The Vertical Spring

Hang a spring from the ceiling. Attach a mass. The spring stretches under gravity until the elastic force balances weight. This new resting position is the new equilibrium.

The key insight: SHM still happens about this new equilibrium, with the same period.

Static extension

At rest, spring force = weight:

$k\delta = mg \implies \delta = \frac{mg}{k}$

$\delta$ is the static extension — how much the spring stretches just from the weight of the mass before any oscillation begins.

SpringMass amplitude 0 k 100 m 2 orientation vertical
Snapshot t 0
Show spring block wall equilibrium
Hide graph-x energy readouts displacement-arrow

The dashed marker shows the natural length $L_0$. The block hangs at $L_0 + \delta$ — the new equilibrium.

Oscillation about the new equilibrium

Displace the block further down and release. It oscillates about the new equilibrium, not the natural length:

SpringMass amplitude 60 k 100 m 2 orientation vertical
Show spring block wall equilibrium displacement-arrow graph-x
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The x-t graph is a perfect cosine. The displacement $x$ is measured from the new equilibrium, not from the ceiling.

Why the period is unchanged

The equation of motion measured from the new equilibrium:

$m\ddot{x} = -k(x + \delta) + mg = -kx - k\delta + mg = -kx$

Since $k\delta = mg$, the gravity term cancels completely. The result is identical to the horizontal case:

$\ddot{x} = -\frac{k}{m}x \implies T = 2\pi\sqrt{\frac{m}{k}}$

Gravity shifts the equilibrium but does not affect the period.

Effect of mass on the static extension

Heavier mass → larger static extension $\delta = mg/k$:

Light mass ($m = 1$ kg):

SpringMass amplitude 0 k 100 m 1 orientation vertical
Snapshot t 0
Show spring block wall equilibrium
Hide graph-x energy readouts displacement-arrow

Heavy mass ($m = 3$ kg, same spring):

SpringMass amplitude 0 k 100 m 3 orientation vertical
Snapshot t 0
Show spring block wall equilibrium
Hide graph-x energy readouts displacement-arrow

Same spring. Heavier mass pulls the equilibrium further down. But once released from the same displacement, both oscillate with periods determined only by $m$ and $k$.

Energy in the vertical spring

The total mechanical energy is still $\frac{1}{2}kA^2$, where $A$ is measured from the new equilibrium. Gravitational PE is absorbed into the redefined potential energy — the mathematics is identical to the horizontal case.

SpringMass amplitude 60 k 100 m 2 orientation vertical
Show spring block wall equilibrium energy KE PE E
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JEE key results

Quantity	Formula
Static extension	$\delta = mg/k$
New equilibrium	$L_0 + \delta$ below ceiling
Period	$T = 2\pi\sqrt{m/k}$ (unchanged)
Amplitude	measured from new equilibrium
Total energy	$\frac{1}{2}kA^2$ (unchanged)

The vertical spring is pedagogically important precisely because gravity appears to complicate things — but cancels out perfectly, revealing the deep structure of SHM.