Superposition, interference, and beats

Start at one fixed point

At one point on an ideal string, two proposed contributions are each +2 mm at t = 0 s. Their sum is +4 mm. At t = 1/2 s, one contribution is −2 mm and the other is +2 mm, so their sum is zero. Adding the signed values explains both results; adding their magnitudes would miss the cancellation.

Choose x along the string, positive to the right, and y as transverse displacement, positive upward from equilibrium. Watch the fixed location x = 0 m as actual time t passes. The graphs below are time records at that location, rather than a picture of the string’s shape at one instant.

Algebra, functions, graphs, and logarithms prepares substitution and graph reading. Geometry, trigonometry, and coordinate systems prepares cosine and phase in radians. The needed addition and trigonometric identity are introduced here. You can first follow the signs and exact table, then return to the formulas.

Units, standards, and physical constants is a published return path for quantity notation. Derivatives, Taylor series, and local approximation is an optional return path for checking the wave equation and the final velocity comparison. No later wave-equation lesson is required for a first reading.

Supply one linear string model

Stipulate a uniform whole-line string with tension T = 0.4 N and mass per length μ = 0.1 kg/m. There is no reflecting endpoint, damping or dispersion in this model. Its rule is y_tt = (T/μ)y_xx: the subscripts mean two derivatives with respect to time or position. The coefficient T/μ = 4 m²/s² gives the positive wave speed c = 2 m/s.

The rule is linear: applying it to y₁ + y₂ gives the sum of the two separate equations. If both fields satisfy this same rule, their sum does too. Their initial displacements and initial transverse velocities must also be added. Superposition means this addition of compatible fields under a linear rule; it is not permission to add arbitrary solutions of different or nonlinear models.

Let A = 0.002 m = 2 mm. Supply y₁(x,t) = A cos(k₁x−ω₁t) and y₂(x,t) = A cos(k₂x−ω₂t), with k₁ = 5π rad/m, k₂ = 6π rad/m, ω₁ = 10π rad/s and ω₂ = 12π rad/s. A wave number k measures phase change per length; angular frequency ω measures phase change per time. Both initial phases are zero.

A trigonometric argument is a dimensionless phase expressed in radians. Both components obey ω² = c²k² and have ω/k = 2 m/s. Keeping kx−ωt fixed therefore moves the phase to the right. Frequency f = ω/(2π) is 5 Hz or 6 Hz; wavelength λ = 2π/k is 0.4 m or 1/3 m. These are exact supplied waves, not measurements of a particular string.

Add values and compare phase

Define τ = t/(1 s), the dimensionless numerical time. At x = 0 m the even cosine function removes the minus sign in each phase: y₁ = (2 mm) cos(10πτ) and y₂ = (2 mm) cos(12πτ). The resultant displacement is y = y₁ + y₂.

Interference describes how the contributions combine. Values of the same sign reinforce the displacement; values of opposite signs partly or completely cancel. “Constructive” and “destructive” refer to that comparison. A zero sum at one point and time does not mean that both component fields vanish everywhere.

At t = 1/4 s the first cosine is cos(5π/2) = 0 and the second is cos(3π) = −1. Thus y = −2 mm. The negative result places the string below equilibrium at this location; it does not tell us that the travelling pattern moves left.

For equal frequencies and wave numbers, a fixed relative phase selects a fixed resultant amplitude. With two 2 mm components, equal phases give amplitude 4 mm; opposite phases differing by π give complete cancellation of the entire supplied fields. Different frequencies instead make their relative phase change with time, even when both began in phase.

Separate the factor from its magnitude

The identity cos a + cos b = 2 cos((a−b)/2) cos((a+b)/2) follows by adding the cosine sum and difference formulas. Applying it at the fixed observer gives y(t) = (4 mm)*cos(pi*tau)*cos(11*pi*tau). The factors reconstruct the original 5 Hz and 6 Hz contributions.

Call B(t) = (4 mm) cos(πτ) the signed modulation factor, and C(t) = cos(11πτ) the dimensionless carrier factor. Then y = BC. B can be negative; a sign change reverses the sign of the carrier contribution. Writing the sum as a product does not turn it into one pure-frequency wave at 5.5 Hz.

Define M(t) = |B(t)| = (4 mm)|cos(πτ)|. This nonnegative modulation magnitude bounds the instantaneous displacement: −M ≤ y ≤ M. It is an envelope bound, not the displacement itself. Equality in |y| ≤ M requires M = 0 or |C| = 1.

For example, at t = 1 s, B = −4 mm and C = −1, giving y = +4 mm and M = 4 mm. A negative factor is consistent with a positive displacement. Squaring M would have units of mm²; no physical intensity, perceived loudness or string-energy formula is being assigned to it here.

Read the three aligned comparisons

The first panel shows the two signed contributions. The second shows their sum between the thin upper and lower bounds +M and −M. The third compares the signed factor B with its magnitude M. All panels use the same time and displacement scales, so their signs and heights can be compared directly.

Circle, square and diamond outlines identify the selected exact states at t = 0, 1/2, 1, 3/2 and 2 s. At coincident component values the smaller square overlaps part of the circle outline; the outer circle rim and square remain distinguishable. The curves use short line segments to approximate the analytic functions. They are supplied model values, not sampled measurements.

The table includes quarter-second states as well as the five marked times. At t = 1/4 s, |y| = 2 mm while M = 2√2 mm. The instantaneous displacement therefore need not reach the envelope. At t = 1/2 s, B = M = y = 0 although the two contributions are nonzero and opposite.

Count beats with a stated convention

Use one beat interval to mean the time between successive maxima of M. Here maxima occur at integer seconds and adjacent maxima are 1 s apart. The beat frequency under this convention is |f₂−f₁| = 1 Hz. The signed factor B takes 2 s to repeat; its magnitude repeats after 1 s because taking the absolute value removes the sign reversal.

The closed display interval 0 ≤ t ≤ 2 s contains three maxima, including both endpoints, but only two complete intervals between them. A half-open interval 0 ≤ t < 2 s contains two maxima. Counting peaks and counting complete peak-to-peak intervals answer different questions.

This particular full waveform also repeats after 1 s: its component phases advance by 10π and 12π, both integer multiples of 2π. One second is its smallest positive period. The exact repetition follows from the supplied 5 Hz and 6 Hz pair; it is not a general consequence of a repeating modulation magnitude.

For a different frequency pair, the fast phase can change from one beat interval to the next. If the ratio of the two nonzero component frequencies is irrational, their finite two-cosine sum has no common positive period. A repeated envelope therefore does not generally imply an exactly repeated full waveform.

Change phase or amplitude

For a separate equal-frequency comparison, keep both amplitudes at 2 mm and both frequencies at 5 Hz, but give the second travelling component an initial phase +π/2. At x = 0 m its contribution is +2 sin(10πτ) mm. The sum is 2[cos(10πτ)+sin(10πτ)] mm = (2√2 mm) cos(10πτ−π/4). Its amplitude is constant and the relative phase does not drift: there is no frequency-difference beat.

For a separate unequal-amplitude comparison, keep the original 5 Hz and 6 Hz frequencies and zero phases, but use amplitudes 3 mm and 1 mm. Let z = 10πτ and δ = 2πτ. The cosine addition formula writes the sum as [3+cos δ] cos z mm − sin δ sin z mm. The two coefficients have a nonnegative magnitude R with R² = [(3+cos δ)²+sin²δ] mm² = (10+6 cos(2πτ)) mm². Thus 2 mm ≤ R ≤ 4 mm: the amplitude difference is the lower bound and the amplitude sum is the upper bound.

A positive lower bound for R prevents complete envelope cancellation, not every zero of the signed displacement. In this unequal case y(0) = 4 mm while y(1/8 s) = −3√2/2 mm. Continuity supplies at least one zero between those times even though R remains at least 2 mm. Keep a modulation magnitude separate from the oscillating real displacement.

Keep the physical meaning and model limits

The propagation speed c = 2 m/s describes how the right-going spatial pattern shifts along x. The transverse velocity y_t describes motion up and down at a fixed x. They are different directions and different quantities. At x = 0 m and t = 1/4 s, differentiating the two components gives y_t = −20π mm/s, not 2 m/s.

The ideal linear small-slope string rule is the premise. The largest possible sum of slope magnitudes is A(k₁+k₂) = 11π/500, approximately 0.0691. That calculation does not supply an experimental error bound for a real string’s nonlinearity, damping, supports or tension changes.

The supplied waves are defined on all real x and all t ≥ 0 s; the figure displays only x = 0 m and 0 ≤ t ≤ 2 s. Whole-line sinusoidal fields do not give a finite total-string energy. No energy-transfer, loudness or intensity conclusion is inferred from these displacement graphs.

The useful sequence is to specify the common linear model, add the compatible signed fields, and then choose the quantity to compare. Instantaneous displacement, a signed factor, a nonnegative modulation magnitude and a counted beat frequency each answer a different question.

Worked examples

Try it yourself

1. Use the main model’s second component, ω₂ = 12π rad/s and k₂ = 6π rad/m. Find its ordinary frequency, wavelength and propagation speed.

   Show the answer and reasoning

   f₂ = 6 Hz, λ₂ = 1/3 m and c = 2 m/s to the right.

   Divide angular frequency by 2π and use λ = 2π/k. Their ratio ω₂/k₂ = 2 m/s is positive in the supplied kx−ωt phase convention.

2. At t = 1 s in the main model, find B, the carrier C, y and M. Explain why y is positive even though B is negative.

   Show the answer and reasoning

   B = −4 mm, C = −1, y = +4 mm and M = 4 mm.

   cos π = −1 and cos 11π = −1. The product of the two negative factors is positive; the absolute magnitude M discards only B’s sign.

3. Count the maxima of M on the closed interval 0 ≤ t ≤ 2 s, then count complete peak-to-peak beat intervals there. State the period of B.

   Show the answer and reasoning

   Three maxima, two complete beat intervals and a 2 s period for B.

   The maxima occur at 0, 1 and 2 s. Consecutive pairs define two intervals. B reverses sign after 1 s and repeats only after 2 s.

4. Replace the main pair by equal amplitudes 2 mm, equal wave numbers 5π rad/m and equal angular frequencies 10π rad/s. Give the second field initial phase π. Is cancellation limited to the observer?

   Show the answer and reasoning

   The supplied resultant field is zero for every real x and all t ≥ 0 s.

   Writing z = kx−ωt gives A cos z + A cos(z+π) = A cos z−A cos z = 0. Identical wave numbers and frequencies, equal amplitudes and the fixed π offset are essential.

5. For the unequal 3 mm and 1 mm comparison, evaluate R at t = 1/2 s and decide whether a nonzero R implies a nonzero signed displacement at every time.

   Show the answer and reasoning

   R(1/2 s) = 2 mm; a nonzero R does not forbid zeros of the signed displacement.

   cos(2π·1/2) = −1 gives R² = 4 mm². The worked sign-change example supplies an instantaneous zero while R ≥ 2 mm.

6. At x = 0 m and t = 1/4 s, calculate y_t for the main pair and compare it with the right-going propagation speed.

   Show the answer and reasoning

   y_t = −20π mm/s, approximately −62.83 mm/s, while c = +2 m/s along the string.

   At the fixed observer y_t = −(20π mm/s) sin(10πτ)−(24π mm/s) sin(12πτ). The sines at τ = 1/4 are 1 and 0. This transverse downward rate differs from motion of the phase pattern along x.

What to carry forward

• Superposition adds compatible signed fields under a stated linear model.
• For the equal-amplitude pair, the signed factor B, its nonnegative magnitude M and the instantaneous displacement y are different quantities.
• Beat counting needs a convention; repeated strength maxima need not imply exact repetition of the entire waveform.
• Phase and amplitude determine cancellation, while propagation speed and transverse velocity retain different physical meanings.