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The crude-oil pumping rig shown is driven at 20 rpm. The inside diameter of the well pipe is 2 in., and the diameter of the pump rod is 0.75 in. The length of the pump rod and the length of the column of oil lifted during the stroke are essentially the same, and equal to $6000 \mathrm{ft}$ During the downward stroke, a valve at the lower end of the pump rod opens to let a quantity of oil into the well pipe, and the column of oil is then lifted to obtain a discharge into the connecting pipeline. Thus, the amount of oil pumped in a given time depends upon the stroke of the lower end of the pump rod. Knowing that the upper end of the rod at $D$ is essentially sinusoidal with a stroke of 45 in. and the specific weight of crude oil is $56.2 \mathrm{lb} / \mathrm{ft}^{3}$, determine (a) the output of the well in $\mathrm{ft}^{3} / \mathrm{min}$ if the shaft is rigid, $(b)$ the output of the well in $\mathrm{ft}^{3} / \mathrm{min}$ if the stiffness of the rod is $2210 \mathrm{N} / \mathrm{m}$, the equivalent mass of the oil and shaft is $290 \mathrm{kg}$, and damping is negligible.

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