Spring-mass system — horizontal, vertical: Worked examples

The scenario matrix

Every spring-mass question is one of these cells. Each worked example below is tagged with the cell it fills.

#	Case class	What makes it special
A	Horizontal, find $T$	Pure $T = 2 π m / k$ , no gravity
B	Vertical, static stretch only	Use $T = 2 π y_{0} / g$ , $m$ and $k$ unknown
C	Energy: amplitude ↔ max speed	$\frac{1}{2} k A^{2} = \frac{1}{2} m v_{ma x}^{2}$
D	Position/velocity at a general instant	Full $x (t) = A cos (ω t + ϕ)$
E	Springs in parallel (stiffer)	$k_{e f f} = k_{1} + k_{2}$
F	Springs in series (softer)	$1/ k_{e f f} = 1/ k_{1} + 1/ k_{2}$
G	Limiting / degenerate inputs	$k \to \infty$ , $k \to 0$ , $m \to 0$ , $A = 0$
H	Real-world word problem	Translate words → symbols
I	Exam twist (combine ideas)	Vertical + energy + amplitude limit

Worked example Example 2 — Cell B: vertical, only the static stretch known

You hang an unknown mass on an unknown spring and it settles $y_{0} = 2.5$ cm below the natural length. Find the period of small vertical bounces.

Forecast: We know neither $m$ nor $k$ separately. Can we still get $T$ ? (Yes — the trick from the parent note.)

Convert units. $y_{0} = 2.5 cm = 0.025$ m. Why this step? SI throughout, else $g$ won't match.
Use the stretch formula. Since $k y_{0} = m g$ , the ratio $m / k = y_{0} / g$ , so $T = 2 π y_{0} / g$ . Why this step? Gravity cancels from the dynamics but hides $m / k$ inside the measured stretch — exactly what we exploit.
Substitute ( $g = 9.8$ m/s²): $T = 2 π 0.025/9.8 = 2 π 2.551 \times 1 0^{- 3}$ .
Evaluate. $2.551 \times 1 0^{- 3} = 0.05051$ , so $T = 2 π (0.05051) = 0.317$ s.

Verify: A $2.5$ cm droop is small → fast bounce → sub-second period ✓. Units: $m / (m/s^{2}) =$ s ✓.

Worked example Example 3 — Cell C: energy, amplitude ↔ max speed

Using Example 1's block ( $m = 2.0$ kg, $k = 50$ N/m), you pull it $A = 0.15$ m and release. Find the maximum speed and the total energy.

Forecast: Where is speed greatest — at the ends or the middle? (Middle, where the spring is relaxed.)

Find $ω$ . $ω = k / m = 50/2.0 = 25 = 5$ rad/s. Why this step? Max speed is $A ω$ ; we need $ω$ first.
Max speed. $v_{ma x} = A ω = 0.15 \times 5 = 0.75$ m/s. Why this step? At $x = 0$ all energy is kinetic (PE $= 0$ ), so speed peaks. See figure below.
Total energy. $E = \frac{1}{2} k A^{2} = \frac{1}{2} (50) (0.15)^{2} = 0.5625$ J. Why this step? At the turning point $x = A$ , KE $= 0$ , so all energy is spring PE $= \frac{1}{2} k A^{2}$ .

Verify: Cross-check via KE at centre: $\frac{1}{2} m v_{ma x}^{2} = \frac{1}{2} (2.0) (0.75)^{2} = 0.5625$ J $= E$ ✓. Energy is conserved as the trade in the figure shows.

Worked example Example 4 — Cell D: position & velocity at a chosen instant

The block of Examples 1&3 ( $ω = 5$ rad/s, $A = 0.15$ m) is released from rest at $x = + A$ at $t = 0$ . Find its position and velocity at $t = 0.20$ s.

Forecast: After $0.20$ s the phase is $ω t = 1.0$ rad $\approx 57°$ — less than a quarter cycle, so it should still be positive but heading toward centre.

Pick the right solution. Released from rest at the far end → $x (t) = A cos (ω t)$ (cosine starts at its max with zero slope). Why this step? "From rest at $x = + A$ " means $x (0) = A$ , $\overset{x}{˙} (0) = 0$ ; cosine matches both, so phase $ϕ = 0$ .
Position. $x = 0.15 cos (5 \times 0.20) = 0.15 cos (1.0) = 0.15 (0.5403) = 0.0810$ m. Why this step? Direct evaluation; angle $1.0$ is in radians.
Velocity. $\overset{x}{˙} = - A ω sin (ω t) = - 0.15 \times 5 \times sin (1.0) = - 0.75 (0.8415) = - 0.631$ m/s. Why this step? Velocity is the derivative; the minus sign says it's moving back toward centre — matching our forecast.

Verify: Energy check: $\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} (2) (0.631)^{2} + \frac{1}{2} (50) (0.081)^{2} = 0.398 + 0.164 = 0.5625$ J $= E$ ✓.

Worked example Example 7 — Cell G: limiting & degenerate inputs

Reason (no big arithmetic) about what happens to $T = 2 π m / k$ in four extreme cases.

Forecast: Guess each before reading the answer.

Infinitely stiff spring, $k \to \infty$ . $T = 2 π m / k \to 0$ . Why this step? Denominator explodes → root → 0. A rigid rod can't oscillate; the "bounce" is infinitely fast (never happens).
Zero stiffness, $k \to 0$ . $T \to \infty$ . Why this step? No restoring force at all → the mass never comes back → infinite period (no oscillation).
Massless block, $m \to 0$ . $T \to 0$ . Why this step? No inertia to overshoot with → the spring snaps it home instantly.
Zero amplitude, $A = 0$ . $T$ is unchanged ( $= 2 π m / k$ ), but the mass simply sits still. Why this step? Period does not depend on $A$ (that's the hallmark of SHM — isochronism). $A = 0$ is the trivial "no motion" solution; the clock still ticks at the same rate.

Verify: All consistent with the formula's structure: $T$ rises with $m$ , falls with $k$ , and is independent of $A$ ✓.

Worked example Example 8 — Cell H: real-world word problem

A car of mass $m = 1200$ kg rests on four identical suspension springs. Each corner carries $1/4$ of the weight and compresses by $y_{0} = 8.0$ cm. Estimate the bounce period of the car body.

Forecast: A big car on soft springs — bounce period around a second? Let's see.

Model it. Treat the four springs as one effective spring supporting the whole mass. The measured droop $y_{0}$ under the full load already encodes $m / k_{e f f}$ . Why this step? "Static stretch under the load it carries" is exactly what $T = 2 π y_{0} / g$ needs — no need to find $k$ or split the mass.
Convert. $y_{0} = 8.0 cm = 0.080$ m.
Apply. $T = 2 π y_{0} / g = 2 π 0.080/9.8 = 2 π 8.163 \times 1 0^{- 3}$ .
Evaluate. $8.163 \times 1 0^{- 3} = 0.09035$ , so $T = 2 π (0.09035) = 0.568$ s.

Verify: About half a second per bounce — realistic for a car (real cars use dampers, see Damped Oscillations, to kill this quickly). Units: s ✓. Notice we never needed the mass at all — the stretch did all the work.

Worked example Example 9 — Cell I: exam twist (vertical + energy + amplitude limit)

A mass hangs on a spring; the static stretch is $y_{0} = 0.05$ m. You start it oscillating with amplitude $A$ . What is the largest amplitude for which the spring never goes slack (i.e. the mass never rises above the natural length)?

Forecast: The spring stays taut only while its extension stays positive. The equilibrium sits at extension $y_{0}$ . So the highest point of the swing must not overshoot the natural length. Guess: $A_{ma x} = y_{0}$ ?

Locate the top of the swing. Oscillation about equilibrium (extension $y_{0}$ ) with amplitude $A$ reaches its highest point at extension $y_{0} - A$ . Why this step? Amplitude is the distance from equilibrium to a turning point; the top turning point is $A$ above equilibrium.
Impose "spring not slack." The spring is at its natural length when extension $= 0$ . To stay taut: $y_{0} - A \geq 0 \Rightarrow A \leq y_{0}$ . Why this step? If $A > y_{0}$ the mass rises past the natural length; the spring would need to push (compress) — a hanging spring can't, so it goes slack and the motion stops being clean SHM.
State the limit and period. $A_{ma x} = y_{0} = 0.05$ m. The period (as long as it stays taut) is $T = 2 π y_{0} / g = 2 π 0.05/9.8 = 2 π (0.07143) = 0.449$ s. Why this step? Same $T = 2 π y_{0} / g$ trick from Energy in SHM/parent — independent of amplitude while SHM holds.

Verify: At $A = A_{ma x} = y_{0}$ the top of the swing exactly grazes the natural length (extension $0$ ) — the boundary case ✓. $T$ : units s, value sub-second, consistent with a $5$ cm stretch ✓.

Recall Quick self-test

Horizontal, $m = 2$ kg, $k = 50$ N/m — period? ::: $2 π 2/50 = 1.257$ s. Two springs $50$ and $30$ N/m in parallel — effective $k$ ? ::: $80$ N/m (they add). Same two in series — effective $k$ ? ::: $18.75$ N/m (reciprocals add). As $k \to \infty$ , the period $T \to$ ? ::: $0$ (an infinitely stiff spring can't oscillate). Largest amplitude before a hanging spring goes slack? ::: $A_{ma x} = y_{0}$ , the static stretch.

Connections

Parent: Spring-mass system — the formulas used here.
Simple Harmonic Motion — why $T$ is independent of amplitude (isochronism, Cell G/I).
Hooke's Law — the $F = - k x$ behind every cell.
Energy in SHM — the $\frac{1}{2} k A^{2}$ used in Cells C, D, I.
Springs in Series and Parallel — Cells E and F.
Simple Pendulum — contrast: its $T$ does depend on $g$ .
Damped Oscillations — the real car (Cell H) is damped, not perpetual.