The Loaded Spring: Time Period

A "LOADED SPRING" refers to a spring of force constant $k$ hung vertically from a fixed support, with a block of mass $m$ attached at its lower, free end, so that the block is free to oscillate up and down under the combined action of the spring force and gravity.\n\nWhen the block is first attached to the hanging spring, the spring stretches from its own natural (unloaded) length by some extension $e_0$, coming to rest at a NEW equilibrium position at which the upward spring force exactly balances the block's downward weight:\n\n$ke_0 = mg$\n\nSuppose the block is now pulled down (or pushed up) by some further distance $x$, measured from this new equilibrium position, and then released. At that instant, the total extension of the spring is $(e_0+x)$, so the NET force on the block, taking the upward direction as positive and measuring $x$ as positive downward from the new equilibrium, works out to be\n\n$F_{\text{net}} = mg - k(e_0+x) = (mg - ke_0) - kx$\n\nBut since $ke_0 = mg$ exactly, by the equilibrium condition established above, the bracketed term $(mg-ke_0)$ is exactly ZERO, leaving simply\n\n$F_{\text{net}} = -kx$\n\nThis is EXACTLY the same S.H.M. restoring-force law derived for the horizontal spring of Section 13.7. In other words, the constant weight of the block does nothing more than shift WHERE the mean position of the oscillation sits -- down by the fixed distance $e_0$ from the spring's own natural, unstretched length -- without changing the FORM of the restoring force about that new mean position in any way at all.\n\nApplying Newton's second law exactly as before, $m\,d^2x/dt^2 = -kx$, therefore gives EXACTLY the same angular frequency and time period as for the horizontal spring-mass system:\n\n$\omega = \sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}}$\n\nNotice a useful practical distinction from the simple pendulum of Section 13.10: the loaded spring's time period depends on the mass $m$ and the force constant $k$, but does NOT depend on $g$ at all -- gravity's only role is to decide WHERE the new equilibrium point lies (through $e_0=mg/k$), not how fast the oscillation abo …