= Solution
Put $M=m_1+m_2$ and let $\mathbf R=(m_1\mathbf r_1+m_2\mathbf r_2)/M$ be the <center of mass>. Applying <Newton's second law> separately to the two <point masses> gives
$$
m_1\ddot{\mathbf r}_1=\mathbf f,\qquad m_2\ddot{\mathbf r}_2=-\mathbf f,\qquad M\ddot{\mathbf R}=0.
$$
Thus the <center of mass> has constant <velocity>, $\mathbf R(t)=\mathbf R(0)+t\dot{\mathbf R}(0)$. For the relative <position> $\mathbf r=\mathbf r_1-\mathbf r_2$, subtraction gives
$$
\ddot{\mathbf r}=\left(\frac1{m_1}+\frac1{m_2}\right)\mathbf f,\qquad \boxed{\mu\ddot{\mathbf r}=\mathbf f},\qquad \mu=\frac{m_1m_2}{m_1+m_2}.
$$
Here $\mu$ is the <reduced mass>; it converts the relative motion into a one-body <equation of motion>.
For the given linear <force>, the relative <position> obeys <simple harmonic motion>, $\ddot{\mathbf r}+\Omega^2\mathbf r=0$, where $\Omega^2=k/\mu$. The initial rest condition gives
$$
\mathbf r(t)=\mathbf r(0)\cos(\Omega t),\qquad |\mathbf r(0)|=d.
$$
For $d>0$, collision first occurs when the relative <position> vanishes, namely when $\Omega t=\pi/2$. Therefore
$$
\boxed{t_{\rm collision}=\frac\pi2\sqrt{\frac{m_1m_2}{k(m_1+m_2)}}}.
$$
The separation $d$ cancels because the <frequency> of this <simple harmonic motion> is independent of its <amplitude>.
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