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Forced Vibrating String: Example 2

Forced Vibrating String: Example 2

Consider a homogeneous string of mass per unit length, m, that is stretched a length, ℓ between two points. At time t = 0, the string is stationary with an initial deflection of ν(x, 0) = sin⎛⎝(πx)/(ℓ)⎞⎠. At t < 0, the string is not loaded, but at time t = 0, two forces are added to the string: a constant downward point force of magnitude F0 at x = ℓ ⁄ 4 and a distributed force F(x, t) = F0sin(4Ωt). Note that the two forces, F0, have equal magnitudes. Find ν(x, t).

Solution

In this example, we have two forces being applied to the string, and we will solve each force separately, beginning with the point force.

1A) Find Ξi for the point force:

The first step is to find Ξi, the generalized force. To do this, we look at the equation for the generalized force:
(1) Ξi = l⌠⌡0F(x, t)φi(x) dx
For a downward point force at x = ℓ ⁄ 4, we write this as
(2) Ξi  = l⌠⌡0 − F01(t)δ⎛⎝x − (ℓ)/(4)⎞⎠φi(x) dx      =  − F01(t)φi(ℓ ⁄ 4)      =  − F01(t)sin(iπ ⁄ 4)
Where 1(t) is the step function (0 for t < 0 and 1 for t ≥ 0).

1B) Using 1A), determine the general solution of ξi for the point force:

Since the point force does not vary with time, we write ξi as
(3) ξi = Aisin(ωit) + Bicos(ωit) + Ci

1C) Using the initial conditions, solve for the unknown coefficients for the point force:

Here, the initial conditions are: no initial velocity and an initial position, which is given by
(4) ν(x, 0) = sin⎛⎝(πx)/(ℓ)⎞⎠
Plugging this in and solving we get
(5) sin⎛⎝(πx)/(ℓ)⎞⎠  = ∞⎲⎳i = 1(Aisin(0) + Bicos(0) + Ci)φi(x)      = (Bi + Ci)φi(x)
(6) ℓ⌠⌡0sin⎛⎝(πx)/(ℓ)⎞⎠sin⎛⎝(jπx)/(ℓ)⎞⎠ dx = ℓ⌠⌡0(Bi + Ci)sin⎛⎝(iπx)/(ℓ)⎞⎠sin⎛⎝(jπx)/(ℓ)⎞⎠ dx
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