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Forced Vibrating String: Example 2

1D) Replace and simplify for ν(x, t) for the point force

Substituting in for all of the constants we get

(15) ν(x, t) = ⎡⎢⎣⎛⎜⎝1 − (2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)⎞⎟⎠cos(ω_it) + (2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)⎤⎥⎦φ_i(x) = ⎡⎢⎣cos(ω_it) + (2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)[1 − cos(ω_it)]⎤⎥⎦φ_i(x) ⎫⎬⎭for i = 1

(16) ν(x, t) = ⎡⎢⎣⎛⎜⎝ − (2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)⎞⎟⎠cos(ω_it) + (2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)⎤⎥⎦φ_i(x) = ⎡⎢⎣(2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)[1 − cos(ω_it)]⎤⎥⎦φ_i(x) ⎫⎬⎭for i ≠ 1

We then use the kronecker delta symbol to show that the cos(ω_it) term only exists for i = 1 and obtain the final solution for the displacement of the string as a result of the point force, and call it ν_pointforce(x, t)

(17) ν_pointforce(x, t) = ^∞⎲⎳_i = 1⎡⎢⎣δ_i1cos(ω_it) + (2F₀1(t)sin⎛⎝(iπ)/(4)⎞⎠)/(mℓω²_i)[1 − cos(ω_it)]⎤⎥⎦φ_i(x)

Now we need to determine the displacement of the string for the distributed force

2A) Find Ξ_i for the distributed force:

(18) Ξ_i = ^l⌠⌡₀F(x, t)φ_i(x) dx

(19) Ξ_i = ^l⌠⌡₀F₀sin(4Ωt)φ_i(x) dx = F₀sin(4Ωt)^l⌠⌡₀sin((iπx)/(ℓ)) dx = (F₀ℓsin(4Ωt))/(iπ)(1 − cos(iπ))

(20) Ξ_i = ⎧⎨⎩ (2F₀ℓsin(4Ωt))/(iπ) for i = odd 0 for i = even

2B) Using 2A), determine the general solution of ξ_i for the distributed force:

Since the forcing function varies as sin(4Ωt), we represent ξ_i as follows

(21) ξ_i = A_isin(ω_it) + B_icos(ω_it) + C_isin(4Ωt)

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1D) Replace and simplify for ν(x, t) for the point force

2A) Find Ξi for the distributed force:

2B) Using 2A), determine the general solution of ξi for the distributed force:

2A) Find Ξ_i for the distributed force:

2B) Using 2A), determine the general solution of ξ_i for the distributed force: