Wave Mechanics Examples: Superposition

This companion article provides exercises for WM09, wave mechanics: Superposition. The exercises are stated first so they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

The set uses only the linear-superposition ideas developed in WM09. In particular,

(1)

and, for two equal-amplitude sinusoidal waves with phase difference Δϕ,

(2)

For unequal amplitudes,

(3)

Superposition is assumed only for a linear wave model. Energy and power are intentionally not inferred from amplitude addition alone; those topics appear later in the series.

How to use this problem set

Attempt all exercises in Part I before consulting Part II. For every numerical result, identify whether you are calculating an instantaneous displacement, an amplitude, a phase difference, or a property of the linear model. These quantities are related, but they are not interchangeable.

Part I: Exercises

Exercise 1: Point-by-point superposition

At one event (x₀,t₀), two disturbances have values

(4)

1. Find the total displacement u at that event.
2. Is the total displacement larger or smaller in magnitude than u₁ alone?
3. Explain why adding the amplitude magnitudes 3.5 + 1.2 would be incorrect here.

Exercise 2: Reading a pointwise sum from a graph

The figure below shows two disturbances and their point-by-point sum at one instant.

From the graph:

1. Estimate u₁(x₀).
2. Estimate u₂(x₀).
3. Use superposition to predict u(x₀).
4. Check that your prediction agrees with the lower panel.

Exercise 3: Limiting interference cases

Two equal sinusoidal waves have amplitude

(5)

Find the resultant amplitude A_R for each phase difference:

1. Δϕ = 0,
2. Δϕ = π∕2,
3. Δϕ = π,
4. Δϕ = 2π.

State whether each case is fully constructive, fully destructive, or partial interference.

Exercise 4: Equal amplitudes with a general phase difference

Two equal waves have amplitude

(6)

and phase difference

(7)

1. Find the resultant amplitude using the WM09 equal-amplitude formula.
2. Compare the result with the maximum possible value 2A.
3. Is the interference more constructive or more destructive than the case Δϕ = π∕2? Explain using the amplitude formula.

Exercise 5: Infer phase difference from a measured resultant amplitude

Two equal-amplitude waves each have

(8)

Their measured resultant amplitude is

(9)

Assume the principal phase difference lies in the interval

(10)

Determine Δϕ.

Exercise 6: Use the phase-resultant graph

The following plot shows the normalized resultant amplitude for two equal waves.

1. At which phase differences is the resultant amplitude maximum?
2. At which phase difference is the resultant amplitude zero?
3. At approximately which phase differences is A_R∕(2A) = 1∕2?
4. Explain why the graph repeats every 2π.

Exercise 7: Unequal amplitudes

Two same-frequency waves have amplitudes

(11)

Find A_R for

1. Δϕ = 0,
2. Δϕ = π∕2,
3. Δϕ = π.

Why can these two waves never cancel completely?

Exercise 8: Pulse overlap

Two equal positive pulses, each with peak amplitude A, move toward one another in a linear medium.

1. What is the peak resultant amplitude during complete overlap?
2. What happens to the component pulses after they separate?
3. Would the same reasoning necessarily remain valid in a strongly nonlinear medium? Explain briefly.

Exercise 9: Opposite-sign pulse overlap

At one instant a positive pulse and a negative pulse completely overlap. Their local shapes are identical except that the first has peak amplitude +5.0 mm and the second has peak amplitude −3.0 mm.

1. What is the resultant peak displacement at complete overlap?
2. Is this complete destructive interference?
3. What would the negative pulse amplitude need to be for complete cancellation?

Exercise 10: Derive the equal-amplitude result

Starting from

u1	= A cos 𝜃,	(12)
u2	= A cos(𝜃 + Δϕ),	(13)

use

(14)

to show that

(15)

Then explain why the physical amplitude is written with an absolute value.

Exercise 11: Linearity and sums of solutions

A linear wave model is written abstractly as

(16)

Suppose

(17)

1. Use linearity to show that u₁ + u₂ is also a solution.
2. More generally, show that au₁ + bu₂ is a solution for constants a and b.
3. Explain why this argument would fail if L were nonlinear.

Exercise 12: Counter-propagating waves

Consider

u1	= A cos(kx − ωt),	(18)
u2	= A cos(kx + ωt).	(19)

Use the cosine-sum identity to show that

(20)

Then answer:

1. Does the resulting expression have the form of a single rigidly translating wave F(x ∓ ct)?
2. Which factor depends only on position?
3. Which factor depends only on time?

Do not yet develop nodes or resonance; those topics belong to the later standing-wave block.

Exercise 13: Diagnose three statements

For each statement, decide whether it is correct. If incorrect, rewrite it accurately.

1. “Superposition means amplitudes are always added as positive numbers.”
2. “If two equal waves differ in phase by π, their instantaneous displacements cancel at every point.”
3. “Whenever two pulses overlap and produce a larger resultant, energy must have been created.”

Exercise 14: Synthesis from an interference measurement

Two same-frequency sinusoidal waves overlap in a region. Their individual amplitudes are

(21)

and the measured resultant amplitude is

(22)

1. Use the unequal-amplitude formula to determine cos Δϕ.
2. Find a principal phase difference in the interval 0 ≤ Δϕ ≤ π.
3. Check that your answer lies between the fully constructive and fully destructive amplitude limits.
4. Explain why the same physical interference state can also be represented by phase differences differing by integer multiples of 2π.

Part II: Complete Worked Solutions

Solution 1: Point-by-point superposition

1. Superposition is algebraic:

   u	= u1 + u2	(23)
   	= 3.5 mm − 1.2 mm	(24)
   	= 2.3 mm .	(25)

2. The magnitude 2.3 mm is smaller than 3.5 mm because the second disturbance has the opposite sign.
3. The numbers 3.5 mm and 1.2 mm are not two positive instantaneous displacements. The second disturbance is negative at the event. Replacing −1.2 by +1.2 would change the physical state being added.

Solution 2: Reading a pointwise sum from a graph

The marked position is x₀ = 1.5 in the plotted coordinate system.

From the upper panel,

(26)

From the middle panel,

(27)

Therefore

(28)

The lower panel passes through approximately 0.5 at the same marked position, confirming the pointwise addition.

Solution 3: Limiting interference cases

For equal amplitudes,

(29)

with A = 4.0 mm.

1. For Δϕ = 0,

   (30)

   This is fully constructive interference.

2. For Δϕ = π∕2,

   AR	= 8.0 mm	(31)
   	= 8.0 mm	(32)
   	= 4 mm	(33)
   	≈ 5.66 mm .	(34)

   This is partial interference.

3. For Δϕ = π,

   (35)

   This is fully destructive interference.

4. For Δϕ = 2π,

   (36)

   The phase difference 2π is equivalent to zero phase difference, so the interference is fully constructive.

Solution 4: Equal amplitudes with a general phase difference

Given

(37)

we obtain

AR	= 2(6.0 mm)	(38)
	= 12.0 mm	(39)
	= 6.0 mm .	(40)

The largest possible resultant is

(41)

Thus the measured resultant is one-half of the fully constructive amplitude.

For Δϕ = π∕2,

(42)

whereas for 2π∕3,

(43)

Therefore 2π∕3 produces the more destructive of the two cases.

Solution 5: Infer phase difference from a measured resultant amplitude

For equal amplitudes,

(44)

Divide by 10.0 mm:

(45)

Because 0 ≤ Δϕ ≤ π, we have

(46)

so the relevant solution is

(47)

Hence

(48)

Solution 6: Use the phase-resultant graph

The plotted function is

(49)

1. The maximum value is 1, occurring at

   (50)

   over the plotted interval.

2. The amplitude reaches zero at

   (51)

3. Set

   (52)

   Over 0 ≤ Δϕ ≤ 2π, this occurs at

   (53)

4. Phase differences that differ by 2π represent the same relative phase state of the two periodic waves. Therefore the interference pattern repeats every 2π.

Solution 7: Unequal amplitudes

Use

(54)

with A₁ = 7.0 mm and A₂ = 3.0 mm.

1. For Δϕ = 0,

   AR	= mm	(55)
   	= mm	(56)
   	= 10.0 mm .	(57)

2. For Δϕ = π∕2,

   AR	= mm	(58)
   	= mm	(59)
   	≈ 7.62 mm .	(60)

3. For Δϕ = π,

   AR	= mm	(61)
   	= mm	(62)
   	= 4.0 mm .	(63)

The minimum possible resultant amplitude is

(64)

not zero. Complete cancellation requires equal component amplitudes.

Solution 8: Pulse overlap

1. During complete overlap, equal positive pulses add:

   (65)

2. In the ideal linear model, the component pulses continue through the overlap and later reappear with their original shapes and directions of propagation.
3. Not necessarily. Strong nonlinear response can make the total response depend on products, powers, or other nonlinear combinations of the disturbances. In that case the simple sum u₁ + u₂ need not remain a valid solution.

Solution 9: Opposite-sign pulse overlap

At complete overlap, the local peak displacement is

(66)

This is destructive interference, but it is not complete cancellation because the magnitudes are unequal.

For complete cancellation, the second pulse would need peak amplitude

(67)

Solution 10: Derive the equal-amplitude result

Starting from

(68)

apply the cosine-sum identity with

(69)

Then

u	= 2A cos cos	(70)
	= 2A cos cos .	(71)

Since cosine is even,

(72)

so

(73)

If the coefficient 2A cos(Δϕ∕2) is negative, that sign can be absorbed into an additional phase shift of π. By convention, amplitude is reported as a nonnegative magnitude. Therefore

(74)

Solution 11: Linearity and sums of solutions

1. Linearity means

   (75)

   Since each term is zero,

   (76)

   Thus u₁ + u₂ is also a solution.

2. More generally,

   L[au1 + bu2]	= aL[u1] + bL[u2]	(77)
   	= 0,	(78)

   so any linear combination with constant coefficients is also a solution.

3. A nonlinear operator generally does not satisfy

   (79)

   For example, if a model contained a term proportional to u², then

   (80)

   and the cross term prevents simple superposition.

Solution 12: Counter-propagating waves

Add the two waves:

(81)

Using the cosine-sum identity with

(82)

we get

u	= 2A cos cos	(83)
	= 2A cos(−ωt) cos(kx)	(84)
	= 2A cos(kx) cos(ωt) .	(85)

1. No. The result separates into a position factor and a time factor rather than appearing as a single rigidly translating function F(x ∓ ct).
2. The purely spatial factor is

   (86)

3. The purely temporal factor is

   (87)

This is the algebraic preview of a standing wave, but node and resonance physics are deferred.

Solution 13: Diagnose three statements

1. Incorrect. Superposition adds the signed instantaneous disturbances:

   (88)

   Amplitudes are nonnegative descriptors of component size and are not simply added without considering phase.

2. Correct for two equal-amplitude, equal-k, equal-ω waves that differ only by a constant phase shift of π. At every event one disturbance is the negative of the other.
3. Incorrect. A larger instantaneous amplitude during constructive interference does not by itself prove that energy has been created. Energy accounting requires the appropriate wave-energy and power expressions, which are treated later in the series.

Solution 14: Synthesis from an interference measurement

Use

(89)

Substitute the measured amplitudes:

(90)

Thus

100	= 64 + 36 + 96 cos Δϕ,	(91)
100	= 100 + 96 cos Δϕ.	(92)

Therefore

(93)

In the principal interval 0 ≤ Δϕ ≤ π,

(94)

The fully constructive limit is

(95)

and the fully destructive limit is

(96)

The measured value

(97)

is therefore physically consistent with partial interference.

Equivalent relative phases differ by integer multiples of 2π:

(98)

They represent the same relative position within the periodic cycle.

Summary of skills practiced

After completing this set, you should be able to:

• add arbitrary disturbances point by point;
• distinguish instantaneous displacement from amplitude;
• identify constructive, destructive, and partial interference;
• calculate the resultant amplitude for equal and unequal component amplitudes;
• infer phase difference from a measured resultant amplitude;
• interpret pulse overlap in a linear model;
• explain why superposition follows from linearity;
• recognize why nonlinear terms can invalidate simple superposition;
• derive the counter-propagating-wave product form that later leads to standing waves;
• avoid making energy claims from amplitude addition alone.

References

References