SHM differential equation — solution - x = A cos(ωt + φ): Worked examples

This page is a worked-example machine for the solution $x=A\cos(\omega t+\phi)$. The equation is solved once; the art is in choosing the two constants $A$ (amplitude) and $ϕ$ (phase) correctly for whatever starting state you are handed. Every possible starting state — sign of position, sign of velocity, starting at the wall, starting at the centre, zero everything — gets its own worked cell below.

Before symbols do any work, recall the two master formulas (both derived in the parent note, never guessed):

The two sign checks (memorise these, use them everywhere)

At $t = 0$ our solution and its derivative read: $x_{0} = A cos ϕ, v_{0} = - A ω sin ϕ .$ Since $A > 0$ and $ω > 0$ :

The picture above is the phase compass: pin $cos ϕ$ 's sign (left/right) and $sin ϕ$ 's sign (up/down) and the arrow lands in one quadrant. Refer back to it in every example.

The scenario matrix

Every SHM "find $x (t)$ " problem is one of these cells. The Example that covers each cell is named. The four nonzero sign combinations of $(x_{0}, v_{0})$ are all present (Cells E, F, G, K).

Cell	Starting position $x_{0}$	Starting velocity $v_{0}$	Which quadrant is $ϕ$ ?	Covered by
A	$+ A$ (at wall)	$0$ (released from rest)	$ϕ = 0$	Ex 1
B	$- A$ (at far wall)	$0$	$ϕ = π$	Ex 2
C	$0$ (centre)	$+$ (moving right)	$ϕ = - π /2$	Ex 3
D	$0$ (centre)	$-$ (moving left)	$ϕ = + π /2$	Ex 4
E	$+$	$+$	Quadrant IV: $- π /2 < ϕ < 0$	Ex 5
F	$+$	$-$	Quadrant I: $0 < ϕ < π /2$	Ex 6
G	$-$	$-$	Quadrant II: $π /2 < ϕ < π$	Ex 7
K	$-$	$+$	Quadrant III: $- π < ϕ < - π /2$	Ex 8
H	$0$ and $0$	$0$	degenerate: no motion	Ex 9
I	word problem (pendulum)	given period + push	full pipeline	Ex 10
J	exam twist (find first time at a position)	—	inverting the cosine	Ex 11

For every example we use the same spring unless told otherwise: $m = 0.5 kg$ , $k = 200 N/m$ , so $ω = k / m = 200/0.5 = 400 = 20 rad/s .$

Example 1 — Cell A: released from rest at the $+$ wall

Worked example Statement

The spring mass is pulled to $x_{0} = + 0.1 m$ and released from rest ( $v_{0} = 0$ ). Find $x (t)$ and the max speed.

Forecast: where in the cosine cycle does "starting stationary at the far right" sit? (Guess $ϕ$ before reading.)

1. Amplitude. $A = x_{0}^{2} + v_{0}^{2} / ω^{2} = 0. 1^{2} + 0 = 0.1 m$ . Why this step? With $v_{0} = 0$ the whole energy is stretch-energy, so the start point is the turning point — that is literally the definition of amplitude.

2. Phase. $cos ϕ$ has the sign of $x_{0} > 0$ (positive) and $sin ϕ$ has the opposite sign of $v_{0} = 0$ (zero). Positive cosine, zero sine ⇒ $ϕ = 0$ . Why this step? $tan ϕ = - v_{0} / (ω x_{0}) = 0$ gives $ϕ = 0$ or $π$ ; the positive-cosine check kills $π$ .

3. Assemble. $x (t) = 0.1 cos (20 t) m$ ; $v_{m a x} = A ω = 0.1 \times 20 = 2 m/s$ . Why this step? Plugging the two constants back into the master solution $x = A cos (ω t + ϕ)$ is the whole point — the constants were the only unknowns left.

Verify: At $t = 0$ : $x = 0.1 cos 0 = 0.1$ ✓, $v = - 0.1 \cdot 20 sin 0 = 0$ ✓. Units of $A ω$ : $m \cdot s^{- 1} = m/s$ ✓.

Example 2 — Cell B: released from rest at the $-$ wall

Worked example Statement

Same spring, pulled to $x_{0} = - 0.1 m$ , released from rest ( $v_{0} = 0$ ). Find $x (t)$ .

Forecast: starting at the left wall must be a half-turn different from Example 1. Which $ϕ$ ?

1. Amplitude. $A = (- 0.1)^{2} + 0 = 0.1 m$ . Why this step? Amplitude is a distance, so the sign of $x_{0}$ drops out — squaring erases it.

2. Phase. $cos ϕ$ takes the sign of $x_{0} < 0$ (negative); $sin ϕ = 0$ . Negative cosine, zero sine ⇒ $ϕ = π$ . Why this step? $tan ϕ = 0$ again offers $0$ or $π$ ; now the negative-cosine requirement forces $π$ . This is exactly why the mistake "always $ϕ = 0$ " fails.

3. Assemble. $x (t) = 0.1 cos (20 t + π) = - 0.1 cos (20 t) m$ . Why this step? We substitute $A = 0.1$ and $ϕ = π$ into $x = A cos (ω t + ϕ)$ ; the identity $cos (θ + π) = - cos θ$ then rewrites it in the cleaner mirror form.

Verify: $x (0) = 0.1 cos π = - 0.1$ ✓. $v (0) = - 0.1 \cdot 20 sin π = 0$ ✓.

Example 3 — Cell C: centre, moving right

Worked example Statement

The mass passes through $x_{0} = 0$ moving in $+ x$ with $v_{0} = + 2 m/s$ . Find $x (t)$ .

Forecast: at the centre it is fastest, so this is the moment of $v_{m a x}$ . That should turn cosine into a sine. Which sign of sine?

1. Amplitude. $A = 0 + v_{0}^{2} / ω^{2} = (2/20)^{2} = 0.1 m$ . Why this step? At the centre all energy is kinetic, so $A = v_{0} / ω$ — the reach is entirely bought by speed.

2. Phase. Here $x_{0} = 0$ , so the formula $tan ϕ = - v_{0} / (ω x_{0})$ is undefined (division by zero) — we drop it and use the direct rule instead. $x_{0} = 0$ ⇒ $cos ϕ = 0$ ⇒ $ϕ = \pm π /2$ . Then $v_{0} > 0$ needs $sin ϕ < 0$ (because $v_{0} = - A ω sin ϕ$ ). Negative sine ⇒ $ϕ = - π /2$ . Why this step? $tan ϕ \to - \infty$ can't distinguish $+ π /2$ from $- π /2$ ; only the velocity sign does.

3. Assemble. $x (t) = 0.1 cos (20 t - π /2) = 0.1 sin (20 t) m$ . Why this step? Substituting the constants and using $cos (θ - π /2) = sin θ$ reveals the pure sine — the signature of a centre-crossing start.

Verify: $x (0) = 0.1 cos (- π /2) = 0$ ✓. $v (0) = - 0.1 \cdot 20 sin (- π /2) = + 2$ ✓.

Example 4 — Cell D: centre, moving left

Example 5 — Cell E: positive position, positive velocity (Quadrant IV of $ϕ$ )

Worked example Statement

The mass is at $x_{0} = + 0.06 m$ and moving outward to the right with $v_{0} = + 1.6 m/s$ . Find $A$ and $ϕ$ .

Forecast: starting on the + side and still speeding outward — it hasn't reached the wall yet, so it's before the turning point in its cycle. Is $ϕ$ positive or negative?

1. Amplitude. $A = 0.0 6^{2} + (1.6/20)^{2} = 0.0036 + 0.0064 = 0.01 = 0.1 m$ . Why this step? Position gives part of the reach, leftover speed gives the rest; Pythagoras adds them because energy (kinetic + potential) adds.

2. Phase. $cos ϕ$ sign = sign of $x_{0} > 0$ (+); $sin ϕ$ sign = opposite of $v_{0} > 0$ ⇒ negative. Positive cosine + negative sine = Quadrant IV, so $- π /2 < ϕ < 0$ . $tan ϕ = - v_{0} / (ω x_{0}) = - 1.6/ (20 \cdot 0.06) = - 1.3 \overline{3}$ ⇒ $ϕ = arctan (- 1.333...) = - 0.9273 rad$ (this lands in Quadrant IV, so it's the correct answer directly). Why this step? Because the raw $arctan$ already returns a Quadrant IV angle, no $\pm π$ correction is needed — the compass confirms it.

3. Assemble. $x (t) = 0.1 cos (20 t - 0.9273) m$ . Why this step? We drop the found constants into the master solution; nothing simplifies further, so we leave $ϕ$ as the numeric radian value.

Verify: $A cos ϕ = 0.1 cos (- 0.9273) = 0.06$ ✓ ( $= x_{0}$ ). $- A ω sin ϕ = - 0.1 \cdot 20 \cdot sin (- 0.9273) = 1.6$ ✓ ( $= v_{0}$ ).

Example 6 — Cell F: positive position, negative velocity (Quadrant I of $ϕ$ )

Worked example Statement

Same numbers as Example 5 but now the mass at $x_{0} = + 0.06 m$ is moving inward (returning), $v_{0} = - 1.6 m/s$ . Find $ϕ$ .

Forecast: it's past the wall and heading home — the later part of the outgoing swing. Should $ϕ$ be positive now?

1. Amplitude. Identical: $A = 0.0 6^{2} + (- 1.6/20)^{2} = 0.1 m$ (velocity squared).

2. Phase. $cos ϕ$ sign = $x_{0} > 0$ (+); $sin ϕ$ sign = opposite of $v_{0} < 0$ ⇒ positive. Positive cosine + positive sine = Quadrant I, so $0 < ϕ < π /2$ . Now $tan ϕ = - v_{0} / (ω x_{0}) = - (- 1.6) / (20 \cdot 0.06) = + 1.3 \overline{3}$ and $arctan (1.333...) = + 0.9273 rad$ — already in Quadrant I. ✓ Why this step? The sign of $v_{0}$ flipped, which flipped the sign of $tan ϕ$ , sending the same-magnitude angle from Quadrant IV (Ex 5) to Quadrant I. Same position, opposite travel ⇒ mirror-image phase.

3. Assemble. $x (t) = 0.1 cos (20 t + 0.9273) m$ . Why this step? Substituting the constants completes the solution; the positive $ϕ$ shifts the curve so the mass is already on its way back at $t = 0$ .

Verify: $A cos ϕ = 0.1 cos (0.9273) = 0.06$ ✓. $- A ω sin ϕ = - 2 sin (0.9273) = - 1.6$ ✓.

Example 7 — Cell G: negative position, negative velocity (arctan lands in the wrong quadrant)

Worked example Statement

The mass is at $x_{0} = - 0.06 m$ moving left with $v_{0} = - 1.6 m/s$ (heading toward the far wall). Find $ϕ$ . Watch the arctan trap.

Forecast: on the − side and moving further −. The compass says cosine negative and sine positive → Quadrant II. Will the raw calculator give that?

1. Amplitude. $A = (- 0.06)^{2} + (- 1.6/20)^{2} = 0.1 m$ .

2. Phase (the trap). $cos ϕ$ sign = $x_{0} < 0$ (−); $sin ϕ$ sign = opposite of $v_{0} < 0$ ⇒ positive. Negative cosine + positive sine = Quadrant II, so $π /2 < ϕ < π$ . But $tan ϕ = - v_{0} / (ω x_{0}) = - (- 1.6) / (20 \cdot (- 0.06)) = 1.6/ (- 1.2) = - 1.3 \overline{3}$ , and a calculator returns $arctan (- 1.333...) = - 0.9273 rad$ — that's Quadrant IV, the wrong place! Add $π$ : $ϕ = - 0.9273 + π = 2.2143 rad$ (Quadrant II ✓). Why this step? $tan$ repeats every $π$ , so $arctan$ can be off by exactly $π$ . The sign compass tells you when to add it — here cosine must be negative, so we push the angle into Quadrant II.

3. Assemble. $x (t) = 0.1 cos (20 t + 2.2143) m$ . Why this step? We substitute the corrected constants; the large positive $ϕ$ places the mass on the negative side heading outward at $t = 0$ .

Verify: $A cos ϕ = 0.1 cos (2.2143) = - 0.06$ ✓. $- A ω sin ϕ = - 2 sin (2.2143) = - 1.6$ ✓.

The figure plots the same magnitude problem for Cells E, F and G on one time axis: identical amplitude, three different phases sliding the curve left/right. That sliding is $ϕ$ .

Example 8 — Cell K: negative position, positive velocity (Quadrant III of $ϕ$ )

Worked example Statement

The mass is at $x_{0} = - 0.06 m$ but moving right (back toward the centre) with $v_{0} = + 1.6 m/s$ . Find $ϕ$ . This is the last missing sign combination.

Forecast: on the − side but heading home. The compass says cosine negative, sine negative → Quadrant III. Will the calculator land there, or must we correct?

1. Amplitude. $A = (- 0.06)^{2} + (1.6/20)^{2} = 0.0036 + 0.0064 = 0.1 m$ (signs squared away).

2. Phase (the trap again). $cos ϕ$ sign = $x_{0} < 0$ (−); $sin ϕ$ sign = opposite of $v_{0} > 0$ ⇒ negative. Negative cosine + negative sine = Quadrant III, so $- π < ϕ < - π /2$ . But $tan ϕ = - v_{0} / (ω x_{0}) = - (1.6) / (20 \cdot (- 0.06)) = - 1.6/ (- 1.2) = + 1.3 \overline{3}$ , and a calculator returns $arctan (1.333...) = + 0.9273 rad$ — that's Quadrant I, wrong! Subtract $π$ : $ϕ = 0.9273 - π = - 2.2143 rad$ (Quadrant III ✓). Why this step? $arctan$ collapsed the two-cosine-signs into one; the compass demands negative cosine, so we shift by $- π$ into Quadrant III. This mirrors Example 7 but with the velocity reversed.

3. Assemble. $x (t) = 0.1 cos (20 t - 2.2143) m$ . Why this step? Substituting the corrected constants finishes the solution; the negative $ϕ$ places the mass on the negative side heading inward at $t = 0$ .

Verify: $A cos ϕ = 0.1 cos (- 2.2143) = - 0.06$ ✓. $- A ω sin ϕ = - 2 sin (- 2.2143) = + 1.6$ ✓.

Example 9 — Cell H: the degenerate case (zero, zero)

Worked example Statement

The mass sits at $x_{0} = 0$ with $v_{0} = 0$ . What is $x (t)$ ?

Forecast: no position, no push — will it ever move?

1. Amplitude. $A = 0 + 0/ ω^{2} = 0$ . Why this step? Amplitude is total reach; with no displacement and no speed there is zero energy to fund any reach.

2. Phase. With $A = 0$ , $ϕ$ does not matter — $0 \cdot cos (anything) = 0$ . The phase becomes meaningless (undefined but irrelevant). Why this step? This is the honest edge case: the two constants collapse to a single trivial solution. Don't hunt for a quadrant that doesn't exist.

3. Assemble. $x (t) = 0$ for all $t$ — the mass stays at equilibrium forever. Why this step? Substituting $A = 0$ into $x = A cos (ω t + ϕ)$ leaves the zero function — the physically obvious "nothing happens" state.

Verify: $\overset{x}{¨} + ω^{2} x = 0 + ω^{2} \cdot 0 = 0$ ✓ — the zero function trivially solves the ODE. Physically: equilibrium is a valid (boring) SHM state.

Example 10 — Cell I: word problem, a pendulum from period + push

Worked example Statement

A simple pendulum swinging through small angles has period $T = 2 s$ . At $t = 0$ its bob is at the bottom ( $x_{0} = 0$ ) and is given a push producing $v_{0} = + 0.15 m/s$ along the arc. Find $ω$ , $A$ , $ϕ$ , and $x (t)$ .

Forecast: bottom of the swing = centre of the motion. Same shape as Example 3, only $ω$ comes from the period instead of from $k / m$ .

0. Legality check. A pendulum only obeys $\overset{x}{¨} = - ω^{2} x$ in the small-angle approximation, where $sin θ \approx θ$ makes the restoring force linear in displacement. Because the push is small, the swing stays small and this linear SHM solution is valid — otherwise the period would depend on amplitude and $x = A cos (ω t + ϕ)$ would not apply. Why this step? The whole matrix assumes true (linear) SHM; we must confirm the system qualifies before using the formulas.

1. Angular frequency. $ω = 2 π / T = 2 π /2 = π \approx 3.1416 rad/s$ . Why this step? The cosine repeats when its argument grows by $2 π$ , so one full period corresponds to $ω T = 2 π$ . We invert that.

2. Amplitude. $A = 0 + v_{0}^{2} / ω^{2} = v_{0} / ω = 0.15/ π \approx 0.04775 m$ . Why this step? Starting at the centre ⇒ all kinetic ⇒ reach $= v_{0} / ω$ , exactly Cell C's logic.

3. Phase. $x_{0} = 0$ makes $tan ϕ$ undefined, so use the direct rule: $cos ϕ = 0$ and $v_{0} > 0$ needs $sin ϕ < 0$ ⇒ $ϕ = - π /2$ .

4. Assemble. $x (t) = \frac{0.15}{π} cos (π t - \frac{π}{2}) = \frac{0.15}{π} sin (π t) m$ . Why this step? Substituting the three found constants and using $cos (θ - π /2) = sin θ$ gives the clean sine of a centre-start swing.

Verify: $x (0) = 0$ ✓. $v (0) = - A ω sin (- π /2) = A ω = (0.15/ π) \cdot π = 0.15$ ✓. Units of $2 π / T$ : $rad / s$ ✓.

Example 11 — Cell J: the exam twist (invert the cosine to find a time)

Worked example Statement

Using Example 1's motion $x (t) = 0.1 cos (20 t)$ : at what first time $t_{1} > 0$ does the mass reach $x = 0.05 m$ ? Also find its speed there.

Forecast: $0.05$ is halfway to the wall. Because cosine is steep near the top and flat later, will $t_{1}$ be more or less than a quarter period?

1. Set up the equation. $0.05 = 0.1 cos (20 t) \Rightarrow cos (20 t) = 0.5$ . Why this step? We know the position and want the time, so we must undo the cosine — that is what $arccos$ is for: "which angle has this cosine?"

2. Invert. $20 t = arccos (0.5) = π /3 rad$ (the smallest positive solution). So $t_{1} = \frac{π /3}{20} = \frac{π}{60} \approx 0.05236 s$ . Why this step? $arccos$ returns the first angle in $[0, π]$ ; dividing by $ω = 20$ converts angle back to seconds. (Later times would be $2 π - π /3$ , etc. — we want the first.)

3. Speed there — use the time-free relation. $v = \pm ω A^{2} - x^{2} = - 20 0. 1^{2} - 0.0 5^{2} = - 20 0.0075 \approx - 1.732 m/s$ . Why this step? The mass started at $+ A$ moving inward, so at $x = + 0.05$ on the way down it moves in $- x$ ⇒ negative sign. Using $v = \pm ω A^{2} - x^{2}$ avoids re-differentiating.

Verify: Period $T = 2 π /20 \approx 0.314 s$ ; a quarter period (to reach centre) is $0.0785 s$ , and $t_{1} \approx 0.0524 s < 0.0785$ ✓ (reaches the halfway point before the centre — cosine falls fast near the top, matching the forecast). Speed magnitude $1.732 < v_{m a x} = 2$ m/s ✓.

Recall One-line recipe for ANY starting snapshot

Compute $A = x_{0}^{2} + v_{0}^{2} / ω^{2}$ ; get the magnitude of $ϕ$ from $tan ϕ = - v_{0} / (ω x_{0})$ (unless $x_{0} = 0$ , where you use $ϕ = \pm π /2$ directly); then set the quadrant using: $cos ϕ$ shares the sign of $x_{0}$ , and $sin ϕ$ is opposite to $v_{0}$ .

Which cell has

ϕ = 0

Released from rest at

+ A

(positive position, zero velocity).

Which cell has

ϕ = π

Released from rest at

- A

(negative position, zero velocity).

Centre, moving right — what is

ϕ

- π /2

(so

x = A sin ω t

Centre, moving left — what is

ϕ

+ π /2

(so

x = - A sin ω t

Negative position, positive velocity — which quadrant is

ϕ

Quadrant III (

- π < ϕ < - π /2

); subtract

π

from the raw arctan.

When must you add

π

to the raw

arctan

for

ϕ

When the required sign of

cos ϕ

(i.e. of

x_{0}

) disagrees with what

arctan

returned — usually

x_{0} < 0

When is

tan ϕ = - v_{0} / (ω x_{0})

unusable?

When

x_{0} = 0

(division by zero); then use

ϕ = \pm π /2

chosen by the sign of

v_{0}

In the zero-zero degenerate case, what is

ϕ

Undefined and irrelevant —

A = 0

so the phase multiplies nothing.