Simple harmonic motion — definition, restoring force F = −kx: Worked examples

Everything here rests on four facts the parent built. Keep them in view:

The scenario matrix

Every SHM problem is one of these cells. The worked examples below are tagged with the cell they hit, and together they cover the whole table.

Cell	What makes it different	Example
A. Period from $m, k$	plug into $T = 2 π m / k$	Ex 1
B. Speed at a position	energy $v = ω A^{2} - x^{2}$	Ex 2
C. Sign of force by quadrant	$x > 0$ vs $x < 0$ vs $x = 0$	Ex 3
D. Is-it-SHM test	linear vs non-linear force	Ex 4
E. Phase / initial conditions	find $A$ and $ϕ$ from start	Ex 5
F. Degenerate & limiting inputs	$x = 0$ , $x = \pm A$ , $k \to 0$ , $m \to \infty$	Ex 6
G. Real-world word problem	translate words → symbols	Ex 7
H. Exam twist (combined springs / energy split)	two ideas at once	Ex 8

Every numeric answer below is machine-checked at the bottom of the page.

Cell A — Period straight from $m$ and $k$

Worked example Example 1 — the plug-in

A block of mass $m = 0.50 kg$ sits on a spring of stiffness $k = 200 N/m$ . Find $T$ , $f$ , and $ω$ .

Forecast: stiffer spring → snappier → shorter $T$ ; heavier block → sluggish → longer $T$ . So a stiff spring with a light block should give a small $T$ . Guess "a fraction of a second".

Step 1 — Compute $ω$ . $ω = \frac{k}{m} = \frac{200}{0.50} = 400 = 20 rad/s .$ Why this step? $ω$ is the natural "spin rate" of the oscillation (the rate the phase angle grows); everything else (T, f) is built from it.

Step 2 — Period is one full turn of phase. $T = \frac{2 π}{ω} = \frac{2 π}{20} = 0.314 s .$ Why this step? One complete cycle happens when the phase $ω t$ advances by $2 π$ radians (one full lap of the imaginary dot), so $ω T = 2 π$ .

Step 3 — Frequency is cycles per second. $f = \frac{1}{T} = \frac{1}{0.314} = 3.18 Hz .$ Why this step? Frequency just inverts period — "how many wobbles per second" vs "seconds per wobble".

Verify: units of $k / m$ are $(N/m) / kg = (kg \cdot m/s^{2} / m) / kg = 1/ s^{2} = 1/ s$ ✓. Matches the parent's Example 1 value $0.314 s$ . ✓

Cell B — Speed at a chosen position

Worked example Example 2 — how fast, and where

Same spring ( $ω = 20 rad/s$ ), amplitude $A = 0.10 m$ . Find the speed at $x = 0.06 m$ , and confirm the maximum speed.

Forecast: $x = 0.06$ is past halfway to the edge $A = 0.10$ , so the block has already slowed from its centre top-speed. Guess a bit below $v_{m a x}$ .

Step 1 — Write energy speed formula. $v = ω A^{2} - x^{2} .$ Why this step? Energy conservation $\frac{1}{2} k A^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}$ solved for $v$ gives exactly this — no need to know the time, only the place.

Step 2 — Substitute numbers. $v = 20 0.1 0^{2} - 0.0 6^{2} = 20 0.01 - 0.0036 = 20 0.0064 = 20 (0.08) = 1.6 m/s .$ Why this step? With the formula already justified, this is the arithmetic — we plug the known $ω$ , $A$ and the target $x$ into it to turn the general rule into one number for this exact position.

Step 3 — Max speed check (at $x = 0$ ). $v_{m a x} = ω A = 20 (0.10) = 2.0 m/s .$ Why this step? Setting $x = 0$ in the formula removes the subtraction — centre is where all energy is kinetic, so speed peaks.

Verify: $1.6 < 2.0$ ✓ (slower than the centre, as forecast). Energy check: $\frac{1}{2} k A^{2} = \frac{1}{2} (200) (0.01) = 1.0 J$ ; and $\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} (0.5) (1. 6^{2}) + \frac{1}{2} (200) (0.0 6^{2}) = 0.64 + 0.36 = 1.0 J$ ✓.

The figure below plots this speed curve. Look at the blue arch: it is a half-ellipse peaking at the centre. The orange dot marks $v_{m a x} = 2.0 m/s$ at $x = 0$ ; the red dot marks our answer $v = 1.6 m/s$ at $x = 0.06$ , sitting lower on the curve; the green dots at $x = \pm A$ sit on the axis where $v = 0$ . The whole shape is the sentence "fastest in the middle, frozen at the edges".

Figure — Simple harmonic motion — definition, restoring force F = −kx

Cell C — Sign of the force in each region

This cell is where readers get burned. The formula $F = - k x$ hides three different situations depending on where the block is.

The figure shows three snapshots stacked vertically. In each, the gray tick is the equilibrium $x = 0$ and the blue dot is the block. Follow the red arrows: in the top row (block on the right) the arrow points left; in the middle row (block at home) there is no arrow, only "F=0"; in the bottom row (block on the left) the arrow points right. Every red arrow aims back at the centre — that is the whole meaning of the minus sign, drawn.

Worked example Example 3 — force direction on the right, at home, and on the left

A block on a spring, $k = 50 N/m$ . Give the force (magnitude and direction) when (a) $x = + 0.20 m$ , (b) $x = 0$ , (c) $x = - 0.20 m$ .

Forecast: the spring always shoves the block back toward the middle. So on the right it pushes left, at the middle it pushes nothing, on the left it pushes right. Magnitudes on the two sides should be equal.

Step 1 — Right side, $x = + 0.20$ . $F = - k x = - (50) (+ 0.20) = - 10 N .$ Why this step? Positive $x$ (right of centre) → the $-$ flips it → $F$ is negative → points left, back to home (top red arrow in the figure). The number's sign is the direction, not "a negative force".

Step 2 — Centre, $x = 0$ . $F = - (50) (0) = 0 N .$ Why this step? At home there is nothing to restore — this is the equilibrium, the only place force vanishes (middle row, no arrow). (But speed is maximum here — don't confuse zero force with rest.)

Step 3 — Left side, $x = - 0.20$ . $F = - (50) (- 0.20) = + 10 N .$ Why this step? Negative $x$ times the $-$ gives a positive force → points right, again toward home (bottom red arrow). Same size as the right side (10 N), opposite direction.

Verify: magnitudes equal ( $10 = 10$ ) ✓; both point inward ✓; force zero only at $x = 0$ ✓. This is the "direction rule" mistake from the parent, resolved by arithmetic.

Cell D — The "is it actually SHM?" test

Worked example Example 4 — two force laws, one imposter

A $2 kg$ mass feels (a) $F = - 8 x$ (N) or (b) $F = - 8 x^{3}$ (N). For each: is it SHM? If yes, find $T$ .

Forecast: SHM demands the force be straight-line in $x$ — exactly proportional, no curves. So (a) should pass; (b), with $x^{3}$ , should oscillate but not with a constant period.

Step 1 — Test (a) for linearity. $F = - 8 x$ has the form $F = - k x$ with $k = 8 N/m$ . Linear ✓ → SHM. Why this step? The single test for SHM is: is $F$ a constant times $- x$ ? Nothing else matters.

Step 2 — Period of (a). $T = 2 π \frac{m}{k} = 2 π \frac{2}{8} = 2 π 0.25 = 2 π (0.5) = 3.14 s .$ Why this step? Once linearity confirms it is SHM, the period formula applies, so we plug the read-off $k = 8$ and $m = 2$ straight in to turn "it's SHM" into an actual timing.

Step 3 — Test (b). $F = - 8 x^{3}$ : the "stiffness" $F / (- x) = 8 x^{2}$ changes with position. Not constant → not SHM. It still oscillates (force always restores), but $T$ would depend on amplitude — the defining SHM property fails. Why this step? If we tried $ω^{2} = k / m$ we'd have no fixed $k$ ; there is no single period, so the theory doesn't apply.

Verify: (a) $T = 3.14 s$ ; slope of $F$ vs $x$ is constant $- 8$ ✓. (b) slope of $F$ is $- 24 x^{2}$ , not constant ✓ (imposter caught).

Cell E — Phase and amplitude from initial conditions

The general SHM solution is $x (t) = A cos (ω t + ϕ)$ , where $ϕ$ (the phase constant, radians) tells us where in the cycle the clock started — it is the phase at $t = 0$ . We need two starting facts — a start position and a start velocity — to pin down the two unknowns $A$ and $ϕ$ .

Worked example Example 5 — released from rest, then given a push

A block has $ω = 4 rad/s$ . (a) Released from rest at $x_{0} = 0.05 m$ . Find $A$ and $ϕ$ . (b) Instead, started at $x_{0} = 0$ moving with $v_{0} = 0.20 m/s$ . Find $A$ and $ϕ$ .

Forecast: (a) starts at the far edge with zero speed → that edge is the amplitude, and cosine already sits at its peak at $t = 0$ , so $ϕ = 0$ . (b) starts dead centre at full speed → it's at the crossing point, so $ϕ = \pm π /2$ , and $A$ comes from the speed.

Step 1 — General position & velocity. $x = A cos (ω t + ϕ), v = \overset{x}{˙} = - A ω sin (ω t + ϕ) .$ Why this step? Two equations at $t = 0$ let us solve for the two unknowns $A, ϕ$ . (The overdot on $x$ just means we differentiated position in time to get velocity.)

Step 2 — Case (a): $x (0) = 0.05$ , $v (0) = 0$ . $v (0) = - A ω sin ϕ = 0 \Rightarrow sin ϕ = 0 \Rightarrow ϕ = 0$ . Then $x (0) = A cos 0 = A = 0.05 m$ . $A = 0.05 m, ϕ = 0.$ Why this step? Starting at rest means the sine (which carries velocity) must vanish; that forces $ϕ = 0$ , and the whole starting displacement is the amplitude.

Step 3 — Case (b): $x (0) = 0$ , $v (0) = + 0.20$ . $x (0) = A cos ϕ = 0 \Rightarrow cos ϕ = 0 \Rightarrow ϕ = \pm \frac{π}{2}$ . $v (0) = - A ω sin ϕ = + 0.20$ . With $ϕ = - π /2$ : $sin (- π /2) = - 1$ , so $v (0) = + A ω = 0.20$ . $A = \frac{0.20}{ω} = \frac{0.20}{4} = 0.05 m, ϕ = - \frac{π}{2} .$ Why this step? Starting at the centre forces cosine to zero, so $ϕ = \pm π /2$ ; we pick the sign that makes the initial velocity point the right way (positive here).

Verify: both give $A = 0.05 m$ (same energy, different starting phase) ✓. Case (b) with $ϕ = - π /2$ : $x = A cos (ω t - π /2) = A sin (ω t)$ , and $\overset{x}{˙} ∣_{0} = A ω = 0.20$ ✓.

Cell F — Degenerate and limiting inputs

You must never be surprised by an edge case. Here we push each quantity to its extreme. One tool we lean on is the acceleration formula $a = - ω^{2} x$ — let's earn it before using it.

Worked example Example 6 — the four edges

Using $E = \frac{1}{2} k A^{2}$ , $v = ω A^{2} - x^{2}$ , $a = - ω^{2} x$ , $T = 2 π m / k$ , describe: (a) $x = 0$ , (b) $x = \pm A$ , (c) $k \to 0$ , (d) $m \to \infty$ . Give numbers where possible with $ω = 10$ , $A = 0.10$ , $k = 100$ , $m = 1$ .

Forecast: centre = fastest & no force; edges = frozen instant but hardest push; a zero-stiffness spring can't oscillate; an infinitely heavy block is too sluggish to ever return.

Step 1 — (a) At the centre $x = 0$ . $v = ω A^{2} - 0 = ω A = 10 (0.10) = 1.0 m/s$ (maximum). $a = - ω^{2} (0) = 0$ . Why this step? All energy is kinetic here → top speed; no displacement → no restoring force.

Step 2 — (b) At the edge $x = A = 0.10$ . $v = ω A^{2} - A^{2} = 0$ . $a = - ω^{2} A = - (1 0^{2}) (0.10) = - 10 m/s^{2}$ (max magnitude). Why this step? The block is momentarily stopped (turning point) yet the force is largest — that maximum force is exactly what reverses it. (Refutes the "acceleration zero at ends" mistake.)

Step 3 — (c) Limit $k \to 0$ . $T = 2 π m / k \to \infty$ ; $ω = k / m \to 0$ . Why this step? A vanishing spring gives a vanishing restoring force → the "period" stretches to infinity → no oscillation at all. The limit warns us the formula degrades gracefully.

Step 4 — (d) Limit $m \to \infty$ . $T = 2 π m / k \to \infty$ ; $ω \to 0$ . Why this step? Infinite inertia can't be turned around in finite time → period diverges. Both degenerate limits ( $k \to 0$ , $m \to \infty$ ) kill the motion, for opposite physical reasons.

Verify: (a) $v_{m a x} = 1.0$ , $a = 0$ ✓; (b) $v = 0$ , $∣ a ∣ = 10$ ✓; (c),(d) $T \to \infty$ (monotone, checked at growing inputs) ✓.

Cell G — Real-world word problem

Worked example Example 7 — a car's suspension

A $1200 kg$ car body rests on four springs. When 4 passengers ( $300 kg$ total) climb in, the body sinks $0.03 m$ . Treating the four springs as one effective spring, find (a) the effective $k$ , (b) the bounce period of the loaded car ( $1500 kg$ ).

Forecast: cars bounce roughly once per second — expect $T$ near $1 s$ .

Step 1 — Get $k$ from the static sag (Hooke's law). Added weight $= m g = 300 \times 9.8 = 2940 N$ compresses the springs by $0.03 m$ : $k = \frac{F}{x} = \frac{2940}{0.03} = 98000 N/m .$ Why this step? At rest the spring force balances the added weight; the sag tells us the stiffness. See Hooke's Law.

Step 2 — Period of the loaded car. $T = 2 π \frac{m}{k} = 2 π \frac{1500}{98000} = 2 π 0.01531 = 2 π (0.1237) = 0.777 s .$ Why this step? Once we have $k$ and the oscillating mass, the SHM period formula does the rest.

Verify: $T = 0.78 s \approx 1 s$ ✓ — matches real suspension "feel". Units: $kg / (N/m) = s^{2} = s$ ✓.

Cell H — Exam twist: combined springs + energy split

Worked example Example 8 — two springs and a "where is KE = PE?" trap

Two identical springs, each $k = 300 N/m$ , are attached in parallel to a $0.75 kg$ block, amplitude $A = 0.08 m$ . (a) Find the effective $k$ and period. (b) At what displacement $x$ is the kinetic energy equal to the potential energy?

Forecast: parallel springs both pull together → effective spring is stiffer (add the $k$ 's), so period is shorter than one spring alone. For (b), the energy is shared 50/50 somewhere between centre and edge — guess around $x = A / 2 \approx 0.7 A$ .

Step 1 — Effective stiffness (parallel). $k_{eff} = k_{1} + k_{2} = 300 + 300 = 600 N/m .$ Why this step? Parallel springs each stretch by the same $x$ and their forces add, so their stiffnesses add.

Step 2 — Period. $T = 2 π \frac{m}{k _{eff}} = 2 π \frac{0.75}{600} = 2 π 0.00125 = 2 π (0.03536) = 0.222 s .$ Why this step? The two springs now behave as one spring of stiffness $k_{eff}$ , so the standard period formula applies — we feed in $k_{eff} = 600$ and the block's mass to get the actual bounce time.

Step 3 — Where KE = PE. Total $E = \frac{1}{2} k_{eff} A^{2}$ . Set $P E = \frac{1}{2} E$ : $\frac{1}{2} k x^{2} = \frac{1}{2} (\frac{1}{2} k A^{2}) \Rightarrow x^{2} = \frac{A ^{2}}{2}$ . $x = \frac{A}{2} = \frac{0.08}{1.4142} = 0.0566 m .$ Why this step? KE = PE means each holds half the total; potential is half the total exactly at $x = A / 2$ . See Energy in SHM.

Verify: at $x = 0.0566$ : $P E = \frac{1}{2} (600) (0.056 6^{2}) = 0.960 J$ ; total $E = \frac{1}{2} (600) (0.0 8^{2}) = 1.92 J$ ; $P E / E = 0.5$ ✓. Stiffer than one spring → $T = 0.222 s < 0.314 s$ from Ex 1 ✓.

Quick Recall

Recall Which cell is each trick? (hide and answer)

Force is $- 8 x^{3}$ : SHM? ::: No — stiffness $8 x^{2}$ isn't constant, so no fixed period (Cell D).
Released from rest at the edge → what is $ϕ$ ? ::: $ϕ = 0$ (cosine peaks at $t = 0$ ), and that edge is the amplitude (Cell E).
Started at centre with speed $v_{0}$ → amplitude? ::: $A = v_{0} / ω$ (Cell E, all energy kinetic there).
Parallel springs combine how? ::: $k_{eff} = k_{1} + k_{2}$ (add), so period gets shorter (Cell H).
KE equals PE at what displacement? ::: $x = A / 2 \approx 0.707 A$ (Cell H).
What happens to $T$ as $k \to 0$ or $m \to \infty$ ? ::: $T \to \infty$ — the motion dies (Cell F).

Connections

Parent: SHM definition & $F=-kx$ — the rules these examples exercise
Energy in SHM — the $\frac{1}{2} k A^{2}$ used in Examples 2, 6, 8
Hooke's Law — the static-sag $k = F / x$ in Example 7
Simple Pendulum — another $F = - k x$ in disguise
Uniform Circular Motion — where $A cos (ω t + ϕ)$ comes from
Damped Oscillations — what changes when a $- b v$ term joins
Taylor Series — why non-linear laws (Ex 4b) still look like SHM for tiny $x$