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Physical Audio Signal Processing
    Elementary String Instruments
       Ideal String Struck by a Mass


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Ideal String Struck by a Mass

In §4.5, we discussed the ideal struck string 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.4.9 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 will consider the more realistic case of an inelastic collision, i.e., where the mass hits the string and sticks to it until something, such as a wave, pushes it away from the string.

For simplicity, let the string length be infinity, and denote its wave impedance by $ R$. Denote the colliding mass by $ m$ and its speed 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.4.20. The connections of the mass impedance with the two semi-infinite string endpoint impedances 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 §L.2 for a fuller discussion of impedances and their parallel/series connection.)

Figure 4.20: Physical model of mass-string collision after time 0.
\begin{figure}\input fig/massstringphy.pstex_t \end{figure}

The equivalent circuit for the mass-string assembly after time zero is shown in Fig.4.21. Note that the string wave impedance 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. Impedances in series can be arranged in any order.

Figure 4.21: Electrical equivalent circuit for the mass and two string endpoints after time zero. The mass is represented by an inductor of $ m$ Henrys which has impedance $ ms$, while each string endpoint is represented by a resistor of $ R$ Ohms (impedance $ R$).
\begin{figure}\input fig/massstringec.pstex_t \end{figure}

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:

$\displaystyle f_m(t) + f_{R}(t) + f_{R}(t) = 0 \protect$ (5.17)

All the signs are ``$ +$'' in this equation because the current (``velocity'' $ v(t)$) flows positively into the $ +$ sign of each impedance. (Note the reference directions indicated by $ +/-$ in Fig.4.21.) Taking the Laplace transform5.9of this equation yields, by linearity of the Laplace transform,

$\displaystyle F_m(s) + 2F_{R}(s) = 0, \protect$ (5.18)

where $ F_m(s)$ denotes the Laplace transform of $ f_m(t)$, and $ F_{R}(s)$ denotes the Laplace transform of $ f_{R}(t)$. 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 [460, Appendix B].5.10The impedance $ ms$ of a mass $ m$ is discussed further in §L.1.

Substituting these relations into Eq.$ \,$(4.18) yields

$\displaystyle m\,s\,V(s) - m\,v_0 + 2R\,V(s) = 0, $

and solving for $ V(s)$ gives

$\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,5.11 we find the velocity of the contact point to be

$\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 §4.5. 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 §4.5.

In a real piano, the hammer, which strikes in an upward 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 Chapter 5 on piano modeling).


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written by 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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