I'm considering an at-rest simple harmonic oscillator (m,k) and want to model the force by a doublet (derivative of dirac delta) at t=0. $$f = \delta'(t)$$ I've already considered the case for a dirac delta, and I'm supposing that a Green's function method will suffice. As normal, $$G = \frac{1}{mw}\sin (w(t-t'))$$ For a dirac delta $f=\delta(t)$, the solution is clearly just $f=\frac{1}{mw}\sin(wt)$ by convolution. I believe this therefore imposes the condition that $\dot x(0^+) = 1/mw$, though I'm not entirely certain.
For doublet, applying integration by parts, I get that $f = \frac{1}{M}\cos(wt)$. However, this clearly cannot be true. In my mind, the idea of $\delta'(t)$ is a spike and a negative spike, meaning the behavior directly after must be $x=0$, since impulses cancel. However, this result clearly conflicts with that. I don't think I have the freedom to impose a coefficient of zero, so I don't know how to proceed.
To simplify the notation, let's set $m=1$. Then, let me show that the solution to $$ x''+w^2x=\delta'(t) \tag{1} $$ such that $x(t)=0$ for $t<0$ is $$ x(t)=\Theta(t)\cos(wt), \tag{2} $$ where $\Theta(\cdot)$ is the Heaviside step function. Indeed, differentiating $x(t)$ twice we find $$ x''(t)=\delta'(t)\cos(wt)-2w\delta(t)\sin(wt)-w^2\Theta(t)\cos(wt). \tag{3} $$ It follows from $(2)$ and $(3)$ that $$ x''+w^2x=\delta'(t)\cos(wt)-2w\delta(t)\sin(wt). \tag{4} $$ To show that the RHS of $(4)$ is equal to $\delta'(t)$, we notice that, for any test function $f\in\cal{D}(\mathbb{R})$, \begin{align} \int_{-\infty}^{\infty}[\delta'(t)\cos(wt)-2w\delta(t)\sin(wt)]f(t)\,dt &=-(\cos(wt)f(t))'-2w\sin(wt)f(t)\big|_{t=0} \\ &=-f'(0)=\int_{-\infty}^{\infty}\delta'(t)f(t)\,dt.\quad{\square} \tag{5} \end{align}
To reconcile this result with your intuition, I suggest that you solve the ODE $$ x''+w^2x=\frac{\delta(t+a)-\delta(t)}{a} $$ and take the limit $a\to 0$ of the solution $$ x(t)=\frac{1}{aw}[\Theta(t+a)\sin(w(t+a))-\Theta(t)\sin(wt)]. $$