From Coordinates to Fields: the Lagrangian Density · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

From beads on a string to the field itself

Science · Forces & Motion · ages 22-24
Name ______________________   Date ____________

  1. True or false: to solve the wave equation on a finite string with fixed ends using d'Alembert's method, you can extend the initial shape outside the string as an odd, periodic function and then apply the infinite-string formula.

    Circle one:   True   False

  2. You pass from finitely many coordinates to a field. What replaces the index that used to label each coordinate?

    • The mass attached to each point
    • The time step of the simulation
    • The position in space itself

  3. d'Alembert's solution is u of x and t equals f of x minus c t plus g of x plus c t. What does this tell you about the motion?

    • Two shapes sliding opposite at speed c, added
    • Each point of the string moves in a circle with radius set by c
    • The shape stays put while its height grows and shrinks

  4. To solve the wave equation on a finite string with fixed ends, you may extend the initial shape outside the string as an odd, periodic function and then use the infinite string method.

    Circle one:   True   False

  5. A string is released from rest, so its motion is u(x, t) = (1/2)[f(x - ct) + f(x + ct)], where f is its initial shape. The wave speed is c = 3 and the initial shape is f(x) = x^2. Find u(1, 2).

    Answer: ______________

  6. A bump with shape f starts centered at 0 and slides as f of x minus c t with c equals 3. Where is the bump centered at t equals 2?

    • At x equals 6
    • At x equals 2
    • At x equals 0

  7. A string of length L is fixed at both ends. Why does the solution only use sin of n pi x over L with whole numbers n?

    • They are the only sine functions that equal zero at both x equals 0 and x equals L
    • They are the only sine functions that can be differentiated twice
    • Cosine functions are banned from every wave problem

  8. A vibrating string satisfies the wave equation, and d'Alembert's solution says u(x, t) = f(x - ct) + g(x + ct). What does this formula tell you about how the string actually moves?

    • The string's shape oscillates in place without anything traveling along it
    • Each point of the string moves in a circle with radius set by c
    • Two shapes sliding opposite at speed c, added
    • The motion is one shape that stays put while its height grows and shrinks

  9. A student models a vibrating string with a single ordinary differential equation in time only. What is wrong with this?

    • Nothing: fields obey ordinary equations
    • The string bends across space too, so the equation must be partial in space and time
    • Wave problems never admit equations at all

  10. A string starts from rest with shape f of x equals x squared and wave speed c equals 2. Its motion is u of x and t equals one half of f of x minus c t plus f of x plus c t. Type u of 2 and 3.

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

From beads on a string to the field itself W1-mt_fRZN3ZByjr-s1

  1. True · An odd extension about each end makes the extension antisymmetric at the endpoints, which forces the solution to stay zero there. Periodicity handles both ends at once, so the infinite-string d'Alembert formula then automatically respects the fixed-end conditions.
  2. The position in space itself · A field has a value at every point, so position takes over the labelling job.
  3. Two shapes sliding opposite at speed c, added · f of x minus c t slides one way, g of x plus c t slides the other, and they add.
  4. True · The odd periodic extension keeps the ends fixed while d'Alembert's formula does the work.
  5. 37 · With zero initial velocity, d'Alembert's solution splits the initial shape into two half-copies traveling opposite ways. Evaluating at x = 1, t = 2 gives (1/2)[f(-5) + f(7)] = (1/2)(25 + 49) = 37.
  6. At x equals 6 · The shape rides along at speed c, so after 2 units of time it has covered 6 units of distance.
  7. They are the only sine functions that equal zero at both x equals 0 and x equals L · Fixed ends demand zero displacement there, and exactly these sines vanish at both ends.
  8. Two shapes sliding opposite at speed c, added · d'Alembert's solution shows the wave equation describes traveling shapes: f(x - ct) is the shape f shifted to the right by ct, and g(x + ct) is the shape g shifted to the left.
  9. The string bends across space too, so the equation must be partial in space and time · A field varies from point to point, so space derivatives belong in its equation of motion.
  10. 40 · Sampling points are 2 minus 6 and 2 plus 6, giving one half of 16 plus 64.
Worksheet · LightMySky