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Converted document Vibrating String: Example 2

Vibrating String: Example 1

For a uniform string attached between two walls, determine the transverse string deflection for an initial string deflection, an initial transverse velocity distribution, and an external force given by:
(1) ν(x, 0) = 0
(2) (∂ν(x, 0))/(∂t) = sin(2πx)/(ℓ)
(3) F1 = Fcos(Ωt)
Where Ω and F are constants. In addition, during the time examined by this analysis, there is an additional constant force F applied in the downward direction at the midpoint of the string. Make sure that you break up the equation into odd, even, and, if necessary, specific mode numbers to see how the coefficients are formulated. Remember that you are asked for ν(x, t).

Solution

From the problem statement, we are given equations for the 2 initial conditions and one of the 2 forcing functions. The other forcing function can be written as:
(4) F2 =  − F∂(x − l ⁄ 2)
It is negative because the force is downward, and ∂(x − l ⁄ 2) indicates that the force is applied at the midpoint.

A) Find Ξi:

The first step is to find Ξi, the generalized force. To do this, we look at the equation for the generalized force:
(5) Ξi = l⌠⌡0F(x, t)φi(x) dx
We then subsitute in our 2 known forces and the mode shape for vibrating strings, which is always sin((iπx)/(l)) (NOTE: only use this assumption for the vibrating string, on other problems you will have to determine the mode shape yourself!), then we solve:
(6) Ξi =  l⌠⌡0(F1 + F2)sin((iπx)/(l)) dx      =  Fcos(Ωt)l⌠⌡0sin((iπx)/(l)) dx − Fl⌠⌡0∂(x − l ⁄ 2)sin((iπx)/(l)) dx      =  (ℓ)/(iπ)Fcos(Ωt)[1 − cos(iπ)] − Fsin((iπ)/(2))
Here we note that cos(iπ) is going to be 0 for even values of i, and 2 for odd values of i. Also, sin((iπ)/(2)) is always 0 for even values of i and alternates from positive to negative for odd values of i. This can be written as:
(7) Ξi = F⎧⎨⎩ (2ℓ)/(iπ)cos(Ωt) − ( − 1)(i − 1) ⁄ 2  if i = odd     0  if i = even
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