The Wave Equation (Term 2)

Question 1

How can we find the total force on an element of length Δx for a graph of u(x,t) against x?

Answer 1

• Take small part of curve (x and x+Δx), and find the angle, which is equal to du/dx since sin(theta) ~= theta for small angles
• Down force and up force equal to -τ du(x)/dx and τ du(x+Δx)/dx respectively
• Total force is these added together

Question 2

What is the total force on an element of length Δx for a graph of u(x,t) against x equal to (using f=ma)?

Answer 2

m = ρΔx, where rho is the mass density, and a = d^2u/dt^2, so make the total force equal to the product of these.

Question 3

How can we rearrange the f=ma version of the total force to get d^2u/dt^2?

Answer 3

• Divide through by τΔx, separate the differential to make it equal to d/dx(du/dx), and rearrange
• This gives d^2u/dt^2 = c^2 d^2u/dx^2, where c^2 = τ/ρ.

Question 4

What is the assumption to start the separation of variables for the wave equation?

Answer 4

u(x,t) = X(x)T(t)

Question 5

How do we separate the variables for the wave equation?

Answer 5

Substitute in the assumed solution, and put the same variables on the same sides, and set this equal to a constant, λ

Question 6

What do we do once we separate the variables?

Answer 6

Try three conditions: λ = 0, λ>0 and λ<0.

Question 7

What happens when you set λ=0?

Answer 7

• Find d^2X/dx^2 = 0, so dX/dx = some constant, say A0
• Integrate again to find X(x) = A0*x + B0
• Same on T side, so T(t) = C0*t + D0
• Since u(x,t) = X(x)T(t), multiply these together.

Question 8

What happens when you set λ > 0?

Answer 8

• Say we set λ = k^2 > 0

- Rearrange T side and find solution T(t) = Cexp(-ckt)+Dexp(ckt)

Question 9

What happens when you set λ < 0?

Answer 9

• Say we set λ = -k^2
• Rearrange the X side this time
• Gives solutions X(x) = Acoskx+Bsinkx
• Rearrange T side and set ω^2 = k^2*c^2, and sub this in
• Find solutions T(t) = Ccosωt + Dsinωt
• Multiply these together to get u(x,t)

Question 10

What is the full general solution for u(x,t)?

Answer 10

Solution for λ=0 and λ<0 added together:

u(x,t) = (A0x + B0)(C0t + D0) + (Acoskx+Bsinkx)(Ccosωt + Dsinωt)

Question 11

How can we find the time average of u?

Answer 11

<u> = 1/t(max) *integral from 0 to t(max) of u(x,t) dt</u>

Question 12

What is <u> equal to?</u>

Answer 12

<u> = (A0x+B0)(C0t+D0), since the other terms integrate to zero.</u>

Question 13

Why is <u> equal to zero? What does u(x,t) now equal?</u>

Answer 13

Because we assume that the string is stationary on average, meaning (A0x+B0)(C0t+D0) = 0.
u(x,t) = (Acoskx+Bsinkx)(Ccosωt + Dsinωt)

Question 14

What are the boundary conditions for a string fixed at x=0 and x=L, and what does this mean for u(x,t)?

Answer 14

That u(x=0,t) = u(x=L,t) = 0

Means (Acosk0+Bsink0) = 0 and (AcoskL+BsinkL) = 0. This gives A = 0, and BsinkL = 0, giving k = nπ/L

Question 15

Using the boundary conditions, what is u(x,t) now equal to?

Answer 15

u(x,t) = sum from n=1-inf of Bsin(nπx/L)*(Ccosωt + Dsinωt), where ω = kc = nπc/L.
So u(x,t) = sum from n=1-inf of sin(nπx/L)*(Cn'cos(cnπt/L) + Dn'sin(cnπt/L))

Question 16

How do we determine the initial shape of u(x,t), if the initial condition is that the shape is up linearly until L/2, and then down linearly until L, peaking at E?

Answer 16

Substitute in t=0 to get the sum of Cn*sin(nπx/L). This is equal to 2Ex/L for 0

Question 17

How do we solve the initial condition stated before?

Answer 17

• Use the orthogonality of sine and cosine
• Multiply both sides by sin(mπx/L) and integrate from 0 to L
• u(x,0) = 2xE/L up to L/2 and 2E(L-x)/L for the rest
• Differentiate general solution found before
• String initially stationary so Dn = 0, giving the final equation with only Cn

Question 18

What is the final equation with only Cn?

Answer 18

u(x,t) = sum of Cn sin(nπx/L)cos(cnπx/L)