ARCHETYPE: Mass Dropped on a Vertical Spring

Lower a bag of flour gently onto a kitchen scale and the pan settles one centimetre down. Now hold the same bag just touching the pan and simply let go — no drop height at all — and the pan dips two centimetres before bouncing back. Drop the bag from half a metre and the dip is eleven centimetres, with the spring underneath momentarily shoving back with eleven times the bag's weight. Nothing about the spring changed; only the history of the landing did. That factor of two — and the quadratic that governs real drops — makes this setup one of the most reliably asked energy problems in the world. CBSE and ISC pose it as a clean conservation exercise; JEE and NEET decorate it with maximum-speed and peak-force follow-ups; AP and A-Level papers dress it up as a design question ("what compression must the buffer allow?"). Every version funnels through the same three-snapshot energy audit you built in Block Meets Spring — with one vertical twist that harvests marks by the thousand: the mass falls $h + x$ , not $h$ , because it keeps descending while the spring squashes. This article drills the archetype to reflex: the master quadratic and its closed form, the equilibrium sag $mg/k$ versus the dynamic maximum, the energy triangle that shows where the kinetic energy peaks, the attached-versus-loose rebound audit, and the SHM lens that turns the whole problem into a circle you already know how to read.

Key takeaways

Drop a mass from height $h$ above a relaxed vertical spring: energy conservation reads $mg(h + x_{\max}) = \tfrac12 k x_{\max}^2$ — the fall distance is $h$ plus the compression itself.
The static (equilibrium) sag is $x_0 = mg/k$ . Released from rest at the spring's top ( $h = 0$ ), the mass overshoots to exactly $x_{\max} = 2x_0$ — twice the gentle-lowering result.
Closed form for any drop: $x_{\max} = x_0 + \sqrt{x_0^2 + 2hx_0}$ .
Maximum speed happens at $x_0$ , where spring force first balances weight — not at first contact, and not at the bottom.
Maximum force (and deceleration) happens at $x_{\max}$ , where the speed is zero: peak spring force $= kx_{\max}$ , peak acceleration $= g\,(x_{\max}/x_0 - 1)$ .
While in contact the motion is simple harmonic about $x_0$ with amplitude $A = \sqrt{x_0^2 + 2hx_0}$ — and $x_{\max} = x_0 + A$ is the same answer wearing SHM clothes.
Key idea: write the energy budget from release point to maximum squash, charging gravity for the full descent $h + x$ — the equilibrium sag $mg/k$ is a landmark on the way down, never the destination.

The core insight

A mass dropped onto a vertical spring converts gravitational potential energy from the entire descent $h + x_{\max}$ into elastic energy $\tfrac12 kx_{\max}^2$ , so the maximum compression solves a quadratic whose closed form $x_{\max} = x_0 + \sqrt{x_0^2 + 2hx_0}$ always overshoots the equilibrium sag $x_0 = mg/k$ — by exactly a factor of two when the drop height is zero.

Where this lives today

Every shipped smartphone survives this article's arithmetic before it ships. Package engineers run drop tests — the product falls from a standard height onto its cushioning, and the cushion is a spring whose $k$ they choose. The trade-off is exactly the $k \to \infty$ limit above: stiff packing squashes little but spikes the peak force ( $F_{\text{peak}} \approx \sqrt{2mghk}$ for deep drops — force grows with the square root of stiffness), while soft packing needs travel the box may not have. Fragility ratings are quoted in g's, and the designer's job is choosing $k$ and allowed $x_{\max}$ so that $g(x_{\max}/x_0 - 1)$ stays below the rating. The same triangle — drop energy, spring travel, peak force — sizes lift buffer springs at the bottom of elevator shafts, aircraft landing-gear oleo struts, and the crash mats under pole vaulters. Bungee operators solve the attached-mass variant for a living: the jumper must stop above the river, which is a statement about $x_0 + \sqrt{x_0^2 + 2hx_0}$ with no margin for the $mgh$ -only error.

The setup

Geometry first, because in an archetype the geometry is the physics. A light vertical spring of constant $k$ stands on the floor, relaxed, its free end at what we will call the origin. A mass $m$ is released from rest a height $h$ above that free end. It free-falls through $h$ , touches the spring at speed $v_c = \sqrt{2gh}$ , and keeps descending as the spring squashes beneath it. Measure the compression $x$ downward from the natural length — that keeps the spring energy formula $\tfrac12 kx^2$ legal with no adjustments. Three landmarks on the way down organize every question in the family:

First touch ( $x = 0$ ): full arrival speed $\sqrt{2gh}$ , spring force zero, weight unopposed — the mass is still gaining speed.

The equilibrium sag ( $x_0 = mg/k$ ): here $kx_0 = mg$ , the net force crosses zero, and the speed peaks. This is where the mass would sit if you lowered it gently — but a dropped mass sails straight through, because passing the balance point at maximum speed is the worst possible moment to stop.

Maximum compression ( $x_{\max}$ ): the speed is zero, the spring force $kx_{\max}$ is the largest of the whole encounter, and the net force points urgently upward. The mass halts here for an instant and is immediately flung back.

Figure

The archetype in three snapshots — release, the balance point, maximum squash

Compare this with the horizontal cousin you already own. In Block Meets Spring, gravity was a spectator: kinetic energy in, spring energy out, done. Here gravity keeps working during the compression — every centimetre of squash is another centimetre of descent, another $mg\,\mathrm{d}x$ of income for the kinetic account. That is the single structural difference, and it is exactly the term students drop: the innocent-looking budget $mgh = \tfrac12 kx^2$ is wrong for every drop, because the true fall distance is $h + x$ .

The master solution pattern

Every question in this family is the same energy audit between the release point and maximum compression. Both endpoints are at rest, so the budget has no kinetic entries at all — gravitational income on one side, elastic storage on the other:

mg\,(h + x_{\max}) = \tfrac12\,k\,x_{\max}^2 \qquad (1)

where: $m$ — the dropped mass; $h$ — drop height above the relaxed spring's top; $x_{\max}$ — maximum compression, measured from natural length; $k$ — spring constant. Read it physically: the left side is gravity's full invoice for a descent of $h + x_{\max}$ ; the right side is the only place that energy can hide when both endpoints are motionless. Rearranged, $(1)$ is a quadratic in $x_{\max}$ — the price of gravity working during the compression. Before solving it, meet the landmark that organizes the answer. If instead you lower the mass gently, your hand bleeds off energy and the mass settles where the forces balance:

x_0 = \frac{mg}{k} \qquad (2)

where: $x_0$ — the static or equilibrium sag. This is a force statement ( $kx_0 = mg$ ), not an energy one. It marks where a dropped mass moves fastest, and it sets the scale for everything else in the problem: compressions come out as multiples of $x_0$ . Solving the quadratic $(1)$ (derivation A) gives the closed form worth memorizing:

x_{\max} = x_0 + \sqrt{x_0^2 + 2hx_0} \qquad (3)

where: the first $x_0$ is the equilibrium sag and the square root is the overshoot beyond it — which the SHM lens will re-derive as the oscillation amplitude. Two readings matter. Set $h = 0$ (released just touching the spring): $x_{\max} = 2x_0$ — the famous factor of two. And for deep drops $h \gg x_0$ the root dominates: $x_{\max} \approx \sqrt{2hx_0} = \sqrt{2mgh/k}$ , which is just the horizontal-archetype result with arrival energy $mgh$ .

question asked	equation to run	answer
gently lowered — where does it settle?	$kx = mg$	$x_0 = mg/k$
released at the top ( $h=0$ ) — max squash?	$mgx = \tfrac12 kx^2$	$x_{\max} = 2x_0$
dropped from $h$ — max squash?	$mg(h+x) = \tfrac12 kx^2$	$x_{\max} = x_0 + \sqrt{x_0^2 + 2hx_0}$
where is the speed maximum?	$kx = mg$	at $x_0$ , value $v_{\max}^2 = 2gh + gx_0$
peak force on the mass / floor?	evaluate at $x_{\max}$	$F_{\text{peak}} = kx_{\max}$

Derivation AThe master quadratic and its closed form

We derive equation (3), the maximum compression for a drop from height $h$ , by energy conservation between two stationary snapshots — release and maximum squash — and then solve the resulting quadratic exactly. The strategy: measure compression $x$ downward from natural length so the elastic term is legally $\tfrac12 kx^2$ , charge gravity for the full descent $h + x$ , and keep the algebra in terms of the landmark $x_0 = mg/k$ so the answer exposes its own anatomy. Watch the discriminant: it is where the drop height enters, and setting $h = 0$ inside it is the fastest route to the factor of two.

Step 1. Energy budget between release (at rest, height $h + x_{\max}$ above the lowest point) and maximum compression (at rest, spring squashed $x_{\max}$ ):

mg(h + x_{\max}) = \tfrac12 k x_{\max}^2

Why both endpoints at rest is the trick: no kinetic term appears anywhere, so the messy speed at first contact never needs computing.

Step 2. Rearrange into standard quadratic form and divide by $k/2$ :

x_{\max}^2 - \frac{2mg}{k}x_{\max} - \frac{2mgh}{k} = 0 \;\;\Rightarrow\;\; x_{\max}^2 - 2x_0\,x_{\max} - 2hx_0 = 0

Why the substitution $x_0 = mg/k$ : it makes every term a length, so dimensional slips become visible instantly.

Step 3. Quadratic formula, keeping the positive root (a compression must be positive):

x_{\max} = x_0 \pm \sqrt{x_0^2 + 2hx_0} \;\;\Rightarrow\;\; x_{\max} = x_0 + \sqrt{x_0^2 + 2hx_0}

Why the negative root is discarded: $\sqrt{x_0^2 + 2hx_0} \ge x_0$ , so the minus sign would give $x_{\max} \le 0$ — the mathematics of an extension the spring never suffers on the way down. (It is, in fact, the highest point of the attached-mass oscillation — see the variants.)

Step 4. Special cases. $h = 0$ : $x_{\max} = x_0 + x_0 = 2x_0 = 2mg/k$ . Deep drop $h \gg x_0$ : $x_{\max} \to \sqrt{2hx_0} = \sqrt{2mgh/k}$ . Verified: numbers from the UG example bank — $m = 2.0$ kg, $h = 0.50$ m, $k = 1960$ N/m, $g = 9.8$ : $x_0 = 1.0$ cm, and $x_{\max} = 0.01 + \sqrt{0.0001 + 0.01} = 0.01 + 0.1005 = 0.111$ m — eleven times the static sag, with peak spring force $kx_{\max} = 217\ \mathrm{N} \approx 11mg$ . Both endpoints check against the budget: $mg(h+x) = 19.6 \times 0.611 = 11.98$ J and $\tfrac12 kx^2 = 980 \times 0.0122 = 11.98$ J.

Go deeper · Derivation B · 1 minWhere the speed peaks — and the SHM reading of the whole flight

Derivation BWhere the speed peaks — and the SHM reading of the whole flight

We derive the location and value of the maximum speed during compression, then show the entire contact phase is simple harmonic motion about $x_0$ — turning equation (3) into "equilibrium plus amplitude" and giving a free cross-check. The strategy: write the kinetic energy as a function of compression $x$ from the running energy budget; it is a downward-opening quadratic in $x$ , so its maximum sits at the vertex — pure algebra, no calculus needed. Watch for the vertex landing exactly on the balance point $x_0$ : that is not a coincidence but a force fact wearing energy clothing.

Step 1. Energy budget from release to a general compression $x$ (speed $v$ there):

\tfrac12 mv^2 = mg(h + x) - \tfrac12 kx^2

Step 2. The right side is a quadratic in $x$ with negative leading coefficient — a downward parabola. Its vertex (maximum) sits at

x = \frac{mg}{k} = x_0

Why this had to happen: speed grows while the net force still points down ( $kx < mg$ ) and shrinks once it points up ( $kx > mg$ ); the changeover — the balance point — is therefore the speed peak. Energy and force agree.

Step 3. Evaluate the budget at $x_0$ to get the peak speed:

\tfrac12 mv_{\max}^2 = mg(h + x_0) - \tfrac12 kx_0^2 = mgh + \tfrac12 mgx_0 \;\;\Rightarrow\;\; v_{\max}^2 = 2gh + gx_0

Why the tidy form: $\tfrac12 kx_0^2 = \tfrac12 mgx_0$ because $kx_0 = mg$ — half the gravitational income over the last stretch is banked in the spring, half stays kinetic.

Step 4. The SHM lens. While in contact, the net force is $mg - kx = -k(x - x_0)$ : a restoring force proportional to displacement from $x_0$ . The motion is SHM about $x_0$ with $\omega = \sqrt{k/m}$ , and the amplitude follows from the speed at the centre: $A = v_{\max}/\omega = \sqrt{(2gh + gx_0)\,m/k} = \sqrt{x_0^2 + 2hx_0}$ .

Step 5. Consistency: the lowest point of an SHM is centre plus amplitude — $x_{\max} = x_0 + A = x_0 + \sqrt{x_0^2 + 2hx_0}$ , exactly equation (3) re-derived without touching the quadratic. Verified: for the anchor numbers ( $x_0 = 1.0$ cm, $h = 50$ cm): $A = \sqrt{1 + 100}\ \mathrm{cm} = 10.05$ cm and $x_0 + A = 11.1$ cm ✓; peak speed $v_{\max} = \sqrt{2(9.8)(0.5) + 9.8(0.01)} = \sqrt{9.898} = 3.15$ m/s, barely above the contact speed $\sqrt{2gh} = 3.13$ m/s — the balance point sits just 1 cm into the squash.

The two derivations draw the same picture from different sides, and one graph holds it all — the energy triangle between gravity's straight line and the spring's parabola:

Figure

The energy triangle — gravity's straight line vs the spring's parabola

For an $h = 0$ release, gravity's cumulative income grows as the straight line $mgx$ while the spring's storage grows as the parabola $\tfrac12 kx^2$ . The parabola starts flatter (near $x = 0$ the spring is feeble), so the line leads and the vertical gap between them is the kinetic energy — widest at $x_0$ , where the slopes match ( $mg = kx$ ), and closed again where the curves recross at $x = 2x_0$ . The whole factor-of-two theorem is the elementary fact that a parabola through the origin needs twice the run to catch a straight line whose slope it matches halfway.

The variants examiners actually pose

Dropped versus lowered (the factor-two interrogation). "A student lowers a block gently onto a spring, then repeats the experiment by releasing it suddenly from the same position. Compare the compressions." Answer: $x_0$ versus $2x_0$ . The gentle hand does negative work, draining the energy that would otherwise carry the mass past the balance point.

Find $k$ (or $h$ , or $m$ ) from a measured squash. Equation (1) rearranged — the design version beloved of A-Level and AP papers. Watch the trap: the given "fall height" is usually to the top of the spring, so the energy line must still read $mg(h + x)$ .

Maximum speed, and where. Balance point $x_0$ , value $v_{\max}^2 = 2gh + gx_0$ (derivation B). The distractor "at first contact" ignores that the net force is still downward for the first $x_0$ of squash.

Peak force and g-force. At the bottom: spring force $kx_{\max}$ , mass's acceleration $a_{\max} = g\,(x_{\max}/x_0 - 1)$ upward, floor pushed down with $kx_{\max}$ . For an $h = 0$ release, $x_{\max} = 2x_0$ gives $a_{\max} = g$ exactly — a lovely zero-cost checkpoint. Deep drops multiply this brutally: the anchor drop ( $x_{\max} = 11.1x_0$ ) decelerates at about $10g$ .

Attached mass — how high does it rebound? If the mass is fastened to the spring, the SHM survives the upswing: it rises to $A - x_0$ above natural length (the discarded root of the quadratic, now physical) and oscillates forever between the two roots with period $2\pi\sqrt{m/k}$ . If it is loose, the spring can push but not pull: the mass separates at natural length with the arrival speed $\sqrt{2gh}$ (energy audit of the return leg) and rises exactly back to its drop height.

Spring on the moon / doubled mass / halved $k$ . Scaling interrogations: $x_0 \propto mg/k$ , and for deep drops $x_{\max} \propto \sqrt{mgh/k}$ — halving $k$ multiplies the deep-drop squash by $\sqrt2$ , not 2.

Limiting cases

case	result	meaning
$h = 0$	$x_{\max} = 2x_0$	released at the top: overshoot doubles the static sag
$h \gg x_0$	$x_{\max} \approx \sqrt{2mgh/k}$	deep drop: arrival KE dominates, gravity-during-squash negligible
$k \to \infty$	$x_{\max} \to 0$ , $F_{\text{peak}} = kx_{\max} \to \sqrt{2mghk} \to \infty$	rigid floor: tiny squash, savage force — why hard landings break things
$k \to 0$	$x_{\max} \to \infty$	pillow-soft spring: enormous travel, gentle force
$g \to 0$	$x_0 \to 0$ , $x_{\max} \to 0$	nothing weighs, nothing squashes
$m$ doubled, $h = 0$	$x_{\max} = 2x_0$ doubles	the factor two is universal; the scale is set by $mg/k$

Common trapThe mass falls h, so mgh equals half k x squared

The single most harvested error in the archetype. The mass does not stop falling when it touches the spring — it descends a further $x_{\max}$ while squashing it, and gravity works the whole way. The correct budget is $mg(h + x_{\max}) = \tfrac12 kx_{\max}^2$ : a quadratic, not a one-liner. The corrupted version $mgh = \tfrac12 kx^2$ silently sets the extra income to zero and underestimates the squash — catastrophically so for shallow drops, where $x_{\max}$ is comparable to $h$ ; check it with $h = 0$ , where the wrong equation predicts no compression at all for a mass visibly bouncing on a spring. Examiners bait the trap by quoting $h$ "above the top of the spring" and offering the $\sqrt{2mgh/k}$ answer among the options; it is only acceptable as an approximation when the problem says $h \gg x_{\max}$ , and even then you should name the approximation.

Worked examples

1 · Board level — dropped versus lowered (CBSE/ISC pattern, 3 marks)

A $0.50\ \mathrm{kg}$ block is (a) lowered gently onto a vertical spring of constant $k = 98\ \mathrm{N/m}$ , and (b) placed just touching the relaxed spring and released from rest. Find the compression in each case. Take $g = 9.8\ \mathrm{m/s^2}$ .

Step 1. Case (a) is a force balance — the hand removes the surplus energy. $kx_0 = mg \Rightarrow x_0 = \dfrac{0.50 \times 9.8}{98} = \boxed{5.0\ \mathrm{cm}}$ .

Step 2. Case (b) is an energy balance between two rest states, with $h = 0$ : $mgx = \tfrac12 kx^2$ .

Step 3. Cancel one power of $x$ (the $x = 0$ root is the release instant): $x_{\max} = \dfrac{2mg}{k} = 2x_0 = \boxed{10.0\ \mathrm{cm}}$ .

Step 4. State the physics for the last mark: released suddenly, the block passes the balance point at maximum speed and overshoots to exactly double the static sag; at $10$ cm the spring pushes with $kx_{\max} = 9.8\ \mathrm{N} = 2mg$ , giving deceleration $g$ at the bottom.

2 · Board level — design the buffer (A-Level/ISC pattern)

A $5.0\ \mathrm{kg}$ crate is dropped from $0.30\ \mathrm{m}$ above a vertical buffer spring and must not compress it beyond $0.20\ \mathrm{m}$ . What minimum spring constant is required, and what peak force does the floor then feel? ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Energy budget with the full descent — $h + x = 0.30 + 0.20 = 0.50$ m: $mg(h+x) = \tfrac12 kx^2$ .

Step 2. Income: $5.0 \times 9.8 \times 0.50 = 24.5$ J. Storage at the limit: $\tfrac12 k (0.20)^2 = 0.020\,k$ .

Step 3. Equate: $k = \dfrac{24.5}{0.020} = \boxed{1225\ \mathrm{N/m}}$ (minimum — a stiffer spring squashes less).

Step 4. Peak force at maximum squash: $F = kx_{\max} = 1225 \times 0.20 = \boxed{245\ \mathrm{N}} = 5mg$ . Sanity: the crate's peak deceleration is $F/m - g = 39.2\ \mathrm{m/s^2} = 4g$ upward — finite and plausible; and the static sag $x_0 = mg/k = 4.0$ cm is comfortably below $x_{\max} = 20$ cm ✓.

3 · Competitive level — the full landmark tour (JEE-Main pattern, ~3 minutes)

A $1.0\ \mathrm{kg}$ ball is dropped from $0.40\ \mathrm{m}$ above a relaxed vertical spring with $k = 100\ \mathrm{N/m}$ . Take $g = 10\ \mathrm{m/s^2}$ . Find (a) the maximum compression, (b) the maximum speed and where it occurs, (c) the peak deceleration.

Step 1. Landmark first: $x_0 = mg/k = 0.10$ m. Budget: $10(0.40 + x) = 50x^2$ , i.e. $5x^2 - x - 0.4 = 0$ .

Step 2. Quadratic formula: $x = \dfrac{1 + \sqrt{1 + 8}}{10} = \dfrac{1+3}{10} = \boxed{0.40\ \mathrm{m}}$ — an engineered clean root; the closed form agrees: $x_0 + \sqrt{x_0^2 + 2hx_0} = 0.1 + \sqrt{0.01 + 0.08} = 0.1 + 0.3$ ✓.

Step 3. Maximum speed at the balance point $x = x_0 = 0.10$ m: $v_{\max}^2 = 2gh + gx_0 = 8 + 1 = 9 \Rightarrow v_{\max} = \boxed{3.0\ \mathrm{m/s}}$ . (Contact speed was $\sqrt{2gh} = 2.83$ m/s — the ball gains speed for the first 10 cm of squash.)

Step 4. Peak deceleration at the bottom: $a = \dfrac{kx_{\max} - mg}{m} = \dfrac{40 - 10}{1} = \boxed{30\ \mathrm{m/s^2}} = 3g$ , directed upward. Timing note: ~3 minutes, and the landmark $x_0$ is the accelerator — computing it first turns parts (b) and (c) into one-line evaluations. If the quadratic's discriminant comes out ugly, suspect an arithmetic slip: setters engineer clean roots here.

4 · Competitive level — attached versus loose (JEE-Advanced flavour, ~4 minutes)

A $2.0\ \mathrm{kg}$ block is dropped from $0.15\ \mathrm{m}$ onto a vertical spring, $k = 200\ \mathrm{N/m}$ , $g = 10\ \mathrm{m/s^2}$ . (a) Find the maximum compression. (b) If the block sticks to the spring, how high above the natural length does it rise, and with what period does it oscillate? (c) If instead it rests loose on the spring, how high does it rebound?

Step 1. Landmarks: $x_0 = mg/k = 0.10$ m; amplitude $A = \sqrt{x_0^2 + 2hx_0} = \sqrt{0.01 + 0.03} = 0.20$ m.

Step 2. (a) $x_{\max} = x_0 + A = \boxed{0.30\ \mathrm{m}}$ . (Check by budget: $mg(h+x) = 20 \times 0.45 = 9.0$ J; $\tfrac12 kx^2 = 100 \times 0.09 = 9.0$ J ✓.)

Step 3. (b) Attached, the block executes SHM about $x_0$ with amplitude $A = 0.20$ m: its top point is $A - x_0 = 0.20 - 0.10 = \boxed{0.10\ \mathrm{m}}$ above natural length — the spring is stretched there, legal only because it can pull an attached block. Period $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.01} = \boxed{0.63\ \mathrm{s}}$ .

Step 4. (c) Loose, the spring releases the block at natural length (a spring cannot pull). Return-leg audit from $x_{\max}$ to natural length: KE there $= \tfrac12 kx_{\max}^2 - mgx_{\max} = 9.0 - 6.0 = 3.0$ J, so $v = \sqrt{3}\ \mathrm{m/s}$ and it rises $v^2/2g = \boxed{0.15\ \mathrm{m}}$ — exactly the drop height, as ideal energy conservation demands. Timing note: ~4 minutes. The attached/loose fork is the whole question: attached ⇒ SHM continues above natural length (top point $A - x_0$ ); loose ⇒ separation at natural length, projectile rules after. Spot the fork in the first read and parts (b)/(c) are 60 seconds each. The elegant audit: loose blocks always return to their release height — if your answer differs, energy leaked somewhere.

Exam traps

Charging gravity only $h$ . The budget line must read $mg(h + x)$ — the flagship error, worth its own warning above. Zero-cost self-check: your equation must survive $h = 0$ with a non-zero answer.

Answering $x_0$ when the paper wants $x_{\max}$ (or vice versa). "Comes to rest momentarily" is the dynamic stop $x_{\max}$ ; "comes to rest finally / hangs in equilibrium / is placed gently" is $x_0$ . Both appear as options in the same MCQ, one exactly half the other for $h = 0$ — the examiners know the confusion sells.

Claiming maximum speed at first contact. For the first $x_0$ of compression the net force still points down; the speed rises until $kx = mg$ . Distractors quote $\sqrt{2gh}$ ; the truth is $\sqrt{2gh + gx_0}$ , always slightly larger.

Treating the bottom as equilibrium. At $x_{\max}$ the speed is zero and the net force is maximal upward — same conflation the horizontal archetype punishes, now with gravity in the mix. Equilibrium lives at $x_0$ , passed long before.

Forgetting the spring cannot pull a loose mass. Rebound questions split at natural length: loose masses separate there (then obey projectile rules); attached masses continue into extension and SHM. Announce which regime you are in.

Keeping the wrong quadratic root. The compression root is the positive one, $x_0 + \sqrt{\cdots}$ . The other root, $x_0 - \sqrt{\cdots} \le 0$ , is not garbage — it is the attached-mass's highest point — but it is never the squash.

Unit mixing. $k$ arrives in N/m, compressions in cm, drop heights in m. Convert before the quadratic: a $k$ in N/cm silently inflates answers a hundredfold, and the discriminant hides the crime.

Which $g$ . CBSE/ISC: $9.8$ unless stated; JEE/NEET: almost always a declared $g = 10$ ; A-Level: $9.81$ . The factor-two theorem is $g$ -proof, but every numeric answer is not — use what the paper states, never mix mid-solution.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · Royal Society, London

Robert Hooke published his spring law in 1676 as the anagram ceiiinosssttuv — insurance while he milked the discovery — unscrambling it two years later as ut tensio, sic vis: as the extension, so the force. He built the law from watch springs and vibrating coils in his role as the Royal Society's Curator of Experiments, a job that had him demonstrate three new experiments a week for decades. The vertical-spring energy problem is pure Hooke arithmetic wearing Newton's dynamics: his linear force law is the reason the storage is a neat parabola $\tfrac12 kx^2$ , the reason dropped masses overshoot to exactly double, and the reason the bounce is simple harmonic. Hooke feuded bitterly with Newton over credit for gravity's inverse square; history sided with Newton on orbits, but every buffer, balance and bathroom scale still runs on Hooke's one-line law.

History

1660

— Hooke discovers the proportionality of spring force and extension, hiding it in an anagram.

1678

— De Potentia Restitutiva publishes the law openly, with the vibrating-spring experiments behind it.

1687

— Newton's Principia supplies the dynamics; energy methods to exploit both mature over the next century.

1840

s — railway buffer stops turn falling-and-arriving masses on giant springs into an engineering specification.

1979

— the Oxford Dangerous Sports Club jumps off Clifton Suspension Bridge on elastic rope: the mass-dropped-on-spring problem, performed on oneself.

Going further

Texts. NCERT Physics Class 11, Ch. 6 (Work, Energy and Power — spring PE and vertical springs) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Morin, Introduction to Classical Mechanics, Ch. 5 (the vertical spring's shifted equilibrium).

Provenance. All results are standard. A verifier should recompute: the closed form (3) from the quadratic in derivation A; the anchor drop ( $m=2.0$ kg, $h=0.50$ m, $k=1960$ N/m, $g=9.8$ ): $x_0 = 1.0$ cm, $x_{\max} = 0.111$ m, peak force $217\ \mathrm{N} \approx 11mg$ , budget $11.98$ J both sides; Example 1's $5.0/10.0$ cm pair; Example 2's $k = 1225$ N/m and $245$ N $= 5mg$ ; Example 3's $x_{\max} = 0.40$ m, $v_{\max} = 3.0$ m/s at $x_0 = 0.10$ m, $a_{\max} = 3g$ ; Example 4's $A = 0.20$ m, top point $0.10$ m above natural length, $T = 0.63$ s, rebound $0.15$ m.

Prerequisites. Block Meets Spring · Potential Energy: Gravitational & Elastic · Conservation of Mechanical Energy. Leads to. Force Method or Energy Method? (technique) · The Spring–Block Oscillator (oscillations chapter) · Gravitational & Elastic Potential Energy (UG deep dive).

Why does suddenly releasing the mass give exactly TWICE the gentle-lowering compression?

Because the spring's storage is a parabola and gravity's income is a straight line. Lowered gently, the mass stops where forces balance: $x_0 = mg/k$ , with your hand having carried away the surplus energy. Released suddenly, nothing removes energy, so the mass must descend until the parabola $\tfrac12 kx^2$ has swallowed the entire line $mgx$ — and a parabola whose slope matches a line at $x_0$ needs exactly $2x_0$ of run to catch its total. The factor two is universal: any mass, any spring, any planet — provided the drop height is zero and the spring is ideal.

Where exactly is the mass in equilibrium during all this?

Only at $x_0 = mg/k$ , the point where $kx = mg$ — and the dropped mass crosses it at maximum speed, which is precisely why it cannot stop there. At maximum compression the mass is momentarily at rest but as far from equilibrium as it ever gets: net force $kx_{\max} - mg$ upward, the encounter's largest. "At rest" and "in equilibrium" are different claims — one about velocity, one about force — and this archetype is the exam's favourite place to test whether you conflate them.

Why does the closed form have a square root, and what is the other root of the quadratic?

The square root is the drop height entering through the discriminant — it is also, beautifully, the amplitude of the SHM that the contact phase secretly is: $x_{\max} = x_0 + A$ . The quadratic's other root, $x_0 - A$ (zero or negative), marks the same energy level on the other side of equilibrium: the highest point an attached mass would reach, a height $A - x_0$ above natural length. For a loose mass that root is unphysical — the spring lets go at natural length — which is why you discard it, with a reason, in written answers.

Does the loose mass really bounce back to exactly its drop height?

In the ideal problem, yes — every ideal store (gravity, spring) repays in full, so the return leg runs the film backward: separation at natural length with speed $\sqrt{2gh}$ , rise of exactly $h$ . Real springs bleed energy into internal friction and sound, so real bounces fall short and shrink. Exams use the ideal version as a free audit: if your computed rebound height exceeds the drop height, energy was created somewhere in your algebra — go find the sign error.

When am I allowed to use x = √(2mgh/k) instead of solving the quadratic?

Only when $h \gg x_0 = mg/k$ , so gravity's income during the squash ( $\sim mgx$ ) is negligible beside the arrival energy $mgh$ — and ideally when the paper signals it ("h is much larger than the compression"). Quantify before trusting: for the anchor drop ( $h = 50$ cm, $x_0 = 1$ cm) the shortcut gives $10.0$ cm against the true $11.1$ cm — a 10% error that MCQ options are spaced to catch. When in doubt, the closed form $x_0 + \sqrt{x_0^2 + 2hx_0}$ costs ten extra seconds and is always right.

Is the motion during contact really simple harmonic — even though gravity acts?

Yes. The net force $mg - kx = -k(x - x_0)$ is a perfect restoring force about the shifted centre $x_0$ : gravity relocates the equilibrium but does not reshape the parabola (the completing-the-square argument from the potential-energy article). So $\omega = \sqrt{k/m}$ exactly as if gravity were absent, and every SHM tool applies — amplitude from speed-at-centre, extremes at centre ± amplitude. The one caveat: for a loose mass the SHM script is valid only while $x > 0$ ; the mass exits the script at natural length, where contact ends.

Why do stiff springs — and hard floors — produce such violent forces if the energy is the same?

The energy $mg(h+x)$ must be absorbed over the travel $x_{\max}$ , and force is roughly energy over distance. Stiff spring ⇒ tiny $x_{\max}$ ⇒ huge force: quantitatively $F_{\text{peak}} = kx_{\max} \approx \sqrt{2mghk}$ for deep drops, growing without bound as $k \to \infty$ . A concrete floor is the limit — micrometres of "compression", thousands of g. This is the entire theory of cushioning: you cannot reduce the energy of a landing, only stretch the distance (and time) over which it is absorbed. Packaging foam, crash mats, bent knees and bungee cords are all purchases of $x_{\max}$ .