Velocity and acceleration in SHM — v = ω√(A² − x²): Worked examples

Before we start, one reminder of what each symbol means, so nothing is unearned:

Definition Cast of symbols (all from the

$A$ = amplitude — the furthest the particle ever gets from the middle, in metres. Picture the two walls it bounces between.
$x$ = displacement — where it is right now, measured from the middle. Positive = one side, negative = the other.
$ω$ (omega) = angular frequency in rad/s — how fast the oscillation "spins" (see Angular frequency ω and time period T). Bigger $ω$ = quicker wobble.
$v$ = velocity (speed with a sign: $+$ moving one way, $-$ the other).
$v_{m a x}$ = maximum speed — the fastest the particle ever goes, which happens at the center; we will show it equals $ω A$ .
$a$ = acceleration — how fast the velocity itself is changing.
$∣ a ∣_{m a x}$ = maximum acceleration magnitude — the strongest push, which happens at the extremes; we will show it equals $ω^{2} A$ .
$T = \frac{2 π}{ω}$ = the time period, seconds for one full there-and-back.

The scenario matrix

Every question about $v$ and $a$ in SHM is one of these cells. We will hit all of them.

Cell	Case class	What's tricky about it	Example
C1	Interior point, outbound ( $0 < x < A$ , $v > 0$ )	choosing the $+$ root	Ex 1
C2	Interior point, inbound ( $0 < x < A$ , $v < 0$ )	same $x$ , opposite sign	Ex 2
C3	Center ( $x = 0$ )	speed max, accel zero	Ex 3
C4	Extreme ( $x = \pm A$ )	speed zero, accel max	Ex 3
C5	Negative side ( $x < 0$ )	sign of $a$ flips to $+$	Ex 4
C6	Degenerate ( $∣ x ∣ = A$ exactly, or $A = 0$ )	square root of zero / no motion	Ex 5
C7	Limiting / ratio (fraction of max speed)	algebra, squaring	Ex 6
C8	Real-world word problem (units, period)	translate words → symbols	Ex 7
C9	Exam twist ( $v$ and $a$ given, find $A, ω$ )	two equations, back-solve	Ex 8

Example 1 — Interior point, outbound (Cell C1)

Worked example Statement

A particle has $A = 0.10 m$ , $ω = 5 rad/s$ . It is at $x = 0.06 m$ , moving away from the center. Find its velocity (with sign) and acceleration.

Forecast: guess first — is the speed closer to its max ( $ω A$ ) or to zero? Is $a$ positive or negative here?

Step 1 — Compute the "leftover room" $A^{2} - x^{2}$ . $A^{2} - x^{2} = 0.1 0^{2} - 0.0 6^{2} = 0.01 - 0.0036 = 0.0064 = 0.08 m$ Why this step? $v = ω A^{2} - x^{2}$ : the root measures how much amplitude is still ahead. At $x = 0.06$ there is $0.08 m$ of "room" left inside the walls.

Step 2 — Pick the sign of the root. Moving away from center on the positive side means moving in the $+ x$ direction, so we take the $+$ root. $v = + ω A^{2} - x^{2} = 5 \times 0.08 = + 0.40 m/s$ Why this step? The $\pm$ isn't a formula quirk — it encodes which way the particle travels (see the outbound/inbound phase idea).

Step 3 — Acceleration from $a = - ω^{2} x$ . $a = - ω^{2} x = - (5)^{2} (0.06) = - 25 \times 0.06 = - 1.5 m/s^{2}$ Why this step? Acceleration in SHM depends only on position, not speed — this is the defining law $a = - ω^{2} x$ from Simple Harmonic Motion — definition a = −ω²x.

Verify: $v_{m a x} = ω A = 5 (0.10) = 0.5 m/s$ , and $0.40 < 0.5$ ✓ (slower than max, as expected off-center). The sign of $a$ is negative while $x$ is positive → it points back toward center, braking the outbound motion. Units: $(rad/s) \cdot m = m/s$ ✓, $(rad/s)^{2} \cdot m = m/s^{2}$ ✓.

The figure below is a bird's-eye view of the straight track the particle slides along. Look at the orange dot — that is the particle at $x = 0.06 m$ . The orange arrow on top is its velocity (pointing outward, $+ 0.40 m/s$ ); the teal arrow underneath is its acceleration (pointing back toward the center, $- 1.5 m/s^{2}$ ). The plum bracket on the right shows the "leftover room" $A^{2} - x^{2} = 0.08 m$ that the velocity formula measures.

Figure — Velocity and acceleration in SHM — v = ω√(A² − x²)

Figure s01 — Ex 1: the particle moves outward (orange) while its acceleration already points home (teal); the plum bracket is the root term.

Example 2 — Same point, inbound (Cell C2)

Worked example Statement

Same particle ( $A = 0.10$ , $ω = 5$ ), same position $x = 0.06 m$ , but now returning toward the center. Find $v$ and $a$ .

Forecast: will the speed change compared to Ex 1? Will the acceleration change?

Step 1 — Speed is identical. $∣ v ∣ = ω A^{2} - x^{2} = 0.40 m/s$ Why this step? The magnitude only depends on where you are, not on which way you go — the particle passes each point at the same speed both ways.

Step 2 — Flip the sign. Moving toward center from the positive side means moving in the $- x$ direction: $v = - 0.40 m/s$ Why this step? This is exactly the "forgetting the $\pm$ " trap from the parent note — every interior point is visited twice per cycle with opposite velocity sign.

Step 3 — Acceleration is unchanged. $a = - ω^{2} x = - 1.5 m/s^{2}$ Why this step? $a$ depends on $x$ alone. Position is the same, so acceleration is the same, regardless of travel direction.

Verify: Ex 1 and Ex 2 have equal $∣ v ∣$ and equal $a$ , differing only in the sign of $v$ . Here $v < 0$ (toward center) and $a < 0$ (also toward center) — the particle is speeding up as it heads home ✓. Compare with Ex 1 where $v > 0$ , $a < 0$ → slowing down going out. Same physics, opposite half of the cycle.

Example 3 — Center and extremes together (Cells C3 & C4)

Worked example Statement

Same particle ( $A = 0.10$ , $ω = 5$ ). Find $v$ and $a$ at (a) the center $x = 0$ , and (b) the extreme $x = + A = 0.10 m$ .

Forecast: at which of these two points is the particle fastest? At which is it "pushed" hardest?

Step 1 — Center, velocity. $v = ω A^{2} - 0^{2} = ω A = 5 (0.10) = 0.5 m/s = v_{m a x}$ Why this step? At $x = 0$ the whole amplitude is "room left," so the root is at its biggest — this is the maximum speed.

Step 2 — Center, acceleration. $a = - ω^{2} (0) = 0$ Why this step? No displacement means no restoring pull. The particle coasts through the middle unaccelerated (all energy is kinetic here).

Step 3 — Extreme, velocity. $v = ω A^{2} - A^{2} = ω 0 = 0$ Why this step? At the wall there is zero room left; the particle has momentarily stopped to reverse.

Step 4 — Extreme, acceleration. $a = - ω^{2} A = - (25) (0.10) = - 2.5 m/s^{2} = ∣ a ∣_{m a x} (directed inward)$ Why this step? Furthest from home = strongest restoring pull. All energy is potential here.

Verify: Center: $v$ max, $a = 0$ . Extreme: $v = 0$ , $∣ a ∣$ max. These are perfect opposites — exactly the anti-phase relationship the parent note's table shows ( $a$ leads $x$ by $180°$ ). Units check: $ω A = 0.5 m/s$ , $ω^{2} A = 2.5 m/s^{2}$ ✓.

The figure below plots both formulas across the whole track at once. The orange curve is speed $v = ω A^{2} - x^{2}$ (solid = outbound $+$ branch, dashed = inbound $-$ branch): notice it peaks at the center and drops to zero at both walls. The teal straight line is $a = - ω^{2} x$ : it is zero at the center and largest in magnitude at the walls, sloping downward because $a$ always opposes $x$ . The two markers pin the two special points from this example.

Figure s02 — Ex 3: speed (orange) is a downward arch peaking at $x = 0$ ; acceleration (teal) is a straight line through the origin, largest at the edges. They are complementary — where one is max, the other is zero.

Example 4 — Negative side, sign of acceleration (Cell C5)

Worked example Statement

Same particle ( $A = 0.10$ , $ω = 5$ ). It is at $x = - 0.06 m$ (the negative side), moving toward the center. Find $v$ and $a$ , paying attention to signs.

Forecast: on the negative side, does the restoring acceleration point in the $+$ or $-$ direction?

Step 1 — Speed magnitude (same as before). $∣ v ∣ = ω A^{2} - x^{2} = 5 0.1 0^{2} - (- 0.06)^{2} = 5 0.0064 = 0.40 m/s$ Why this step? $x$ is squared inside the root, so $x = - 0.06$ gives the same value as $x = + 0.06$ . Position's sign doesn't affect speed magnitude.

Step 2 — Velocity sign. Moving from the negative side toward center means moving in the $+ x$ direction: $v = + 0.40 m/s$ Why this step? "Toward center" from $x < 0$ points the same way as increasing $x$ .

Step 3 — Acceleration sign flips. $a = - ω^{2} x = - (25) (- 0.06) = + 1.5 m/s^{2}$ Why this step? Because $x$ is negative, $- ω^{2} x$ becomes positive — the pull is toward the center, which is now in the $+$ direction. This is the whole point of the minus sign in $a = - ω^{2} x$ : it always aims home, whichever side you're on.

Verify: On the positive side (Ex 1) $a$ was $- 1.5$ ; on the negative side it is $+ 1.5$ . Both point toward $x = 0$ ✓. The restoring nature holds on both sides, confirming Simple Harmonic Motion — definition a = −ω²x.

Example 5 — Degenerate inputs (Cell C6)

Worked example Statement

(a) A particle with $A = 0.10$ , $ω = 5$ is reported to be at $x = 0.10 m$ . What is $v$ ? (b) A "particle" is set up with $A = 0$ . Describe its motion.

Forecast: can the particle ever be at $x > A$ ? What does $A = 0$ even mean physically?

Step 1 — Exactly at the amplitude. $v = ω A^{2} - A^{2} = ω 0 = 0$ Why this step? $0 = 0$ : the particle is frozen at the turning point. This is the boundary case — $x$ can equal $A$ but never exceed it, because then $A^{2} - x^{2} < 0$ and the square root of a negative number is not real. The formula itself forbids $∣ x ∣ > A$ .

Step 2 — The forbidden zone. If someone claims $x = 0.12 > A$ : $A^{2} - x^{2} = 0.01 - 0.0144 = - 0.0044 < 0$ . No real speed exists → the situation is impossible for this amplitude. The maths is telling you the walls are at $\pm A$ .

Step 3 — Zero amplitude. With $A = 0$ : for all $x$ , we must have $x = 0$ (the only allowed point), and $v = ω 0 - 0 = 0$ , $a = - ω^{2} (0) = 0$ . The particle sits still at the center forever — a degenerate "oscillation" with no oscillating.

Verify: (a) $v = 0$ at $∣ x ∣ = A$ agrees with Ex 3. (b) $A = 0 \Rightarrow$ no motion is physically sensible: nothing displaces it, so nothing restores it. Both edge cases are consistent with $v = ω A^{2} - x^{2}$ having a real value only when $∣ x ∣ \leq A$ ✓.

Example 6 — Fraction of maximum speed (Cell C7)

Worked example Statement

At what displacement $x$ (as a fraction of $A$ ) is the speed exactly one third of its maximum? Then evaluate for $A = 0.10$ , $ω = 5$ : find $x$ and the accompanying acceleration.

Forecast: will $x$ be closer to the center or to the edge? (Recall half-max already happened at $0.866 A$ .)

Step 1 — Set up the ratio. $ω A^{2} - x^{2} = \frac{1}{3} ω A$ Why this step? $v_{m a x} = ω A$ ; "one third of max" means the left side equals $\frac{1}{3} ω A$ . The $ω$ 's will cancel — the fraction of speed doesn't depend on how fast it oscillates.

Step 2 — Cancel $ω$ and square both sides. $A^{2} - x^{2} = \frac{1}{3} A \Rightarrow A^{2} - x^{2} = \frac{1}{9} A^{2}$ Why this step? Squaring removes the square root so we can solve for $x^{2}$ . (Squaring is safe here: both sides are positive lengths.)

Step 3 — Solve for $x$ . $x^{2} = A^{2} - \frac{1}{9} A^{2} = \frac{8}{9} A^{2} \Rightarrow x = \frac{8}{9} A = \frac{2 2}{3} A \approx 0.943 A$ Why this step? To drop speed all the way to a third, you must be very near the wall — the speed collapses steeply near the edge (this matches the parent's half-speed-at- $0.866 A$ result).

Step 4 — Numbers. $x \approx 0.943 \times 0.10 = 0.0943 m, a = - ω^{2} x = - 25 (0.0943) \approx - 2.36 m/s^{2}$ Why this step? Once $x$ is known, acceleration is immediate from $a = - ω^{2} x$ .

Verify: Check the speed: $v = 5 0.1 0^{2} - 0.094 3^{2} = 5 0.01 - 0.008889 = 5 0.001111 \approx 5 (0.03333) = 0.1667 m/s$ , and $\frac{1}{3} v_{m a x} = \frac{1}{3} (0.5) = 0.1667 m/s$ ✓. The $ω$ cancelled, so the fraction $0.943 A$ is universal for any SHM.

Example 7 — Real-world word problem (Cell C8)

Worked example Statement

A loudspeaker cone oscillates in SHM with a period $T = 0.004 s$ and total peak-to-peak travel of $2.0 mm$ . Find (a) its maximum speed, and (b) its speed when it is $0.5 mm$ from the center.

Forecast: speakers vibrate fast — do you expect a max speed of centimetres/s or metres/s?

Step 1 — Translate words into symbols. Peak-to-peak $= 2 A$ , so amplitude $A = \frac{1}{2} (2.0 mm) = 1.0 mm = 1.0 \times 1 0^{- 3} m$ . Angular frequency from the period: $ω = \frac{2 π}{T} = \frac{2 π}{0.004} = 500 π \approx 1570.8 rad/s$ . Why this step? "Peak-to-peak" is the full swing = twice the amplitude — a classic word-problem trap. And $ω = 2 π / T$ is the bridge from a measured time period to our formulas (see Angular frequency ω and time period T).

Step 2 — Maximum speed. $v_{m a x} = ω A = 1570.8 \times 1.0 \times 1 0^{- 3} \approx 1.571 m/s$ Why this step? Max speed is at the center, $v = ω A$ . Despite tiny amplitude, the huge $ω$ makes it fast — that's why speakers move air audibly.

Step 3 — Speed at $x = 0.5 mm$ . $v = ω A^{2} - x^{2} = 1570.8 (1.0 \times 1 0^{- 3})^{2} - (0.5 \times 1 0^{- 3})^{2}$ $= 1570.8 (1.0 - 0.25) \times 1 0^{- 6} = 1570.8 0.75 \times 1 0^{- 3} \approx 1570.8 (0.8660) (1 0^{- 3}) \approx 1.360 m/s$ Why this step? $x = 0.5 mm = \frac{1}{2} A$ , so we plug into the position form. No time needed.

Verify: At $x = \frac{1}{2} A$ , $v / v_{m a x} = 1 - (1/2)^{2} = 3 /2 \approx 0.866$ , and indeed $1.360/1.571 \approx 0.866$ ✓. Units: $rad/s \times m = m/s$ ✓. Answer is metres/second, as forecast for a fast small oscillation.

Example 8 — Exam twist: back-solve for $A$ and $ω$ (Cell C9)

Worked example Statement

An examiner tells you: at a certain instant a SHM particle has speed $v = 3 m/s$ and acceleration magnitude $∣ a ∣ = 16 m/s^{2}$ at displacement $x = 0.04 m$ . Find $ω$ , then the amplitude $A$ .

Forecast: you have two measured quantities and two unknowns — which formula gives $ω$ directly without needing $A$ ?

Step 1 — Get $ω$ from acceleration (it doesn't involve $A$ ). $∣ a ∣ = ω^{2} ∣ x ∣ \Rightarrow ω^{2} = \frac{∣ a ∣}{∣ x ∣} = \frac{16}{0.04} = 400 \Rightarrow ω = 20 rad/s$ Why this step? The law $a = - ω^{2} x$ links only $a$ , $ω$ , $x$ — three quantities that are either known or wanted, with no $A$ in sight. Taking magnitudes removes the sign, and dividing isolates $ω^{2}$ . Solve for $ω$ first to unlock everything else.

Step 2 — Use velocity to find $A$ . $v = ω A^{2} - x^{2} \Rightarrow A^{2} - x^{2} = \frac{v}{ω} = \frac{3}{20} = 0.15$ Why this step? Now that $ω$ is known, the velocity formula has a single unknown, $A$ . Divide both sides by $ω$ to strip it off the root.

Step 3 — Square and solve for $A$ . $A^{2} - x^{2} = 0.1 5^{2} = 0.0225 \Rightarrow A^{2} = 0.0225 + 0.0 4^{2} = 0.0225 + 0.0016 = 0.0241$ $A = 0.0241 \approx 0.1552 m$ Why this step? Squaring undoes the square root (safe: both sides positive), then we add $x^{2}$ across and take the final root to get the amplitude.

Verify: Plug the answers straight back into both master formulas. Acceleration: $∣ a ∣ = ω^{2} x = 400 (0.04) = 16 m/s^{2}$ ✓. Velocity: $v = 20 0.0241 - 0.0016 = 20 0.0225 = 20 (0.15) = 3 m/s$ ✓. Sanity: $∣ x ∣ = 0.04 < A = 0.155$ , so the reported point is legitimately inside the range ✓. As a bonus, the time period is $T = 2 π / ω \approx 0.314 s$ .

Recall Quick self-test on the matrix

Which cell is "same $x$ , opposite travel direction"? ::: C2 — speed identical, velocity sign flipped. On the negative side, what is the sign of the acceleration when $x < 0$ ? ::: Positive, because $a = - ω^{2} x$ and $x$ is negative — still points toward center. What does the formula do if you feed it $∣ x ∣ > A$ ? ::: It demands the square root of a negative number → physically impossible; the walls are at $\pm A$ . At what fraction of amplitude is speed one-third of maximum? ::: $x = \frac{2 2}{3} A \approx 0.943 A$ . In a word problem, "peak-to-peak = 2 mm" gives what amplitude? ::: $A = 1 mm$ (half of peak-to-peak).

Connections

Velocity and acceleration in SHM — v = ω√(A² − x²) (parent)
Simple Harmonic Motion — definition a = −ω²x
Energy in SHM — kinetic and potential
Phase and phase difference
Reference circle (projection of uniform circular motion)
Angular frequency ω and time period T
Pythagorean identity sin² + cos² = 1