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Wave Mechanics: Deriving the 1D String Wave Equation from Newton's Second Law (Topic)

Wave Mechanics: Deriving the 1D String Wave Equation from Newton’s Second Law

WM13 introduced the derivative structure

|------2---|
utt =-c-uxx-
(1)

and showed that translating profiles such as F(x ct) satisfy it. That was a kinematic result: it described a mathematical relation obeyed by a shape-preserving traveling disturbance.

WM14 asks the dynamical question:

|------------------------------------------------------|
Why   should a real stretched string obey that equation?|
--------------------------------------------------------
(2)

The answer comes directly from Newton’s second law. A curved string has slightly different Tension directions at neighboring points. The difference between those tension directions produces a transverse force. That force is proportional to the local curvature of the string, and Newton’s second law then turns curvature into transverse acceleration. Under the ideal assumptions developed below, this leads to

|------------|
-μutt =-Tuxx--
(3)

or

|------------|
|utt = T-uxx.|
-------μ-----|
(4)

Comparison with the standard one-dimensional wave equation gives

|----∘-----|
|       T  |
|c =    -. |
--------μ--|
(5)

This derivation is standard in treatments of waves on stretched strings and is a canonical example of how a continuum partial differential equation emerges from Newtonian mechanics [1235].

1 The physical model

Consider a thin flexible string stretched primarily along the x direction. Let

u(x,t)
(6)

denote its transverse displacement from equilibrium.

The derivation uses several assumptions. They are not merely mathematical conveniences; they define the physical model.

  • The string is continuous and perfectly flexible, with negligible bending stiffness.
  • The string has uniform linear mass density μ.
  • The equilibrium tension magnitude T is approximately constant along the string.
  • motion is transverse; longitudinal motion is neglected to first order.
  • The slope is small:
    |  |
||∂u||
|∂x| ≪  1.
    (7)

  • Damping, gravity, and distributed external transverse forces are neglected.

The most important approximation is the small-slope condition. The displacement itself need not be zero, but neighboring pieces of the string must make only small angles with the equilibrium x direction. For a sinusoidal wave, the small-slope requirement is roughly controlled by the dimensionless quantity Ak.

These assumptions produce the linear string wave equation. If the slopes become large, the tension varies strongly, the string stretches significantly, or bending stiffness matters, additional nonlinear or higher order terms appear.

2 Choose a short string element

Select a small piece of string extending from x to x + Δx.

If μ is the mass per unit equilibrium length, then the mass of this element is approximately

|------------|
Δm---=-μ-Δx.--
(8)

The element is pulled by the rest of the string at both ends. Because an ideal string can sustain tension but not bending moment, the tension force at each end acts tangent to the local string direction.

PIC

Figure. A short string element between x and x + Δx. The neighboring string pulls tangentially on the two ends with approximately equal tension magnitude T, but the directions differ because the string is curved.

Let the local tangent angles be 𝜃L at the left end and 𝜃R at the right end.

3 Geometry connects tangent angle to spatial slope

At any point on the string, the slope of the tangent line is

tan 𝜃 = ∂u-.
        ∂x
(9)

This relation is geometric and exact for the graph u(x,t) at a fixed time.

The small-slope assumption gives

|𝜃 | ≪ 1,
(10)

so the standard small-angle approximations apply:

sin𝜃 ≃  tan 𝜃 ≃ 𝜃.
(11)

Therefore

|------------|
|sin𝜃 ≃  ∂u. |
---------∂x--|
(12)

This is the step that linearizes the force law.

A more exact relation would be

sin𝜃 =  ∘--ux---,
          1 + u2x
(13)

which reduces to sin 𝜃 ux when |ux|≪ 1.

4 Resolve the tension forces

The right-hand tension contributes a transverse component

+T sin 𝜃 ,
        R
(14)

while the left-hand tension contributes

− T sin 𝜃L.
(15)

The net transverse force on the element is therefore

Fu = T sin𝜃R −  T sin 𝜃L.
(16)

Using the small-slope approximation,

F  ≃  T [(u )  − (u )  ].
 u        x R     x L
(17)

In coordinate form,

Fu ≃  T [ux (x + Δx, t) − ux(x, t)].
(18)

The horizontal components are

Fx  = T cos𝜃R − T cos 𝜃L.
(19)

For small slopes,

cos𝜃 ≃  1,
(20)

so the horizontal components cancel to first order. This is consistent with the model assumption that the leading motion is transverse and that the tension magnitude can be treated as constant.

5 Curvature produces a transverse force

Rewrite the transverse force as

       [                        ]
         ux(x-+-Δx,-t) −-ux(x,t)
Fu ≃ T             Δx             Δx.
(21)

As the element becomes arbitrarily short,

ux (x + Δx, t) − ux(x, t)
----------------------- − → uxx (x,t).
          Δx
(22)

Thus

|---------------|
Fu-≃--T-uxxΔx.---
(23)

This equation contains the central physical idea:

|--------------------------------------------------|
curvature creates a transverse imbalance  of tension.
----------------------------------------------------
(24)

If the string is locally straight, uxx = 0, the tension directions balance in the transverse direction and there is no transverse net force from tension. If the string is curved, the two tension vectors do not cancel transversely.

PIC

Figure. For a concave-down string element, uxx < 0. The tension imbalance points downward, so the transverse acceleration is also negative. The signs of curvature and acceleration therefore agree in the linear string model.

6 Apply Newton’s second law

The transverse acceleration of the material element is

au =  utt.
(25)

The element mass is

Δm  = μ Δx.
(26)

Newton’s second law in the transverse direction is

Fu = (Δm  )au.
(27)

Substitute the force and mass expressions:

T uxxΔx  = μ Δx utt.
(28)

Cancel the nonzero element length Δx:

|------------|
μu   = T u  .|
---tt-----xx--
(29)

Finally divide by μ:

|------------|
|      T     |
|utt = --uxx.|
-------μ-----
(30)

This is the one-dimensional linear wave equation for an ideal stretched string.

PIC

Figure. The derivation chain from string geometry to Newton’s second law. A difference in local slope creates a transverse tension imbalance; in the continuum limit that slope difference becomes uxx.

7 Identify the wave speed

The standard one-dimensional wave equation has the form

utt = c2uxx.
(31)

The string equation is

utt = T-uxx.
      μ
(32)

Therefore

 2   T
c  = --
     μ
(33)

and

|----∘-----|
|       T  |
|c =    -. |
--------μ--|
(34)

The positive root is used for the speed magnitude. Direction is carried by the traveling-wave form F(x ct) or G(x + ct) rather than by assigning a negative value to the speed magnitude.

This result says:

  • increasing the tension makes waves travel faster;
  • increasing the mass per unit length makes waves travel slower.

More precisely,

---------
|    √ --|
-c ∝---T-|
(35)

when μ is fixed, and

|--------|
|     1  |
|c ∝ √---|
-------μ--
(36)

when T is fixed.

PIC

Figure. Wave speed on an ideal string scales with the square root of tension and with the inverse square root of linear mass density.

8 Dimensional check

A correct physical equation must have consistent units.

Tension has units of force:

          kg-m-
[T ] = N =   s2 .
(37)

Linear mass density has units

      kg
[μ] = --.
      m
(38)

Therefore

[T ]
 --
 μ = kg m ∕s2
--------
  kg∕m (39)
= m2
-s2. (40)

Taking the square root gives

[∘  --]
    T     m
    --  = --,
    μ      s
(41)

which is the correct dimension for speed.

The differential equation is also dimensionally consistent. If u is a displacement,

       m-
[utt] = s2
(42)

and

[T    ]   m2  1    m
 --uxx  = --2 --=  -2.
 μ         s  m    s
(43)

9 The equation expresses a local feedback law

The string equation can be read physically as

|------------------------------------|
-local acceleration-∝-local curvature.
(44)

A region that is locally concave upward has

uxx > 0,
(45)

so the transverse acceleration is upward. A region that is locally concave downward has

uxx < 0,
(46)

so the transverse acceleration is downward.

At an inflection point,

uxx = 0,
(47)

so the string has no transverse acceleration from the local tension imbalance at that instant, even though the displacement or velocity at that point may be nonzero.

This is more informative than viewing the wave equation as a purely symbolic relation between second derivatives. The equation is a local Newtonian law: curvature produces force, force produces acceleration, and the resulting motion changes the curvature at neighboring points.

10 Why the equation is linear

The equation

μu  =  Tu
  tt      xx
(48)

is linear in the field u. If u1 and u2 are solutions for the same constant T and μ, then any linear combination

au1 + bu2
(49)

is also a solution.

This mathematical linearity is the reason the superposition principle from WM09 works for the ideal string model.

The linearity comes from the modeling assumptions. In particular, the approximation

sin 𝜃 ≃ tan 𝜃 ≃ ux
(50)

replaces the exact geometric force relation by one that is linear in the slope. If the slope is not small, that simplification fails and nonlinear effects can appear.

11 Consistency with a translating disturbance

WM13 showed that a sufficiently smooth right-moving profile

u (x,t) = F (x − ct)
(51)

obeys

       2
utt = cuxx.
(52)

The mechanical derivation now says that the stretched string obeys

      T
utt = --uxx.
      μ
(53)

The two are consistent when

|--------|
|c2 = T-.|
------μ--|
(54)

Thus the traveling-wave speed is no longer just a parameter in a chosen function. The mechanical properties of the string determine it.

A sinusoidal wave

u(x,t) = A cos(kx − ωt + ϕ )
(55)

therefore satisfies

  2   T  2
ω   = --k ,
      μ
(56)

or

|--------|
-ω-=-ck.-|
(57)

For the ideal string this is a nondispersive relation: all sinusoidal components have the same phase speed ω∕k = c within the model.

12 The role of initial and boundary conditions

The wave equation does not by itself determine one unique motion. WM12 made that point explicit.

For a finite string we still need initial data such as

u(x,0) = f(x )
(58)

and

u (x,0) = g(x ),
 t
(59)

plus boundary conditions such as

u(0,t) = 0,    u(L, t) = 0
(60)

for a fixed-fixed string.

The local differential law

μutt = Tuxx
(61)

tells us how the interior evolves. The initial and boundary conditions select the particular physical solution.

13 What changes outside the ideal-string assumptions?

The derivation also shows where more complicated models come from.

If the tension varies with position, a more general linearized string equation has the structure

           ∂
μ(x )utt = ---(T (x)ux),
          ∂x
(62)

rather than simply Tuxx.

If bending stiffness is important, as in a beam or stiff wire, higher spatial derivatives appear. If damping is important, velocity-dependent terms appear. If slopes become large, geometric nonlinearities appear. If external forcing acts along the string, a forcing term appears on the right-hand side.

Thus

|-----------|
u  =  T-u   |
|tt   μ  xx |
-------------
(63)

should be understood as the governing equation of a specific idealized physical system, not as a universal equation for every one-dimensional object that can vibrate.

14 Worked example 1: compute wave speed

A string is under tension

T = 180 N
(64)

and has linear mass density

μ = 0.012 kg/m.
(65)

The wave speed is

c = ∘ ---
  T-
  μ (66)
= ∘ ------
  -180--
  0.012 m/s (67)
= √ ------
  15000 m/s (68)
122 m/s . (69)

15 Worked example 2: scaling with tension and density

Suppose an ideal string initially has speed c0.

If the tension is increased by a factor of four while μ stays fixed,

    ∘  ----
       4T0-
c =     μ  = 2c0.
(70)

So quadrupling tension doubles wave speed.

If instead the linear density is increased by a factor of four while T stays fixed,

    ∘ ----
c =   -T--=  c0.
      4μ0    2
(71)

So quadrupling linear density halves wave speed.

16 Worked example 3: find the required tension

A string has

μ = 0.020 kg/m
(72)

and must support waves at speed

c = 75 m/s.
(73)

From

     T
c2 = --,
     μ
(74)

we obtain

T = μc2 (75)
= (0.020)(75)2 N (76)
= 112.5 N . (77)

17 Worked example 4: connect the wave equation to k and ω

Consider

u(x,t) = 0.005m  cos(8x − 240t).
(78)

The Wavenumber and angular frequency are

k = 8 rad/m,      ω =  240 rad/s.
(79)

Hence

    ω-   240-
c =  k =  8   = 30 m/s.
(80)

If the string has

μ = 0.010 kg/m,
(81)

then the required tension is

T = μc2 (82)
= (0.010)(30)2 N (83)
= 9.0 N . (84)

The same conclusion follows from matching the second derivatives:

         2                2
utt = − ω u,     uxx = − k u.
(85)

Substitution into

μutt = Tuxx
(86)

gives

μω2 =  Tk2,
(87)

which is equivalent to

T-   (ω-)2
μ =    k   .
(88)

18 Common mistakes

  • Mistake: using displacement itself as the restoring-force variable. For an ideal stretched string, the local tension imbalance is controlled by curvature uxx, not directly by u.
  • Mistake: assuming ux is the wave speed. It is spatial slope.
  • Mistake: assuming ut is the propagation speed. For string displacement it is the transverse material velocity at fixed x.
  • Mistake: forgetting that the mass of the small element is μΔx.
  • Mistake: adding the two vertical tension components instead of taking their signed difference.
  • Mistake: using c = T∕μ. The correct speed is the square root c = ∘ -----
  T∕ μ.
  • Mistake: treating constant tension as exact for arbitrary large slopes. It is part of the linear ideal-string approximation.
  • Mistake: thinking the wave equation alone fixes the motion. Initial and boundary conditions are still required.

19 What WM14 adds to the wave-mechanics language

The earlier lessons built the mathematical structure of a wave. WM14 supplies the mechanical cause for that structure in an ideal string.

The chain is

|----------------------------------------------------------------------|
-curvature-−→--tension-imbalance-−-→--transverse-force-−→--acceleration.-|
(89)

Quantitatively,

|------------|
|μutt = Tuxx |
--------------
(90)

and therefore

                 |---------|
------------     |    ∘ ---|
|      2    |    |      T- |
utt-=-c-uxx-,    |c =   μ .|
                 -----------
(91)

This is the first point in the series where the one-dimensional wave equation has been obtained from a physical law rather than merely recognized as a relation satisfied by a chosen traveling waveform.

20 References

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]   William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1, OpenStax, 2016, Section 16.3, “Wave Speed on a Stretched String.”

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

[5]   Massachusetts Institute of Technology, 8.03SC Physics III: Vibrations and Waves - The Physics of Waves, Fall 2016, MIT OpenCourseWare, material on the one-dimensional wave equation and transverse waves on a string.


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Classification:
Physics Classification46.40.Cd (Mechanical wave propagation (including diffraction, scattering, and)
 46.40.-f (Vibrations and mechanical waves )
 02.30.Jr (Partial differential equations)
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