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Forced Vibrating String: Example 2

For all i ≠ j ≠ 1, we get a trivial solution (0 = 0). The other cases to consider are for i = j = 1 and i = j ≠ 1.

(7) (ℓ)/(2) = (B_i + C_i)(ℓ)/(2) B_i = 1 − C_i ⎫⎬⎭for i = j = 1

(8) 0 = (B_i + C_i)(ℓ)/(2) B_i = − C_i ⎫⎬⎭for i = j ≠ 1

Now we look to the second initial condition

(9) (∂ν(x, 0))/(∂t) = 0 = ξ_i̇(t)φ_i(x)

and take the derivative of ξ_i(t) and divide by φ_i(x)

(10) ξ_i̇ = 0 = A_iω_icos(0) − B_iω_isin(0) = A_iω_i 0 = A_i

We now have A_i. B_i and C_i, however, are still in terms of each other. In order to resolve this, we look to the Generalized Equations of Motion:

(11) M_i(ξ_ï + ω²_iξ_i) = Ξ_i

Knowing that A_i = 0, we know that

(12) ξ_i = B_icos(ω_it) + C_i ξ_i̇ = − B_iω_isin(ω_it) ξ_ï = − B_iω²_icos(ω_it)

Plugging into the generalized equation of motion, we have

(13) (mℓ)/(2)[ − B_iω²_icos(ω_it) + ω²_i(B_icos(ω_it) + C_i)] = F₀1(t)sin((iπ)/(4))

Noting that the cos terms on the left cancel, we simplify the expression and solve for C_i

(14) (mℓ)/(2)[C_iω²_i] = F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠ C_i = (2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)