Discussion
If \(u(t)\) is a unit step function, the solution of the differential equation \(m\dfrac{d^{2}x}{dt^{2}} + kx = u(t)\) in Laplace domain is
Answer: A
The Laplace-domain solution is \(\dfrac{1}{s(ms^{2} + k)}\).
Take the Laplace transform with zero initial conditions. The second derivative transforms to \(s^{2}X(s)\) and the unit step to \(1/s\):
\[ \begin{aligned} ms^{2}X(s) + kX(s) &= \frac{1}{s} \\ X(s) &= \frac{1}{s(ms^{2} + k)} \end{aligned} \]
\(1/(ms^{2} + k)\) is the transfer function, which is the response to a unit impulse, not a step. \(s/(ms^{2} + k)\) and \(1/[s^{2}(ms^{2} + k)]\) correspond to the derivative of the impulse and to a ramp input respectively.
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