A steel string of length

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 steel string of length L=1.2 mL = 1.2\,\mathrm{m}, wave speed c=300 m s−1c = 300\,\mathrm{m\,s^{-1}}, is fixed at x=0x=0 and x=Lx=L. The string is undisturbed, y(x,0)=0y(x,0)=0 everywhere, when a hammer strikes it uniformly over 0≤x≤L/80 \le x \le L/8, imparting an initial transverse velocity ∂y/∂t(x,0)=v0=2.0 m s−1\partial y/\partial t(x,0) = v_0 = 2.0\,\mathrm{m\,s^{-1}} there and leaving the rest of the string at rest. Air resistance and internal damping are ignored, so the string obeys ∂2y/∂t2=c2 ∂2y/∂x2\partial^2y/\partial t^2 = c^2\,\partial^2y/\partial x^2. (a) Explain why the coefficients AnA_n of cos⁡ωnt\cos\omega_nt are all zero in y=∑nsin⁡(nπx/L)[Ancos⁡ωnt+Bnsin⁡ωnt]y=\sum_n\sin(n\pi x/L)[A_n\cos\omega_nt+B_n\sin\omega_nt]. (b) Derive a formula for BnB_n from the initial velocity profile, and evaluate B1B_1, B2B_2 and B3B_3. (c) State the fundamental frequency f1f_1 and confirm ω3=3ω1\omega_3=3\omega_1. (d) Using only the three terms found, evaluate the three-term partial sum for the transverse displacement at the midpoint x=L/2x=L/2 at t=2.00 mst = 2.00\,\mathrm{ms}. (e) Summed to convergence, the series gives 5.0×10−4 m5.0\times10^{-4}\,\mathrm{m} at that point and time. Explain, from how BnB_n depends on nn, why three terms recover less than half of this, and estimate the mode number beyond which the coefficients begin to fall off as 1/n21/n^2.

Browse the problem library