Wave hitting a boundary with a mass on it? If we have a transverse wave that is infinite in the $-x$ direction and terminate by a mass $m$, that is allowed to move in the $y$ direction at $x=0$ as shown in the diagram below: 
I think we can use the following (approximation) for our boundary conditions:
$$-T \frac{\partial y}{\partial x}=m\frac{\partial^2 y}{\partial t^2}$$ (evaluated at $x=0$)
Where $T$ is the tension in the string. This however gives a amplitude releflection coefficent of:
$$r=e^{i\phi}$$
Where $\phi$ is some phase. 
This means that the average energy of the reflected wave is the same as the indicdent wave. But this is breaking the conservation of energy since the mass $m$ also has a non-zero average energy. This means the total average energy of the mass and the reflected wave will be greater then that of the incident wave. Please can someone explain this to me, i.e. why it is or is not valid? 
 A: I don't think you are doing anything wrong. You're simply forgetting that your reflexion co-efficient, calculated as a phase, assumes sinusoidal excitation and thus, tacitly, you've assumed that the wave system is of infinite extent in time. In particular, the system has reached a steady state. What energy is going into the mass is periodically coming out again as the reflected wave. As the energy leaves the mass to join the reflected wave, a balancing amount is brought to the mass by the incident wave. There is periodic shuttling of energy just as in a lossless LC resonant circuit. You might try exploring a general exitation to get a deeper grasp of things: let's assume that $f(z-c\,t)$ is the functional form of the incident wave, $g(z+c\,t)$ that of the reflected wave and that the mass is at $z=0$. From your correct equation $-T \frac{\partial y}{\partial x}=m\frac{\partial^2 y}{\partial t^2}$ I get:
$$T\,c\,\left(\dot{f}(-c\,t) -\dot{g}(c\,t)\right) = m\,\ddot{y}(t)\tag{1}$$
From the continuity equation (the end of the string's motion must be the same as that of the  lumped mass) I get:
$$g(c\,t) +f(-c\,t) =y(t)\tag{2}$$
Whence:
$$m\,c^2\left(\ddot{g}(c\,t) +\ddot{f}(-c\,t)\right) =m\,\ddot{y}(t)\tag{3}$$
so that you can get rig of $y$ by combining (1) and (3) and find the differential equation for $g$ in terms of $f$. If you set some excitations up in, say, Mathematica, you should see the progression of energy from the incident wave, to the mass and then to the reflected wave.
