The Spring–Block Oscillator

Pull a block on a spring 2 cm to the side and release it; time the round trip. Now pull it 8 cm — four times as far — and time it again. Same answer. The big swing and the small wiggle complete their cycles in exactly the same time, a property called isochronism, and it is strange enough that it took humanity until the 1650s to notice it, trust it, and build the first accurate clocks around it. Every mechanical watch made since is a standing bet that this equality never fails. The spring–block oscillator is where that bet is easiest to understand — and it is the single most important worked case of SHM you will meet. The previous modules defined SHM by $a = -\omega^2 x$ and unpacked its timetable; here, for the first time, a real system hands you its $\omega$ : put Hooke's law $F = -kx$ into Newton's second law and out comes $\omega = \sqrt{k/m}$ , fixed by the stiffness and the mass before you ever touch the block. How hard you pull sets the amplitude and the top speed; it gets no vote on the timing. This module extracts the two headline numbers — the period $T = 2\pi\sqrt{m/k}$ and the maximum speed $v_{\max} = \omega A$ — and then earns the exam mileage: the vertical spring (gravity shifts the home position but not the clock), the stopwatch-to-spring-constant inversion, the added-mass period shift, and the drop-a-load-on-it amplitude puzzles. Master this system and you own the template every other oscillator in the chapter is measured against.

Key takeaways

A block of mass $m$ on an ideal spring of stiffness $k$ feels $F = -kx$ , so $a = -(k/m)x$ : SHM with $\omega = \sqrt{k/m}$ , chosen by the hardware alone.
The period $T = 2\pi\sqrt{m/k}$ is independent of amplitude (isochronism) and independent of gravity — the same on the Moon, in a lift, or in orbit.
Maximum speed $v_{\max} = \omega A$ at the equilibrium crossing; maximum acceleration $a_{\max} = \omega^2 A$ at the turning points — one power of $\omega$ apart.
Hanging the spring vertically shifts the equilibrium down by $x_0 = mg/k$ and changes nothing else: same $k$ , same $T$ , oscillation now centred on the sagged point.
Adding mass slows the clock as $\sqrt{m}$ : $T_2/T_1 = \sqrt{m_2/m_1}$ — the ratio form that kills most MCQs.
Key idea: the spring–block oscillator ticks at $T = 2\pi\sqrt{m/k}$ — stiffness races the clock, inertia drags it, and neither amplitude nor gravity can touch it.

The core insight

Hooke's law inside Newton's second law makes the block's acceleration $a = -(k/m)x$ — SHM with $\omega = \sqrt{k/m}$ and period $T = 2\pi\sqrt{m/k}$ , fixed by hardware; amplitude sets how far and how fast ( $v_{\max} = \omega A$ ), never how long.

Where this lives today

The spring–block oscillator is the most manufactured physical system in history — you own dozens. Every quartz watch and phone carries a tiny tuning-fork crystal vibrating at exactly 32,768 Hz; digital logic halves that fifteen times to one tick per second, and the timekeeping survives battery sag precisely because the period ignores amplitude. MEMS accelerometers — the chips that flip your screen and fire airbags — are literal micromachined spring–mass systems reading force from the displacement of a proof mass. Atomic force microscopes feel individual atoms through the frequency shift of a vibrating cantilever, a spring–block whose $\sqrt{k/m}$ moves when atomic forces add stiffness. At the other end of the scale, the 660-tonne tuned mass damper hanging in Taipei 101 is a building-sized oscillator deliberately tuned to the tower's sway period so it drinks energy out of typhoon-driven oscillations. And aboard the ISS, astronauts strap into a chair on a calibrated spring and time its oscillation to measure their own body mass — $m = kT^2/4\pi^2$ , this module's equation (2) run backwards, as a medical instrument in a place where scales are useless.

The physical picture

Watch one full cycle. At the right turning point, $x = +A$ , the block is momentarily at rest but the stretched spring is pulling its hardest — maximum force, maximum acceleration, zero speed. The block accelerates left, gaining speed as the pull weakens, and at the natural length the force has died to zero — but the block is now at top speed and inertia carries it straight through. On the far side the compressed spring pushes back, braking the block to a stall at $x = -A$ , and the story runs in reverse. The motion is an endless argument between the spring's eagerness to restore and the mass's reluctance to stop — and neither ever wins.

Figure

One cycle of the spring-block oscillator — force, speed and energy trade places

Notice what each snapshot pairs together: maximum force with zero speed (the turning points) and maximum speed with zero force (the centre). Speed and force peak half a cycle out of step — the single most tested fact about this system. If you have the SHM-kinematics module fresh, you can also see the whole motion as the shadow of a point circling at rate $\omega = \sqrt{k/m}$ : the circle analogy supplies every timetable formula on demand, and this module supplies the $\omega$ to feed it.

The math: hardware chooses the frequency

Put the block on a frictionless horizontal track and measure $x$ from the spring's natural length. The only horizontal force is the spring's, $F = -kx$ , so Newton's second law reads

a = -\frac{k}{m}\,x \qquad (1)

where: $k$ — the spring constant (N/m), the stiffness; $m$ — the block's mass. Compare with the definition of SHM from two modules back, $a = -\omega^2 x$ : the match is exact, with one identification. The equation contains a single knob — stiffness over inertia — and no amplitude anywhere, which is isochronism guaranteed before anything is solved:

\omega = \sqrt{\frac{k}{m}}, \qquad T = 2\pi\sqrt{\frac{m}{k}}, \qquad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \qquad (2)

where: $\omega$ — angular frequency (rad/s); $T$ — period; $f$ — frequency in Hz. Read the structure: stiffness sits on top (a stiffer spring ticks faster), inertia sits below (a heavier block ticks slower), and both live under a square root — quadrupling either one only doubles its effect. This inertia-over-stiffness square root is the master pattern of the whole chapter; the technique module at the chapter's end drills it across every system you will meet. With $\omega$ in hand, the entire SHM toolkit from the kinematics module comes along free:

v_{\max} = \omega A, \qquad a_{\max} = \omega^2 A, \qquad v = \pm\,\omega\sqrt{A^2 - x^2} \qquad (3)

where: $A$ — the amplitude you chose at release. One power of $\omega$ separates $v_{\max}$ from $a_{\max}$ — given both in a problem, divide to get $\omega = a_{\max}/v_{\max}$ instantly. And the energy module's budget applies verbatim: $E = \tfrac12 kA^2 = \tfrac12 mv_{\max}^2$ .

given	find	route
$m$ , $k$	$T$	$T = 2\pi\sqrt{m/k}$ directly
stopwatch ( $N$ cycles in $t$ s)	$k$	$T = t/N$ , then $k = 4\pi^2 m/T^2$
$v_{\max}$ and $a_{\max}$	$\omega$ , $A$	$\omega = a_{\max}/v_{\max}$ , $A = v_{\max}^2/a_{\max}$
added mass, $T_1 \to T_2$	$m$ or $\Delta m$	$(T_2/T_1)^2 = (m + \Delta m)/m$
vertical spring, sag $x_0$	$T$	$k = mg/x_0$ , so $T = 2\pi\sqrt{x_0/g}$

Derivation AFrom Hooke's law to the period

We derive $T = 2\pi\sqrt{m/k}$ by matching the spring's force law against the defining equation of SHM, then confirm the solution honestly by substitution. The strategy: write Newton's second law, recognise the $a \propto -x$ shape, identify $\omega^2$ by comparison, and let the kinematics module's timetable do the rest. Watch the two things that never enter: the amplitude and gravity.

Step 1. Newton's second law on the track. The spring is the only horizontal force: $ma = -kx$ , so $a = -(k/m)x$ . The acceleration is proportional to displacement and oppositely directed — the SHM signature, with no approximation (an ideal spring is exactly linear).

Step 2. Identify the frequency. The definition of SHM is $a = -\omega^2 x$ . Comparing coefficients: $\omega^2 = k/m$ . There is nothing to solve — the hardware constants walk straight into the frequency slot.

Step 3. Borrow the timetable. SHM kinematics established that $a = -\omega^2 x$ is executed by $x(t) = A\cos(\omega t + \phi)$ , with $A$ and $\phi$ set by how the motion starts. Check it against this system by substitution: $v(t) = -\omega A\sin(\omega t + \phi)$ , $a(t) = -\omega^2 A\cos(\omega t + \phi) = -\omega^2 x$ ✓ — the spring equation is satisfied at every instant.

Step 4. Read off the period. The cosine repeats when its argument grows by $2\pi$ : $\omega T = 2\pi$ , so $T = 2\pi/\omega = 2\pi\sqrt{m/k}$ . No $A$ anywhere.

Step 5. Why the amplitude cancels — the physical audit. Double the pull: every force along the path doubles, so every acceleration doubles — but the block also has twice the distance to cover, and for a linear force law the two doublings cancel exactly. Bend the law even slightly (a rubber band, a large-angle pendulum) and the cancellation breaks: the period starts to drift with amplitude and the system stops being a good clock. Verified: dimensions $\sqrt{m/k} = \sqrt{\mathrm{kg \cdot m/N}} = \sqrt{\mathrm{s^2}} = \mathrm{s}$ ✓; limits $k \to \infty$ gives $T \to 0$ (a rigid link doesn't oscillate, it jerks) and $m \to \infty$ gives $T \to \infty$ ✓; matches the standard result in NCERT Ch. 14 and HC Verma Ch. 12.

The vertical spring — gravity moves home, not the clock

Hang the same block from the same spring and something both new and reassuring happens. Gravity stretches the spring to a new equilibrium, the sag where spring pull balances weight:

kx_0 = mg \qquad \Longrightarrow \qquad x_0 = \frac{mg}{k} \qquad (4)

where: $x_0$ — the equilibrium extension. Now displace the block a distance $y$ below that sagged point. The spring force grows by $ky$ while the weight stays fixed, so the net force is $-ky$ : exactly Hooke's law about the new centre, with the same $k$ . The constant $mg$ has been absorbed into relocating the home position, and everything else — $\omega$ , $T$ , $v_{\max}$ , the energy parabola — is untouched:

T_{\text{vertical}} = 2\pi\sqrt{\frac{m}{k}} = T_{\text{horizontal}} \qquad (5)

where: the deeper reading is that a spring clock is gravity-blind. Take it to the Moon and $x_0$ shrinks to a sixth, but $T$ does not move — which is precisely why astronauts measure body mass in orbit by strapping into a chair on a calibrated spring and timing its oscillation: $m = kT^2/4\pi^2$ works where a bathroom scale reads zero. (Contrast the pendulum, next module, whose $T$ is built from $g$ — the pair of behaviours is a favourite exam discriminator.) One practical gift of equation (4): the sag itself measures the hardware, $g/x_0 = k/m = \omega^2$ , so a spring that sags 2.5 cm under its load must oscillate at $T = 2\pi\sqrt{x_0/g} = 2\pi\sqrt{0.025/9.8} = 0.32$ s — no force meter needed.

Figure

Vertical spring — the equilibrium shifts by mg/k, the period does not shift at all

Limiting cases

case	result	meaning
$A \to 4A$	$T$ unchanged, $v_{\max} \to 4v_{\max}$	isochronism: amplitude buys speed and energy, never time
$k \to 4k$	$T \to T/2$	stiffness accelerates the clock — but only as $\sqrt{k}$
$m \to 4m$	$T \to 2T$	inertia drags the clock — as $\sqrt{m}$
vertical, on the Moon	same $T$ , sag $x_0 \to x_0/6$	the spring clock is gravity-blind; the sag is not
in a freely falling lift	same $T$ , oscillation centres at natural length	gravity gone: $x_0 = 0$ , dynamics unchanged
spring cut in half	$k' = 2k$ , so $T' = T/\sqrt2$	preview of the cut-and-combined-springs archetype

Common trap“Pull it farther and each cycle takes longer”

The most tempting wrong intuition about oscillators. A bigger pull means more distance to cover — but it also means proportionally bigger forces and speeds all along the path, and for a linear force the two effects cancel to the last decimal: $T = 2\pi\sqrt{m/k}$ contains no $A$ . Watch the companion traps in the same breath: $v_{\max} = \omega A$ but $a_{\max} = \omega^2 A$ — one power of $\omega$ apart, endlessly swapped in MCQs — and maximum speed happens at $x = 0$ where the force is zero, not at the turning points where the force is largest. Speed and force peak half a cycle out of step; any option pairing "maximum speed" with "maximum force" is wrong on sight.

Worked examples

1 · Board level — the standard setup (CBSE/ISC/AP pattern)

A $0.50\ \mathrm{kg}$ block on a frictionless table is attached to a spring of constant $k = 200\ \mathrm{N/m}$ , pulled $8.0\ \mathrm{cm}$ from equilibrium and released. Find $\omega$ , $T$ , $v_{\max}$ , and the block's speed at $x = 4.0\ \mathrm{cm}$ .

Step 1. Hardware first: $\omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20\ \mathrm{rad/s}$ .

Step 2. Period: $T = 2\pi/\omega = 2\pi/20 = \boxed{0.314\ \mathrm{s}}$ — about three cycles per second, whether the pull was 8 cm or 2 cm.

Step 3. Top speed at the centre: $v_{\max} = \omega A = 20 \times 0.080 = \boxed{1.6\ \mathrm{m/s}}$ .

Step 4. Speed at half amplitude: $v = \omega\sqrt{A^2 - x^2} = 20\sqrt{(0.080)^2 - (0.040)^2} = 20 \times 0.0693 = \boxed{1.39\ \mathrm{m/s}}$ — still 87% of top speed at half the displacement (the energy module's ¾ landmark: $v = v_{\max}\sqrt3/2$ ✓).

2 · Board level — reading the hardware off a stopwatch (A-Level/IB pattern)

A $1.2\ \mathrm{kg}$ block hanging from a spring completes 15 oscillations in $9.0\ \mathrm{s}$ with amplitude $5.0\ \mathrm{cm}$ . Find the spring constant, the maximum acceleration, and the equilibrium sag. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Period from timing many cycles (the standard accuracy trick): $T = 9.0/15 = 0.60\ \mathrm{s}$ , so $\omega = 2\pi/T = 10.5\ \mathrm{rad/s}$ .

Step 2. Invert for stiffness: $k = m\omega^2 = 1.2 \times (10.5)^2 = \boxed{132\ \mathrm{N/m}}$ . That the spring hangs vertically is irrelevant — gravity shifts the equilibrium, not the period.

Step 3. Maximum acceleration: $a_{\max} = \omega^2 A = (10.5)^2 \times 0.050 = \boxed{5.5\ \mathrm{m/s^2}}$ — over half a $g$ , at the turning points.

Step 4. The sag: $x_0 = mg/k = 1.2 \times 9.8/132 = \boxed{8.9\ \mathrm{cm}}$ . Audit via the shortcut: $T = 2\pi\sqrt{x_0/g} = 2\pi\sqrt{0.089/9.8} = 0.60$ s ✓.

3 · Competitive level — the added-mass ratio (JEE-Main/NEET pattern, ~2 minutes)

A block of unknown mass $m$ hanging from a spring oscillates with period $2.0\ \mathrm{s}$ . When a $2.0\ \mathrm{kg}$ load is gently attached to it, the period becomes $3.0\ \mathrm{s}$ . Find $m$ .

Step 1. The ratio kills $k$ before it is ever needed: $\dfrac{T_2}{T_1} = \sqrt{\dfrac{m + 2}{m}}$ .

Step 2. Square: $\left(\dfrac{3}{2}\right)^2 = \dfrac{m+2}{m}$ , so $9m = 4m + 8$ and $m = \boxed{1.6\ \mathrm{kg}}$ .

Step 3. Sense check: mass grew by a factor $3.6/1.6 = 2.25$ , and $\sqrt{2.25} = 1.5 = T_2/T_1$ ✓. Timing note: ~90 s. Never compute $k$ — the ratio is the whole method, and the numbers are chosen so the squares are clean. Distractor watch: solving $T_2/T_1 = (m+2)/m$ without the square root gives $m = 4$ kg, planted among the options every time.

4 · Competitive level — JEE-Advanced stretch: where you drop the load matters (~4 minutes)

A $3.0\ \mathrm{kg}$ block on a smooth floor oscillates on a spring with amplitude $8.0\ \mathrm{cm}$ . A $1.0\ \mathrm{kg}$ lump of putty is dropped vertically onto the block and sticks. Find the new amplitude if the putty lands (a) as the block crosses the mean position, (b) as the block sits momentarily at the extreme.

Step 1. Case (a) — landing at the mean. The block is at top speed $v_{\max} = \omega A$ . The vertical drop-and-stick is a horizontal-momentum event: $mv_{\max} = (m + \Delta m)v'$ , so $v' = \tfrac{3}{4}\omega A$ . Kinetic energy is lost to the splat — momentum, not energy, crosses a sticking collision.

Step 2. The new system has $\omega' = \sqrt{k/(m+\Delta m)} = \omega\sqrt{3/4}$ . The landing happened at the centre, so $v'$ is the new maximum speed: $A' = \dfrac{v'}{\omega'} = \dfrac{(3/4)\,\omega A}{\omega\sqrt{3/4}} = A\sqrt{\dfrac{3}{4}}$ . Numerically $A' = 8.0 \times 0.866 = \boxed{6.9\ \mathrm{cm}}$ .

Step 3. Case (b) — landing at the extreme. The block is at rest; sticking to it changes no velocity and steals no energy. The spring's stored energy $\tfrac12 kA^2$ is untouched, so the turning point stays where it was: $A' = \boxed{8.0\ \mathrm{cm}}$ , unchanged. Only the clock changes: $T' = T\sqrt{4/3}$ in both cases.

Step 4. The architecture: one event, two conservation regimes. At the mean, the collision happens at full speed — momentum bookkeeping shrinks the amplitude by $\sqrt{m/(m+\Delta m)}$ . At the extreme, the collision happens at zero speed — nothing dynamical happens at all. Between the two, examiners have a full menu of variants. Timing note: ~3.5–4 min. Tag the landing position before writing anything — it decides the whole solution. Distractor watch: applying energy conservation across the sticky landing in case (a) gives $A' = 8\sqrt{3}/2 \approx 6.9$ … by luck-shaped coincidence? No: it gives $A' = A\sqrt{3/4}$ only if you (wrongly) conserve energy and (wrongly) scale — the honest route is momentum, then energy after. Write the momentum line explicitly; JEE marks the method.

Exam traps

Amplitude in the period. $T = 2\pi\sqrt{m/k}$ — no $A$ . Any option in which the period responds to the pull is wrong, and any lab question about "does amplitude affect $T$ " wants no, for small ideal-spring oscillations with the isochronism argument.

Gravity in the period. Also absent. Vertical, horizontal, Moon, lift — the spring clock reads the same. What gravity moves is the centre (sag $mg/k$ ); questions that show a vertical spring and offer $T = 2\pi\sqrt{m/k \pm \text{something}\,g}$ are testing exactly this reflex. (The pendulum behaves oppositely — keep the two clocks separate in your head.)

$\sqrt{}$ discipline in ratios. $T \propto \sqrt{m}$ and $T \propto 1/\sqrt{k}$ : quadruple the mass to double the period; "double the mass doubles the period" is the planted error. Ratio questions want the square-then-solve routine of Example 3.

$v_{\max}$ vs $a_{\max}$ . $\omega A$ versus $\omega^2 A$ . Given both, $\omega = a_{\max}/v_{\max}$ and $A = v_{\max}^2/a_{\max}$ — two of the fastest inversions in the chapter, and two of the most swapped.

Measuring $x$ from the natural length in vertical problems. All SHM formulas want displacement from the sagged equilibrium. From the natural length you must carry gravity separately, and mixing the two references double-counts $mg$ — the energy module's headline warning, and the most common wrong setup in vertical-spring energy parts.

Cut springs jumping early. "The spring is cut in half, so $k$ halves" — backwards: half the coils share the same pull, each stretches half as much, so the half-spring is twice as stiff, $k' = 2k$ and $T' = T/\sqrt2$ . The full cut-and-combined story is the next archetype; here just refuse the reflex.

Which $g$ ? Only the sag uses it: CBSE/JEE/NEET papers usually declare $g = 10\ \mathrm{m/s^2}$ (then $x_0 = mg/k$ comes out clean); IB uses 9.8, CIE 9.81. The period needs no $g$ at all — if your $T$ formula has $g$ in it, re-read the question to check it isn't a pendulum.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · London

Hooke was the irascible polymath of the early Royal Society — microscopist, architect of post-Fire London, coiner of the biological word "cell" — and the man who found the force law that makes this module tick. He published it in 1676 as an anagram, ceiiinosssttuv, unscrambled two years later to ut tensio, sic vis: as the extension, so the force. He understood immediately what linearity buys: a spring's oscillations keep time regardless of how hard they swing, so a coiled spring could replace the pendulum in a portable clock. His spring-regulated watch designs collided with Huygens' balance-spring patent in 1675 and ignited one of science's nastiest priority feuds — a feud only settled by history calling the law "Hooke's" and the clock "Huygens'". Every mechanical watch since carries both men's physics in one oscillating spring.

History

1581

— Galileo, timing a swinging chandelier in Pisa against his pulse, notices that wide and narrow swings keep the same time.

1656

— Huygens builds the first pendulum clock; timekeeping error collapses from minutes to seconds per day.

1675

— Huygens invents the spiral balance-spring watch regulator; Hooke claims priority and the quarrel echoes for decades.

1676

— Hooke publishes the linear law of elasticity as an anagram; ut tensio, sic vis makes spring oscillation exactly harmonic.

1687

— Newton's Principia supplies

F = ma

, turning Hooke's force law into the equation of harmonic motion.

1927

— Marrison and Horton build the first quartz clock: a vibrating crystal replaces the spring, isochronism now electronic.

Going further

Texts. NCERT Physics Class 11, Ch. 14 (oscillations of a loaded spring) · HC Verma, Concepts of Physics Vol. 1, Ch. 12 · Resnick–Halliday–Walker Ch. 15 · The university-track spring articles carry the mass-of-spring correction and the full differential-equation treatment.

Provenance. All results are standard. A verifier should recompute: Example 1's $\omega = 20$ rad/s, $T = 0.314$ s, $v_{\max} = 1.6$ m/s, $v(4\ \mathrm{cm}) = 1.39$ m/s; Example 2's $k = 132$ N/m, $a_{\max} = 5.5\ \mathrm{m/s^2}$ , $x_0 = 8.9$ cm with the $T$ – $x_0$ audit; Example 3's $m = 1.6$ kg; Example 4's $A' = A\sqrt{3/4} = 6.9$ cm (mean) and $8.0$ cm (extreme).

Prerequisites. Periodic Motion & the Definition of SHM · SHM Kinematics · Spring Force & Hooke's Law · Energy in SHM. Leads to. The Simple Pendulum · Cut & Combined Springs (archetype) · Damped & Forced Oscillations.

Why must the motion be a sine or cosine — couldn't the block move some other way?

The force law leaves it no choice. $a = -\omega^2 x$ demands a function whose curvature is proportional to minus itself, and sinusoids are the only bounded functions with that property. You can verify rather than derive it at this level: substitute $x = A\cos(\omega t + \phi)$ , differentiate twice (conceptually: the slope of the slope), and watch the equation balance for any $A$ and $\phi$ provided $\omega^2 = k/m$ . Two free constants matching two facts about the start — where the block was, how fast it moved — is exactly enough freedom, so this family covers every possible motion of the system.

If gravity stretches the vertical spring, why doesn't it change the period?

Because gravity is a constant force, and constants can't oscillate anything — they only relocate the balance point. About the sagged equilibrium, every extra newton of spring force from displacement $y$ is $-ky$ , exactly as on the table; the $mg$ and the $kx_0$ cancel each other line by line. The test that this is real physics and not bookkeeping: the ISS body-mass chair keeps perfect spring time in free fall, where "down" has lost all meaning. Contrast the pendulum, whose restoring force is gravity — remove $g$ and it stops swinging entirely.

What actually decides the phase constant φ?

The starting snapshot. Release from rest at $+A$ : the cosine must start at its crest, $\phi = 0$ . Kick the block through equilibrium at $t = 0$ : the motion is a pure sine, $\phi = -\pi/2$ in the cosine convention. In general $x_0 = A\cos\phi$ and $v_0 = -\omega A \sin\phi$ pin both $A$ and $\phi$ from the initial position and velocity. Exams at this level nearly always choose one of the two clean starts; write "released from rest at the extreme ⇒ $x = A\cos\omega t$ " or "kicked from the mean ⇒ $x = A\sin\omega t$ " as your first line and the phase never troubles you again.

Does the spring's own mass matter?

The ideal-spring formulas assume it doesn't, and at board level that assumption is always granted. In reality the coils near the block move almost as fast as the block while the coils near the wall barely move, so the spring contributes part of its mass to the inertia — for a uniform spring, one-third of it: $T = 2\pi\sqrt{(m + m_s/3)/k}$ . JEE-Advanced has asked for this as a stated-formula application, and the lab version explains why measured periods run slightly long on light blocks. If the problem doesn't mention spring mass, take it as zero — silence is the licence.

Two identical springs — does connecting them to the block side by side really differ from end to end?

Enormously: side by side (parallel) they share the load, each supplying force for the full displacement, so stiffnesses add ( $2k$ , period shrinks by $\sqrt2$ ); end to end (series) they share the stretch, each feeling the full force, so compliances add ( $k/2$ , period grows by $\sqrt2$ ). Same two springs, a factor of 2 apart in effective stiffness and a factor 2 in $T^2$ . The full machinery — including the cut spring that becomes stiffer, $k' = nk$ — is the next module's archetype; the sanity anchor to carry there: more spring paths = stiffer, longer spring chain = softer.

The block on my lab track slows down and stops after a minute. Which formula was wrong?

None — the model was idealised. Friction and air drag steadily tax the energy budget ( $E \propto A^2$ , so the amplitude decays), and the oscillation dies. What is remarkable, and worth noticing in the data, is that the period barely changes while the amplitude collapses: isochronism holds to the end, each ever-smaller cycle taking the same time. The systematic story of that decay — and of what happens when you push back in rhythm — is the damped and forced oscillations module, one step ahead.