Derive

Question 1

Derive the wave equation

Answer 1

draw string

F(net) = T tan{θ2} - T tan{θ1}

F(perp) = T [dy/dx|xo+δx/2 - dy/dx|xo-δx/2]

F = ma

m = pdx

a = d^2y/dx^2|x0

Question 2

Derive potential energy of a string element

Answer 2

δs = √((δx)^2 + (δy)^2)

~ δx[1+1/2(δy/δx)^2]

Δs = δs - δx

giving Epot on the formula sheet

Question 3

Derive the characteristic impedance

Answer 3

F(perp) = -T dy/dx|x(0)

dy/dt = -c dy/dx

Z = T/C

Question 4

Derive the reflection at boundaries

Answer 4

Ψ(1) = Aexp{i(ωt-k(1)x)} + Bexp{i(ωt+k(1)x)}

Ψ(2) = Cexp{i(ωt-k(2)x)}

Ψ(1) = Ψ(2) at x = 0

then

T(Ψ’(1)) = T(Ψ’(2)) at x = 0

solve for B over A

Question 5

Derive the transmission at boundaries

Answer 5

T = 1 - Reflection coefficient

i.e. T = 1 - R

= 2Z1/(Z1+Z2)

Question 6

Derive the reflection at boundaries for a mass

Answer 6

Ψ(1) = Aexp{i(ωt-k(1)x)} + Bexp{i(ωt+k(1)x)}

Ψ(2) = Cexp{i(ωt-k(2)x)}

Ψ(1) = Ψ(2) at x = 0

then

T(Ψ’(2)-Ψ’(1)) = M(Ψ’‘(2)) at x = 0

solve for B over A

Question 7

Show Pi = Pr + Pt

Answer 7

Pi = 1/2 A^2 ω^2 Z(1)

Pr = 1/2 B^2 ω^2 Z(1)

Pt = 1/2 C^2 ω^2 Z(2)

Question 8

Derive the normal modes on a string when both ends are fixed.

Answer 8

y = X(x)T(t)

wave equation

T(t) = T(0)sin(ωt+θ)
X(t) = X(0)sin(kx+φ)

sub for y = X(x)T(t)

y = 0 at x = 0 => φ = 0

y = 0 at x = L

k = nπ/L

Question 9

Derive the wave equation in a solid bar

Answer 9

F(net) = YA[ε(x+δx/2)-ε(x-δx/2)]

F(net) = ma

m = Aδxp

Question 10

derive the dispersion relation

Answer 10

Fnet = K(δa(n+1)-2δa(n)+δa(n-1))

δa(n) = Aexp{i(wt-ka(n))}

F = ma

leads to dispersion equation

Question 11

Derive the 2D wave equation

Answer 11

vertical restoring force = T [δy (dz/dx)|x0+δx/2 - δy (dz/dx)|x0-δx/2]

horizontal restoring force = T [δx (dz/dy)|y0+δy/2 - δx (dz/dy)|y0-δy/2]

= Tδyδx (d^2z/dx^2) + Tδyδx (d^2z/dy^2)

F = ma

leads to the 2D wave equation

Question 12

Derive the general solution of the 2D wave equation

Answer 12

z = Ae{i(ωt-[xk(x)+yk(y)]}) + Be{i(ωt-[xk(x)-yk(y)]})

z = 0 at y = 0

z = 0 at y = b

Question 13

The general solution of the 2D wave equation can be rewritten as

Answer 13

z = A {sin(xk(x))}{sin(yk(y))}{sin(ωt))}
{cos(xk(x))}{cos(yk(y))}{cos(ωt))}

Question 14

Derive the period equation

Answer 14

y(1) = y(0)sin(kx)cos(kct+φ)
y(2) = y(0)sin[k(L-x)]cos(kct+φ)

at x = 0 y(1) = 0
at x = L/2 y(1)(L/2,t)=y(2)(L/2,t)
at x = L y(2) = 0

F = ma

T( dy(2)/dx|L/2 - dy(1)/dx|L/2) = M d^2y(1)/dt^2|L/2