Mass Termination Model

The previous discussion solved for the motion of an ideal mass after hitting an ideal string of infinite length. We now investigate the model from the string's point of view. As before, we will be interested in a digital waveguide model (sampled traveling-wave model) of the string, for efficiency's sake, and we therefore will need to know what the mass ``looks like'' at the end of each string segment.

Let's number the string segments to the left and right of the mass by 1 and 2, respectively. Then Eq.$\,$(4.17) above may be written

$$f_m(t) + f_{1m}(t) + f_{2m}(t) = 0, \protect$$ (5.19)

where $f_{1m}(t)$ denotes the force applied by string-segment 1 to the mass (defined as positive in the ``up'', or positive-$y$ direction), $f_{2m}(t)$ is the force applied by string-segment 2 to the mass (again positive upwards), and $f_m(t)$ denotes the inertial force applied by the mass to both string endpoints (where again, a positive force points up).

To derive the traveling-wave relations in a digital waveguide model, we want to use the force-wave variables $f_1=f^{{+}}_1+f^{{-}}_1$ and $f_2=f^{{+}}_2+f^{{-}}_2$ that we defined for vibrating strings in §4.1.5; i.e., we defined $f\isdeftext -Ky'$, where $K$ is the string tension and $y'$ is the string slope.

A moment's thought (or careful reading of §G.7.2) reveals that under our definition, string force acts to the right, irrespective of how the sloped string segment might be moving. That is, as shown in Fig.4.22, a negative string slope pulls ``up'' to the right. The same sloped string segment pulls ``down'' when viewed as acting to the left.

Figure 4.22: Depiction of a string segment with negative slope (center), pulling up to the right and down to the left. (Horizontal force components are neglected.)

In the present problem, considering the physical sum of forces at the mass-string junction in Fig.4.20, we must have

$$m\dot v(t) + K\,y'_1(t,0) - K\, y'_2(t,0) = 0, \protect$$ (5.20)

which translates to the force-wave relations

$$f_m(t) - f_1(t) + f_2(t) = 0. \protect$$ (5.21)

The force relations can be checked individually. For string 1,

$$m\dot v(t) + K\,y'_1(t,0) = 0$$

states that a positive slope in the string-segment to the left of the mass corresponds to a negative acceleration of the mass by the endpoint of that string segment. Similarly, for string 2,

$$m\dot v(t) - K\,y'_2(t,0) = 0$$

says that a positive slope on the right accelerates the mass upwards. Similarly, a string segment with negative slope pulls ``up'' to the right and ``down'' to the left, as shown in Fig.4.22.

Now that we have expressed the string forces in terms of string force-wave variables, we can derive digital waveguide models by performing the traveling-wave decompositions $f_1=f^{{+}}_1+f^{{-}}_1$ and $f_2=f^{{+}}_2+f^{{-}}_2$ and using the Ohm's law relations $f^{{+}}_i=Rv^{+}_i$ and $f^{{-}}_i=-Rv^{-}_i$ for $i=1,2$(introduced above near Eq.$\,$(4.6)).

Let's first consider how the mass looks from the viewpoint of string 1, assuming string 2 is at rest. In this situation (no incoming wave from string 2), string 2 will appear to string 1 as a simple resistor (or dashpot) of $R$ Ohms in series with the mass impedance $ms$. In a traveling-wave model, an incoming force wave will reflect from the mass, causing some motion of the mass. The mass motion is also the motion of both string endpoints. On string 2, we see a ``dispersive transmitted wave'', and on string 1 we see a ``wake'' following the reflection from the mass.

The traveling-wave counterpart of Eq.$\,$(4.21) above, with no incoming wave from string 2, is given by performing the traveling-wave decompositions $f_1(t,x)=f^{{+}}_1(t-x/c) + f^{{-}}_1(t+x/c)$ and $f_2(t,x)=f^{{+}}_2(t+x/c) + f^{{-}}_2(t-x/c)=f^{{-}}_2(t-x/c)$. Thus, we have, at time $t$ and at the collision position $x=0$,

$$f_m(t) - [f^{{+}}_1(t) + f^{{-}}_1(t)] + f^{{+}}_2(t) = 0.$$

We will now omit the common ``(t)'' arguments and write simply $f_m + f^{{+}}_1 + f^{{-}}_1 + f^{{+}}_2=0$. In the Laplace domain, dropping the common ``(s)'' arguments, we have

$$F_m - (F^{+}_1 + F^{-}_1) + F^{+}_2 = 0.$$

Substituting the Ohm's law relations for the string waves and using the impedance relation $F_m = ms$ for the mass (now assuming a zero initial mass velocity $v_0=0$), we obtain

$$msV - RV^{+}_1 + RV^{-}_1 + RV^{+}_2 = 0.$$

Since the incoming wave on string 2 is $V^{-}_2=0$, we have $V^{+}_2=V$, because it must always be true that $V=V^{+}_1+V^{-}_1=V^{+}_2+V^{-}_2$ (the mass and string endpoints are in series and therefore have a common velocity). We can consequently write

$$\begin{eqnarray*} && (R+ms)V - RV^{+}_1 + RV^{-}_1 \;=\; 0\\ \Rightarrow \; && (R+ms)(V^{+}_1+V^{-}_1) - RV^{+}_1 + RV^{-}_1 \;=\; 0. \end{eqnarray*}$$

We can now solve this equation for the velocity reflection transfer function (or velocity reflectance) of the mass, as seen from string 1:

$$\rho^v_1 \isdef \frac{V^{-}_1}{V^{+}_1} = -\frac{ms}{ms+2R}$$

It is always good to check that our answers make physical sense in limiting cases. For this problem, easy cases to check are $m=0$ and $m=\infty$. When the mass is $m=0$, the reflectance goes to zero (no reflected wave at all). When the mass goes to infinity, the reflectance approaches $\rho^v_1=-1$, corresponding to a rigid termination, which also makes sense.

By physical symmetry, the mass looks the same from string 2. Therefore,

$$\rho^v_2 \isdef \frac{V^{+}_2}{V^{-}_2} = \rho^v_1 = -\frac{ms}{ms+2R}.$$

In other words, the reflectance of the mass is $-ms/(ms+2R)$ from either string segment.

The results of this section can be more quickly obtained as a special case of the main result of §G.10, by choosing $N=2$ waveguides meeting at a load impedance $R_J(s)=ms$.