6 · Lissajous Figures — Theory + 5 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 five 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. $\dfrac{\omega_x}{\omega_y} = \dfrac{f_x}{f_y} = \dfrac{N_x}{N_y}$ — $x$-frequency uses $x$-touches ($N_x$ = touches on a vertical side, $N_y$ = touches on a horizontal side).

Worked example: figure-8 lying sideways ($\infty$), with $x = a\sin 2\omega t$, $y = b\sin\omega t$. Left wall touched twice ($N_x = 2$), top wall touched once ($N_y = 1$). 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)

202220232024×3 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 $f_x/f_y = N_x/N_y$ (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. $\boxed{\text{Path of perpendicular SHMs};\ \text{ratio }f_x/f_y = N_x/N_y;\ \text{CRO X-Y demo}}$.

Exam tip

Asked ×3 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?

20232024×2 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 clockwise (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. Asked ×2.

S6-Q3

Find the resultant of two perpendicular simple harmonic vibrations and discuss Lissajous figures.

2019202220232024×4 Recall

Superposition: the spot is at $(x(t), y(t))$ with $x = a\sin(\omega_1 t+\phi)$, $y = b\sin(\omega_2 t)$. Eliminate $t$ to get the path. 8-mark question: derivation + phase cases + frequency-ratio cases + uses.

Steps

Write $x = a\sin(\omega_1 t+\phi)$, $y = b\sin(\omega_2 t)$ (reason: superposition — each instant gives one spot).
For $\omega_1 = \omega_2 = \omega$: expand $x/a = \sin\omega t\cos\phi + \cos\omega t\sin\phi$, put $\sin\omega t = y/b$, square to kill the root (reason: §2 derivation).
Resultant: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{2xy}{ab}\cos\phi = \sin^2\phi$ — always an ellipse for 1:1 (reason: Formula 1).
Discuss phase: $\phi = 0$ → straight line $y = (b/a)x$; $\phi = \pi/2$ (equal $a$) → circle $x^2+y^2 = a^2$; $\phi = \pi$ → other diagonal (reason: phase table §2).
Discuss frequency ratio: $1:2$ → figure-8; $2:1$ → sideways figure-8; in general $f_x/f_y = N_x/N_y$ by counting wall touches (reason: Formula 2, tangency method).
Uses: unknown-frequency measurement, phase measurement, CRO calibration (reason: §1 list).

Answer

Resultant of perpendicular SHMs: $\boxed{\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} - \dfrac{2xy}{ab}\cos\phi = \sin^2\phi}$ (ellipse for 1:1). Lissajous figures are these paths; shape encodes frequency ratio ($f_x/f_y = N_x/N_y$) and phase. Uses: frequency/phase measurement, CRO calibration.

Exam tip

Asked ×4 — the most repeated 8-marker in this unit. Write the derivation, then the phase table, then one ratio example (1:2 → figure-8), then two uses. That covers all 8 marks.

S6-Q4

What happens when two perpendicular vibrations of slightly different time periods are superimposed on a particle?

20222023×2 Recall

Slightly different periods → frequencies nearly but not exactly equal ($\omega_1 \approx \omega_2$). Phase gap $\phi(t) = (\omega_1-\omega_2)t$ drifts slowly with time.

Steps

Write $x = a\sin(\omega_1 t+\phi_0)$, $y = b\sin(\omega_2 t)$ with $\omega_1 \approx \omega_2$ but not equal (reason: "slightly different time periods").
The effective phase gap $\phi(t) = (\omega_1-\omega_2)t + \phi_0$ grows slowly (reason: beat-like phase drift).
At any instant the spot follows the 1:1 ellipse of Formula 1 with that $\phi$ — but $\phi$ keeps changing (reason: Formula 1 with time-varying phase).
So the figure slowly morphs: line → ellipse → circle → ellipse → line, cycling continuously (reason: phase table §2, $\phi$ sweeping $0$ to $2\pi$).
On a CRO screen this looks like a slowly rotating/breathing figure — a "drifting" Lissajous figure (reason: same drift that §4 warns about when frequencies are not exactly locked).

Answer

$\boxed{\text{The 1:1 Lissajous figure slowly morphs through all phases}}$ — line → ellipse → circle → ellipse → line, repeating. The phase drift $(\omega_1-\omega_2)t$ makes the figure appear to rotate/breathe on screen.

Exam tip

Asked ×2. Key phrase: "phase difference changes slowly with time". Contrast with exact 1:1 (still figure).

S6-Q5

What are Lissajous figures? Show one such figure where ratio of frequencies is $1:2$.

2023 Recall

Lissajous figure = path of $(x(t), y(t))$ for perpendicular SHMs. Ratio $f_x:f_y = 1:2$ means $x = a\sin\omega t$, $y = b\sin 2\omega t$. Tangency: $f_x/f_y = N_x/N_y$.

Steps

Define Lissajous figures: paths from perpendicular SHM superposition, shape fixed by frequency ratio and phase (reason: §1).
For $1:2$: $x = a\sin\omega t$, $y = b\sin 2\omega t = 2b\sin\omega t\cos\omega t$ (reason: double-angle).
Eliminate $t$: with $\sin\omega t = x/a$, $y = \pm\dfrac{2bx}{a}\sqrt{1-x^2/a^2}$ — a figure-8 standing upright (reason: $y = 0$ at $x = 0$ and $x = \pm a$, crossing at the origin).
Check by tangency: touches each vertical wall once ($N_x = 1$), each horizontal wall twice ($N_y = 2$), so $f_x/f_y = 1/2$ ✓ (reason: Formula 2).
Sketch: upright "8" — top loop and bottom loop meeting at the origin (reason: the curve crosses itself at $(0,0)$).

Answer

$\boxed{1:2 \Rightarrow \text{upright figure-8}}$: $x = a\sin\omega t$, $y = b\sin 2\omega t$; touches give $N_x:N_y = 1:2$. Sketch an "8" crossing at the origin.

Exam tip

Draw the "8" upright (not sideways — that is $2:1$). Mark $N_x = 1$, $N_y = 2$ on your sketch to prove the ratio.

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