To solve part (b), we will start from the given solution for $u(x, t)$ and apply the boundary condition at $x = 0$.

The solution is given as:

$u(x, t) = e^{-ikx - i\omega t} + r e^{ikx - i\omega t}.$

The boundary condition is:

$m \frac{\partial^2 u}{\partial t^2}(0, t) = T \frac{\partial u}{\partial x}(0, t).$

1. Step 1: Compute the derivatives

   Compute $\frac{\partial^2 u}{\partial t^2}$:

   $\frac{\partial u}{\partial t} = -i\omega \left(e^{-ikx - i\omega t} + r e^{ikx - i\omega t}\right),$

   $\frac{\partial^2 u}{\partial t^2} = -\omega^2 \left(e^{-ikx - i\omega t} + r e^{ikx - i\omega t}\right).$

   At $x = 0$:

   $\frac{\partial^2 u}{\partial t^2}(0, t) = -\omega^2 \left(1 + r\right)e^{-i\omega t}.$

   Compute $\frac{\partial u}{\partial x}$:

   $\frac{\partial u}{\partial x} = -ik e^{-ikx - i\omega t} + rik e^{ikx - i\omega t}.$

   At $x = 0$:

   $\frac{\partial u}{\partial x}(0, t) = -ik \left(1 - r\right)e^{-i\omega t}.$

2. Step 2: Apply the boundary condition

   Substituting into the boundary condition:

   [ m \cdot \left[-\omega^2 (1 + r)e^{-i\omega t}\right] = T \cdot \left[-ik (1 - r)e^{-i\omega t}\right]. ]

   Simplify by canceling $e^{-i\omega t}$ and the negative sign:

   $m\omega^2 (1 + r) = T k (1 - r).$

3. Step 3: Solve for $r$

   Expand and collect terms involving $r$:

   $m\omega^2 + m\omega^2 r = Tk - Tk r.$

   Rearrange:

   $r \left(m\omega^2 + Tk\right) = Tk - m\omega^2.$

   Solve for $r$:

   $r = \frac{Tk - m\omega^2}{m\omega^2 + Tk}.$

4. Step 4: Express $k$ using $v = \frac{\omega}{k}$

   From $v = \frac{\omega}{k}$, we have $k = \frac{\omega}{v}$. Substituting this into the expression for $r$:

   $r = \frac{T\frac{\omega}{v} - m\omega^2}{m\omega^2 + T\frac{\omega}{v}}.$

   Simplify:

   $r = \frac{\frac{T\omega}{v} - m\omega^2}{m\omega^2 + \frac{T\omega}{v}}.$

This is the expression for $r$.