There are 2 conditions:

- Geometrical boundary condition: the displacement at x = 0 must be continuous at all times.

- Dynamical boundary condition: transverse force at x = 0 must be continuous.

Suppose 2 different strings which are joined at x = 0 so the tension cannot vary between them, only rhô can differ:

- Incident wave: yi = A*exp(omega*t - k1*x),

- Reflected wave: yr = B*exp(omega*t + k1*x) at the boundary,

- Transmitted wave: yt = C*exp(omega*t - k2*x) at the boundary.

Here we have different wave numbers but same angular frequency as rhô varies from one string to the other.

Relation between A, B and C for the geometrical boundary:

Because of what it implies, at all times we have:

yi(x = 0) + yr(x = 0) = yt(x = 0),

So A + B = C.

Relation between A, B and C for the dynamical boundary:

The only element we know/have is that the transverse force at x = 0 must be continuous. As we have seen before, at the boundary, the transverse force component is given by F = T2*sin theta2 - T1*sin theta1.
Remember small angle approximation turns sin into tan and tan is the gradient of the tension. Thus, for the general case, there is no net force at a massless boundary so:

T2*dyt/dx (at x = 0) = T1*d(yi+yr)/dx (at x = 0).

For our simple case, assume T1 = T2 = T.

So dynamical boundary can be re-written in this case:

- The product of T and the gradient must be continuous,

- If T is the same in both strings, only the string gradient has to be continuous.

As we have an expression for yi, yr and yt, can calculate the gradient dyi/dx, dyr/dx and dyt/dx and re-write the boundary condition at x = 0:

dyi/dx + dyr/dx = dyt/dx and calculate when x = 0.

We get: -k1*A + k1*B = -k2*C, using geometrical boundary condition:

-k1*A + k1*B = -k2(A + B).

If we want B/A, we need to eliminate C like we just have.

Multiply both sides by T so that later on we can use impedance. Use k = omega/c, c = sqrt(T/rhô) and sqrt(T*rhô) = Z too:

-Z1*A + Z1*B = -Z2*A - Z2*B, giving

B(Z1 + Z2) = A(Z1 - Z2)

So B/A = (Z1 - Z2)/(Z1 + Z2).

When you add a point mass at the join, the geometrical boundary condition doesn't change but the dynamical one does. It changes because this time there will be a net transverse force at the location of the mass (that's the force that causes the mass to accelerate). Therefore, the gradient doesn't have to be continuous, there is a possibility of a kink (petit défaut) at the join.

The dynamical boundary condition for 2 strings with an added mass is gotten from F = m*a. You get the same equation as before but mass is M.

We have the wave equations yi, yr and yt, so we can calculate the gradients and second derivative, d^2yt/dt^2. remember lastly the unconstrained case where T1 = T2.

Remember k*T = omega*Z.

Finally, we find that B/A = (Z1 - Z2 - i*M*omega)/(Z1 + Z2 + i*M*omega).

	Created by Sia N about 9 years ago

Physics (Waves and Diffraction) Note on Waves and Diffraction, created by Sia N on 29/10/2017.