ARCHETYPE: Losing Contact in SHM

After a violent earthquake, seismologists walking the field sometimes find boulders sitting beside fresh craters — rocks that were thrown clear of their own sockets. No hand lifted them. The ground under them accelerated downward faster than gravity could make the boulders follow, and for a moment the planet simply left them behind, airborne. That field observation is used to this day as a lower bound on how hard the ground shook: "the ground's downward acceleration exceeded $g$ here." This archetype is that physics on a lab bench: a block resting — not attached — on a platform that oscillates vertically in SHM. The platform can push the block (normal force) but can never pull it. On the way through the bottom of the stroke the platform presses extra hard, and the block "weighs" more. Near the top of the stroke the platform is accelerating downward, and it needs less and less push to carry the block along — until, if the shaking is vigorous enough, the required push hits zero. The instant the ledger demands a negative normal force, the contract breaks: the block departs on its own private projectile arc. The whole family — "maximum amplitude before the coin leaves the piston," "maximum frequency for contact," "where exactly does it fly off," and the horizontal cousin "when does the block skid" — is governed by one comparison: $\omega^2 A$ versus $g$ . JEE has dressed this comparison in fifty costumes; this module teaches you to undress it in three lines.

Key takeaways

A block resting on a vertically oscillating platform feels $N = m(g - \omega^2 y)$ , where $y$ is the displacement above the mean: the platform presses hardest at the bottom of the stroke, lightest at the top.
Contact survives while $N \geq 0$ everywhere, i.e. while $\omega^2 A \leq g$ — the block leaves the platform the moment the platform's maximum downward acceleration $\omega^2 A$ exceeds $g$ .
Thresholds to quote: $A_{\max} = g/\omega^2$ at fixed frequency; $\omega_{\max} = \sqrt{g/A}$ (so $f_{\max} = \frac{1}{2\pi}\sqrt{g/A}$ ) at fixed amplitude.
If $\omega^2 A > g$ , separation happens on the way up at height $y^* = g/\omega^2$ above the mean — where the platform's downward acceleration first equals $g$ — not at the extreme.
Horizontal cousin: a block on a horizontally oscillating table skids when the needed force exceeds friction's ceiling — slip when $\omega^2 A > \mu g$ .
Key idea: a surface can push but never pull — contact in vertical SHM survives exactly while $\omega^2 A \leq g$ , and the crisis always comes at (or on the way to) the top of the stroke.

The core insight

For a block riding a vertically oscillating platform, $N = m(g - \omega^2 y)$ : the normal force runs lightest at the top of the stroke, so contact demands $\omega^2 A \leq g$ ; beyond that the block lifts off at $y^* = g/\omega^2$ , where the platform first accelerates downward at $g$ .

Where this lives today

Engineering runs this archetype daily, in both directions. Shake tables — building-sized oscillating platforms — qualify everything from satellites to hospital wings by driving them through prescribed $\omega^2 A$ schedules; test engineers speak in "g's of excitation," literally this module's comparison. Earthquake codes bound the anchoring of rooftop water tanks, server racks and museum statues using the $N \to 0$ condition: anything unbolted becomes a projectile when the ground's downward acceleration passes $g$ , which modern accelerographs have recorded repeatedly. The same physics is exploited on purpose: vibratory conveyors and paint mixers run super-critical so their contents repeatedly lift off and re-land (that is what does the mixing and the marching); ultrasonic cleaning baths toss contaminant particles off surfaces thousands of times a second; and loudspeaker engineers watch dust dance off a woofer cone the moment its drive crosses $\omega^2 A = g$ — a desktop demonstration of the boulder in the field.

The setup

A platform (piston, shake table, loudspeaker cone) executes vertical SHM with amplitude $A$ and angular frequency $\omega$ : displacement $y$ measured upward from the mean position. A block of mass $m$ rests on it — no glue, no bolts, contact only. Questions the exam asks: the normal force at a given point of the cycle; the apparent weight at top and bottom; the largest $A$ (or largest $f$ ) for which the block never leaves; and, past the threshold, where the block lifts off and how high it flies. Two forces act on the block while contact lasts: gravity $mg$ down, normal force $N$ up. Riding with the platform, the block shares the platform's acceleration $a_y = -\omega^2 y$ (SHM's defining kinematics: directed toward the mean, proportional to displacement). Newton's second law, taking up as positive:

N - mg = m\,a_y = -\,m\,\omega^2 y \quad\Longrightarrow\quad N = m\left(g - \omega^2 y\right) \qquad (1)

where: $y$ runs from $-A$ (bottom of stroke) to $+A$ (top). Read the two extremes off equation (1) at sight:

N_{\text{bottom}} = m(g + \omega^2 A) \qquad N_{\text{top}} = m(g - \omega^2 A) \qquad (2)

where: at the bottom the platform must both hold the block up and fling it upward — the block "weighs" more (a bathroom scale on the platform would read high, exactly the lift-accelerating-up story). At the top the platform is dropping away beneath the block — the scale reads light. The whole archetype is the observation that $N_{\text{top}}$ can reach zero:

\text{contact everywhere} \iff \omega^2 A \leq g \qquad (3)

where: $\omega^2 A$ — the platform's maximum acceleration, the one number to compare against $g$ . Equality is the critical shake: the block goes momentarily weightless at the top, touching the platform with zero force, and any increase in amplitude or frequency breaks the contract.

Figure

The stroke and its two crises — pressed at the bottom, abandoned at the top

The master solution pattern

Derivation AThe normal-force ledger, the threshold, and the lift-off point

We derive $N(y)$ , extract the contact condition $\omega^2 A \leq g$ , and locate the lift-off point $y^* = g/\omega^2$ for a super-critical shake. Strategy: Newton's second law on the block alone, with the platform's SHM kinematics imposed as long as contact lasts; then interrogate the one inequality a passive surface must satisfy, $N \geq 0$ . Watch that the threshold is mass-blind — the block's $m$ divides out of the contract.

Step 1. Forces on the block during contact: $N$ up, $mg$ down. Contact means the block copies the platform's motion, so its acceleration is the SHM value $a_y = -\omega^2 y$ (toward the mean). Second law (up positive): $N - mg = -m\omega^2 y$ , giving equation (1): $N = m(g - \omega^2 y)$ .

Step 2. Audit the extremes. Bottom ( $y = -A$ ): $N = m(g + \omega^2 A) > mg$ — consistent with the lift-accelerating-up experience; no crisis possible. Top ( $y = +A$ ): $N = m(g - \omega^2 A)$ — the cycle's minimum. The normal force oscillates about $mg$ with "amplitude" $m\omega^2 A$ , at the platform's own frequency.

Step 3. Impose the surface's one law: $N \geq 0$ (push, never pull). The worst case is the top, so contact survives the whole cycle iff $m(g - \omega^2 A) \geq 0$ , i.e. $\omega^2 A \leq g$ — equation (3). The mass has cancelled: a coin and an anvil leave the same platform at the same setting. Rearranged as design limits: $A_{\max} = g/\omega^2$ , $\omega_{\max} = \sqrt{g/A}$ , $f_{\max} = \frac{1}{2\pi}\sqrt{g/A}$ .

Step 4. Super-critical shake: where does it leave? For $\omega^2 A > g$ , set $N = 0$ in (1): $y^* = g/\omega^2$ , a height below the top ( $y^* < A$ exactly because $\omega^2 A > g$ ). Rising through $y^*$ , the platform's downward acceleration demand passes $g$ ; gravity is the only downward force available to a free block, so the block cannot follow — it detaches while still moving upward, with the SHM speed at that height, $v^* = \omega\sqrt{A^2 - y^{*2}}$ (SHM kinematics' speed-position law).

Step 5. After lift-off, the block is a projectile: it rises a further $v^{*2}/2g$ , then falls back and lands on the (still oscillating) platform — the landing timing is a hard problem the exams do not ask; the rise height they do (Example 4). Verified: threshold dimensions $[\omega^2 A] = \mathrm{s^{-2} \cdot m} = \mathrm{m/s^2}$ ✓ comparable to $g$ ; critical case $\omega^2 A = g$ puts $y^* = A$ (leaves exactly at the top with $v^* = 0$ — grazing weightlessness) ✓; $\omega \to 0$ gives $N \to mg$ everywhere (a parked platform) ✓.

The horizontal cousin. Lay the oscillation flat — block on a table shaking horizontally with the same $A$ , $\omega$ — and the vertical ledger goes quiet ( $N = mg$ always; no lift-off) while friction inherits the contract. The block needs horizontal force $m\omega^2 A$ (at the extremes) to keep up, and static friction can supply at most $\mu N = \mu mg$ :

\text{no skidding} \iff \omega^2 A \leq \mu g \qquad (4)

where: the same comparison with $g$ replaced by $\mu g$ — friction is a weaker god than gravity by the factor $\mu$ . Past the threshold the block doesn't fly; it slips, and the stacked-blocks module's who-drags-whom machinery takes over. One threshold pattern, two orientations: this pairing is a JEE-Main favourite.

Figure

The normal force against position — a straight line that must not cross zero

Limiting cases

case	result	meaning
$\omega^2 A \ll g$	$N \approx mg$ always	gentle shaking: the block barely notices
$\omega^2 A = g$	$N_{\text{top}} = 0$ , grazing	critical: momentary weightlessness at the top, contact just survives
$\omega^2 A > g$	lift-off at $y^* = g/\omega^2 < A$	departure on the way up, then free projectile
$\omega \to$ large at fixed $A$	$f_{\max} = \frac{1}{2\pi}\sqrt{g/A}$	high-frequency shaking tosses everything: $\omega$ enters squared
horizontal table, friction $\mu$	slip when $\omega^2 A > \mu g$	same law, friction's ceiling instead of gravity's
block glued / attached	no separation ever; glue tension $= m(\omega^2 y - g)$ when demanded	changing "resting" to "attached" changes the problem completely

Common trap“The block flies off at the bottom, where the platform pushes hardest”

Upside down. The bottom of the stroke is where the block is most secure — the platform must push up with $m(g + \omega^2 A)$ , well above the weight, to fling the block upward; a scale there reads heavy, never zero. The danger zone is the top: there the platform accelerates downward at $\omega^2 A$ , and the most gravity can do to a free block is accelerate it down at $g$ . If the platform demands more, the surface would have to pull the block down — and surfaces cannot pull. $N$ hits zero and the block departs. Twin traps in the same family: separation happens at the top only in the critical case; for vigorous shaking it happens below the top at $y^* = g/\omega^2$ , on the way up. And the threshold is mass-independent — "heavier blocks hold on longer" is bait, since $m$ multiplies both the demand and gravity's supply.

Worked examples

1 · Board level — maximum amplitude at a set frequency (JEE-Main early slot)

A horizontal platform vibrates vertically in SHM at $f = 2.0\ \mathrm{Hz}$ . A small block rests on it. What is the largest amplitude for which the block never leaves the platform? ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Angular frequency: $\omega = 2\pi f = 4\pi = 12.57\ \mathrm{rad/s}$ , so $\omega^2 = 158.0\ \mathrm{s^{-2}}$ .

Step 2. Contact condition at the top: $\omega^2 A \leq g$ , so $A_{\max} = \dfrac{g}{\omega^2} = \dfrac{9.8}{158.0} = \boxed{6.2\ \mathrm{cm}}$ .

Step 3. Sense check: at 2 Hz, a 6 cm stroke already produces accelerations of a full $g$ — vigorous shaking by any human standard, which is why loose objects survive most machinery but not paint mixers.

2 · Board level — maximum frequency at a set amplitude (NEET-style single comparison)

A coin rests on a piston oscillating vertically with amplitude $5.0\ \mathrm{cm}$ . Up to what frequency does the coin stay in contact throughout? ( $g = 10\ \mathrm{m/s^2}$ .)

Step 1. Threshold: $\omega_{\max} = \sqrt{g/A} = \sqrt{10/0.050} = \sqrt{200} = 14.1\ \mathrm{rad/s}$ .

Step 2. Convert: $f_{\max} = \dfrac{\omega_{\max}}{2\pi} = \dfrac{14.1}{6.283} = \boxed{2.25\ \mathrm{Hz}}$ .

Step 3. Audit by the reverse route: at $2.25$ Hz, $\omega^2 A = 200 \times 0.05 \times \ldots$ — directly, $\omega^2 A = (14.1)^2 \times 0.050 = 10.0 = g$ ✓, the grazing case exactly.

3 · Competitive level — the scale on the piston (JEE-Main pattern, ~2.5 minutes)

A $0.50\ \mathrm{kg}$ block rests on a platform oscillating vertically with amplitude $10\ \mathrm{cm}$ at $\omega = 8.0\ \mathrm{rad/s}$ . Find (a) the normal force at the highest and lowest points, (b) the apparent weight (scale reading) as the platform passes the mean position, (c) whether the block ever leaves. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. The one comparison first: $\omega^2 A = 64 \times 0.10 = 6.4\ \mathrm{m/s^2} < g$ — contact holds everywhere; part (c) answered before any force is computed.

Step 2. Top: $N_{\text{top}} = m(g - \omega^2 A) = 0.50 \times (9.8 - 6.4) = \boxed{1.7\ \mathrm{N}}$ — the block "weighs" about a third of its true weight.

Step 3. Bottom: $N_{\text{bottom}} = m(g + \omega^2 A) = 0.50 \times 16.2 = \boxed{8.1\ \mathrm{N}}$ — 65% overweight.

Step 4. Mean position: $y = 0$ , so $N = mg = \boxed{4.9\ \mathrm{N}}$ — true weight, twice per cycle, at maximum speed. The scale needle swings between 1.7 and 8.1 N around its honest reading. Timing note: ~2–2.5 min. Run the threshold comparison first, always — it classifies the problem and sometimes ends it. Distractor watch: swapping top and bottom (the "flies off at the bottom" reflex), and reading "at the mean" as zero force rather than zero extra force.

4 · Competitive level — JEE-Advanced stretch: lift-off point and flight ceiling (~4 minutes)

A block rests on a platform oscillating vertically with $A = 10\ \mathrm{cm}$ and $\omega = 14\ \mathrm{rad/s}$ ( $\omega^2 \approx 196\ \mathrm{s^{-2}}$ , so $\omega^2 A = 19.6\ \mathrm{m/s^2} = 2g$ ). Find (a) the height above the mean at which the block leaves the platform, (b) its speed there, (c) the maximum height above the mean it reaches in free flight. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Lift-off where the demanded downward acceleration first equals $g$ (on the way up): $y^* = \dfrac{g}{\omega^2} = \dfrac{9.8}{196} = 0.050\ \mathrm{m} = \boxed{5.0\ \mathrm{cm}}$ — halfway to the top, not at the top.

Step 2. Speed at departure, from SHM's speed-position law: $v^* = \omega\sqrt{A^2 - y^{*2}} = 14\sqrt{(0.10)^2 - (0.05)^2} = 14 \times 0.0866 = \boxed{1.21\ \mathrm{m/s}}$ , upward.

Step 3. Free flight from $y^*$ : additional rise $\Delta h = \dfrac{v^{*2}}{2g} = \dfrac{1.47}{19.6} = 0.075\ \mathrm{m}$ . Ceiling above the mean: $y^* + \Delta h = 0.050 + 0.075 = \boxed{12.5\ \mathrm{cm}}$ .

Step 4. Read the punchline: the block's ceiling (12.5 cm) is above the platform's own top (10 cm) — the block genuinely gets thrown, like the earthquake boulders. The platform meanwhile turns at 10 cm and dives; block and platform will meet again somewhere on the way down, with a thud whose timing is (mercifully) beyond the syllabus. Timing note: ~3.5–4 min. The architecture: SHM until $N = 0$ , projectile after — one clean handover at $y^*$ . The three reflexes: $y^* = g/\omega^2$ (not $A$ ), speed from $\omega\sqrt{A^2 - y^2}$ (not $\omega A$ ), and projectile bookkeeping from the departure point, not the mean. Distractor watch: "leaves at the top with speed zero" (gives ceiling 10 cm) and using $v_{\max} = \omega A = 1.4$ m/s at departure (gives 15 cm) — both on the option list.

Exam traps

The crisis is at the top. Minimum $N$ at $y = +A$ ; maximum at the bottom. Any option placing separation at the lowest point, or claiming the scale reads zero at the bottom, dies on equation (2).

One comparison decides everything. Compute $\omega^2 A$ and set it against $g$ (vertical) or $\mu g$ (horizontal). Run this first; half the MCQs end there.

$\omega$ enters squared. Doubling the frequency quarters the permissible amplitude ( $A_{\max} = g/\omega^2$ ). Linear-in- $\omega$ reasoning ("twice the frequency, half the amplitude") is the planted wrong scaling.

Mass never enters the threshold. $m$ multiplies both the demand ( $m\omega^2 A$ ) and the supply ( $mg$ ): coins, blocks and anvils depart together. Mass does enter the normal-force values — read what the question asks.

Super-critical lift-off is below the top. $y^* = g/\omega^2 < A$ whenever separation happens at all, and the block leaves moving upward with $v^* = \omega\sqrt{A^2 - y^{*2}}$ . "Leaves at the extreme" is only the grazing case $\omega^2 A = g$ .

"Resting on" versus "attached to". Attached blocks (glued, bolted, or the platform is a spring the block hangs from) can be pulled — no separation, ever; the glue or spring simply goes into tension. The archetype's whole drama lives in the words "rests on"; underline them in the stem.

Which $g$ ? JEE/NEET default to $10\ \mathrm{m/s^2}$ here (thresholds come out clean: $A_{\max} = 10/\omega^2$ ); CBSE-style derivation parts use 9.8. The critical-case audits ( $\omega^2 A \stackrel{?}{=} g$ ) inherit whichever you chose — never mix within a problem.

Origins · 1 minJohn Milne and how this was discovered

John Milne

1850–1913 · Tokyo & the Isle of Wight

Milne was the mining engineer who turned earthquake shaking into a measurable signal. Hired by the Imperial College of Engineering in Tokyo in 1876, he arrived to find earthquakes plentiful and instruments nonexistent; with colleagues Ewing and Gray he built the first practical seismographs — swinging and hanging masses whose inertia lets them stand still while the ground moves, the exact inverse of this module's block that must be forced to follow the ground. His horizontal-pendulum seismograph of 1880 became the world standard, and the global network he later ran from his home on the Isle of Wight was the first planet-wide instrument of any kind. Milne also read the crude field evidence — overturned lanterns, tossed stones, the boulder-out-of-its-socket signature — as acceleration data, arguing that ground shaking near great earthquakes can exceed $g$ : nature running this archetype's super-critical case, with continents for platforms.

History

1880

— Milne, Ewing and Gray build the first practical seismographs in Tokyo: suspended masses that read the shaking ground.

1906

— The San Francisco earthquake's field evidence (thrown objects, shifted monuments) is mined for acceleration estimates against g.

1933

— The first strong-motion accelerographs are installed in California, born from the Long Beach earthquake.

1971

— The San Fernando earthquake delivers the first instrumental recording of ground acceleration exceeding 1 g.

2011

— Japan's dense K-NET array records peak accelerations near 3 g in the Tōhoku earthquake — platforms outrunning gravity, on the record.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1, Ch. 12 (block-on-piston problems) · DC Pandey SHM chapter, contact-separation drill set · Resnick–Halliday–Walker Ch. 15 · The lifts-and-apparent-weight module for the $N = m(g \pm a)$ ledger this archetype cycles.

Provenance. All results are standard JEE canon. A verifier should recompute: Example 1's $A_{\max} = 9.8/158.0 = 6.2$ cm; Example 2's $f_{\max} = 2.25$ Hz with the grazing audit; Example 3's 1.7/4.9/8.1 N triple; Example 4's $y^* = 5.0$ cm, $v^* = 1.21$ m/s, ceiling 12.5 cm.

Prerequisites. SHM Kinematics · Weight, Normal Force & Apparent Weight · The Spring–Block Oscillator. Leads to. Time-Period Pattern Recognition (technique).

Why can't the normal force just go negative for a moment and then recover?

Because a normal force is not a spring or a rope — it is the electromagnetic refusal of two surfaces to interpenetrate, and it exists only while they press. It can push with any strength but the instant "holding the block down" is required, there is simply no mechanism: nothing tethers the block. So the mathematics $N = m(g - \omega^2 y)$ is valid only while its output is non-negative; the moment the formula asks for $N < 0$ , the premise (shared motion) fails, contact ends, and the block's true acceleration becomes just $-g$ . This "inequality constraint" style of reasoning — solve assuming contact, then check the assumption — is the same logic as string-goes-slack and car-leaves-the-road problems.

The threshold has no mass in it — does that mean a feather and an anvil leave together?

For the rigid-body idealisation, yes, and it is worth internalising why: the platform must supply the block's whole weight plus the extra $m\omega^2 y$ ; both the demand and gravity's contribution scale with $m$ , so the crossover $\omega^2 A = g$ is mass-blind — the same cancellation that makes free-fall universal. Real feathers complicate life with air drag and real anvils with platform flexing, but exam blocks are ideal. Where mass does matter is in every force question: $N_{\text{top}}$ and $N_{\text{bottom}}$ are proportional to $m$ , so scale-reading problems need it.

How is N at the top being small related to feeling light in a descending lift?

It is the identical ledger. The lift module's apparent weight is $N = m(g + a)$ with $a$ the upward acceleration; an oscillating platform is just a lift whose acceleration cycles sinusoidally, $a = -\omega^2 y$ . Top of stroke = lift accelerating downward (light, $N = m(g - \omega^2 A)$ ); bottom = lift accelerating upward (heavy). The platform's grazing case $N_{\text{top}} = 0$ is the snapped-cable lift: momentary true weightlessness. One formula, three costumes — lift, platform, and (in the next chapters) orbit.

After the block leaves, does it land back at the same point of the platform's cycle?

Almost never, and this is why exams stop at the flight ceiling. From lift-off, the block flies as a projectile (parabolic in time), while the platform continues its cosine; their reunion is where a parabola meets a sinusoid — a transcendental equation with no clean solution, solved numerically in engineering practice (vibratory conveyor designers do exactly this to set hop lengths). What is clean, and asked: the departure point $y^* = g/\omega^2$ , the departure speed $\omega\sqrt{A^2 - y^{*2}}$ , and the rise $v^{*2}/2g$ . If a problem seems to demand the landing instant, re-read it — it is asking one of the clean three.

In the horizontal case, does the block fly off the table when it slips?

No — it skids, which is a gentler failure. Vertically, losing contact means losing the only connection, so the block goes ballistic. Horizontally, the normal force ( $N = mg$ , steady) survives; what fails at $\omega^2 A = \mu g$ is static friction's grip, so the block starts sliding relative to the table while kinetic friction ( $\mu_k mg$ ) still drags it about. The subsequent motion — block lagging, slipping forward and back — is the stacked-blocks module's relative-sliding machinery on an oscillating base. For MCQs, the asked quantity is nearly always just the threshold amplitude $\mu g/\omega^2$ or frequency $\frac{1}{2\pi}\sqrt{\mu g/A}$ .

Is this the same physics as objects being thrown upward by earthquakes?

Precisely, at geological scale. When the ground under a boulder accelerates downward faster than $g$ , the ground leaves the boulder — from the ground's frame, the boulder "jumps." Field seismologists treat displaced-but-unbroken objects (stones out of sockets, coffins lifted in vaults, unbolted tanks walked across floors) as crude accelerometers: their displacement proves $a_{\text{ground}} > g$ locally. Instrumental records have confirmed super-g shaking many times since 1971. The lab version is your Example 4; the code-book version is why hospitals bolt down everything heavier than a stapler.