Boundary conditions and standing waves

Start with two fixed ends

A supplied string has equilibrium length L = 2 m. Its ends stay at transverse displacement zero. At t = 0 s, the point labelled x = 0.5 m is 0.01 m above equilibrium and the point labelled x = 1.5 m is 0.01 m below it. Both points initially have zero transverse velocity. A chosen smooth profile between the ends will make this example exact.

Use one fixed inertial frame. The equilibrium label x increases to the right; y is signed transverse displacement, positive upward. A material point keeps its x label in this model while its y changes. An endpoint condition describes what an end must do throughout time. Initial data describe how the whole string starts.

Algebra, functions, graphs, and logarithms and Geometry, trigonometry, and coordinate systems prepare substitution, graphs and sine/cosine phases. Units, standards, and physical constants is a published return path for quantity notation. The needed definitions are introduced here, so first follow the points and exact table.

Simple harmonic motion is an optional return path for periodic motion. Derivatives, Taylor series, and local approximation and Partial differential equations and Green functions are optional return paths for the derivative checks. The wave rule is defined locally; another wave lesson is not required for a first reading.

Supply the string rule

Stipulate a uniform ideal string with tension T = 4/5 N and mass per equilibrium length ρ = 1/5 kg/m. It is flexible, has small transverse slopes and has no longitudinal motion, damping or external drive. The model is defined for 0 ≤ x ≤ 2 m and t ≥ 0 s. Its rule is y_tt = (T/ρ)y_xx; subscripts mean two derivatives with respect to the indicated quantity.

T/ρ = 4 m²/s², so the positive wave speed is c = √(T/ρ) = 2 m/s. The equation gives acceleration in m/s² on both sides: y_xx has units 1/m and c² has units m²/s². The tension is not a spring stiffness and ρ is not mass per volume.

These supplied quantities and equations define a teaching model. Small slope motivates its transverse approximation; it does not certify a measured error for a real string. We will check two smooth fields directly, rather than infer a unique field from the finite points in a drawing.

Fit a pattern to fixed ends

For fixed ends the boundary data are y(0,t) = y(L,t) = 0. Try a nonzero spatial pattern sin(kx), where k is the wave number in rad/m. The left value is already zero. The right value stays zero when sin(kL) = 0, so the positive choices are k_n = nπ/L with n = 1, 2, 3, … . The zero field is a separate trivial case.

Select n = 2. Then k = π/m and ω = ck = 2π/s. Supply the field y(x,t) = (0.01 m) sin((π/m)x) cos((2π/s)t). Each phase is dimensionless and measured numerically in radians. The angular frequency ω gives frequency f = ω/(2π) = 1 Hz and period P = 1 s.

Its stationary nodes are x = 0 m, 1 m and 2 m: their displacement and transverse velocity remain zero at every time. The selected antinodes x = 0.5 m and 1.5 m have displacement amplitude 0.01 m. The two lobes have opposite signs and their material motions are opposite in phase.

A single normal mode has one fixed spatial pattern multiplied by one oscillating time factor. The pattern changes sign in place; its nodes do not travel along the string. Other allowed indices describe other patterns, not a requirement that this release contain every one.

Choose how the pattern starts

The chosen cosine factor supplies the initial shape y(x,0) = (0.01 m) sin((π/m)x) and initial velocity y_t(x,0) = 0 m/s. Endpoint conditions alone do not specify either amplitude or this release. For example, replacing the time cosine by sin((2π/s)t) keeps both fixed-end conditions but starts with zero displacement and generally nonzero velocity.

Check the rule before substituting a special point: y_tt = −ω²y and y_xx = −k²y. Since ω = ck, y_tt − c²y_xx = 0 everywhere on the declared interval, including all nodes and instants when y is zero. This direct identity does not divide by a vanishing sine or cosine.

We have verified this supplied field and its data. We have not expanded an arbitrary initial pluck into infinitely many modes or proved a general completeness or existence theorem. Those are additional mathematical questions.

Separate a flat snapshot from rest

At t = 1/4 s, cos((2π/s)t) = 0, so every point has displacement zero: a spatial photograph is a straight profile. That statement alone says nothing about transverse velocity at that instant.

Differentiate first: y_t(x,t) = −(π/50 m/s) sin((π/m)x) sin((2π/s)t). At the same quarter-second instant, the left antinode has velocity −π/50 m/s and the right one has +π/50 m/s. They pass through equilibrium in opposite directions.

A stationary node has zero spatial sine coefficient at every time; a flat snapshot has a zero time cosine only at a selected instant. At x = 1 m both displacement and transverse velocity remain zero. Away from the nodes, a material point can cross equilibrium while moving.

Change only the right boundary

Keep the left end fixed, but use a different ideal right-end arrangement: a negligibly massive end can slide transversely without an applied transverse force. In this linear model its zero-force condition is T y_x(L,t) = 0. Because T > 0, this is y_x(L,t) = 0. The tension is still 4/5 N; zero transverse force does not mean zero tension.

For a nonzero sine pattern the right condition becomes cos(kL) = 0. Write the positive allowed choices as k_n = (2n+1)π/(2L), with n = 0, 1, 2, … . Select n = 0: k = π/(4 m) and ω = π/(2 s). The supplied rest-release field is y(x,t) = (0.01 m) sin((π/(4 m))x) cos((π/(2 s))t).

At x = L and t = 0 s this field has displacement +0.01 m and slope zero. Its horizontal tangent meets the force condition, while the end is not fixed at y = 0. Its selected period is 4 s. This ideal constrained end is not a prediction for an arbitrary real end that is suddenly released.

Compare the allowed selections

The speed c = 2 m/s is unchanged because the tension and linear density are unchanged. The permitted spatial pattern changes because the right-end constraint changes. Therefore the selected angular frequency changes too. The two rows use different explicit index conventions: fixed-fixed n = 2 and fixed/force-free n = 0.

Changing tension would be a different comparison. With the fixed-fixed pattern held at n = 2, c is proportional to √T at fixed density, and ω = ck. This predicts a frequency within the supplied ideal model; it is not a driven-resonance peak or an observed frequency of a real apparatus.

Compare profiles with material motion

The upper and middle panels are spatial snapshots of the two different boundary models at t = 0 s. The upper short-dashed curve also shows its complete zero profile at t = 1/4 s. The lower panel follows two fixed material labels from that upper model as actual time changes.

The table uses those same labels, x_L = 0.5 m and x_R = 1.5 m. Displacement and velocity occupy different columns with different units. Both histories reach displacement zero together at quarter-period times, while their velocities have opposite signs.

All three panels share the vertical displacement scale. The spatial panels and time panel use metres and seconds horizontally. Vertical magnification makes the spatial slopes look much larger than their physical values; the fixed model has |y_x| ≤ π/100, and the force-free model has |y_x| ≤ π/400. The drawing is not a true-angle picture.

Worked examples

Try it yourself

1. Keep the original length, tension and density but choose fixed-fixed n = 1. Find k, ω and P, and check the endpoint pattern.

   Show the answer and reasoning

   k = π/(2 m), ω = π/s and P = 2 s. The shape sin((π/(2 m))x) is zero at x = 0 m and 2 m. It is a different allowed fixed-end pattern, not the original n = 2 release.

   Use k_n = nπ/L with n = 1, then ω = ck and P = 2π/ω. The endpoint sine values must be checked for this changed index.

2. For the original n = 2 field at t = 1/8 s, compare x = 1 m with the two antinodes. Is the middle node only momentarily at equilibrium?

   Show the answer and reasoning

   At x = 1 m the spatial sine is zero, so y and y_t are zero throughout time. At this instant the antinode displacements are +√2/200 m and −√2/200 m, since cos(π/4) = √2/2. The middle point is a stationary node, not just a zero-time-factor snapshot.

   The node has a zero spatial sine coefficient, while the antinodes have coefficients +1 and −1. At t = 1/8 s the time cosine is √2/2; a zero spatial factor and a selected time factor have different meanings.

3. Find displacement and transverse velocity at x = 0.5 m, t = 1/4 s in the original fixed-fixed field. Explain the different conclusions.

   Show the answer and reasoning

   y = 0 m because cos(π/2) = 0; y_t = −π/50 m/s because the differentiated time factor is −sin(π/2). The point is crossing equilibrium downward, not at rest.

   The velocity is the time derivative of the field, so it contains a sine time factor. Substituting into displacement alone cannot decide whether the point is at rest.

4. With the same fixed-fixed endpoints, test y_trial = (0.01 m) cos((π/m)x) cos((2π/s)t) at both ends when t = 0 s.

   Show the answer and reasoning

   Both end values are +0.01 m because cos(0) = cos(2π) = 1. The trial does not satisfy y(0,t) = y(2 m,t) = 0. Satisfying the differential equation alone would not repair this boundary mismatch.

   A field must meet both the rule and the specified endpoint data. Here the nonzero end values already disqualify the trial from the chosen fixed-fixed boundary problem.

5. At x = 2 m, t = 0 s in the mixed-boundary model, give y, T and T y_x. Explain which quantity is constrained to zero.

   Show the answer and reasoning

   y = +0.01 m, T = 4/5 N and T y_x = 0 N. The zero quantity is the transverse boundary force; the model retains nonzero tension and permits nonzero end displacement.

   At the mixed right end the slope vanishes. Multiplying that dimensionless slope by retained tension gives a zero transverse force, while the displacement is unrestricted by a fixed-value condition.

6. Keep L = 2 m, ρ = 1/5 kg/m and fixed-fixed n = 2, but stipulate T = 16/5 N. Find the new c, ω and P. Is this a prediction of a driven resonance peak?

   Show the answer and reasoning

   T/ρ = 16 m²/s², so c = 4 m/s. The allowed k remains π/m, giving ω = 4π/s and P = 1/2 s. This is the selected free mode in a changed ideal model; no drive or resonance-response measurement has been supplied.

   The changed tension is four times the original value, so the positive speed doubles at fixed density. The spatial pattern is selected by the unchanged endpoints; ω changes through c, without any supplied drive.

What to carry forward

• Boundary data constrain the ends throughout time; initial displacement and velocity specify the release.
• A stationary node is different from a complete zero-profile snapshot; differentiate before deciding whether a material point is moving.
• Fixed displacement and zero transverse boundary force select different patterns. Zero transverse force can coexist with nonzero tension and nonzero end displacement.