6 · Lissajous Figures — Theory + 2 Solved PYQs

Perpendicular SHM superposition, ellipse/circle/line by phase, tangency frequency ratio, CRO demo. Easy theory then solved questions.

This page covers superposition of two perpendicular SHMs. It has four short theory sections, one figure, and two solved PYQs.

1. What are Lissajous figures? Formation

Shake a light spot left-right and up-down at the same time. The spot draws a looping pattern on the screen. That pattern is a Lissajous figure. Its shape tells the frequency ratio and phase gap of the two shakes.

For $x = a\sin(\omega_1 t+\phi)$, $y = b\sin(\omega_2 t)$, the spot $(x,y)$ traces a curve fixed by ratio $\omega_1:\omega_2$ and phase $\phi$.

Graphical method

Plot $x(t)$ and $y(t)$ sine strips on the top and side of a square. At each instant $t$, drop a line from the $x$-strip and run a line from the $y$-strip; their crossing point is the spot. Join the crossings over one full period. Changing $\phi$ slides the side strip and morphs the figure.

Uses

Measure unknown frequency: feed known $f$ to one axis, unknown to the other, read the ratio from the figure.
Measure phase gap between two signals.
Calibrate a CRO (check amplifiers and time base).
Study vibrations (turntable wobble, tuning-fork comparison).

2. Analytical method: 1:1 case

Remove time $t$ from the two equations. For equal frequencies the path is always an ellipse; phase picks circle or line.

For $x = a\sin(\omega t+\phi)$, $y = b\sin\omega t$, expand $x/a = \sin\omega t\cos\phi + \cos\omega t\sin\phi$, use $\sin^2+\cos^2 = 1$ and square to kill the root:

Formula 1. $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{2xy}{ab}\cos\phi = \sin^2\phi$ — an ellipse (tilted unless $\phi = \pi/2$).

Phase table (1:1)

$\phi = 0$: $(x/a - y/b)^2 = 0$, so $y = (b/a)x$ — straight line through origin.
$\phi = \pi/2$, equal $a$: $x^2+y^2 = a^2$ — circle.
$\phi = \pi$: line $y = -(b/a)x$ — other diagonal.

Lissajous circle for 1:1 with phase pi-by-2 and figure-8 for 2:1

Fig 1. Left — 1:1, $\phi = \pi/2$, equal amplitudes gives a circle. Right — 2:1 gives a figure-8.

3. Frequency ratio by tangency method

Count touches on the bounding box. $N_x$ = touches on a vertical side, $N_y$ = touches on a horizontal side.

Formula 2. $\dfrac{\omega_x}{\omega_y} = \dfrac{f_x}{f_y} = \dfrac{N_y}{N_x}$ (crossed ratio — $x$-frequency uses $y$-touches).

Worked example: figure-8 lying sideways ($\infty$). Left wall touched once ($N_x = 1$), top wall touched twice ($N_y = 2$). Ratio $f_x:f_y = 2:1$. So if the $y$-plates get $50$ Hz, the $x$-plates get $100$ Hz.

4. Electrical demonstration (CRO)

Wiring: Generator 1 → CRO X-input (horizontal plates). Generator 2 → CRO Y-input (vertical plates). Set CRO to X-Y mode (no internal time base; the spot position is directly $(V_x, V_y)$). Common ground all three chassis. Start near $50$ Hz with equal gains. Adjust one generator until the figure stands still — drifting figure means frequencies not in exact ratio.

Solved PYQs

S6-Q1

What are Lissajous figures? (two practical applications / demonstration by electrical experiment)

20232024×2 Recall

Lissajous figures are the paths traced by a particle under two perpendicular SHMs $x = a\sin(\omega_1 t+\phi)$, $y = b\sin\omega_2 t$. Shape is fixed by the ratio $\omega_1:\omega_2$ and phase $\phi$.

Steps

Define: the path of $(a\sin(\omega_1 t+\phi), b\sin\omega_2 t)$ from perpendicular SHM superposition (reason: each instant gives one spot).
Shape fixed by frequency ratio + phase (reason: Eqs. 1–2 above).
Application 1: unknown-frequency measurement by tangency ratio $f_x/f_y = N_y/N_x$ (reason: count box-touches).
Application 2: phase measurement or CRO calibration using a still figure (reason: shape encodes phase).
CRO wiring: two signal generators → X-input and Y-input, CRO in X-Y mode, common ground, still figure means locked ratio (reason: no time base; spot is $(V_x, V_y)$).

Answer

Lissajous figures are the looping paths traced by superposition of two perpendicular SHMs. Shape depends on the frequency ratio and phase. $\boxed{\text{Path of perpendicular SHMs};\ \text{ratio }f_x/f_y = N_y/N_x;\ \text{CRO X-Y demo}}$.

Exam tip

Asked ×2 with "applications" explicit in 2024. Always write two uses plus a short CRO wiring note.

S6-Q2

Two SHMs of same frequency and amplitude, phase difference $\pi/2$, in perpendicular directions. What is the shape of the resultant motion?

2023 Recall

For equal $\omega$ and equal $a$, phase $\phi = \pi/2$ gives Formula 1 with $\cos\phi = 0$, $\sin^2\phi = 1$: a circle of radius $a$.

Steps

$y = a\sin(\omega t+\pi/2) = a\cos\omega t$ (reason: $\sin(x+\pi/2) = \cos x$).
$x = a\sin\omega t$ gives $\sin\omega t = x/a$; $y = a\cos\omega t$ gives $\cos\omega t = y/a$.
Square and add: $x^2/a^2 + y^2/a^2 = \sin^2\omega t + \cos^2\omega t = 1$ (reason: identity $\sin^2 + \cos^2 = 1$ kills $t$).
Multiply by $a^2$: $x^2 + y^2 = a^2$.

Answer

$\boxed{x^2 + y^2 = a^2}$ — a circle of radius $a$, traced anticlockwise (the $y$-motion leads the $x$-motion by $90°$).

Exam tip

State all three circle conditions (1:1 ratio, equal amplitudes, $90°$ phase). The examiner checks you know it is not always a circle.

Reference. Subrahmanyam & Brij Lal — Waves & Oscillations Ch-5 (graphical and analytical methods, uses).