the impedances of the two strings being matched
72
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