Mass Reflectance from Either String

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 Ohms in series with the mass impedance . (This observation will be used as the basis of a rapid solution method in §9.3.1 below.)

When a wave from string 1 hits the mass, it will cause the mass to
move. This motion carries both string endpoints along with it.
Therefore, both the reflected and transmitted waves include this mass
motion. We can say that we see a ``dispersive transmitted wave'' on
string 2, and a dispersive reflection back onto string 1. Our object
in this section is to calculate the transmission and reflection
*filters* corresponding to these transmitted and reflected waves.

By physical symmetry the *velocity* reflection and transmission
will be the same from string 1 as it is from string 2. We can say the
same about force waves, but we will be more careful because the sign
of the transverse force flips when the direction of travel is
reversed.
^{10.12}Thus, we expect a scattering junction of the form shown in
Fig.9.17 (recall the discussion of physically
interacting waveguide inputs in §2.4.3). This much invokes
the superposition principle (for simultaneous reflection and transmission),
and imposes the expected symmetry: equal reflection filters
and
equal transmission filters
(for either force or velocity waves).

Let's begin with Eq.(9.12) above, restated as follows:

The traveling-wave decompositions can be written out as

where a ``+'' superscript means ``right-going'' and a ``-'' superscript means ``left-going'' on either string.

Let's define the mass position to be zero, so that Eq.(9.14) with the substitutions Eq.(9.15) becomes

Let's now omit the common ``(t)'' arguments and write more simply

Let denote the common velocity of the mass and string endpoints. Then we have for the mass (as discussed above at Eq.(9.13)), and the Ohm's-law relations for the string are and . Making these substitutions gives

In the Laplace domain, dropping the common ``(s)'' arguments,

To compute the reflection coefficient of the mass seen on string 1, we may set , which means , so that we have

Also, since , we can substitute and solve for the common velocity to get

From this, the reflected velocity is immediate:

The transmitted velocity is of course . We have thus found the

By physical symmetry, we also have ,

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

The results of this section can be more quickly obtained as a special case of the main result of §C.12, by choosing waveguides meeting at a load impedance . The next section gives another fast-calculation method based on a standard formula.

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