Factors affecting the vibration of tennis racquets

Abstract

Measurements and calculations are presented on the vibration of the handle of a tennis racquet. It is shown that the vibration amplitude depends primarily on the stiffness of the racquet, but it also depends on the mass of the racquet, the mass distribution, the vibration frequency, the impact location, the impact duration, the stiffness of the ball and the stiffness of the string plane. Mass added at the tip of a racquet usually reduces handle vibrations when the ball impacts near the tip, but it increases handle vibrations for impacts in the throat region. Mass added to the handle usually reduces handle vibrations, regardless of the impact point. Mass added at a vibration node point has no effect on the vibration frequency, only a small effect on amplitude, and can usefully be added to very light and stiff racquets to increase racquet power without affecting the low vibration amplitude of these racquets. The amplitude of the fundamental vibration mode decreases as the impact duration increases, decreasing to zero at all impact points when the impact duration is 1.5 times longer that the fundamental vibration period.

Appendices

Appendix A: Fourier transform of half-sine waveform

Any function $v(t)$ that is symmetrical about $t = 0$ can be expressed in the form $v(t) = \int _{0} ^{ \infty } a(\omega ) \cos (\omega t)\, \mathrm{d}\omega$ where $a(\omega ) = (1/\pi ) \int _{-\infty } ^{ \infty } v(t) \cos (\omega t) \, \mathrm{d}t$ is the Fourier transform of $v(t)$. A symmetric half-cos function is $v(t) = v_{0} \cos (\omega _{0} t)$ when $-\tau _{0} < t < \tau _{0}$ and $v(t) = 0$ when $t < -\tau _{0}$ or $t > \tau _{0}$ where $\tau _{0} = \pi /(2\omega _{0})$. The Fourier transform of this function is

$$\begin{aligned} a(\omega ) &= \frac{v_{0}}{\pi } \int _{-\tau _{0}} ^{ \tau _{0}} \cos (\omega _{0} t) \cos (\omega t) \, {\mathrm{d}t}\\ &= \frac{v_{0}}{\pi } \int _{0} ^{ \tau _{0}} [ \cos (\omega _{0} + \omega )t + \cos (\omega _{0} - \omega ) t] \, {\mathrm{d}}t \end{aligned}$$

which is easily integrated to give

$$\begin{aligned} a(\omega ) = \frac{ v_{0} \cos (\omega \tau _{0})}{\tau _{0} (\omega _{0}^{2} - \omega ^{2}) } \, . \end{aligned}$$

The duration of $v(t)$ is $\tau = 2\tau _{0}$. $a(\omega )$ is not zero when $\omega \tau _{0} = \pi /2$ since then $\omega _{0} = \omega$. The first zero of $a(\omega )$ is at $\omega \tau _{0} = 1.5 \pi$ and then $f = \omega / (2\pi ) = 1.5/\tau$. Typical results relevant to racquets are shown in Fig. 13.

Fig. 13

Fourier transforms for half-sine waveforms of duration $\tau = 5$, 6 and 7 ms

Appendix B: Three-element beam model

Consider a rectangular beam of mass $M$ and length $L$ constructed from three uniform sections each of length $L/3$ and with masses $m_{1}$, $m_{2}$ and $m_{3}$. The total mass is $M = m_{1}+m_{2}+m_{3}$ and the balance point is located at distance $B = (m_{1} + 3m_{2} + 5m_{3})L/(6M)$ from the $m_{1}$ end. The moment of inertia about the $m_{1}$ end is $I = (m_{1} + 7m_{2} + 19m_{3})L^{2}/27$. Measurements of $M$, $B$ and $I$ can therefore be used to determine $m_{1}$, $m_{2}$ and $m_{3}$.