A string of length

PHYSICS · Subjective · Fermi University Physics · Oscillations, Waves & Acoustics · Coupled oscillators, normal modes and Fourier analysis · Coupled oscillators, normal modes and Fourier analysis

Modes, Packets and Media, Ch The continuum limit, Fourier series and the spectrum of a measured signal

Problem

A string of length LL, mass per unit length μ\mu and tension τ\tau is fixed at x=0x=0 and x=Lx=L. A point mass MM is attached at the midpoint x=L/2x=L/2; the string carries no mass of its own exactly at that point, so all the inertia there belongs to MM. Away from the midpoint the string obeys the usual equation for a uniform string, −τϕ′′=ω2μϕ-\tau\phi'' = \omega^2\mu\phi. At the midpoint the displacement is continuous, and Newton's second law applied to the point mass, pulled transversely by the tension on each side, gives the slope jump τ[ϕn′(L/2+)−ϕn′(L/2−)]=−Mωn2 ϕn(L/2)\tau\left[\phi_n'(L/2^+) - \phi_n'(L/2^-)\right] = -M\omega_n^2\,\phi_n(L/2) for any normal mode ϕn\phi_n of angular frequency ωn\omega_n. Fixed ends give ϕn(0)=ϕn(L)=0\phi_n(0)=\phi_n(L)=0. (a) Using two integrations by parts on each side of the midpoint, separately for modes ϕm\phi_m and ϕn\phi_n of different frequencies (ωm≠ωn\omega_m\ne\omega_n), show that both give the same expression for τ∫0Lϕm′ϕn′ dx\tau\int_0^L\phi_m'\phi_n'\,\mathrm dx once the slope-jump condition is used, apart from a factor of ωn2\omega_n^2 on one side and ωm2\omega_m^2 on the other. (b) Hence show that ∫0Lμ ϕmϕn dx+M ϕm(L/2)ϕn(L/2)=0.\int_0^L\mu\,\phi_m\phi_n\,\mathrm dx + M\,\phi_m(L/2)\phi_n(L/2) = 0.

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