Hang a mass from a spring and it stretches, then bounces about its new resting point. Does gravity change the motion that The Equation of Simple Harmonic Motion described? Only where it is centered.
Why it works
Take $y$ downward from the spring's natural length. Newton's second law: $m\ddot{y} = mg - ky$. At equilibrium $\ddot{y} = 0$, so the stretch there is $d = mg/k$. Now measure from that point: $y = d + u$. Then $$m\ddot{u} = mg - k(d + u) = (mg - kd) - ku = -ku.$$ The constant force of gravity is exactly cancelled by the extra stretch $kd$, leaving the same equation as a horizontal spring.
Key idea
A vertical spring oscillates about its stretched equilibrium $d = mg/k$ with the same period as without gravity: $$T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{d}{g}}.$$ Gravity shifts the center of the motion; it doesn't change how stiff the motion is.
Worked example: Stretch and period
A $0.5\ \text{kg}$ mass hangs from a spring with $k = 20\ \text{N/m}$. Find the equilibrium stretch and the period.
- $d = \dfrac{mg}{k} = \dfrac{0.5(9.8)}{20} = 0.245\ \text{m}$.
- $T = 2\pi\sqrt{0.5/20} = 0.993\ \text{s}$, the same as on a horizontal track.
- Also $2\pi\sqrt{d/g} = 2\pi\sqrt{0.245/9.8} = 0.993\ \text{s}$: the stretch alone tells you the period.
Answer: $d = 0.245\ \text{m}$; $T = 0.993\ \text{s}$.
Worked example: Let go at the natural length
The same mass is hooked onto the unstretched spring and released from rest. How far does it fall, and what is the largest spring force?
- It starts at rest $d = 0.245\ \text{m}$ above equilibrium, so its amplitude is $A = d$.
- It turns around $d$ below equilibrium: a total drop of $2d = 0.49\ \text{m}$.
- There the stretch is $2d$, so the spring pulls with $k(2d) = 2mg = 9.8\ \text{N}$, twice the weight.
Answer: It drops $0.49\ \text{m}$; the spring force peaks at $9.8\ \text{N}$.
Energy tells the same story. Measured from the stretched equilibrium, the spring energy and the gravitational energy together add up to $\tfrac{1}{2}ku^2$ plus a constant (the linear terms in $u$ cancel because $kd = mg$). So the energy sloshes between $\tfrac{1}{2}mv^2$ and $\tfrac{1}{2}ku^2$, exactly as on a horizontal spring.
Common mistake
Thinking gravity makes a hanging spring oscillate faster or slower. Gravity is the same at every point of the motion, so it can't supply a restoring force; only the spring's extra stretch or squeeze does. Check: the period formula $2\pi\sqrt{m/k}$ has no $g$ in it. (A pendulum is different: there gravity is the restoring force, so $g$ appears.)