Question

derive the classical wave equation. Show the relationship between this equation and the electromagnetic wave equation.

Answer 1

wave equation for waves on a taut string, stretched between  x=0 and  x=L, tension T newtons, density ρ kg/meter. Assuming the string’s equilibrium position is a straight horizontal line (and, therefore, ignoring gravity), and assuming it oscillates in a vertical plane, we use f(x,t) to denote its shape at instant t, so f(x,t) is the instantaneous upward displacement of the string at position x. We assume the amplitude of oscillation remains small enough that the string tension can be taken constant throughout.

The wave equation is derived by applying F=ma to an infinitesimal length dx of string (see the diagram below). We picture our little length of string as bobbing up and down in simple harmonic motion, which we can verify by finding the net force on it as follows.

At the left hand end of the string fragment, point x, say, the tension T is at a small angle df(x)/dx to the horizontal, since the tension acts necessarily along the line of the string. Since it is pulling to the left, there is a downward force component Tdf(x)/dx.  At the right hand end of the string fragment there is an upward force Tdf(x+dx)/dx.    

Putting f(x+dx)=f(x)+(df/dx)dx, and adding the almost canceling upwards and downwards forces together, we find a net force T(d²f/dx²)dx  on the bit of string. The string mass is ρdx, so F=ma becomes

T(∂²f(x,t)/∂x²)dx=ρdx(∂²f(x,t)∂t²)

giving the standard wave equation

∂²f(x,t)/∂x²=1/v²(∂²f(x,t)∂t²) ..................(i)

with wave velocity given by  v²=T/ρ;

Now since we have a differential equation of second order we can use the method of operation of variables.

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

Substituting (ii) in (i)

T(t)⋅∂²X(x)/∂x²=X(x)/v²⋅∂²T(t)/∂t²

which simplifies to

1/X(x)⋅∂²X(x)/∂x²=1/T(t)v²⋅∂²T(t)/∂t²=K

where K is called the "separation constant". By setting each side equal to K, two 2^nd order homogeneous ordinary differential equations are made.

d²X(x)/dx²−KX(x)=0 ................(iii)

and

d²T(t)/dt²−Kv²T(t)=0 ..........(iv)

We are particularly interested in the example with specific boundary conditions (the wave has zero amplitude at the ends).

u(x=0,t)=0

u(x=L,t)=0

The general solution of the homogenous second order linear DE will be of eq (iii) can be

X(x)=A⋅cos(ax)+B⋅sin(bx)

Using the boundary conditions;

X(x)=B⋅sin(nπx/ℓ)

From which we find the separation constant k that

K=−(nπ/ℓ)²

Using the separation constant k and Solving for the ODE of eq.(iv)

T(t)=Dcos(nπνt/ℓ)+Esin(nπνt/ℓ)

where DD and EE are constants and nn is an integer (>1>1), which is shared between the spatial and temporal solutions.

Unfortunately, we do not have the boundary conditions like with the spatial solution to simplify the expression of the general temporal solutions in eq(iii). However, these solutions can be simplified with basic trigonometry identities to

T_n(t)=A_ncos(nπνt/ℓ+ϕ_n)

where A_n is the maximum displacement of the string (as a function of time), commonly known as amplitude, and ϕ_n is the phase and nn is the number from required to establish the boundary conditions.

Now. since we had

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

=> u_n=A_ncos(ω_nt+ϕ_n)sin(nπx/ℓ)

Both equations have same type of equation. In the em wave equation we have two superposition of classical wave equations of magnetic field and electric field mutually perpendicular to each other.