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Forced Vibrating String: Example 2

2D) Replace and simplify for ν(x, t) for the distributed force

(29) ν(x, t) = ^∞⎲⎳_i = oddξ_iφ_i

We have two cases, one for i = 1 and one for i ≠ 1

(30) ν(x, t) = ⎧⎨⎩ ( − 4Ω)/(ω_i)⎛⎝(4F₀)/(iπm(ω²_i − 16Ω²))⎞⎠sin(ω_it) + cos(ω_it) + (4F₀)/(iπm(ω²_i − 16Ω²))sin(4Ωt) for i = 1 ( − 4Ω)/(ω_i)⎛⎝(4F₀)/(iπm(ω²_i − 16Ω²))⎞⎠sin(ω_it) + (4F₀)/(iπm(ω²_i − 16Ω²))sin(4Ωt) for i ≠ 1

Noticing that the cos(ω_it) term is the only difference between the two, we can use the kronecker delta symbol to show this, and say

(31) ν(x, t) = ^∞⎲⎳_i = odd⎛⎝(4F₀)/(iπm(ω²_i − 16Ω²))⎛⎝ − (4Ω)/(ω_i)sin(ω_it) + sin(4Ωt)⎞⎠ + δ_i1cos(ω_it)⎞⎠φ_i(x)

The final step is to combine the displacements from both forces to obtain the final solution:

(32) ^∞⎲⎳_i = 1⎡⎢⎣δ_i1cos(ω_it) + (2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)[1 − cos(ω_it)]⎤⎥⎦φ_i(x) + ^∞⎲⎳_i = odd⎛⎝(4F₀)/(iπm(ω²_i − 16Ω²))⎛⎝ − (4Ω)/(ω_i)sin(ω_it) + sin(4Ωt)⎞⎠ + δ_i1cos(ω_it)⎞⎠φ_i(x)

where φ_i(x) = (iπx)/(ℓ) and ω_i = (iπ)/(ℓ)√((T)/(m)).