Wave Mechanics Examples: Deriving the 1D String Wave Equation

This companion article provides exercises for WM14, wave mechanics: Deriving the 1D String wave equation from Newton’s Second Law. All exercises are stated first so that they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

The central mechanical result of WM14 is

(1)

or, equivalently,

(2)

Comparison with the standard one-dimensional wave equation

(3)

gives the wave speed

(4)

The purpose of this exercise set is not merely to use the final speed formula. The problems repeatedly return to the physical chain

(5)

that makes the wave equation a consequence of Newtonian mechanics [1, 2, 3, 5].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. Keep the distinction between the following quantities explicit:

• displacement u,
• slope u_x,
• curvature u_xx,
• transverse material velocity u_t,
• transverse acceleration u_tt,
• Tension T, and
• linear mass density μ.

Also keep track of which steps are exact geometry and which steps depend on the small-slope approximation.

Part I: Exercises

Exercise 1: Identify the ideal-string assumptions

For each statement below, say whether it is part of the ideal linear string model used in WM14. If it is not part of the model, explain what kind of additional physics it would introduce.

1. The string is perfectly flexible and has negligible bending stiffness.
2. The linear mass density μ is uniform.
3. The tension magnitude T is approximately constant.
4. The string can have arbitrarily large slope without changing the derivation.
5. Damping is neglected.
6. Longitudinal motion is neglected to first order.
7. Gravity and distributed transverse forcing are neglected.

Why is the small-slope condition

(6)

especially important for obtaining a linear wave equation?

Exercise 2: Read the force diagram

The following figure shows a short curved string element.

Use the figure to answer:

1. What is the approximate mass of the element?
2. Why do the tension forces point away from the element at both ends?
3. Write the signed transverse component of the tension force at the left end.
4. Write the signed transverse component at the right end.
5. Write the net transverse force before making any small-angle approximation.
6. Explain why equal tension magnitudes do not imply zero net transverse force.

Exercise 3: From tangent angle to slope

At a point on the string,

(7)

1. Which part of this relation is geometry rather than approximation?
2. Under the small-slope assumption, write the approximation that replaces sin 𝜃 by u_x.
3. The exact relation can be written as

   (8)

   Evaluate the exact value and the linear approximation when u_x = 0.10.

4. Repeat part (c) for u_x = 0.80.
5. Which case better justifies the linear approximation, and why?

Exercise 4: Estimate the curvature force from neighboring slopes

At a particular instant, a short element of string has

(9)

The measured slopes at the two ends are

(10)

Using the linearized force expression,

(11)

find:

1. the net transverse force,
2. the sign of the transverse acceleration,
3. the finite-difference estimate of u_xx,
4. the same force using F_u ≃ Tu_xxΔx.

Exercise 5: Complete the Newton’s-law derivation

Starting with

(12)

and

(13)

carry out the following steps.

1. Multiply and divide the slope difference by Δx.
2. Take the limit as Δx → 0 to identify u_xx.
3. Write Newton’s second law in the transverse direction.
4. Substitute the force, mass, and acceleration expressions.
5. Cancel the common factor and obtain the governing PDE.
6. Identify c² and derive c.

Exercise 6: Dimensional consistency

Use

(14)

and

(15)

to show:

1. [T∕μ] = m²∕s²,
2. [] = m∕s,
3. [u_tt] = m∕s² when u is displacement,
4. [(T∕μ)u_xx] = m∕s².

Why would the formula c = T∕μ be dimensionally wrong?

Exercise 7: Compute the wave speed

A string has tension

(16)

and linear mass density

(17)

1. Compute the ideal wave speed.
2. If a disturbance travels 8.0 m, how long does it take to travel that distance?
3. If the tension is unchanged but the linear density is doubled, what is the new speed?

Exercise 8: Read the square-root scaling laws

The following figure summarizes the dependence of wave speed on tension and linear mass density.

Answer the following.

1. If tension is multiplied by 9 at fixed μ, by what factor does c change?
2. If tension is reduced to one-quarter of its original value, by what factor does c change?
3. If μ is multiplied by 9 at fixed T, by what factor does c change?
4. By what factor must the tension change to triple the wave speed at fixed μ?
5. By what factor must μ change to double the wave speed at fixed T?

Exercise 9: Find the required tension or density

1. A string with μ = 0.015 kg/m must support waves at c = 90 m/s. Find the required tension.
2. A string under tension T = 150 N supports waves at c = 100 m/s. Find its linear mass density.
3. A uniform string has total mass m = 0.060 kg and length L = 3.0 m. Find μ and then find the wave speed if T = 120 N.

Exercise 10: Check a sinusoidal wave against the string equation

A transverse displacement is

(18)

The string has linear mass density

(19)

1. Identify k and ω.
2. Find the propagation speed c = ω∕k.
3. Find the tension required for this wave to satisfy the ideal-string equation.
4. Compute u_xx and u_tt in terms of u.
5. Verify directly that μu_tt = Tu_xx.

Exercise 11: Interpret curvature and acceleration

The next figure shows three local string geometries.

For each panel, state the signs of u_xx and u_tt. Then answer:

1. Can a point with u_xx = 0 still have nonzero displacement?
2. Can a point with u_xx = 0 still have nonzero transverse velocity?
3. Does u_xx = 0 imply that the entire string is straight?
4. Why does the sign of acceleration match the sign of curvature in the ideal model?

Exercise 12: Linearity and superposition

Suppose u₁(x,t) and u₂(x,t) both satisfy

(20)

for the same constant T and μ.

1. Show by direct substitution that

   (21)

   also satisfies the equation for arbitrary constants a and b.

2. Explain how this result connects WM14 back to the superposition principle from WM09.
3. Explain why the same conclusion need not hold if the exact nonlinear force relation is retained instead of the small-slope approximation.

Exercise 13: Diagnose physical and mathematical mistakes

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

1. “The transverse restoring force is proportional directly to displacement u.”
2. “If a string segment is locally straight, the ideal tension force gives no transverse acceleration at that instant.”
3. “The wave speed is T∕μ.”
4. “A larger linear mass density makes waves propagate faster if tension is fixed.”
5. “The small-slope approximation is what makes the ideal string equation linear.”
6. “Once the PDE is known, no initial or boundary conditions are needed.”

Exercise 14: Synthesis - reconstruct the string wave equation from measurements

An experimenter measures a uniform string with

(22)

A sinusoidal disturbance on the string has wavelength

(23)

1. Compute the linear mass density μ.
2. Compute the wave speed from T and μ.
3. Compute the Wavenumber k.
4. Use ω = ck to compute the angular frequency.
5. Compute the ordinary frequency f.
6. Write the governing string PDE with the numerical value of T∕μ.
7. Write one right-moving sinusoidal solution with amplitude A = 3.0 mm and zero phase constant.
8. Explain in one sentence why satisfying the PDE does not by itself guarantee that this sinusoid is the actual motion of a finite string.

Part II: Complete Worked Solutions

Solution 1: Identify the ideal-string assumptions

1. Yes. Perfect flexibility and negligible bending stiffness are part of the model. If bending stiffness matters, higher spatial derivatives enter the governing equation.
2. Yes. WM14 assumes uniform μ. A spatially varying density would produce a more general equation with μ(x).
3. Yes. The tension magnitude is treated as approximately constant. If T varies appreciably with position or deformation, the simple form Tu_xx must be replaced by a more general expression.
4. No. Arbitrarily large slope violates the approximation that linearizes the tension components. Large slopes introduce geometric nonlinearity.
5. Yes. Damping is neglected. Including damping would introduce a velocity-dependent term.
6. Yes. Longitudinal motion is neglected to first order so that the leading motion is transverse.
7. Yes. Gravity and distributed transverse forcing are omitted in the basic derivation. Including them would add forcing terms to the PDE.

The small-slope condition is crucial because

(24)

This replaces the exact nonlinear geometric relation by one that is linear in u_x. After differentiating once more in space, the force becomes proportional to u_xx, which yields a linear PDE in u.

Solution 2: Read the force diagram

1. The approximate mass is

   (25)

2. Tension is a pulling force. The neighboring string pulls each end of the selected element away from the element along the local tangent direction.
3. The left transverse component is

   (26)

4. The right transverse component is

   (27)

5. Therefore the exact signed transverse force for equal tension magnitudes is

   (28)

6. Equal magnitudes do not imply cancellation because force is a vector. If the tangent directions are different, the vertical components differ. Curvature changes the directions even when the tension magnitudes are the same.

Solution 3: From tangent angle to slope

1. The exact geometric relation is

   (29)

2. For small slope,

   (30)

3. For u_x = 0.10,

   sin 𝜃	=	(31)
   	≈ 0.0995.	(32)

   The linear approximation gives

   (33)

   The error is very small.

4. For u_x = 0.80,

   sin 𝜃	=	(34)
   	=	(35)
   	≈ 0.625.	(36)

   The linear approximation would give 0.80, which is substantially different.

5. The u_x = 0.10 case is much better because |u_x|≪ 1, exactly the regime assumed by the linear model.

Solution 4: Estimate the curvature force from neighboring slopes

The data are

(37)

and

(38)

1. Fu	≃ T	(39)
   	= 60(0.010 − 0.040) N	(40)
   	= −1.8 N .	(41)

2. Because the net transverse force is negative, Newton’s second law gives negative transverse acceleration.
3. The finite-difference estimate is

   uxx	≃ m−1	(42)
   	= −1.5 m−1 .	(43)

4. Then

   Fu	≃ TuxxΔx	(44)
   	= 60(−1.5)(0.020) N	(45)
   	= −1.8 N ,	(46)

   which agrees with part (a).

Solution 5: Complete the Newton’s-law derivation

Start with

(47)

1. Insert Δx∕Δx:

   (48)

2. In the continuum limit,

   (49)

   Therefore

   (50)

3. Newton’s second law is

   (51)

4. Substitute

   (52)

   and the force expression:

   (53)

5. Cancel Δx:

   (54)

   Equivalently,

   (55)

6. Compare with

   (56)

   Thus

   (57)

   and

   (58)

Solution 6: Dimensional consistency

1. 	=	(59)
   	= .	(60)

2. Therefore

   (61)

3. If u has units of meters, then differentiating twice with respect to time gives

   (62)

4. Since

   (63)

   we have

   (64)

The expression T∕μ has units of speed squared, not speed. Therefore c = T∕μ would have the wrong dimensions. The square root is required.

Solution 7: Compute the wave speed

Given

(65)

1. c	= m/s	(66)
   	= m/s	(67)
   	≈ 110.7 m/s .	(68)

2. The travel time is

   t	=	(69)
   	= s	(70)
   	≈ 0.0723 s .	(71)

3. Doubling μ reduces speed by 1∕:

   (72)

Solution 8: Read the square-root scaling laws

From

(73)

at fixed μ and

(74)

at fixed T:

1. Multiplying tension by 9 multiplies speed by

   (75)

2. Reducing tension to T₀∕4 gives

   (76)

3. Multiplying μ by 9 gives

   (77)

4. To triple c,

   (78)

   Squaring gives

   (79)

5. To double c at fixed T,

   (80)

   Squaring gives

   (81)

   so

   (82)

Solution 9: Find the required tension or density

1. From

   (83)

   we obtain

   T	= (0.015)(90)2 N	(84)
   	= 121.5 N .	(85)

2. Rearranging gives

   (86)

   Therefore

   μ	= kg/m	(87)
   	= 0.015 kg/m .	(88)

3. The linear mass density is

   (89)

   Then

   c	= m/s	(90)
   	= m/s	(91)
   	≈ 77.5 m/s .	(92)

Solution 10: Check a sinusoidal wave against the string equation

The wave is

(93)

1. (94)

2. (95)

3. T	= μc2	(96)
   	= (0.025)(30)2 N	(97)
   	= 22.5 N .	(98)

4. For a cosine wave,

   (99)

   and

   (100)

5. The left-hand side is

   (101)

   The right-hand side is

   (102)

   Therefore

   (103)

   as required.

Solution 11: Interpret curvature and acceleration

For the ideal string,

(104)

Because T∕μ > 0, u_tt and u_xx always have the same sign.

The left panel has

(105)

The middle panel has

(106)

The right panel has

(107)

1. Yes. A point may have nonzero displacement while its local curvature is zero.
2. Yes. Zero instantaneous acceleration does not imply zero velocity.
3. No. It only says the string is locally straight to second order at that point. Other parts of the string may be curved.
4. The sign agreement follows directly from multiplication by the positive coefficient T∕μ.

Solution 12: Linearity and superposition

Suppose

(108)

and

(109)

Let

(110)

Then

(111)

and

(112)

Therefore

μutt	= aμ(u1)tt + bμ(u2)tt	(113)
	= aT(u1)xx + bT(u2)xx	(114)
	= Tuxx.	(115)

Thus

(116)

is also a solution.

This is the mathematical foundation for superposition in the ideal string model. It connects the physical derivation in WM14 directly to the superposition principle introduced in WM09.

If the exact nonlinear geometry is retained, the force no longer depends linearly on u_x and its derivatives. In that case the sum of two solutions need not be another solution.

Solution 13: Diagnose physical and mathematical mistakes

1. Incorrect. For the ideal stretched string, the local restoring force is controlled by curvature u_xx, not directly by displacement u.
2. Correct within the ideal model. If u_xx = 0, the local transverse tension imbalance vanishes at that instant.
3. Incorrect. The correct speed is

   (117)

4. Incorrect. At fixed tension, increasing μ makes the wave speed smaller:

   (118)

5. Correct. The small-slope approximation converts the geometric force law into one linear in the slope, leading to a linear PDE.
6. Incorrect. The PDE gives the local evolution law, but initial and boundary conditions are still needed to select the actual motion of a finite string.

Solution 14: Synthesis - reconstruct the string wave equation from measurements

The measured quantities are

(119)

and

(120)

1. The linear mass density is

   (121)

2. The wave speed is

   c	= m/s	(122)
   	= m/s	(123)
   	≈ 63.25 m/s .	(124)

3. (125)

4. ω	= ck	(126)
   	≈ (63.25)(5π) rad/s	(127)
   	≈ 993.5 rad/s .	(128)

5. (129)

   The same result follows from f = c∕λ.

6. (130)

   Therefore the governing PDE is

   (131)

7. With amplitude

   (132)

   one right-moving solution is

   (133)

8. Satisfying the PDE is not sufficient because the actual finite-string motion must also satisfy the prescribed initial conditions and boundary conditions.

Common mistakes

• Mistake: treating tension as a vertical force rather than a force tangent to the string.
• Mistake: adding the two transverse tension components without signs.
• Mistake: using u instead of u_xx as the quantity controlling the local restoring force.
• Mistake: forgetting that the small element mass is μΔx.
• Mistake: confusing slope u_x with curvature u_xx.
• Mistake: using c = T∕μ instead of c = .
• Mistake: assuming the linear wave equation remains exact for large slopes.
• Mistake: concluding that u_xx = 0 implies zero displacement or zero velocity.
• Mistake: forgetting that the PDE still requires initial and boundary data to determine a unique physical solution.

What WM14E1 reinforces

The central lesson is the mechanical origin of the string wave equation. For a short element,

(134)

while the transverse tension imbalance becomes

(135)

Newton’s second law then gives

(136)

and therefore

(137)

The equation should be read physically as

(138)

That interpretation is the bridge from elementary Newtonian mechanics to continuum wave dynamics.

References