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[parent] Wave Mechanics Examples: Wavenumber (Example)

Wave Mechanics Examples: Wavenumber

This companion entry provides self-study exercises for WM05, Wavenumber. All exercises are stated first so that they can be attempted before the worked solutions are read. Part II then develops each solution step by step and identifies common mistakes where useful.

The notation follows the angular-wavenumber convention used throughout this wave mechanics series and in standard wave-mechanics texts:

    2π
k = ---.
     λ
(1)

Thus k measures spatial phase change per unit distance. The unqualified word “wavenumber” is used differently in some spectroscopic contexts, where it commonly denotes reciprocal wavelength 1∕λ; Exercise 11 addresses this important distinction [1235].

The problems remain deliberately within the WM05 spatial description. They do not require propagation speed or a time-dependent field u(x,t). Those ideas begin in WM06 and WM07.

How to use this problem set

For every calculation, ask what k means physically before substituting numbers: it is an angular spatial rate. Track whether a quantity is a length, a reciprocal length, a phase, or a displacement. Whenever a cosine or sine appears, verify that its argument is dimensionless, with phase understood in radians.

WM05 relations used in this set:
k =  2π,     λ = 2-π,     𝜃(x) = kx + ϕ,
     λ            k
u (x) = A cos(kx + ϕ),     Δ𝜃 =  kΔx.
For angular wavenumber, k is commonly described in radians per meter, while its SI dimensional unit is inverse length. One wavelength corresponds to a phase advance of 2π.

Part I: Exercises

Exercise 1: Wavelength to angular wavenumber

A spatial sinusoid has wavelength

λ = 0.80 m.

Find:

  1. the angular wavenumber k in exact form;
  2. the numerical value of k in radians per meter;
  3. the phase advance over a distance of 0.20 m.

Exercise 2: Angular wavenumber to wavelength

A periodic spatial profile has angular wavenumber

k = 12.0rad/m.

Determine:

  1. its wavelength in meters;
  2. its wavelength in centimeters;
  3. the distance required for a phase advance of π radians.

Exercise 3: Interpreting radians per meter

A profile has

k = 4π rad/m.

  1. How much phase is accumulated over 0.25 m?
  2. What fraction of a complete spatial cycle is this?
  3. Determine the wavelength.
  4. Explain in words what 4π rad/m means.

Exercise 4: Reading wavelength and wavenumber from a graph

The following graph shows a spatial sinusoid.

PIC

Figure. A spatial profile for Exercise 4. Read the amplitude and wavelength from the geometry before calculating wavenumber.

Determine:

  1. the amplitude A;
  2. the wavelength λ;
  3. the angular wavenumber k in exact and numerical form;
  4. the reciprocal wavelength 1∕λ in cycles per meter;
  5. the total phase advance from x = 0 to x = 2.40 m.

Exercise 5: Mapping position into spatial phase

For a pattern with

k = π rad/m,     ϕ =  0,

the spatial phase is 𝜃(x) = kx.

PIC

Figure. position and phase for Exercise 5. The wavelength is 2.00 m, so quarter-wavelength steps produce phase increments of π∕2.

  1. Verify the phase values shown at x = 0, 0.50, 1.00, 1.50, and 2.00 m.
  2. What phase is reached at x = 2.50 m?
  3. Reduce the phase from part (b) to an equivalent phase between 0 and 2π.
  4. What is the displacement u∕A at x = 1.50 m for u = A cos(kx)?

Exercise 6: Evaluating a spatial sinusoid

Consider

                  (        )
u(x) = 5.0mm   cos  πx +  π- ,
                    2     3

where x is measured in meters.

  1. Identify A, k, and ϕ.
  2. Determine the wavelength.
  3. Evaluate u(0).
  4. Evaluate u(23 m).
  5. Evaluate u(43 m).

Exercise 7: Phase difference between two positions

A profile has

k = 4π rad/m.

Compare the spatial phase at

x1 = 0.20 m,     x2 = 0.95 m.

  1. Find Δx = x2 x1.
  2. Find Δ𝜃 = kΔx.
  3. Reduce the phase difference to an equivalent value between 0 and 2π.
  4. Determine the wavelength and express the separation Δx in wavelengths.
  5. For a cosine profile, what does the equivalent phase difference say about the displacement values at the two positions?

Exercise 8: Distance corresponding to a phase change

A spatial pattern has

k =  8.0 rad/m.

Find the distance required for a phase change of:

  1. π∕2;
  2. π;
  3. 2π.

Then verify that the answer to part (c) agrees with the wavelength calculated from λ = 2π∕k.

Exercise 9: Phase constant as a spatial shift

Compare

u1(x) = A cos(πx )

and

              (      π )
u2 (x ) = A cos πx +  -- ,
                     2

where x is in meters.

PIC

Figure. The two profiles have the same amplitude and wavenumber. Their phase constants shift the location of corresponding maxima without changing wavelength.

  1. Determine the common wavelength.
  2. Find the maximum of u1 nearest the origin.
  3. Find the maximum of u2 nearest the origin.
  4. Determine the spatial shift between those maxima.
  5. Does a positive phase constant shift this cosine pattern toward positive or negative x?

Exercise 10: Comparing two wavenumbers

Two spatial profiles have the same amplitude but different wavenumbers:

                          5π-
kA = 2π rad/m,      kB =  2  rad/m.

  1. Find λA and λB.
  2. Which profile has the shorter wavelength?
  3. Which profile accumulates phase more rapidly with increasing x?
  4. How many complete wavelengths of each profile fit in a 4.0 m interval?

Exercise 11: Angular wavenumber versus spectroscopic wavenumber

A monochromatic spatial pattern has wavelength

λ = 500 nm.

In this Wave Mechanics series, angular wavenumber means

    2π
k = ---.
     λ

In spectroscopy, the term “wavenumber” commonly denotes reciprocal wavelength

     1
^ν =  -,
     λ

often reported in cm1 [5].

Find:

  1. k in radians per meter;
  2. ν in inverse meters;
  3. ν in inverse centimeters;
  4. the factor relating k and ν.

Exercise 12: Dimensional audit

Assume x and λ have dimensions of length, k has dimensions of inverse length, and ϕ is a phase angle. For each expression, state whether it can serve as a valid trigonometric argument and explain why.

  1. kx + ϕ
  2. k + x
  3. x∕λ
  4. 2πx∕λ

Which of the dimensionally valid expressions automatically measures phase in the same angular convention used for cos(kx + ϕ)?

Exercise 13: Inferring wavenumber from measured phase change

An experiment shows that the spatial phase increases by

      3π
Δ 𝜃 = ---
       2

over a distance

Δx  = 0.30 m.

Assume the phase varies linearly with position.

  1. Determine k in exact form.
  2. Give the numerical value of k in radians per meter.
  3. Determine the wavelength.
  4. Predict the phase advance over 0.10 m.

Exercise 14: Challenge—reconstruct a spatial sinusoid

A measured sinusoidal profile has amplitude

A  = 6.0mm.

Consecutive maxima occur at

x = − 0.10 m     and     x =  0.70 m.

Model the profile as

u(x) = A cos(kx + ϕ ).

Choose the phase constant in the interval 0 ϕ < 2π.

  1. Determine the wavelength.
  2. Determine the angular wavenumber k.
  3. Use the maximum at x = 0.10 m to determine ϕ.
  4. Write a complete equation for u(x).
  5. Evaluate u(0.10 m) and u(0.30 m).
  6. Determine the reciprocal wavelength in cycles per meter.

Part II: Complete Worked Solutions

Solution 1: Wavelength to angular wavenumber

Given

λ = 0.80 m,

use k = 2π∕λ.

  1.               |----------|
    --2π---   |5π-       |
k = 0.80 m =  |2  rad/m  .
              -----------
  2.      |----------|
k ≈  7.85-rad/m--.
  3. Over Δx = 0.20 m,
                  ( 5π)          |π-|
Δ 𝜃 = k Δx =    --- (0.20) = |--.
                2            -2--

    So one quarter of a wavelength corresponds to one quarter of a 2π phase cycle.

Common error. Using k = 1∕λ would calculate reciprocal wavelength, not angular wavenumber.

Solution 2: Angular wavenumber to wavelength

Given

k = 12.0rad/m,

  1.      2π     2π    |--------|
λ =  ---=  ---- = -0.524m--|
      k    12.0

    to three significant figures.

  2.           |--------|
0.524m  = -52.4cm--.
  3. A phase advance of π is half of a complete 2π cycle, so the required distance is half a wavelength:
          π     π     |--------|
Δx  = --=  ---- ≈ -0.262m--.
      k    12.0

Solution 3: Interpreting radians per meter

For

k = 4π rad/m,

  1.                           |--|
Δ 𝜃 = kΔx  = (4π )(0.25) = -π-.
  2. Since one full cycle is 2π, a phase advance of π is
    |--------------|
|one-half cycle .
---------------
  3.            |------|
λ =  2π-=  0.50 m .
     4π    --------
  4. The value 4π rad/m means that increasing position by one meter advances the spatial phase by 4π radians, corresponding to two full phase cycles per meter.

Solution 4: Reading wavelength and wavenumber from a graph

From the graph,

  1. The vertical extrema are ±3.0 mm, so
    |------------|
A--=-3.0mm---.
  2. Adjacent maxima occur at x = 0 and x = 1.20 m, so
    |-----------|
-λ-=-1.20-m-.
  3.            |----------|
    -2π-   |5π-       |  |-----------|
k = 1.20 = | 3 rad/m  |≈ -5.24rad/m--.
           -----------
  4.               |--------------|
1-= ---1---=  0.833 cycles/m .
λ   1.20 m    ----------------
  5. The interval 2.40 m contains two wavelengths, hence two complete angular cycles:
                             ----
Δ 𝜃 = kΔx  =  5π(2.40) = |4π .
              3          ----|

Solution 5: Mapping position into spatial phase

Here

k = π rad/m,      𝜃(x ) = πx.

  1. Substituting the marked positions gives
       𝜃(0) = 0,
          π-
𝜃(0.50) = 2 ,
𝜃(1.00) = π,

𝜃(1.50) = 3π-,
           2
𝜃(2.00) = 2π.

    These match the figure.

  2.                     |---|
                    |5π-|
𝜃(2.50) = π(2.50) = | 2 .
                    -----
  3. Subtracting one complete cycle 2π,
               |--|
5π- − 2π = |π-.
 2         -2--
  4. At x = 1.50 m,
             (   )
 u         3π     |-|
-- = cos   ---  = -0 .
A          2

Solution 6: Evaluating a spatial sinusoid

The profile is

                  (        )
u(x) = 5.0mm   cos  πx +  π- .
                    2     3

  1. Comparing with u = A cos(kx + ϕ),
                      |-------------|     |------|
|------------|    |    π-       |     |    π-|
A--=-5.0mm---,    |k = 2 rad/m  ,     ϕ =  3 .
                  --------------      --------
  2.             |-----|
λ =  -2π-=  4.0-m-.
     π∕2
  3. At x = 0,
                 ( π)    |-------|
u(0) = 5.0 cos  3  =  2.5-mm--.
  4. At x = 23 m,
         π ( 2)    π   2 π
𝜃 =  --  -- +  --= ---,
     2   3     3    3

    so

     (  )           (   )    |---------|
u  2-  = 5.0cos   2π-  = |− 2.5 mm .
   3              3      -----------
  5. At x = 43 m,
           (  )
     π- 4-    π-
𝜃 =  2  3   +  3 = π,

    so

      ( 4)                |---------|
u   --  = 5.0cos(π) = -−-5.0-mm--.
    3

Solution 7: Phase difference between two positions

Given

k = 4π rad/m,

  1.                    |-------|
Δx =  0.95 −  0.20 = -0.75-m-.
  2.                            |--|
Δ 𝜃 = kΔx  =  (4π)(0.75 ) = 3π .
                           ----
  3. Modulo 2π,
              |--|
3π − 2π = -π-.
  4.      2π    |------|
λ =  ---=  0.50-m-.
     4π

    Thus

    Δx     0.75   |----------------|
-λ--=  0.50 = -1.5-wavelengths--.
  5. A phase difference of π reverses the cosine value:
    cos(𝜃 + π ) = − cos𝜃.

    Therefore the two displacements have equal magnitude and opposite sign for the same-amplitude cosine profile.

Solution 8: Distance corresponding to a phase change

Using

                      Δ 𝜃
Δ𝜃 =  kΔx,     Δx  =  ---,
                       k

with k = 8.0 rad/m:

  1.                   |--------|
Δx =  π∕2-=  π--≈ |0.196 m .
      8.0    16   ---------
  2.       -π-   |--------|
Δx =  8.0 ≈ -0.393m--.
  3.        2π    π    |-------|
Δx  =  ---=  --≈  0.785-m-.
       8.0    4

    The wavelength computed independently is

         2π-   2π-
λ =  k  =  8.0 ≈ 0.785m,

    so the results agree.

Solution 9: Phase constant as a spatial shift

The common wavenumber is

k = π rad/m.

  1. λ =  2π-= |2.0m--.
     π    -------|
  2. For u1 = A cos(πx), a maximum occurs when
    πx = 2πn.

    The maximum nearest the origin is

    |------|
-x-=-0-.
  3. For u2, a maximum satisfies
          π-
πx +  2 = 2 πn.

    Choosing n = 0 gives

                   |------------|
πx =  − π,     |x = − 0.50 m .
        2      --------------
  4. The corresponding maxima are separated by
    |-------------|
-Δx--=-0.50-m-.
  5. The positive phase constant moves the corresponding maximum from x = 0 to x = 0.50 m, so this cosine pattern shifts toward

    negative x .

Common sign error. In cos(kx + ϕ), a positive ϕ produces a negative spatial shift. Solving a constant-phase condition is safer than guessing from the sign.

Solution 10: Comparing two wavenumbers

  1. For profile A,
          2π   |-------|
λA =  2π-= -1.00m--.

    For profile B,

           2π    |-------|
λB =  -----= -0.80m--.
      5π∕2
  2. Profile B has the shorter wavelength.
  3. Since kB > kA, profile B accumulates phase more rapidly with increasing x.
  4. In L = 4.0 m,
           L--   4.0-   |-|
NA  =  λ  =  1.00  = -4 ,
        A

    and

           L--   4.0-   |-|
NB  =  λB =  0.80 = -5 .

Solution 11: Angular wavenumber versus spectroscopic wavenumber

First convert

                   −7
500nm  =  5.00 × 10   m.

  1.                    |-----------------|
k = -----2π-----≈  |1.26 × 107 rad/m  .
    5.00 × 10− 7   ------------------
  2.                    |--------------|
^ν =  -----1------= |2.00 × 106 m −1|.
     5.00 × 10 −7   ----------------
  3. Since 1 m = 100 cm,
                      |---------------|
         6  − 1   |        4   − 1|
2.00 × 10  m   =  2.00-×-10--cm--- .
  4. |--------|
k-=--2π^ν-.

    Thus angular wavenumber and reciprocal wavelength differ by a factor of 2π.

Solution 12: Dimensional audit

A trigonometric argument must be dimensionless, with angular phase interpreted in radians.

  1. kx+ϕ is valid. The product kx is dimensionless and can be added to the dimensionless phase ϕ.
  2. k + x is not valid. An inverse length cannot be added to a length.
  3. x∕λ is dimensionless, so it can serve mathematically as a trigonometric argument. By itself, however, it measures a fraction of a cycle, not angular phase in radians.
  4. 2πx∕λ is dimensionless and directly represents angular phase in radians. It is equivalent to kx when k = 2π∕λ.
  5. is dimensionless. For the angular-wavenumber definition, = 2π, so it represents one complete angular cycle.

The expressions that automatically match the angular convention used in cos(kx + ϕ) are kx + ϕ, 2πx∕λ, and—when interpreted as a specific complete cycle—.

Solution 13: Inferring wavenumber from measured phase change

Given

      3π
Δ 𝜃 = -2-,    Δx  = 0.30 m,

  1.     Δ 𝜃    3π∕2    |---------|
k = ----=  -----=  5π-rad/m--.
    Δx     0.30
  2.      |----------|
k ≈  15.7-rad/m--.
  3.      2π    |------|
λ =  ---=  0.40-m-.
     5π
  4. Over 0.10 m,
                              |π-|
Δ𝜃 =  kΔx  = (5π)(0.10) = |--.
                          -2--

Solution 14: Challenge—reconstruct a spatial sinusoid

The measured amplitude is

A  = 6.0mm,

and consecutive maxima occur at 0.10 m and 0.70 m.

  1. Consecutive corresponding maxima are one wavelength apart:
                         |-------|
λ = 0.70 − (− 0.10) =-0.80m--.
  2.             |---------|
     2π     |5π       |
k =  ----=  |---rad/m |.
     0.80    -2---------
  3. At a cosine maximum,
    kx + ϕ = 2πn.

    Using the maximum at x = 0.10 m and choosing n = 0,

    ( 5π )
  ---  (− 0.10) + ϕ = 0.
   2

    Therefore

      π              |----π--|
− --+ ϕ =  0,    |ϕ = -- .
  4              ------4-|

    This lies in the requested interval 0 ϕ < 2π.

  4. A complete model is
    |------------------(----------)--|
|                    5π-    π-   |
|u(x) = 6.0mm   cos   2 x + 4   ,|
---------------------------------

    where x is measured in meters.

  5. At x = 0.10 m,
         5π-        π-   π-
𝜃 =  2 (0.10) + 4 =  2,

    so

    |--------------|
|u(0.10m ) = 0 .
---------------

    At x = 0.30 m,

         5π         π
𝜃 =  --(0.30) + --=  π,
     2          4

    so

    |---------------------|
|u(0.30m ) = − 6.0mm  .
-----------------------
  6. 1      1      |--------------|
--=  -------= -1.25cycles/m--.
λ    0.80m

Summary of skills practiced

After completing this set, you should be able to:

  • convert between wavelength and angular wavenumber;
  • interpret k as spatial phase advance per unit distance;
  • calculate phase changes from spatial separations and vice versa;
  • read wavelength and wavenumber from a spatial graph;
  • evaluate u(x) = A cos(kx + ϕ) at specified positions;
  • determine how a phase constant shifts a spatial profile;
  • distinguish angular wavenumber k from reciprocal wavelength 1∕λ;
  • check the dimensional consistency of spatial-phase expressions;
  • infer k, λ, and ϕ from measured spatial data.

WM06 next introduces a disturbance that moves through space. The fixed spatial phase kx + ϕ developed here will later combine with temporal phase ωt to form a traveling-wave phase.

References

[1]   A. P. French, Vibrations and Waves, M.I.T. Introductory Physics Series, W. W. Norton & Company, 1971.

[2]   Frank S. Crawford, Jr., Waves, Berkeley Physics Course, Volume 3, McGraw-Hill, 1968.

[3]   Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 1, OpenStax, 2016, Chapter 16, especially Section 16.2, “Mathematics of Waves.”

[4]   I. M. Mills, B. N. Taylor, and A. J. Thor, “Definitions of the Units Radian, Neper, Bel, and Decibel,” National Institute of Standards and Technology, 2001.

[5]   International Union of Pure and Applied Chemistry, “wavenumber,” Compendium of Chemical Terminology (the Gold Book), 5th ed., online version 5.0.0, 2025, doi:10.1351/goldbook.W06664.


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Physics Classification46.40.-f (Vibrations and mechanical waves )
 45.20.Dd (Newtonian mechanics)
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