Figure di Lissajous en

From "Physics, Waves, Music": a website about the physics of waves and sound, acoustics of musical instruments, musical scales, harmony and music.

AI-translated from Italian. Original page: Figure_di_Lissajous

Lissajous figures

We do not present Lissajous figures here in their rigorous mathematical definition, but only as a tool for the analysis of oscillatory phenomena.

Suppose we have to study two simple harmonic oscillators and that we have measured experimentally their laws of motion x1(t) and x2(t), which we can plot on two graphs of the type x(t).

If we now want to deduce from these graphs alone the ratio between the frequencies of oscillation and the relative phase shift, without knowing the exact analytical solution in advance, a mere comparison of the two laws of motion can be difficult (see the figure on the left).

A much more effective representation of the properties we want to determine is a diagram in which time does not appear explicitly and the two quantities x1(t) and x2(t) are plotted one against the other: at each fixed time we take x1(t) and x2(t), which become, respectively, the abscissa and the ordinate of the points in the new graph. In the particularly simple case of pure harmonic oscillations these curves are called "Lissajous figures" (figure on the right). Lissajous figures are particularly useful because:

  • they represent the trajectory actually described by a point whose motion is the superposition of two harmonic motions in perpendicular directions;
  • from the Lissajous figures, and this is our aim, it is immediate to determine the ratio between the frequencies when it is a rational number;
  • from the Lissajous figures it is also immediate to determine the phase shift between the two oscillators for particularly simple angles (e.g. 30°, 60°, etc.).

Let us illustrate the simple graphical procedure:


Draw two straight lines parallel to the axes, so that they do not pass through the points where the branches of the curve cross each other. Now count the intersections of these lines with the curve. In our case we have 8 intersections along the line parallel to the x2 axis (red line) and 10 along the line parallel to the x1 axis (green line). The ratio between the number of intersections with the red line and the number with the green line is equal to the ratio between the frequencies of the oscillators x1 and x2. In this case: 810=f1f2=45

When the frequencies are equal (ratio 1:1) the figure is an ellipse and the phase shift can be read on the x2 axis. If x1=A1sin⁡(ωt) and x2=A2sin⁡(ωt+ϕ), at the instant when x1=0 we have x2=±A2sin⁡ϕ: calling ±x2,0 the points where the ellipse crosses the x2 axis, we therefore have

sin⁡ϕ=x2,0A2.

For ϕ=0 the ellipse reduces to a segment; for ϕ=90∘ it crosses the axis at the points of maximum amplitude (and, if the amplitudes are equal, it is a circle).


Overview

Some Lissajous figures obtained for simple frequency ratios, keeping the phase shift between the oscillators equal to 0.

1:1 1:2 1:3 1:4
2:1 3:1 4:1

Some Lissajous figures obtained keeping the frequencies of the oscillators equal and varying their phase shift.

φ=0° φ=30° φ=60° φ=90°
φ=180° φ=150° φ=120°

The only way to distinguish positive from negative phase shifts is to watch how the figures form in time.

φ=+60° φ=-60°

If the ratio between the frequencies is almost, but not exactly, a ratio of two whole numbers, the figure does not stay still. For example, with 440 Hz and 441 Hz the phase shift grows slowly and the figure goes through all the shapes of the previous table, repeating every 1/|f1−f2| seconds, that is every second. On an oscilloscope this makes it possible to tune two sounds: one of the two frequencies is adjusted until the figure stops.

Check your understanding

Problem. In a still Lissajous figure, a line parallel to the x2 axis crosses the curve at 6 points, a line parallel to the x1 axis at 4 points. What is the ratio between the frequencies? What musical interval do the two sounds form? Solution. Following the procedure described above, f1/f2 = 6/4 = 3/2. The ratio 3/2 corresponds to a fifth: for example 660 Hz and 440 Hz.

Problem. Two oscillations of the same frequency give an ellipse that crosses the x2 axis at half its maximum amplitude. What is the phase shift? Solution. From the relation sin⁡ϕ=x2,0/A2 = 0.5 we get φ = 30° or φ = 150°. The two cases are distinguished by the orientation of the ellipse: as in the table, for 30° it is elongated along the diagonal of the 1st and 3rd quadrants, for 150° along that of the 2nd and 4th.

Problem. On an oscilloscope a 440 Hz tuning fork is compared with an almost equal sound. The figure changes shape and becomes the same again every half second. What frequency can the second sound have? Solution. The figure repeats every 1/|f1−f2| seconds, so |f1−f2| = 2 Hz: the second sound is at 438 Hz or at 442 Hz. To find out which, just change its frequency slightly and see whether the figure slows down or speeds up.


"Physics, Waves, Music": a website about the physics of waves, acoustics of musical instruments, musical scales, harmony and music.