A uniform string of length under tension gives wave speed and is fixed at both ends,

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Oscillations, Waves & Acoustics · Superposition, boundaries and interference

Mathematical Methods for Physics, Ch Separation of variables in Cartesian, cylindrical and spherical geometry

Problem

A uniform string of length L=1.0 mL=1.0\,\mathrm{m} under tension gives wave speed c=250 m s−1c=250\,\mathrm{m\,s^{-1}} and is fixed at both ends, y(0,t)=y(L,t)=0y(0,t)=y(L,t)=0. A hammer strikes it so that at t=0t=0 the string has zero displacement everywhere but a uniform transverse velocity ∂y∂t∣t=0={2.0 m s−1,0.40 m≤x≤0.50 m0,elsewhere\left.\frac{\partial y}{\partial t}\right|_{t=0}=\begin{cases}2.0\,\mathrm{m\,s^{-1}}, & 0.40\,\mathrm m\le x\le0.50\,\mathrm m\\ 0, &\text{elsewhere}\end{cases} Its displacement obeys the wave equation ∂2y/∂t2=c2∂2y/∂x2\partial^2y/\partial t^2=c^2\partial^2y/\partial x^2. (a) Write the general fixed-fixed solution y(x,t)=∑nsin⁡(nπx/L)[Ancos⁡ωnt+Bnsin⁡ωnt]y(x,t)=\sum_n\sin(n\pi x/L)\big[A_n\cos\omega_nt+B_n\sin\omega_nt\big] with ωn=nπc/L\omega_n=n\pi c/L, and use the two initial conditions to show An=0A_n=0 for every nn and Bn=2Lωn∫0L∂y∂t∣t=0sin⁡(nπx/L) dx.B_n=\frac{2}{L\omega_n}\int_0^L\left.\frac{\partial y}{\partial t}\right|_{t=0}\sin(n\pi x/L)\,\mathrm dx. (b) Carry out the integral for the given strike, showing that Bn=2v0Ln2π2c[cos⁡(nπx1/L)−cos⁡(nπx2/L)],B_n=\frac{2v_0L}{n^2\pi^2c}\Big[\cos(n\pi x_1/L)-\cos(n\pi x_2/L)\Big], with v0=2.0 m s−1v_0=2.0\,\mathrm{m\,s^{-1}}, x1=0.40 mx_1=0.40\,\mathrm m, x2=0.50 mx_2=0.50\,\mathrm m. (c) Evaluate B1B_1 and B2B_2 in millimetres.

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