By thinking about the balance of kinetic energy, we can get some useful information about the bandwidth that can be excited when a bouncing hammer hits a vibrating structure. It is convenient to think about the problem in idealised form, as sketched in Fig. 1. A point mass $m$ hits a chosen position on the system, approaching at speed $v_1$ and rebounding with speed $v_2$. After the impact, the structure is vibrating with some velocity $v(t)$ at the impacted point.

The contact force may be a single pulse or a cluster of multiple pulses, but in either case once the rebound process is complete the total impulse imparted to the structure must be given by the jump in momentum of the bouncing mass:

$$I=m(v_1+v_2). \tag{1}$$

We can write the force waveform as $I f(t)$ so that

$$\int{f(t) dt} = 1 \tag{2}$$

where the integral is taken over the entire duration of the pulse(s). The response velocity $v(t)$ is then given by the standard convolution integral (see section 2.2.8)

$$v(t)=I \int_{-\infty}^t{g(t-\tau) f(\tau) d \tau} \tag{3}$$

in terms of the impulse response

$$g(t) = \sum_n{\dfrac{\cos \omega_n t}{M_n}} \mathrm{~~~for~~} t\ge 0 \tag{4}$$

where mode $n$ has resonance frequency $\omega_n$ and effective mass at the impacted point $M_n$. This effective mass is related to the mode shape at the impacted point, which we can write in a shorthand way as $u_n(x_0)$: then

$$M_n=\dfrac{1}{u_n^2(x_0)} \tag{5}$$

where the mode shape is assumed to have been normalised in the usual way with respect to the mass distribution of the structure: see section 2.2.5, where the normalisation condition was stated in discrete form in equation (12). Note that the influence of damping is ignored in equation (4), because an impact event is too short for damping to have a significant influence.

Suppose first that the impact is an ideal, instantaneous one: then

$$f(t) = \delta(t), \tag{6}$$

the Dirac delta function (see section 2.2.8). Equations (3) and (4) then reduce to the simple form

$$v(t) = I \sum_n{\dfrac{\cos \omega_n t}{M_n}} . \tag{7}$$

The expression $I \cos \omega_n t /M_n$ is thus the velocity of the effective mass $M_n$ representing mode $n$, and since we know that the mass matrix is always diagonalised when we use modal coordinates, we can write a simple equation representing the kinetic energy of the system before and after impact:

$$\frac{1}{2}m v_1^2 = \frac{1}{2}m v_2^2 +\frac{1}{2} \sum_n{M_n \dfrac{I^2}{M_n^2}} \tag{8}$$

where the right-hand side has been evaluated immediately after the impact, at $t=0+$ so that all the cosine factors are simply equal to 1. Notice that equation (8) ignores any actual loss of energy during the collision: the aim here is to find an upper limit on the post-impact vibration.

Substituting from equation (1) and writing

$$v_2= \lambda v_1 \tag{9}$$

we obtain

$$1- \lambda^2 = (1+\lambda)^2 \sum_n{\dfrac{m}{M_n}} . \tag{10}$$

Cancelling a factor $(1+ \lambda)$ leaves

$$1-\lambda = (1+\lambda)K \tag{11}$$

where

$$K=\sum_n{\dfrac{m}{M_n}} \tag{12}$$

and so

$$\lambda = \dfrac{1-K}{1+K}. \tag{13}$$

But now we hit a snag. The speed ratio $\lambda$ must be positive for a rebound, and for that to happen equation (13) requires $K \le 1$. The limiting case would have $K=1$, which is the condition for all the kinetic energy from the original moving mass to turn into vibration in the structure, leaving nothing for a rebound speed. But for a structure with an infinite number of modes, $K \rightarrow \infty$ because nearly all modes have a value of $M_n$ which is of the order of the total mass of the structure.

We can immediately make a crude estimate based on this result. For any strongly-excited mode, the modal mass $M_n$ would be expected to be of the order of 1/3 of the total mass of the vibrating structure. So $K \le 1$ means that the maximum number of strongly-excited modes like this must be less than about 1/3 of the ratio of the structure mass to the hammer mass.

To go beyond this crude estimate, we can try a more realistic (but still idealised) case. We can use the contact force waveform found in section 2.2.6 and reprised in section 12.1: a half-cycle of sine wave, which is the result of hitting a rigid surface through a linear contact spring. The contact resonance frequency $\Omega$ is now a variable, which we can use to describe impacts of different durations. Specifically, we can write

$$f(t)=A \cos \Omega t \mathrm{~~~for~~} -T \le t \le T \tag{14}$$

and zero outside that range, where the contact duration is $2T$. Since this duration is a half-cycle at the frequency $\Omega$, we have

$$T=\dfrac{\pi}{2 \Omega}. \tag{15}$$

The pulse is sketched in Fig. 2. The constant $A$ is determined by the unit-area condition (2), with the result

$$A=\dfrac{\pi}{4T}. \tag{16}$$

Now we can evaluate the convolution integral (3) with the impulse response (4). After a little algebraic manipulation, the result for the velocity $v(t)$ at time $t=T$ immediately after the pulse finishes is

$$v(T)=\dfrac{I}{2} \sum_n{\dfrac{1}{M_n} \left\lbrace \dfrac{\Omega^2}{\Omega^2 – \omega_n^2} (1 + \cos \omega_n \pi/\Omega) \right\rbrace} . \tag{17}$$

The effective velocity of the modal mass $M_n$ just after the pulse finishes is thus

$$v_n=\dfrac{I}{M_n} \dfrac{\Omega^2}{\Omega^2 – \omega_n^2} \dfrac{(1 + \cos \omega_n \pi/\Omega)}{2} \tag{18}$$

and the kinetic energy of the vibrating structure is $(1/2)\sum_n{M_n v_n^2}$. Substituting this into the equivalent of equation (8), we can obtain an equation identical to equation (11) if we replace $K$ by

$$K^\prime (\Omega) = \sum_n{\dfrac{m}{M_n} \left\lbrace \dfrac{\Omega^2}{\Omega^2 – \omega_n^2} \dfrac{(1 + \cos \omega_n \pi/\Omega)}{2} \right\rbrace^2} . \tag{19}$$

It is reassuring to see that we recover the delta-function version (12) in the limit $T \rightarrow 0$ so that $\Omega \rightarrow\infty$, as we would expect. But with a finite value of $\Omega$ the problem of $K$ being infinite has gone away. Once the modal frequency $\omega_n$ grows much bigger than $\Omega$, the term inside the curly brackets dies away: the finite-length impulse has the effect of a low-pass filter on the modal sum.

We can see the effect most clearly by an example. Figure 3 shows the function $K^\prime(\Omega)$ for the particular case of a rectangular plate of total mass 200 g and lowest resonance frequency 100 Hz being impacted by a “hammer” with mass 5 g. The plate is the same one used for some of the examples in section 12.1, with details given in section 12.1.1. The function $K^\prime$ satisfies $K^\prime(0)=0$, and we have already noted that it tends to infinity as $\Omega \rightarrow \infty$, so there is always at least one value of $\Omega$ for which $K^\prime=1$. In practice, as in the example in Fig. 3, there usually seems to be just one such value, which represents the shortest possible impact (of the chosen form (14)) for which a rebound might occur without multiple contacts. It is also the condition for maximum energy transfer from the moving mass to the vibrating plate.

The interpretation is that if you want to hit this plate with a hammer of this particular mass, you should equip it with a sufficiently soft tip that the contact resonance is below the threshold value (around 3.5 kHz) if you want even a fighting chance of tapping with a single impact. In reality this condition gives an upper limit, but it should give a guide to the order of magnitude and the trend. In real impacts it is of course possible that multiple contacts could occur at significantly lower values of the contact stiffness — some computed examples are shown in section 12.1, and we will see others in later sections of this chapter.

In Fig. 4, the threshold value of contact resonance frequency is plotted for a wide range of hammer masses. The same plate is being tapped in all cases. Figure 5 shows the same data in a different form, plotting the threshold contact time as a function of hammer mass.

Finally, we should note that this approximate calculation can be easily extended to include the effect of the vibration modes of the beater, or drumstick. The same force that drives the structure also acts (in the opposite direction) on the beater. In a simple model, such as the pinned-free beam model that is used to represent a drumstick in section 12.1, the mass we have so far considered corresponds to a rigid-body mode of the beater. If the beater also has modes with non-zero frequency, they will acquire kinetic energy from the impact in exactly the same way as discussed above for the modes of the structure. So all we need to do to extend the analysis to this case is include these extra modes of the beater, suitably mass-normalised, into the expressions for $K(\Omega)$ or $K^\prime (\Omega)$ in equations (12) and (19).