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Physical Audio Signal Processing
    Virtual Musical Instruments
       String Excitation
          Ideal String Struck by a Mass

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Ideal String Struck by a Mass

In §6.6, the ideal struck string was modeled as a simple initial velocity distribution along the string, corresponding to an instantaneous transfer of linear momentum from the striking hammer into the transverse motion of a string segment at time zero. (See Fig.6.10 for a diagram of the initial traveling velocity waves.) In that model, we neglected any effect of the striking hammer after time zero, as if it had bounced away at time 0 due to a so-called elastic collision. In this section, we consider the more realistic case of an inelastic collision, i.e., where the mass hits the string and remains in contact until something, such as a wave, or gravity, causes the mass and string to separate.

For simplicity, let the string length be infinity, and denote its wave impedance by $ R$. Denote the colliding mass by $ m$ and its speed prior to collision by $ v_0$. It will turn out in this analysis that we may approximate the size of the mass by zero (a so-called point mass). Finally, we neglect the effects of gravity and drag by the surrounding air. When the mass collides with the string, our model must switch from two separate models (mass-in-flight and ideal string), to that of two ideal strings joined by a mass $ m$ at $ x=0$, as depicted in Fig.9.12. The ``force-velocity port'' connections of the mass and two semi-infinite string endpoints are formally in series because they all move together; that is, the mass velocity equals the velocity of each of the two string endpoints connected to the mass (see §7.2 for a fuller discussion of impedances and their parallel/series connection).

Figure 9.12: Physical model of mass-string collision after time 0.
\includegraphics[width=\twidth]{eps/massstringphy}

The equivalent circuit for the mass-string assembly after time zero is shown in Fig.9.13. Note that the string wave impedance $ R$ appears twice, once for each string segment on the left and right. Also note that there is a single common velocity $ v(t)$ for the two string endpoints and mass. LTI circuit elements in series can be arranged in any order.

Figure 9.13: Electrical equivalent circuit for the mass and two string endpoints after time zero. The mass is represented by an inductor of $ m$ Henrys, while each string endpoint is represented by a resistor of $ R$ Ohms (impedance $ R$).
\includegraphics{eps/massstringec}

From the equivalent circuit, it is easy to solve for the velocity $ v(t)$. Formally, this is accomplished by applying Kirchoff's Loop Rule, which states that the sum of voltages (``forces'') around any series loop is zero:

$\displaystyle 0 \eqsp f_m(t) + f_{R}(t) + f_{R}(t) \protect$ (10.8)

All of the signs are `$ +$' in this equation because the ``current'' (velocity $ v(t)$) flows into the `$ +$' sign of each element. These reference directions indicated by `$ +/-$' on each element in Fig.9.13 may be chosen arbitrarily, but it is convenient to adopt the convention that $ v(t)$ flows into the `$ +$' sign for each reaction force, and into the `$ -$' sign for each action force (or driving force or voltage/current source, etc.). With this convention, Kirchoff's Loop Rule effectively states ``the sum of all action forces equals the sum of all reaction forces,'' where the forces are understood to act on a common point. This is essentially Newton's third law of motion ``for every action there is an equal and opposite reaction'' (§B.1). In our mass-string example, all three forces are defined to be reaction forces because there is no external driving force on the mass, and no incoming waves from the string segments. External driving forces on the mass-string junction will be formulated later below.10.8 We could equally well have defined the mass inertial force as a driving force on the two string segments, or the string segments could comprise a pair of driving forces for the mass. Ultimately, any force polarities may be assigned and consistently handled.

Taking the Laplace transform10.9of Eq.$ \,$(9.8) yields, by linearity,

$\displaystyle 0 \eqsp F_m(s) + 2F_{R}(s), \protect$ (10.9)

where $ F_m(s)$ and $ F_{R}(s)$ denote the Laplace transforms of $ f_m(t)$ and $ f_{R}(t)$, respectively. As discussed above, the impedance relation for each string endpoint is given by

$\displaystyle f_{R}(t) = R\,v(t) \quad\longleftrightarrow\quad F_{R}(s) = R\,V(s), $

where $ V(s)$ denotes the Laplace transform of $ v(t)$.

For the mass, we have

$\displaystyle f_m(t) = m\,a(t)\;=\; m\,\frac{d}{dt} v(t) \quad\longleftrightarrow\quad F_m(s) = m\left[s\,V(s) - v_0\right], $

where we used the differentiation theorem for Laplace transforms [449, Appendix D].10.10Note that the mass is characterized by its impedance
$ F_m(s)/V(s) = ms$ when the initial velocity $ v_0$ is zero (§7.1).

Substituting these relations into Eq.$ \,$(9.9) yields

$\displaystyle m\,v_0 \eqsp m\,s\,V(s) + 2R\,V(s). \protect$ (10.10)

We see that the initial momentum $ m\,v_0$ of the mass effectively provides an impulsive external driving force

$\displaystyle f_{\mbox{ext}}(t)\eqsp m\,v_0\,\delta(t). $

That is, an equivalent problem formulation is to start with the mass at rest and in contact with the string, followed by striking the mass with an ideal hammer (impulse) that imparts momentum $ m\,v_0$ to the mass at time zero. This formulation is diagrammed in Fig.9.14.

Figure: Electrical equivalent circuit for the mass and two string endpoints after time zero, using an impulsive driving force in place of a nonzero initial velocity. (Cf. Fig.9.13.)
\includegraphics{eps/massstringecimp}

An advantage of the external-impulse formulation is that the system has a zero initial state, so that an impedance description (§7.1) is complete. In other words, the system can be fully described as a series combination of the three impedances $ ms$, $ R$ (on the left), and $ R$ (on the right), driven by an external force-source $ f_{\mbox{ext}}(t)$.

Solving Eq.$ \,$(9.10) for $ V(s)$ yields

$\displaystyle V(s) = \frac{m\,v_0}{ms + 2R}. $

Since the Laplace transform of $ e^{-at}u(t)$ is $ 1/(s+a)$, where $ u(t)$ denotes the Heaviside unit step function,10.11 the velocity of the contact point is

$\displaystyle v(t) = v_0\, e^{-{\frac{2R}{m}t}}, \quad t\ge 0. $

We see that at time zero the mass velocity is $ v_0$, as it must be, and after that it decays exponentially to zero with time-constant $ m/2R$. The decay rate depends on the ratio of the mass to the string wave impedance. In particular, the heavier the mass, the slower the mass velocity decays to zero. Since $ R=\sqrt{K\epsilon }$, the greater the string tension $ K$ or mass-density $ \epsilon $, the faster the mass velocity decays to zero.

The displacement of the string at $ x=0$ is given by the integral of the velocity:

$\displaystyle y(t,0) = \int_0^t v(\tau)\,d\tau = v_0\,\frac{m}{2R}\,\left[1-e^{-{\frac{2R}{m}t}}\right] $

where we defined the initial transverse displacement as $ y(0,0)=0$. The final displacement of the string is

$\displaystyle y(\infty,0) = v_0\,\frac{m}{2R}. $

Thus, the final string displacement is proportional to both the ``hammer mass'' and the initial striking velocity; it is inversely proportional to the string wave impedance $ R$.

The momentum of the mass before time zero is $ mv_0$, and after time zero it is

$\displaystyle m\,v(t) = m\,v_0\, e^{-{\frac{2R}{m}t}}. $

The force applied to the two string endpoints by the mass is given by $ f_m(t) = 2Rv(t)$. From Newton's Law, $ f=ma=m\dot v$, we have that momentum $ mv$, delivered by the mass to the string, can be calculated as the time integral of applied force:

$\displaystyle \int_0^t f_m(\tau)\,d\tau = 2R\int_0^t v(\tau)\,d\tau = 2Rv_0\fr... ...eft(1-e^{-{\frac{2R}{m}t}}\right) = m\,v_0\left(1-e^{-{\frac{2R}{m}t}}\right). $

Thus, the momentum delivered to the string by the mass starts out at zero, and grows as a relaxing exponential to $ mv_0$ at time infinity. We see that an ideal string struck inelastically by a mass does not at all correspond to an instantaneous momentum transfer, as considered in §6.6. Instead, the mass's momentum is transferred over a period of time (in this case infinite time). This is why it is possible to approximate the mass width by zero in this analysis, unlike in the excitation-by-initial-velocity in §6.6.

In a real piano, the hammer, which strikes in an upward (grand) or sideways (upright) direction, falls away from the string a short time after collision, but it may remain in contact with the string for a substantial fraction of a period (see §9.4 on piano modeling).


Subsections
Previous: String Excitation
Next: Mass Termination Model

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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