Impedance Matching

Question 1

Power in terms of Impedance

Answer 1

Pi = 1/2 Z₁A²ω²
Pr = 1/2Z₁A²ω²
Pt = 1/2Z₂A²ω²

Question 2

When there is no mass at the boundary

B/A =

C/A =

Answer 2

B/A = (Z₁ - Z₂)/(Z₁ + Z₂)
C/A = 2Z₁/(Z₁ + Z₂)

Question 3

All energy arriving at the boundary with the incident wave leaves with

Answer 3

the reflected and transmitted wave. Because energy is conserved

Question 4

When there is no mass at the boundary

Answer 4

energy is conserved

Question 5

The simple joined string shows

Answer 5

no reflection if Z2 = Z1

Question 6

A sinusoidal wave incident on a fixed or free end gives

Answer 6

rise to a reflected wave, and that when the incident and reflected waves interact, we get standing waves

Question 7

λn =

Answer 7

2π/kn = 2π/nπ/L

= 2L/n

Question 8

fn =

Answer 8

c/λn

= n * c/2L

Question 9

Each normal mode satisfies the

Answer 9

boundary condition of the string being fixed at each end

Question 10

The n=1 mode is called the

Answer 10

fundamental mode or the 1st harmonic

Question 11

When do you get perfect nodes

Answer 11

if incident and reflected waves have identical amplitudes, this happens only if the incident and reflected wave energies are the same. In which case no energy is lost at the boundaries and so the total net energy flux is 0.

Question 12

plucked strings are

Answer 12

pulled aside and released with zero velocity

Question 13

consider two different strings joined at x = 0 their yi , yr and yt can be defined as

Answer 13

yi = Aoe^(i{wt-k1x})
yr = Aoe^(i{wt+k1x})
yt = A2e^(i{wt-k2x})

Question 14

To solve two different strings joined at x = 0 with a mass M

Answer 14

1) yi+yr = yt

2) Apply boundary conditions for
a) yi+yr = yt x = 0

2) Consider the forces acting on the mass M at x = 0
T d(yi+yr)/dx - Td(yt)/dx = M d^2(yt)/dx^2

3) substitute progressive transverse wave into 2)

simplify and solve

Question 15

for a progressive transverse wave y(x,t) =

Answer 15

y(x ± ct)

dy/dt = ± c dy/dx

dy/dx = 1/c dy/dt

Question 16

To solve two different strings joined at x = 0 (unconstrained) with no mass

Answer 16

1) yi+yr = yt

2) take the derivatives of yi+yr = yt

3) Apply boundary conditions for
a) the derivatives of yi+yr = yt x = 0

simplify and solve

Question 17

if a string is unconstrained x = 0 it is the

Answer 17

gradient that is continuous

Question 18

Two requirements to prevent energy/power being reflected

Answer 18

Z2 = √Z1Z3

and

L = λ/4

Question 19

General Q involves

Answer 19

1) yi + yr = yt

2) evaluate yi+yr = yt at the conditions usually x = 0 (or x = L)

3) find derivatives of yi + yr = yt and multiply by each component by tension T

4) evaluate at boundary condition

5) Simplify and solve.

Question 20

standing waves on a stretched string

Answer 20

y(x,t) = yosin{kx+Φ}sin{kct+θ}

Question 21

undisplaced from equilibrium

Answer 21

t = 0

Question 22

displaced from the equilibrium

Answer 22

dy/dt = 0