Superposition of SHMs

A tabla player tunes the left drum, the bayan, against a tanpura drone. The drone puts out a steady note — a single frequency humming through the room. The bayan is close, but not quite right. As both sound together you hear not two notes but one note that throbs — loud, soft, loud, soft — at a slow, steady beat. Tighten the tabla's tuning ring a fraction of a turn and the throbbing slows. Tighten it a touch more and the beats vanish: the two notes have locked into the same frequency and become one.

Those beats are a superposition of simple harmonic motions. Each sound wave at your eardrum is driving it in SHM. Two SHMs with nearly equal frequencies, added together, produce the slow amplitude throb you hear as a beat. Two SHMs with exactly equal frequencies add to another SHM — a single pure tone. And if two SHMs act on the same particle along perpendicular axes — as electron beams do in a cathode-ray oscilloscope, or as atoms do in a two-dimensional optical trap — the particle does not move along a line at all. It traces an ellipse, a circle, or a tangled looping curve called a Lissajous figure, and the shape of that curve tells you exactly what the two oscillations are doing.

Superposition is the single most useful idea in waves and oscillations. It says: when two SHMs act together, the displacements just add, point by point, moment by moment. Nothing else enters. No new physics, no coupling term, no hidden interaction. Two motions on one particle become one motion, found by adding vectors. That is it. The consequences — beats, interference patterns, Lissajous figures, Fourier analysis, standing waves — all fall out of that one idea.

What "same line" and "perpendicular" mean

The two cases you need to distinguish up front:

• Same line. Both SHMs push the particle back and forth along the same axis. A mass at the end of a spring being nudged by two different sinusoidal forces is this case. So is your eardrum — each sound wave pushes it along the ear-canal direction, and two sound waves add along that one direction.
• Perpendicular directions. One SHM pushes the particle along x, another along y. The particle moves in the xy-plane, and its path is a curve, not a line. The classic example is the cathode-ray tube oscilloscope: the horizontal deflection plates run one SHM, the vertical plates run another, and the electron beam traces out whatever curve the two motions create on the screen.

Same-line superposition gives you beats and resultant SHMs. Perpendicular superposition gives you Lissajous figures. Both are governed by the same rule — displacements add — but the geometry is different.

Same line, same frequency — the phasor method

Start with the cleanest case: two SHMs of the same frequency \omega along the same line.

The phase difference \varphi is what distinguishes them — x_2 leads x_1 by \varphi radians. The question is: what is x_1 + x_2?

You could expand \sin(\omega t + \varphi) = \sin(\omega t)\cos\varphi + \cos(\omega t)\sin\varphi, collect terms, and crunch trigonometry. That works, but there is a faster and more physical method: phasors.

A phasor is a rotating arrow. An SHM of amplitude A and phase \alpha can be represented by an arrow of length A that makes angle \alpha + \omega t with the x-axis. The SHM itself is the projection of the arrow onto a fixed axis (say the vertical axis). As the arrow rotates, the projection oscillates sinusoidally — that is an SHM.

The phasor method works because two phasors rotating at the same rate stay at a fixed angle to each other. Their vector sum is another arrow that also rotates at \omega — which means the resultant is another SHM at the same frequency \omega. You only need to find the length and phase of the summed arrow. That is vector addition.

Deriving the resultant amplitude and phase

Take two phasors at t = 0: one of length A_1 along the x-axis (representing x_1 = A_1\sin(\omega t), which has phase 0), and one of length A_2 at angle \varphi above the x-axis (representing x_2 = A_2\sin(\omega t + \varphi)).

Step 1. Resolve A_2 into components along and perpendicular to A_1.

Why: the resultant phasor is the vector sum A_1\hat{x} + A_2\cos\varphi\,\hat{x} + A_2\sin\varphi\,\hat{y}. To add vectors, you add components.

Step 2. Sum the components.

Why: A_1 contributes only along x. A_2 contributes A_2\cos\varphi along x and A_2\sin\varphi along y.

Step 3. The resultant length is the Pythagorean combination of the components.

Why: the length of a vector is the square root of the sum of the squares of its components — the Pythagorean theorem in disguise.

Step 4. Expand the first square.

Step 5. Use \cos^2\varphi + \sin^2\varphi = 1 to collapse the last two terms.

Why: this is the law of cosines applied to the parallelogram. The resultant amplitude depends on the two individual amplitudes and on the phase difference. If \varphi = 0 the arrows point the same way and add head-to-tail; if \varphi = \pi they point opposite ways and subtract.

Step 6. The resultant phase \alpha (the angle the sum makes with A_1) comes from the component ratio.

Why: any vector with components (R_x, R_y) makes angle \arctan(R_y/R_x) with the x-axis. The resultant SHM is therefore x(t) = A\sin(\omega t + \alpha).

Two SHMs of the same frequency add to a third SHM of the same frequency. The only question is how big it is and where the peaks land, and those are set by the phasor parallelogram.

The three key phase differences

The amplitude formula hides three famous cases.

• In phase (\varphi = 0). \cos\varphi = 1, so A^2 = A_1^2 + A_2^2 + 2A_1 A_2 = (A_1 + A_2)^2 and A = A_1 + A_2. The amplitudes add. This is constructive interference along a line: two sources pushing in step at the same moment.
• Out of phase (\varphi = \pi). \cos\varphi = -1, so A^2 = A_1^2 + A_2^2 - 2A_1 A_2 = (A_1 - A_2)^2 and A = |A_1 - A_2|. The amplitudes subtract. This is destructive interference — and if A_1 = A_2, the two SHMs cancel out perfectly. Silence. A flat line.
• Quadrature (\varphi = \pi/2). \cos\varphi = 0, so A^2 = A_1^2 + A_2^2. The amplitudes add in quadrature, just as the legs of a right triangle combine to the hypotenuse.

These three are the workhorses of interference, diffraction, and AC circuit analysis.

Same line, nearly equal frequencies — beats

Now loosen the condition: let the two SHMs have nearly but not exactly equal frequencies. Say

with \omega_1 \approx \omega_2 and equal amplitudes for simplicity. Add them.

Step 1. Use the sum-to-product identity \cos P + \cos Q = 2\cos\!\left(\frac{P+Q}{2}\right)\cos\!\left(\frac{P-Q}{2}\right).

Why: this identity is the sum-to-product formula you saw in trigonometry. It rewrites the sum of two cosines as a single cosine multiplied by a slowly varying amplitude. The identity itself is derived from the cosine addition formula — expand \cos(X+Y) + \cos(X-Y) = 2\cos X\cos Y and set X = (P+Q)/2, Y = (P-Q)/2.

Step 2. Label the two frequencies that appear.

• \omega_{\text{fast}} = (\omega_1 + \omega_2)/2 — the average of the two frequencies.
• \omega_{\text{slow}} = (\omega_1 - \omega_2)/2 — half the difference.

Why: because \omega_1 \approx \omega_2, the average is close to either frequency — this is the tone you hear. The difference is small — this is the slow envelope that rises and falls.

Step 3. Interpret the formula.

The motion is an oscillation at the fast frequency, modulated by a slowly varying amplitude 2A\cos(\omega_{\text{slow}}\,t). The envelope itself oscillates, and each time it passes through zero the combined sound drops to silence.

Step 4. Count the beats per second.

The envelope's magnitude |2A\cos(\omega_{\text{slow}}\,t)| peaks twice per period of \omega_{\text{slow}}, because cosine hits its largest magnitude both at its positive peak and at its negative peak. The audible beat frequency — how many loud-moments per second your ear hears — is therefore twice \omega_{\text{slow}}/2\pi:

Why: one full period of |\cos| is half the period of \cos, so the beat frequency is 2 \times (\omega_{\text{slow}}/2\pi) = (\omega_1 - \omega_2)/2\pi = f_1 - f_2 in magnitude.

This is the tabla tuning recipe. You strike the bayan and the tanpura drone together. You count beats per second. You tighten the string until the beats slow down, then vanish. When they vanish, f_1 = f_2 and the tabla is in tune with the drone.

Perpendicular directions, same frequency — the ellipse

Now switch geometries. Let the two SHMs act on the same particle along perpendicular axes:

The particle has coordinates (x, y) at each moment, and you want the shape of the curve the particle traces in the xy-plane. Eliminate t between the two equations.

Step 1. Solve the x equation for \sin(\omega t).

Why: you need both \sin(\omega t) and \cos(\omega t) to plug into the y equation after expanding the phase-shifted sine.

Step 2. Expand the y equation.

Why: the sine addition formula splits \sin(\omega t + \varphi) into a part that depends on \sin(\omega t) = x/A (which you already know) and a part that depends on \cos(\omega t) (which you still need to eliminate).

Step 3. Isolate the \cos(\omega t) term.

Why: move the known piece to the left so only the unknown \cos(\omega t) remains on the right.

Step 4. Square both sides and replace \cos^2(\omega t) = 1 - \sin^2(\omega t) = 1 - x^2/A^2.

Why: squaring kills the \pm that came with the square root, and the Pythagorean identity eliminates \cos(\omega t) in favour of x.

Step 5. Expand the left side and divide through by B^2.

Step 6. Collect the x^2/A^2 terms on the left.

Step 7. Use \cos^2\varphi + \sin^2\varphi = 1.

Why: this is the general equation of an ellipse (or line, or circle) inscribed in the rectangle |x|\le A, |y|\le B. The shape depends on \varphi — the phase difference between the two perpendicular SHMs. Different phase differences give different ellipse orientations.

The three key shapes

Plug in the famous values of \varphi and watch the ellipse deform into its special cases.

Case 1: \varphi = 0 (in phase). Then \sin\varphi = 0, \cos\varphi = 1, and the equation becomes

A straight line through the origin, slope B/A. Both coordinates pass through zero together and reach their extremes together.

Case 2: \varphi = \pi/2 (quadrature). Then \sin\varphi = 1, \cos\varphi = 0, and the cross-term vanishes.

An ellipse with axes aligned to the coordinate axes. If additionally A = B you get a circle of radius A. The particle swings in a circle at constant angular rate — like a particle moving along a radius of a carousel viewed from above.

Case 3: \varphi = \pi (out of phase). Then \sin\varphi = 0, \cos\varphi = -1.

A straight line with negative slope. Same line, reversed direction.

For phase differences between these (say \varphi = \pi/4, or \varphi = 2\pi/3), you get tilted ellipses — the axes no longer line up with x and y. The farther from 0 or \pi, the rounder the ellipse. At \pi/2 it is fattest (and un-tilted); at \pi/4 it is a narrow diagonal oval.

Watch the ellipse being traced

The animation below shows the particle driven by x(t) = \cos(2\pi t) and y(t) = \cos(2\pi t - \pi/2) = \sin(2\pi t). With equal amplitudes and \varphi = \pi/2 (quadrature), this traces a circle of radius 1. The red dot is the particle; the trail shows the curve it lays down.

Changing the phase difference to \varphi = 0 collapses this circle into a diagonal line, and changing it to \varphi = \pi/4 tilts it into a narrow ellipse. The next animation shows the \varphi = \pi/4 case — the particle still moves smoothly, but the path is now a slanted oval.

Perpendicular directions, unequal frequencies — Lissajous figures

Let the two frequencies differ. Write

Eliminating t is no longer simple — the curve is in general not an ellipse but a Lissajous figure. The key result is about closure.

Theorem. The Lissajous figure closes (the particle returns to its starting point and retraces) if and only if \omega_x/\omega_y is a rational number p/q in lowest terms. The closed figure has p lobes along the y-axis and q lobes along the x-axis.

Why it closes if the ratio is rational. Write \omega_x = p\Omega and \omega_y = q\Omega for some common frequency \Omega (this works whenever \omega_x/\omega_y = p/q). Then x has period 2\pi/(p\Omega) and y has period 2\pi/(q\Omega). The joint motion repeats whenever both oscillations simultaneously return to their starting states — every T = 2\pi/\Omega (since T contains exactly p periods of x and q periods of y). Hence the curve closes with period T.

Why it does not close if the ratio is irrational. If \omega_x/\omega_y is irrational, there is no common \Omega. The two motions never return to their starting state simultaneously, and the curve never exactly repeats. It traces denser and denser, eventually filling the rectangle |x|\le A, |y|\le B — a result that in ergodic theory is called equidistribution.

Why lobe counting: in one joint period, x goes through p cycles, so it hits its extreme value x = A a total of p times. Each time it reaches that extreme, the curve touches the vertical boundary of the rectangle — that is a lobe along the y-axis. Similarly, y reaches B a total of q times, giving q lobes along the x-axis.

Watch a Lissajous figure being traced

The animation below shows a particle driven by x(t) = \sin(2\pi \cdot 2 t) and y(t) = \sin(2\pi t + \pi/2) — a frequency ratio \omega_x/\omega_y = 2 and a quarter-period phase offset. The result is the classic figure-eight.

Changing the ratio changes the shape entirely. A 3:2 ratio gives three vertical lobes and two horizontal lobes — a much more intricate weaving. Here is that animation.

On a real cathode-ray oscilloscope — the kind still used in school and college physics labs across India — this is exactly how you measure an unknown frequency. Feed the unknown signal to the y-plates and a reference signal from a calibrated oscillator to the x-plates. Tune the reference until the figure on the screen stops rotating and locks into a stationary Lissajous pattern. Count the lobes: p vertical, q horizontal. The unknown frequency is (p/q) \times f_{\text{reference}}.

Worked examples

Common confusions

• "Adding two SHMs always gives another SHM." Only if they have the same frequency and act along the same line. Same frequency along perpendicular lines gives an ellipse (a closed curve, but not a 1D SHM). Different frequencies along the same line gives beats (not a simple sinusoid). Different frequencies along perpendicular lines gives a Lissajous figure.

• "Beats are destructive interference." They involve both constructive and destructive interference — alternating every half-envelope-period. The loud moments are constructive (the two SHMs momentarily in phase), the silent moments are destructive (momentarily out of phase). Beats are interference unfolding in time.

• "The beat frequency is (\omega_1 - \omega_2)/2." That is the angular frequency of the envelope, but the audible beat frequency is twice that, because |\cos| has twice as many peaks per unit time as \cos. The useful formula is f_{\text{beat}} = |f_1 - f_2|.

• "A Lissajous figure with ratio 1:1 and \varphi = 0 is a circle." No — at \varphi = 0 it is a straight line along y = (B/A)x. You need \varphi = \pi/2 (with A = B) for the circle. At \varphi = \pi/2 with A \ne B, it is an axis-aligned ellipse.

• "An irrational-ratio Lissajous figure eventually closes." It does not. The path is dense in the bounding box (comes arbitrarily close to every point), but never exactly returns to its starting state. Real oscilloscope traces look "almost stationary" only because they are actually close to a rational ratio and the drift is slow.

• "Phasors are a separate subject from SHM." They are the same subject in different clothing. A phasor is the rotating arrow whose projection is the SHM. Every manipulation you do on phasors (scaling, adding, rotating) is a valid manipulation on SHMs.

Where this leads next

• The Wave Equation — once SHMs travel through space, the equation y(x, t) = A\sin(kx - \omega t) takes over, and the superposition principle promotes from a particle-level rule to a field-level rule.
• Principle of Superposition — the broader statement that applies to waves of any kind, including light and sound, and leads into interference and diffraction.
• Forced Oscillations and Resonance — a driven SHM is itself the sum of two SHMs (transient + steady state). Superposition tells you their behaviour near resonance.
• Simple Pendulum — the cleanest physical realisation of a single SHM, and the starting block for every superposition problem.
• Energy in SHM — the energy of a sum of SHMs is not the sum of energies, because energy is quadratic in amplitude. The cross term is the whole physics of interference.