Wave Mechanics Review: Numerical and Conceptual Review of the 1D String Wave Equation

WM14 derived the one-dimensional ideal-string wave equation from Newton’s second law,

(1)

or equivalently,

(2)

The corresponding wave speed is

(3)

WM14E1 concentrated on the mechanical derivation itself. WM14E2 has a different purpose: use the equation as a working physical model. The problems ask the reader to predict wave speeds, infer string properties from measurements, test candidate solutions, interpret local curvature and acceleration, evaluate the small-slope approximation, and decide when the ideal model is or is not appropriate.

The central physical reading remains

(4)

Because T∕μ > 0, the signs of u_xx and u_tt must agree in the ideal linear string model [1, 2, 3, 5].

How to use this review

All exercises are stated first. Complete worked solutions appear in Part II. The calculations are intentionally mixed with conceptual questions so that numerical fluency does not become detached from the underlying mechanics.

Useful relations include

(5)

and for a sinusoidal traveling wave,

(6)

with

(7)

For that wave to satisfy the ideal string equation,

(8)

Part I: Exercises

Exercise 1: Compare four strings

Four ideal strings have the following Tensions and linear mass densities:

1. Compute the wave speed on each string.
2. Rank the strings from fastest to slowest.
3. Which comparison best demonstrates that tension alone does not determine wave speed?
4. Which quantity actually controls the speed in this model?

Exercise 2: Infer string properties from time of flight

A pulse is detected by two sensors on the same stretched string.

The sensor positions and arrival times are

x1	= 1.20 m,	t1	= 0.020 s,	(9)
x2	= 4.80 m,	t2	= 0.065 s.	(10)

The string tension is 72 N.

1. Find the measured wave speed.
2. Infer the linear mass density μ.
3. What mass would a 2.00 m length of this string have?
4. State one reason why using two separated sensors is preferable to estimating the speed from a single snapshot.

Exercise 3: Square-root scaling

A reference string has wave speed

(11)

Use the scaling law

(12)

to predict the new speed for each modification.

1. The tension is multiplied by 9 while μ is unchanged.
2. The density is multiplied by 4 while T is unchanged.
3. The tension is multiplied by 9 and the density by 4.
4. Both tension and density are doubled.
5. Explain why doubling the tension does not double the wave speed.

Exercise 4: Design the required tension

A string has

(13)

1. What tension is required to obtain a wave speed of 90 m/s?
2. If the maximum safe tension is only 120 N, what is the maximum ideal-model wave speed?
3. By what percentage would the desired 145.8 N tension exceed the safe limit?
4. Name one physical reason why the ideal formula should not be extrapolated beyond the safe operating range of a real string.

Exercise 5: Connect material parameters to wave parameters

A string is held at

(14)

and has

(15)

A sinusoidal traveling wave on the string has frequency

(16)

Find:

1. the wave speed c,
2. the wavelength λ,
3. the Wavenumber k,
4. the angular frequency ω,
5. the numerical value of ω∕k, and
6. the numerical value of T∕μ.

Explain how parts (e) and (f) are related.

Exercise 6: Read the PDE locally

For an ideal string,

(17)

At three different events, the spatial curvature has the values

event A: uxx	= −0.60 m−1,	(18)
event B: uxx	= +0.20 m−1,	(19)
event C: uxx	= 0.	(20)

1. Compute T∕μ.
2. Find u_tt at each event.
3. State the direction of transverse acceleration at A and B.
4. At event C, does u_tt = 0 imply that the displacement is zero? Does it imply that the transverse velocity is zero?

Exercise 7: Infer the string from a measured sinusoid

A measured wave is well approximated by

(21)

with x in meters and t in seconds.

1. Identify k and ω.
2. Determine the propagation speed.
3. What value of T∕μ is required for this wave to satisfy the string wave equation?
4. If μ = 0.025 kg/m, find T.
5. Evaluate u, u_xx, and u_tt at x = 0, t = 0.
6. Verify numerically at that event that u_tt = (T∕μ)u_xx.

Exercise 8: Which functions satisfy the wave equation?

Consider the PDE

(22)

For each candidate below, determine whether it satisfies the PDE for all x and t.

1. u = sin(x − 20t)
2. u = sin(2x − 20t)
3. u = sin x cos(20t)
4. u = x² + 400t²

For every function that does satisfy the PDE, state whether it looks like a single traveling sinusoidal wave, a Standing Wave, or neither.

Exercise 9: Use linearity without re-solving the PDE

Two functions are

u1(x,t)	= 0.002 cos(5x − 150t),	(23)
u2(x,t)	= 0.001 sin(8x − 240t).	(24)

1. Find the propagation speed associated with each component.
2. Show that both components satisfy the same wave equation.
3. Without differentiating the sum directly, explain why

   (25)

   also satisfies that equation.

4. Evaluate the total displacement at x = 0, t = 0.
5. What assumption about the physical string makes this superposition argument valid?

Exercise 10: Test the small-slope approximation

The exact transverse geometric factor associated with a local slope u_x is

(26)

The linear string derivation replaces this by

(27)

1. For u_x = 0.10, compute the exact factor and the linear approximation.
2. Compute the percentage by which the linear value exceeds the exact value.
3. Repeat parts (a) and (b) for u_x = 0.80.
4. Which slope is more consistent with the ideal linear string model?
5. For a sinusoid u = A cos(kx − ωt), why is the dimensionless product Ak useful when judging the small-slope assumption?

Exercise 11: Compare theory with a simple experiment

A string has independently measured parameters

(28)

A pulse travels a measured distance of 2.00 m. Three timing trials give

(29)

1. Predict the ideal wave speed from T and μ.
2. Find the mean measured travel time.
3. Find the measured wave speed from the mean time.
4. Compute the percentage difference between measured and predicted speed, using the predicted value in the denominator.
5. Infer μ from the measured speed and the known tension.
6. Give two plausible experimental or modeling reasons for a small discrepancy.

Exercise 12: Diagnose conceptual statements

For each statement, decide whether it is correct in the ideal linear string model. If it is incorrect, rewrite it accurately.

1. Doubling the tension doubles the wave speed.
2. If u_xx = 0 at one event, then u_tt = 0 at that event.
3. If u_tt = 0 at one event, then the string must be at equilibrium there.
4. Two ideal strings with the same ratio T∕μ have the same wave speed.
5. In the linear model, changing the wave amplitude changes the propagation speed.
6. The PDE alone determines one unique motion of a finite string.

Exercise 13: Reconstruct the model from observed wave data

A 3.00 m uniform string has total mass

(30)

A traveling sinusoidal wave on the string has

(31)

Assume the ideal linear string model is valid.

1. Find μ.
2. Find c.
3. Infer the tension T.
4. Write the governing PDE in the form u_tt = Cu_xx and determine C.
5. Find k and ω.
6. Compute the maximum slope magnitude Ak.
7. Based on part (f), comment on whether the small-slope assumption appears reasonable.
8. Write one right-moving sinusoidal solution with zero phase constant.

Exercise 14: Decide when the ideal model should be modified

For each physical change below, identify which assumption of the WM14 model is being challenged and state qualitatively what kind of modification you would expect in a more complete model.

1. The string has appreciable bending stiffness.
2. The transverse motion experiences strong viscous damping.
3. The linear mass density varies significantly with position.
4. The slope is no longer small.
5. The tension changes substantially as the string stretches.
6. A distributed transverse force acts along the string.

The goal is not to derive each extended equation, but to identify why the simple constant-coefficient linear PDE can no longer be expected to be exact.

Part II: Complete Worked Solutions

Solution 1: Compare four strings

For every string,

(32)

For string A,

cA	=	(33)
	=	(34)
	= 60.0 m/s .	(35)

For string B,

cB	=	(36)
	=	(37)
	≈ 63.25 m/s .	(38)

For string C,

cC	=	(39)
	=	(40)
	≈ 83.67 m/s .	(41)

For string D,

cD	=	(42)
	=	(43)
	≈ 31.62 m/s .	(44)

Therefore

(45)

A useful comparison is A versus B. String B has the larger tension, but it also has the larger linear mass density, so the speed increase is much smaller than a tension-only argument would suggest. The controlling quantity is the ratio

(46)

not T alone.

Solution 2: Infer string properties from time of flight

The sensor separation is

Δx	= x2 − x1	(47)
	= 4.80 − 1.20	(48)
	= 3.60 m.	(49)

The travel-time difference is

Δt	= t2 − t1	(50)
	= 0.065 − 0.020	(51)
	= 0.045 s.	(52)

Thus

(53)

From

(54)

we obtain

(55)

Therefore

μ	=	(56)
	=	(57)
	= 0.01125 kg/m .	(58)

For a 2.00 m length,

(59)

Using two sensors turns the measurement into a direct time-of-flight experiment. It avoids having to infer velocity from a single frozen picture and reduces sensitivity to uncertainty in the launch time of the pulse.

Solution 3: Square-root scaling

The normalized scaling law is

(60)

1. If T∕T₀ = 9 and μ∕μ₀ = 1,

   (61)

   Therefore

   (62)

2. If T∕T₀ = 1 and μ∕μ₀ = 4,

   (63)

   so

   (64)

3. If T∕T₀ = 9 and μ∕μ₀ = 4,

   (65)

   Thus

   (66)

4. If both are doubled,

   (67)

   so

   (68)

5. Tension enters under a square root. Doubling T multiplies c by

   (69)

   not by 2.

Solution 4: Design the required tension

From

(70)

solve for tension:

(71)

For the desired 90 m/s,

T	= (0.018)(90)2	(72)
	= (0.018)(8100)	(73)
	= 145.8 N .	(74)

If T_max = 120 N,

cmax	=	(75)
	≈ 81.65 m/s .	(76)

The requested tension exceeds the safe limit by

× 100%

= 21.5% .

(77)

A real string can yield, stretch, change its tension-density relation, or fail mechanically. The ideal formula is a model inside a valid operating regime, not a license to extrapolate past material limits.

Solution 5: Connect material parameters to wave parameters

The speed is

c	=	(78)
	=	(79)
	≈ 63.25 m/s .	(80)

The wavelength is

λ	=	(81)
	=	(82)
	≈ 0.506 m .	(83)

The wavenumber is

k	=	(84)
	≈ 12.42 rad/m .	(85)

The angular frequency is

ω	= 2πf	(86)
	= 250π	(87)
	≈ 785.4 rad/s .	(88)

Then

(89)

Also,

(90)

The relation between these results is

(91)

The left side is c² from wave kinematics, while the right side is c² from string mechanics.

Solution 6: Read the PDE locally

The coefficient is

(92)

The PDE gives

(93)

At event A,

(94)

The acceleration is downward if positive u is upward.

At event B,

(95)

so the acceleration is upward.

At event C,

(96)

Zero instantaneous acceleration does not imply zero displacement, and it does not imply zero transverse velocity. It only says that the ideal-model transverse force balance is momentarily zero at that event.

Solution 7: Infer the string from a measured sinusoid

From

(97)

we identify

(98)

The speed is

(99)

Therefore

(100)

If μ = 0.025 kg/m,

(101)

For this sinusoid,

(102)

and

(103)

At x = 0, t = 0,

(104)

Thus

(105)

and

(106)

Finally,

(107)

which agrees with u_tt exactly.

Solution 8: Which functions satisfy the wave equation?

The PDE is

(108)

so the characteristic wave speed is

(109)

1. For

   (110)

   the ratio ω∕k = 20∕1 = 20, so it satisfies the PDE. It is a single right-moving sinusoidal wave.

2. For

   (111)

   the ratio is 20∕2 = 10, not 20. Equivalently,

   (112)

   These are not equal, so this function does not satisfy the PDE.

3. For

   (113)

   we have

   (114)

   Thus

   (115)

   It satisfies the PDE and has standing-wave form.

4. For

   (116)

   we have

   (117)

   Since

   (118)

   the function mathematically satisfies the PDE. It is neither a bounded traveling sinusoid nor a standing sinusoidal mode. This is a useful reminder that a PDE admits a broader family of mathematical solutions than the familiar harmonic waves.

Solution 9: Use linearity without re-solving the PDE

For u₁,

(119)

For u₂,

(120)

Thus both satisfy

(121)

The differential operator

(122)

is linear. Therefore, if it gives zero when acting on u₁ and zero when acting on u₂, it also gives zero when acting on any linear combination. Hence

(123)

satisfies the same PDE.

At x = 0, t = 0,

u1(0, 0)	= 0.002,	(124)
u2(0, 0)	= 0,	(125)

so

(126)

The superposition argument depends on the linearized ideal-string model. Large-slope geometric nonlinearities or other nonlinear material effects would generally destroy simple superposition.

Solution 10: Test the small-slope approximation

The exact geometric factor is

(127)

while the linear approximation is

(128)

For u_x = 0.10,

sexact	=	(129)
	≈ 0.09950 ,	(130)

while

(131)

The relative overestimate is

(132)

For u_x = 0.80,

sexact	=	(133)
	=	(134)
	≈ 0.6247 ,	(135)

while

(136)

The relative overestimate is

(137)

Therefore |u_x| = 0.10 is much more consistent with the small-slope approximation.

For

(138)

we have

(139)

so the maximum slope magnitude is

(140)

Thus the dimensionless product Ak is a direct measure of the largest slope and is a useful model-validity indicator.

Solution 11: Compare theory with a simple experiment

The ideal prediction is

cpred	=	(141)
	=	(142)
	≈ 94.87 m/s .	(143)

The mean measured travel time is

t	=	(144)
	≈ 0.02103 s .	(145)

The measured speed is therefore

cmeas	=	(146)
	≈ 95.09 m/s .	(147)

The percentage difference is

× 100%

≈ 0.23% .

(148)

Using the measured speed to infer density,

μinferred	=	(149)
	=	(150)
	≈ 0.00896 kg/m .	(151)

This is close to the independently measured 0.0090 kg/m.

Possible discrepancy sources include timing resolution, uncertainty in sensor spacing, imperfect knowledge of tension, slight nonuniformity in μ, damping, stiffness, or nonideal pulse detection. The very small discrepancy here is consistent with the ideal model being a good approximation for this experiment.

Solution 12: Diagnose conceptual statements

1. Incorrect. Doubling tension multiplies the speed by if μ is unchanged.
2. Correct. In the ideal model,

   (152)

   so u_xx = 0 implies u_tt = 0 at that event.

3. Incorrect. Zero instantaneous acceleration does not require zero displacement. A point can have nonzero displacement and zero acceleration in a general wave field if the local curvature is zero. It also need not have zero velocity.
4. Correct. Equal T∕μ means equal c² and therefore equal positive propagation speed c.
5. Incorrect within the linear ideal-string model. The speed depends on T and μ, not on wave amplitude. At sufficiently large amplitude or slope, however, the assumptions behind the linear model can fail.
6. Incorrect. The PDE specifies the local evolution law but does not by itself select one unique finite-string motion. Initial conditions and boundary conditions are also required.

Solution 13: Reconstruct the model from observed wave data

The linear mass density is

(153)

The wave speed is

(154)

The tension is

T	= μc2	(155)
	= (0.030)(60.0)2	(156)
	= 108 N .	(157)

The PDE coefficient is

(158)

Therefore

(159)

The wavenumber is

(160)

The angular frequency is

(161)

With

(162)

the maximum slope magnitude is

Ak	= (0.0020)(15.71)	(163)
	≈ 0.0314 .	(164)

Because this maximum slope is much smaller than 1, the small-slope assumption appears very reasonable.

One right-moving solution is

(165)

Solution 14: Decide when the ideal model should be modified

1. Appreciable bending stiffness violates the perfectly flexible string assumption. A more complete model generally includes higher-order spatial derivatives associated with bending resistance.
2. Strong viscous damping violates the no-damping assumption. A damping term involving a time derivative, often proportional to u_t, would be expected.
3. Significant variation of μ with position violates the uniform-density assumption. The governing equation becomes variable-coefficient rather than having one constant value of T∕μ everywhere.
4. Large slope invalidates the small-angle linearization. The exact geometric relation between tangent angle and slope introduces nonlinear terms, so the simple superposition principle generally fails.
5. If tension changes significantly with stretching, the constant-T approximation is no longer valid. Tension may become coupled to the local deformation, producing nonlinear dynamics.
6. A distributed transverse force violates the force-free interior assumption. The PDE would acquire a forcing term on the right-hand side or, equivalently, an added force-per-unit-length term in the Newton-law balance.

The common theme is that

(166)

is the governing equation of a specific idealized model. Its power comes from knowing both what it predicts and the assumptions under which those predictions are expected to be accurate.

Common review mistakes

• Mistake: treating T itself as the speed. The speed is .
• Mistake: forgetting to square the measured speed when solving for T or μ.
• Mistake: assuming zero curvature means zero displacement.
• Mistake: checking only c = fλ and forgetting that the mechanical model also requires c² = T∕μ.
• Mistake: assuming every mathematical solution of the PDE must look like one sinusoidal traveling wave.
• Mistake: using superposition outside the linear regime without checking the model assumptions.
• Mistake: treating agreement with theory as exact proof of the model. Experimental agreement supports a model over the tested regime; it does not make the approximations universally exact.

What WM14E2 reinforces

WM14E2 turns the derivation into a working set of diagnostic tools. The same compact equation

(167)

can be read in several complementary ways:

• locally, it relates curvature to acceleration;
• globally, it supports waves traveling at c = ;
• experimentally, it lets measured travel time determine string properties;
• spectrally, it requires ω = ck for sinusoidal solutions; and
• as a model, it is valid only under the assumptions used in the Newton-law derivation.

That combination of mechanics, calculus, numerical prediction, and model checking is the main payoff of the 1D string wave equation.

References

References