76 the wave equation we consider vibrations of a

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7.6 The Wave Equation We consider vibrations of a guitar string (or a similar elastic string). We assume that the motion of the string is transverse , so that it goes only up and down (and not sideways). Let ?(?, ?) denote the displacement of the string at a point ? and time ? , and our goal is to calculate ?(?, ?) . The motion of an element of the string (?, ? + Δ?) is governed by Newton’s second law of motion ?𝑎 = ?. (7.6.1) The acceleration 𝑎 = ? ?? (?, ?) . If 𝜌 denotes the density of the string, then the mass of the element is ? = 𝜌Δ? . (The string is assumed to be uniform , so that 𝜌 > 0 is a constant.) We also assume that the internal tension is the only force acting on this element and that the magnitude of the tension ? is constant throughout the string. Our final assumption is that the slope of the string ? ? (?, ?) is small, for all ? and ? . Observe that ? ? (?, ?) = tan 𝜃 , the slope of the tangent line to the function ? = ?(?) , with ? fixed. The vertical (acting) component of the force at the right end point of our element is ? sin 𝜃 ≈ ? tan 𝜃 = ?? ? (? + Δ?, ?), because for small angles 𝜃 , sin 𝜃 ≈ tan 𝜃 ≈ 𝜃 . At the left end point, the vertical compo- nent of the force is ?? ? (?, ?) , and the equation (7.6.1) becomes 𝜌Δ?? ?? (?, ?) = ?? ? (? + Δ?, ?) − ?? ? (?, ?). Divide both sides by 𝜌Δ? , and denote ?/𝜌 = ? 2 : ? ?? (?, ?) = ? 2 ? ? (? + Δ?, ?) − ? ? (?, ?) Δ? . Letting Δ? → 0 , we obtain the wave equation ? ?? (?, ?) = ? 2 ? ?? (?, ?).
284 Chapter 7. The Fourier Series and Boundary Value Problems - 6 ? ? > 9 ? 𝜃 ? ? ? + Δ? Forces acting on an element of a string. We consider now the vibrations of a string of length ? , which is fixed at the end points ? = 0 and ? = ? , with given initial displacement ?(?) and given initial velocity ?(?) : ? ?? = ? 2 ? ?? for 0 < ? < ? and ? > 0, ?(0, ?) = ?(?, ?) = 0 for ? > 0, ?(?, 0) = ?(?) for 0 < ? < ?, ? ? (?, 0) = ?(?) for 0 < ? < ?. Perform separation of variables, by setting ?(?, ?) = ?(?)?(?) and obtaining ?(?)? (?) = ? 2 ? (?)?(?), ? (?) ? 2 ?(?) = ? (?) ?(?) = −?. Using the boundary conditions gives ? + ?? = 0, ?(0) = ?(?) = 0, ? + ?? 2 ? = 0. Nontrivial solutions of the first problem occur at ? = ? ? = ? 2 𝜋 2 ? 2 , and they are ? ? (?) = sin ?𝜋 ? ? . The second equation then takes the form ? + ? 2 𝜋 2 ? 2 ? 2 ? = 0. Its general solution is ? ? (?) = 𝑏 ? cos ?𝜋? ? ? + ? ? sin ?𝜋? ? ?, where 𝑏 ? and ? ? are arbitrary constants. The function ?(?, ?) = ?=1 ? ? (?)? ? (?) = ?=1 (𝑏 ? cos ?𝜋? ? ? + ? ? sin ?𝜋? ? ?) sin ?𝜋 ? ?
7.6. The Wave Equation 285 satisfies the wave equation and the boundary conditions. It remains to satisfy the initial conditions. Compute ?(?, 0) = ?=1 𝑏 ? sin ?𝜋 ? ? = ?(?), which requires that 𝑏 ? = 2 ? ? 0 ?(?) sin ?𝜋 ? ? ??, (7.6.2) and ? ? (?, 0) = ?=1 ? ? ?𝜋? ? sin ?𝜋 ? ? = ?(?), which implies that ? ? = 2 ?𝜋? ? 0 ?(?) sin ?𝜋 ? ? ??. (7.6.3) Answer. ?(?, ?) = ?=1 (𝑏 ? cos ?𝜋? ? ? + ? ? sin ?𝜋? ? ?) sin ?𝜋 ? ?, (7.6.4) with the 𝑏 ? ’s computed by (7.6.2), and the ? ? ’s by (7.6.3). The last formula shows that the motion of the string is periodic in time, similarly to the harmonic motion of a spring (the period is 2𝜋? 𝜋? ). This is understandable, because we did not account for the dissipation of energy in our model of vibrating string. Example 7.6.1. Find the displacements ? = ?(?, ?) : ? ?? = 9? ?? for 0 < ? < 𝜋 and ? > 0, ?(0, ?) = ?(𝜋, ?) = 0 for ? > 0, ?(?, 0) = 2 sin ? for 0 < ? < 𝜋, ? ? (?, 0) = 0 for 0 < ? < 𝜋. Here ? = 3 and ? = 𝜋 . Because ?(?) = 0 , all ? ? = 0 , while the 𝑏 ? ’s are the Fourier sine series coefficients of 2 sin ? on the interval (0, 𝜋) , so that 𝑏 1 = 2 and all other 𝑏 ? = 0 . Using (7.6.4) gives the answer: ?(?, ?) = 2 cos 3? sin ? . To interpret this answer, we use a trigonometric identity to write ?(?, ?) = sin(? − 3?) + sin(? + 3?).

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