Solution
= Solution
Let $x=0.400\,\mathrm m$ be the compression and $m_t=0.550\,\mathrm{kg}$ the combined <mass>. Starting from rest at the top, it falls through $H+x$ before maximum compression. Hence
$$
\frac12kx^2=m_tg(H+x),\qquad\boxed{k=94.5\,\mathrm{N\,m^{-1}}.}
$$
The gravitational drop during compression must be included. Equating $kx$ to the weight would instead describe a static equilibrium, not the turning point after a fall.