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[parent] Wave Mechanics Examples: Oscillation in Space (Example)

Wave Mechanics Examples: Oscillation in Space

This companion entry provides self-study exercises for WM04, Oscillation in Space. The exercises are stated first so that they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

The problems remain deliberately within the WM04 spatial description. They use spatial profiles u(x), wavelength, amplitude, and spatial periodicity. Wavenumber k, propagation, and the full space-time field u(x,t) are reserved for later lessons.

How to use this problem set

For every graph, first identify the horizontal variable. If the horizontal axis is position x, then horizontal separations are distances, not times. When measuring a wavelength, use two corresponding points in adjacent cycles: crest to crest, trough to trough, or the same displacement with the same local direction of change with x.

WM04 relations used in this set:
u = u(x ),     u(x + λ) = u (x ),    A =  max |u(x)|.
For a simple sinusoidal spatial profile,
             (   x)
u(x) = A cos  2π -- .
                 λ
The wavelength λ is the smallest positive spatial distance over which the complete pattern repeats.

Part I: Exercises

Exercise 1: Time description or space description?

For each expression or statement, identify whether it describes variation in time, variation in space, or neither.

  1. u(t)
  2. u(x)
  3. “the displacement at several positions in one photograph”
  4. “the displacement at one sensor as the seconds pass”
  5. u(x + λ) = u(x)
  6. u(t + T) = u(t)

For parts (e) and (f), state the physical meaning of the horizontal repeat quantity.

Exercise 2: Spatial repetition

A spatially periodic profile has wavelength

λ = 0.80 m.

If

u(0.35m ) = 2.1mm,

what can you say about the values of u(x) at

1.15 m,     1.95m,      − 0.45 m?

Explain why the result follows from spatial periodicity rather than from any assumption about motion.

Exercise 3: Reading amplitude and wavelength from a graph

The following graph shows a spatial profile.

PIC

Figure. A periodic spatial profile for Exercise 3. The horizontal axis is position, not time.

Determine:

  1. the equilibrium value;
  2. the amplitude A;
  3. the wavelength λ;
  4. the peak-to-peak vertical range;
  5. the number of complete spatial cycles between x = 0 and x = 4.0 m.

Exercise 4: Counting wavelengths in a known length

A periodic pattern has wavelength

λ = 0.25 m.

How many complete wavelengths fit into each distance?

  1. 1.00 m
  2. 2.50 m
  3. 0.75 m

Then state the general relation between a length L, wavelength λ, and the number N of complete wavelengths when L contains an integer number of cycles.

Exercise 5: Length occupied by several cycles

A spatial pattern repeats every

λ = 1.20cm.

Find the total length occupied by:

  1. 3 wavelengths;
  2. 12 wavelengths;
  3. 25 wavelengths.

Give the answer to part (c) in both centimeters and meters.

Exercise 6: Equal displacement is not enough

The graph below marks two positions P and Q with the same displacement.

PIC

Figure. Points P and Q have equal displacement but opposite local slope. They are not corresponding points of adjacent cycles.

For the marked points,

     λ            5λ
xP = --,    xQ  = ---.
      6            6

Answer the following.

  1. Show that uP = uQ for the profile
                 (   x )
u (x ) = A cos 2π -- .
                 λ
  2. Find the separation xQ xP as a fraction of λ.
  3. Is that separation one wavelength?
  4. Explain why equal displacement alone cannot be used to measure λ.

Exercise 7: Unit conversions for wavelength

Convert each wavelength to the requested unit.

  1. 0.035 m to centimeters;
  2. 4.8 mm to meters;
  3. 2.5 km to meters;
  4. 650 nm to meters.

Which of these quantities is largest physically, regardless of the numerical value written before the unit conversion?

Exercise 8: Same amplitude, different wavelength

Two spatial profiles have the same amplitude,

A  = 5.0mm,

but their wavelengths are

λ  = 2.0m,     λ  =  0.50 m.
 1               2

Determine:

  1. which pattern repeats more frequently in space;
  2. how many cycles of each pattern fit into 4.0 m;
  3. whether the shorter wavelength implies a larger amplitude;
  4. the peak-to-peak vertical range of each profile.

Exercise 9: Reading a sinusoidal spatial equation

Consider

                (   --x- )
u(x ) = 0.040 cos  2π0.60  ,

where x and u are measured in meters.

Find:

  1. the amplitude;
  2. the wavelength;
  3. u(0);
  4. u(0.15 m);
  5. u(0.30 m);
  6. u(0.60 m).

Exercise 10: Verify the spatial period algebraically

For

             (   x-)
u (x ) = A cos 2π λ  ,

show directly that

u (x + λ) = u(x).

Then explain why the same proof also shows

u(x + n λ) = u(x)

for every integer n.

Exercise 11: Spatial landmarks within one wavelength

For

             (     )
u (x ) = A cos 2π x- ,
                 λ

find the displacement at

x =  0,    x =  λ,     x = λ-,    x =  3λ,     x = λ.
                4          2           4

Describe how these five points correspond to one complete spatial cycle.

Exercise 12: Infer wavelength from measured crest positions

Successive crests of a spatial pattern are observed at

0.42 m,     1.17 m,     1.92m,      2.67 m.

Assuming these are consecutive crests:

  1. determine the wavelength;
  2. determine the distance from the first to the fourth crest;
  3. determine how many complete wavelengths lie between the first and fourth crest;
  4. explain why four listed crests do not span four wavelengths.

Exercise 13: Dimensional and conceptual audit

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

  1. “The wavelength is 0.50 s.”
  2. “The period is 1.2 m.”
  3. “A shorter wavelength must have a smaller amplitude.”
  4. “A picture of u(x) by itself proves that the pattern is moving.”
  5. “If u(x + λ) = u(x), then 2λ is also a repeat distance.”
  6. “The wavelength normally means the smallest positive repeat distance.”

Exercise 14: Challenge—reconstruct a spatial pattern from observations

A laboratory photograph shows a periodic string profile. The string reaches maximum upward displacement at

x =  0.20m,     0.70 m,     1.20 m,     1.70m.

The maximum displacement is +6.0 mm and the minimum is 6.0 mm.

Determine:

  1. the amplitude;
  2. the wavelength;
  3. the peak-to-peak range;
  4. how many complete wavelengths lie between x = 0.20 m and x = 1.70 m;
  5. the next crest position to the right of 1.70 m;
  6. whether these observations alone determine that the pattern is traveling to the right, traveling to the left, or stationary.

Part II: Complete Worked Solutions

Solution 1: Time description or space description?

  1. u(t) is a temporal description: the independent variable is time.
  2. u(x) is a spatial description: the independent variable is position.
  3. A photograph showing several positions at one instant is a spatial description.
  4. One sensor observed as seconds pass gives a temporal description.
  5. u(x + λ) = u(x) describes spatial periodicity. The repeat quantity λ is a distance.
  6. u(t+T) = u(t) describes temporal periodicity. The repeat quantity T is a time interval.

Key distinction. Period and wavelength are analogous repeat measures, but they belong to different independent variables and have different physical units.

Solution 2: Spatial repetition

The wavelength is 0.80 m. The listed positions differ from 0.35 m by integer multiples of the wavelength:

1.15 − 0.35 = 0.80m  = λ,

1.95 − 0.35 = 1.60 m =  2λ,

and

− 0.45 − 0.35 = − 0.80m  = − λ.

Therefore

|----------------------------------------------|
|u(1.15 m ) = u(1.95m ) = u(− 0.45m ) = 2.1mm   .
-----------------------------------------------

This conclusion follows only from the repeating spatial profile. No time variable or propagation assumption is needed.

Solution 3: Reading amplitude and wavelength from a graph

  1. The profile is centered on
    |------|
-u-=-0-.
  2. The graph reaches +3 mm and 3 mm, so
    |----------|
A  = 3 mm  .
------------
  3. Adjacent crests occur at x = 0 and x = 2.0 m, so
    |----------|
-λ-=-2.0m--.
  4. The peak-to-peak vertical range is
    2A =  6mm.

    Thus

    |---------------------------|
|peak-to-peak range =  6mm  .
-----------------------------
  5. The interval from 0 to 4.0 m contains
          4.0 m    |-----------------|
N  =  ------= |2 complete cycles.
      2.0 m    -------------------

Solution 4: Counting wavelengths in a known length

When a length contains an integer number of complete cycles,

     L-
N =  λ .

With λ = 0.25 m:

  1.      1.00   |--|
N  = ---- = -4-.
     0.25
  2.              |--|
N  =  2.50-=  10-.
      0.25
  3.      0.75   |--|
N  = ---- = -3-.
     0.25

Thus the general relation is

|--------|
-L-=-N-λ--

or equivalently N = L∕λ.

Solution 5: Length occupied by several cycles

Use

L  = N λ.

With λ = 1.20 cm:

  1.                   |-------|
L =  3(1.20 cm ) = -3.60-cm-.
  2.                   |--------|
L = 12(1.20 cm ) =-14.4cm--.
  3.                   |--------|
L = 25(1.20 cm ) =-30.0cm--.

    Since 100 cm = 1 m,

              ---------
30.0cm  = |0.300 m  .
          ---------|

Solution 6: Equal displacement is not enough

The profile is

             (   x )
u (x ) = A cos 2π -- .
                 λ

  1. At xP = λ∕6,
               (   1 )         ( π)    A
uP =  A cos  2π--  = A  cos  -- =  --.
               6             3     2

    At xQ = 5λ∕6,

               (     )         (    )
uQ =  A cos  2π 5- =  A cos  5π-  =  A.
                6             3      2

    Hence

    |--------------|
|           A- |
|uP = uQ =  2  .
---------------
  2.                       |---|
xQ  − xP =  5λ-− λ- = |2λ-.
            6     6   -3---
  3. No. The separation is 2λ∕3, not λ.
  4. The same displacement can occur at more than one location within a single cycle. To measure wavelength using repeated displacement values, the points must also occupy corresponding locations in the pattern, including the same local direction of change with x.

Common error. Two equal values of u need not be separated by one wavelength.

Solution 7: Unit conversions for wavelength

  1.         (       )
          100cm--    |------|
0.035 m    1 m     = -3.5cm--.
  2.        (    1m    )    |------------|
4.8mm    ---------   = -4.8-×-10−-3m-.
         1000 mm
  3.        (        )   -------------
         1000m--    |        3   |
2.5km     1km     = -2.5 ×-10-m--.
  4.        (   −9   )   |--------------|
650nm    10---m-  = |6.50 × 10−7m  .
          1nm       ---------------

The largest physical wavelength is

|------|
2.5km  .
--------

Solution 8: Same amplitude, different wavelength

The profiles have

A =  5.0mm,      λ1 = 2.0 m,     λ2 = 0.50 m.

  1. Profile 2 has the shorter wavelength, so it repeats more frequently in space.
  2. Over 4.0 m,
         4.0    |-|           4.0    |-|
N1 = --- =  2-,    N2  = ---- =  8-.
     2.0                 0.50
  3. No. Wavelength controls horizontal repetition distance, while amplitude controls vertical size. The two profiles were explicitly given the same amplitude.
  4. For both profiles,
                       |--------|
2A  = 2(5.0mm  ) = |10.0 mm  .
                   ----------

Solution 9: Reading a sinusoidal spatial equation

The equation is

                (        )
u(x ) = 0.040 cos  2π--x-  .
                    0.60

Comparing with

             (   x )
u (x ) = A cos 2π -- ,
                 λ

gives

|------------|     |-----------|
-A-=-0.040-m-,     λ-=--0.60-m--.

Now evaluate the profile.

  1.                     |--------|
u(0) = 0.040 cos0 = -0.040-m--.
  2. Since 0.15 = λ∕4,
                       (  )   |--|
u(0.15) = 0.040cos  π-  = |0 .
                    2     ---
  3. Since 0.30 = λ∕2,
                            |---------|
u (0.30 ) = 0.040 cosπ = − 0.040 m .
                        -----------
  4. Since 0.60 = λ,
                              |-------|
u(0.60) = 0.040 cos(2 π) = 0.040-m-.

The values trace one complete cosine cycle in space.

Solution 10: Verify the spatial period algebraically

Start with

             (     )
u (x ) = A cos 2π x- .
                 λ

Then

u(x + λ) = A cos (   x + λ )
  2π------
      λ (1)
= A cos (          )
 2π x-+ 2π
    λ (2)
= A cos (   x-)
 2π λ (3)
= u(x). (4)

Therefore

-----------------
|                |
u-(x +-λ) =-u(x)-.

For any integer n,

                  (            )
u (x + n λ) = A cos  2πx- + 2πn  .
                      λ

Cosine is unchanged by any integer multiple of 2π, so

|-----------------|
-u(x-+-nλ-) =-u(x) .

Solution 11: Spatial landmarks within one wavelength

Evaluate the cosine argument at each position.

       |       |
---x---|2πx∕-λ-|u(x)-
   0   |  0    | A
  λ∕4  | π∕2   | 0
       |       |
  λ∕2  |  π    |− A
 3λ ∕4 |3 π∕2  | 0
   λ   |  2π   | A

Thus the spatial profile moves through

|------------------------|
|A →  0 →  − A →  0 →  A |
-------------------------

as x increases through one wavelength. The final point is equivalent to the initial point and completes one full spatial repetition.

Solution 12: Infer wavelength from measured crest positions

The crest positions are

0.42,   1.17,   1.92,  2.67m.

Neighboring separations are

1.17 − 0.42 = 0.75m,

1.92 − 1.17 = 0.75m,

2.67 − 1.92 = 0.75m.

Therefore

|-----------|
|λ = 0.75 m .
------------

From the first crest to the fourth crest,

2.67 − 0.42 = 2.25m.

Hence

|------------|
2.25-m-=--3λ-.

There are three complete wavelengths between four consecutive crests because four marked points contain only three intervals.

Common counting error. Count the intervals between repeated features, not only the number of marked features.

Solution 13: Dimensional and conceptual audit

  1. Incorrect. Wavelength is a distance, so it must have a length unit, for example “the wavelength is 0.50 m.”
  2. Incorrect. Period is a time interval, so a valid example would be “the period is 1.2 s.”
  3. Incorrect. Amplitude and wavelength are independent geometric features of a spatial profile.
  4. Incorrect. A single function u(x) or one photograph gives only a spatial profile. Motion requires time information.
  5. Correct. If λ is a repeat distance, then any integer multiple such as 2λ is also a repeat distance.
  6. Correct. The wavelength normally means the smallest positive repeat distance of the complete spatial pattern.

Solution 14: Challenge—reconstruct a spatial pattern from observations

The consecutive maxima are located at

0.20,   0.70,   1.20,  1.70m.

  1. The extremes are +6.0 mm and 6.0 mm, so
    |------------|
A--=-6.0mm---.
  2. Consecutive maxima are separated by
    0.70 − 0.20 = 0.50m.

    The later pairs give the same result, so

    |-----------|
-λ-=-0.50-m-.
  3. The peak-to-peak range is
          |--------|
2A =  12.0-mm--.
  4. The distance from the first to the fourth crest is
    1.70 − 0.20 = 1.50m.

    Therefore

         1.50   |------------------------|
N  = ---- = |3 complete wavelengths  .
     0.50   -------------------------
  5. The next crest is one wavelength farther to the right:
                 |-------|
1.70 + 0.50 = -2.20m--.
  6. The photograph gives only spatial information at one instant. It cannot determine whether the pattern is moving right, moving left, or not moving at all. Time-dependent information would be required.

What this set prepares you for

WM04E1 has treated wavelength as a spatial repeat distance without introducing any spatial angular-rate symbol. WM05 will package the factor

2π-
 λ

into the wavenumber k. This will create the direct spatial analogue of the temporal relation ω = 2π∕T.

Summary of skills practiced

After completing this set, you should be able to:

  • distinguish a spatial profile u(x) from a time history u(t);
  • apply the spatial periodicity relation u(x + λ) = u(x);
  • identify amplitude and wavelength from a graph;
  • count complete wavelengths within a known distance;
  • convert wavelength units correctly;
  • distinguish amplitude from wavelength;
  • recognize why equal displacement alone does not establish one wavelength;
  • evaluate a simple sinusoidal spatial profile at key positions;
  • verify spatial periodicity algebraically;
  • infer wavelength from repeated spatial features;
  • distinguish spatial structure from propagation.

"Wave Mechanics Examples: Oscillation in Space" is owned by bloftin.
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Keywords:  wave mechanics, spatial oscillation, spatial periodicity, wavelength, spatial profile, amplitude, sinusoidal spatial pattern, periodic function, exercises, worked solutions

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Cross-references: function, trace, equilibrium, motion, relations, position, graph, field, space-time, WM04

This is version 1 of Wave Mechanics Examples: Oscillation in Space, born on 2026-09-11.
Object id is 1155, canonical name is WaveMechanicsExamplesOscillationInSpace.
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Classification:
Physics Classification46.40.-f (Vibrations and mechanical waves )
 45.20.Dd (Newtonian mechanics)
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