ω, T, f relationships: Worked examples — Physics deep dive

This is the drill page for the ω–T–f subtopic. We march through every kind of question the three rhythm quantities can throw at you — normal cases, weird limits, degenerate zeros, word problems, an exam trap, and even a negative $ω$ — and we solve each one from the ground up.

Before we start, a one-second reminder of the only three tools we ever use here (built in the parent note):

A few symbols show up in the examples below, so let's build every one of them before we ever use it:

Definition The symbols inside

$x (t) = A cos (ω t + ϕ)$

$t$ is time, measured in seconds — the reading on your stopwatch. Everything on this page is a function of this one ticking variable.
$x (t)$ is the position at time $t$ — where the oscillating thing sits (e.g. how far the mass is from its rest point), measured in a length unit like cm or m.
$A$ is the amplitude: the largest distance from the rest point the motion ever reaches. Because $cos$ swings between $- 1$ and $+ 1$ , the position $x$ swings between $- A$ and $+ A$ . It is a size, not a rhythm — it never affects $ω$ , $f$ , or $T$ .
The phase is the whole quantity inside the cosine, $ω t + ϕ$ . Picture the merry-go-round point: the phase is the angle (in radians) it has turned through so far. As $t$ grows, $ω t$ grows — that's the rhythm ticking forward.
$ϕ$ (Greek letter "phi") is the starting phase: the angle the point had already turned to at $t = 0$ , before the stopwatch started. It just shifts the whole wave left or right in time; it does not change $ω$ , $f$ , or $T$ .

More on phase in Phase and Phase Difference; more on amplitude and this whole form in Simple Harmonic Motion.

The scenario matrix

Every question this topic can ask lands in one of these cells. Each worked example below is tagged with its cell so you can see the coverage is complete.

Cell	What makes it distinct	Example
A. Forward chain	Given $T$ → find $f$ and $ω$	Ex 1
B. Backward chain	Given $ω$ → find $f$ and $T$	Ex 2
C. Read-off from equation	Given $x (t) = A cos (ω t + ϕ)$ → extract $ω, f, T$	Ex 3
D. Very large $f$ (limit)	High frequency → tiny period, huge $ω$	Ex 4a
E. Very small $f$ (limit)	Low frequency → huge period, tiny $ω$	Ex 4b
F. Degenerate: $f = 0$	No oscillation at all — what happens to $T, ω$ ?	Ex 5
G. Degenerate: $ω = 0$ inside cosine	The "motion" that never moves	Ex 5
H. Unit-trap	Frequency given in rpm / minutes, not Hz	Ex 6
I. Scaling / proportion	"If $f$ triples, what happens to $T, ω$ ?"	Ex 7
J. Real-world word problem	Merry-go-round / wheel, geometry involved	Ex 8 (figure)
K. Exam twist	Wrong-pairing trap ( $T = 2 π f$ ?) resolved by units	Ex 9
L. Sign / direction	Negative $ω$ — what does $ω < 0$ mean?	Ex 10

Worked Examples

Ex 1 — Forward chain (Cell A)

Worked example Given the period, find everything

A mass on a spring completes one full up-and-down cycle in $T = 0.25 s$ . Find $f$ and $ω$ .

Forecast: Before reading on — guess: is $f$ bigger or smaller than 1? Is $ω$ bigger or smaller than $f$ ?

Step 1 — $f = 1/ T = 1/0.25 = 4 Hz$ . Why this step? $f$ counts how many cycles fit in one second. If one cycle eats $0.25 s$ , four of them fit in a second. That reciprocal is the definition of frequency.

Step 2 — $ω = 2 π f = 2 π (4) = 8 π \approx 25.13 rad/s$ . Why this step? $ω$ measures the same rhythm but in radians. Each cycle sweeps $2 π$ radians, so 4 cycles/s means $4 \times 2 π$ rad/s. That's why $ω$ is always the bigger number here — it carries the $2 π$ crown.

Forecast resolved: $f = 4$ is bigger than 1 (short period packs many cycles in), and $ω = 8 π \approx 25.1$ is bigger than $f$ — the $2 π$ always inflates it. Both guesses were "bigger".

Verify: Loop back: $T = 2 π / ω = 2 π / (8 π) = 0.25 s$ ✔. Units: $[Hz] = s^{- 1}$ ✔, $[ω] = rad/s$ ✔.

Ex 2 — Backward chain (Cell B)

Worked example Given angular frequency, walk it back

An oscillator has $ω = 15 rad/s$ . Find $T$ and $f$ .

Forecast: Will $T$ be more or less than 1 second? Will $f$ be more or less than $ω$ ?

Step 1 — $T = 2 π / ω = 2 π /15 \approx 0.4189 s$ . Why this step? $T$ is the time for the angle to sweep one full turn, $2 π$ . Dividing the total angle ( $2 π$ ) by the rate of sweeping angle ( $ω$ ) gives the time. That's just distance ÷ speed, but for angle.

Step 2 — $f = ω / (2 π) = 15/ (2 π) \approx 2.387 Hz$ . Why this step? We strip off the $2 π$ . One cycle is $2 π$ radians of turning, so dividing a radians-per-second rate by the pure number $2 π$ tells you how many whole $2 π$ -turns (i.e. cycles) happen per second — that's Hz. $f$ is smaller than $ω$ for exactly the same reason $ω$ is bigger in Ex 1.

Forecast resolved: $T \approx 0.419$ is less than 1 second, and $f \approx 2.39$ is less than $ω = 15$ — dividing by $2 π$ shrank it. Both guesses were "less".

Verify: $f = 1/ T = 1/0.4189 \approx 2.387 Hz$ ✔ — the two independent routes agree.

Ex 3 — Read-off from the equation (Cell C)

Worked example Pull the rhythm straight out of

$x (t)$ An object moves as $x (t) = 5 cos (6 π t + \frac{π}{3}) cm$ . Find $ω$ , $f$ , $T$ .

Forecast: Which number in that equation is $ω$ — is it the $6 π$ , the $5$ , or the $π /3$ ?

Step 1 — $ω = 6 π rad/s$ . Why this step? Compare with the standard SHM form $A cos (ω t + ϕ)$ , whose symbols we defined above. The thing multiplying $t$ is $ω$ by definition. The $5$ is the amplitude $A$ (in cm) — the biggest distance reached; the $π /3$ is the starting phase $ϕ$ — the angle already turned at $t = 0$ — and neither of those affects the rhythm.

Step 2 — $f = ω / (2 π) = 6 π / (2 π) = 3 Hz$ . Why this step? Convert radian-rate to cycle-rate; the $π$ cancels cleanly, which is the tell-tale sign you did it right.

Step 3 — $T = 1/ f = 1/3 \approx 0.3333 s$ . Why this step? Straight reciprocal, the cheapest possible last step.

Forecast resolved: $ω$ is the $6 π$ (the coefficient of $t$ ) — not the amplitude $5$ and not the phase $π /3$ . If you guessed $6 π$ , you had it.

Verify: Phase must advance by $2 π$ in one period: $ω T = 6 π \times \frac{1}{3} = 2 π$ ✔.

Ex 4 — Limiting values: very fast and very slow (Cells D & E)

Worked example Push frequency to extremes

(a) [Cell D — very fast.] A guitar string vibrates at $f = 440 Hz$ (musical note A). Find $T$ and $ω$ . (b) [Cell E — very slow.] A slow ocean swell has $f = 0.05 Hz$ . Find $T$ and $ω$ .

Forecast: For the fast string, is $T$ big or tiny? For the slow swell, is $T$ big or tiny?

Step 1a (Cell D) — $T = 1/ f = 1/440 \approx 0.002273 s$ . Why this step? High frequency = many cycles per second = each cycle is squeezed short. As $f \to$ large, $T = 1/ f \to 0$ .

Step 2a (Cell D) — $ω = 2 π f = 2 π (440) = 880 π \approx 2764.6 rad/s$ . Why this step? Fast rhythm ⇒ the angle sweeps enormously fast. $ω$ grows without bound as $f$ grows.

Step 1b (Cell E) — $T = 1/ f = 1/0.05 = 20 s$ . Why this step? Low frequency = few cycles per second = each cycle is stretched long. As $f \to 0^{+}$ , $T = 1/ f \to \infty$ .

Step 2b (Cell E) — $ω = 2 π f = 2 π (0.05) = 0.1 π \approx 0.3142 rad/s$ . Why this step? Slow rhythm ⇒ the angle barely creeps forward.

Forecast resolved: Fast string → $T \approx 0.0023 s$ is tiny; slow swell → $T = 20 s$ is big. High $f$ shrinks $T$ , low $f$ stretches it.

Verify: (a) $ω T = 880 π \times 0.002273 \approx 2 π$ ✔. (b) $ω T = 0.1 π \times 20 = 2 π$ ✔. The product $ω T = 2 π$ is always true — that's the invariant guarding every case.

Ex 5 — Degenerate inputs: $f = 0$ and $ω = 0$ (Cells F & G)

Worked example What if there is no oscillation?

(a) A "system" has $f = 0$ . What are $T$ and $ω$ ? (b) What does $x (t) = A cos (0 \cdot t + ϕ)$ describe? (Recall $A$ = amplitude, $t$ = time, $ϕ$ = starting phase.)

Forecast: Can the period of a non-oscillating thing be a real number? What curve does a zero- $ω$ cosine trace?

Step 1a — $ω = 2 π f = 2 π (0) = 0 rad/s$ . Why this step? No cycles per second ⇒ no radians swept per second. The angle never moves.

Step 2a — $T = 1/ f = 1/0 \to \infty$ (undefined / infinite). Why this step? If you never complete a cycle, the "time for one cycle" is infinite. So $T$ is not a finite number here — the reciprocal breaks down exactly when there is nothing to repeat. This is the boundary case you must flag, not force a number onto.

Step 1b — $x (t) = A cos (0 + ϕ) = A cos ϕ = constant$ . Why this step? With $ω = 0$ , the argument never changes as $t$ ticks, so cosine is frozen. The "motion" is a stationary point sitting at height $A cos ϕ$ (a fixed fraction of the amplitude $A$ ) — a flat horizontal line on the $x$ – $t$ graph. It is the degenerate limit of SHM: an oscillation with infinite period is just not moving.

Forecast resolved: No — the period is not a finite real number; it is infinite. And the zero- $ω$ cosine traces a flat horizontal line (a constant), not a wave.

Verify: Check the invariant softly: $ω T = 2 π$ would need $0 \cdot \infty$ , an indeterminate form — consistent with $T$ being undefined, not with any finite value. So the only self-consistent reading is $ω = 0, f = 0, T = \infty$ ✔.

Ex 6 — Unit trap: rpm not Hz (Cell H)

Worked example Frequency hidden in minutes

A record player spins at $33 \frac{1}{3}$ revolutions per minute (rpm). Find $f$ (Hz), $T$ , and $ω$ .

Forecast: Is $f$ going to be roughly $33$ , or roughly $0.5$ ? Watch the "per minute".

Step 1 — Convert to Hz: $f = \frac{33.333 rev}{60 s} = 0.5556 Hz$ . Why this step? Hz means cycles per second, but we were handed cycles per minute. Divide by $60 s/min$ so the "minutes" cancel and seconds appear. Skipping this is the classic unit blunder.

Step 2 — $T = 1/ f = 1/0.5556 = 1.8 s$ . Why this step? Reciprocal. Sanity: one full turn should take a bit under 2 seconds — matches a slow turntable feel.

Step 3 — $ω = 2 π f = 2 π (0.5556) \approx 3.491 rad/s$ . Why this step? Standard conversion once $f$ is finally in Hz.

Forecast resolved: $f \approx 0.56 Hz$ — the roughly $0.5$ guess wins. The " $33$ " was cycles per minute, so it must shrink by 60 to become Hz.

Verify: $T = 60/33.333 = 1.8 s$ directly (60 s ÷ revs-per-minute) ✔, and $ω T = 3.491 \times 1.8 \approx 2 π$ ✔.

Ex 7 — Scaling and proportion (Cell I)

Ex 8 — Real-world word problem with geometry (Cell J)

Worked example Merry-go-round rider

A child sits at radius $r = 2 m$ on a merry-go-round that makes one full turn every $T = 4 s$ . Find $f$ , $ω$ , and the child's linear speed $v$ around the circle.

Forecast: Which is bigger, the child's speed in m/s or the angular frequency in rad/s?

Step 1 — $f = 1/ T = 1/4 = 0.25 Hz$ . Why this step? One turn every 4 s ⇒ a quarter turn each second.

Step 2 — $ω = 2 π / T = 2 π /4 = π /2 \approx 1.571 rad/s$ . Why this step? Uniform Circular Motion is the birthplace of $ω$ : one lap sweeps $2 π$ radians, so the sweep-rate is $2 π$ divided by the lap time. Look at the blue arc labelled "angle sweeps at rate omega" in the figure — that's the angle growing at rate $ω$ .

Step 3 — Linear speed $v = ω r = (π /2) (2) = π \approx 3.142 m/s$ . Why this step? This is where geometry enters. A point at radius $r$ moving through angle $Δ θ$ travels an arc-length $r Δ θ$ (definition of the radian!). Dividing by time turns $Δ θ / time = ω$ into speed $= r ω$ . The yellow arrow labelled "v = omega r" in the figure is this tangential velocity. Note it depends on $r$ , while $ω$ does not — a child further out moves faster but sweeps the same angle-rate.

Forecast resolved: Here $v = π \approx 3.14 m/s$ is bigger than $ω = π /2 \approx 1.57 rad/s$ — because the radius $r = 2$ (bigger than 1) magnifies $ω$ into $v$ . (At $r < 1$ it would flip, so the answer genuinely depends on $r$ .)

Verify: Whole circumference $= 2 π r = 4 π m$ , covered in $T = 4 s$ , so $v = 4 π /4 = π m/s$ ✔ — matches $ω r$ .

Ex 9 — Exam twist: the wrong-pairing trap (Cell K)

Worked example Which formula is legal?

A student is asked: "Given $f = 5 Hz$ , a classmate writes $T = 2 π f$ and $ω = 2 π / f$ . Are these right? Give the correct $T$ and $ω$ ."

Forecast: Trust your gut — does a $2 π$ belong with $T$ , or only with $ω$ ?

Step 1 — Kill the wrong pairing using units. Why this step? $T$ must come out in seconds. The factor $2 π$ is just a pure number (radians are dimensionless), so $2 π f$ still has the units of $f$ , namely per-second — that's a frequency, not a time. A formula for a time that produces a per-second quantity is impossible, so $T = 2 π f$ is dead on arrival. The $2 π$ belongs with $ω$ only because $ω$ is defined to count radians per second, not because $2 π$ carries any unit.

Step 2 — Correct period: $T = 1/ f = 1/5 = 0.2 s$ . Why this step? $T$ and $f$ are pure reciprocals; no $2 π$ ever touches them. ("f Tlips T.")

Step 3 — Correct angular frequency: $ω = 2 π f = 2 π (5) = 10 π \approx 31.42 rad/s$ . Why this step? $ω = 2 π / f$ would give $\approx 1.257$ , far too small for a 5 Hz motion — another sanity red flag ( $1/ f$ is just seconds, the size of a period, not a fast radian-rate). The right form multiplies by $2 π$ .

Forecast resolved: The $2 π$ belongs only with $ω$ . Both of the classmate's formulas were wrong; $T = 1/ f$ and $ω = 2 π f$ are the legal pairings.

Verify: $ω T = 10 π \times 0.2 = 2 π$ ✔. The student's $T = 2 π f = 10 π \approx 31.4$ "s" is absurdly long for 5 cycles-per-second — clearly wrong, confirmed.

Ex 10 — Sign and direction: what does $ω < 0$ mean? (Cell L)

Worked example A negative angular frequency

A rotating point is described by $x (t) = A cos (ω t)$ with $ω = - 4 π rad/s$ . What are $f$ and $T$ , and what does the minus sign physically mean?

Forecast: Does a negative $ω$ make the period negative too? And does the rhythm get faster or slower than a $+ 4 π$ oscillator?

Step 1 — The rhythm depends only on the size $∣ ω ∣$ . Why this step? $cos$ is an even function: $cos (- θ) = cos (θ)$ . So $cos (- 4 π t) = cos (4 π t)$ — the position-vs-time graph is identical to the $+ 4 π$ case. The sign never changes how fast the wave repeats; only $∣ ω ∣$ does.

Step 2 — $f = ∣ ω ∣/ (2 π) = 4 π / (2 π) = 2 Hz$ . Why this step? Frequency counts cycles per second — a count, which cannot be negative. We feed in the magnitude. Likewise $T = 2 π /∣ ω ∣ = 2 π / (4 π) = 0.5 s$ ; a time-per-cycle is never negative.

Step 3 — What the minus does mean. Why this step? In Uniform Circular Motion, $ω$ is the underlying spinning point's angular velocity, and its sign is direction: $ω > 0$ = counter-clockwise, $ω < 0$ = clockwise. So $ω = - 4 π$ is the same speed, opposite spin direction. The oscillation's shadow looks the same; the wheel behind it turns the other way. Direction lives in the sign, magnitude lives in $∣ ω ∣$ .

Forecast resolved: $T$ is not negative — it is $+ 0.5 s$ , using $∣ ω ∣$ . And the rhythm is the same speed as $+ 4 π$ ; the minus only flips the direction of rotation, not the pace.

Verify: $∣ ω ∣ T = 4 π \times 0.5 = 2 π$ ✔, and $cos (- 4 π (0.5)) = cos (- 2 π) = 1 = cos (2 π)$ — the graphs coincide ✔.

Recall Quick self-test across the matrix

$T = 0.25$ s → $f$ ? ::: $4$ Hz $ω = 15$ rad/s → $f$ ? ::: $\approx 2.387$ Hz $x = 5 cos (6 π t + π /3)$ → $T$ ? ::: $1/3 \approx 0.333$ s $f = 0$ → $T$ ? ::: infinite / undefined (no cycle to complete) $ω = 0$ cosine → what motion? ::: a frozen constant, no oscillation $33 \frac{1}{3}$ rpm → $T$ ? ::: $1.8$ s Triple $f$ → $T$ becomes? ::: one-third Merry-go-round $T = 4$ s, $r = 2$ m → $v$ ? ::: $π \approx 3.14$ m/s Is $T = 2 π f$ ever valid? ::: never — units force $T = 1/ f$ $ω = - 4 π$ → $T$ and meaning of the sign? ::: $T = 0.5$ s (uses $∣ ω ∣$ ); the minus means clockwise rotation, same speed

Connections

ω, T, f relationships — the parent this page drills.
Uniform Circular Motion — Ex 8's arc and $v = ω r$ , and Ex 10's sign-as-direction come from here.
Simple Harmonic Motion — Ex 3, Ex 5, Ex 10 read $ω$ (and amplitude $A$ ) out of $A cos (ω t + ϕ)$ .
Phase and Phase Difference — the argument $ω t + ϕ$ and the starting phase $ϕ$ we defined.
Wave Speed v = fλ — next step: convert this time-rhythm into spatial wavelength.
Springs and Pendulums — where the numerical $T$ and $ω$ actually come from physically.