Exercises — ω, T, f relationships

The phrase "lasts $0.6\text{s}$" is a duration of one cycle — seconds per cycle. That is the definition of the period $T$. $T=0.6\text{s}$ It is not $f$ (that would be "cycles per second", a count) and not $\omega$ (that would be "rad/s"). Units decide it: a plain "seconds" answer is always $T$.

Compare with the standard form $x=A\cos (\omega t+\varphi )$. The angular frequency $\omega$ is literally the number multiplying $t$ inside the cosine. $\omega =6\text{rad/s}$ Nothing to calculate — you just pattern-match. ($A=5\text{cm}$, $\varphi =0$.)

Step 1 — get $f$. Frequency is cycles per second, the reciprocal of seconds-per-cycle: $f=\frac{1}{T}=\frac{1}{0.25}=4\text{Hz}.$ Why reciprocal? If one cycle takes a quarter second, four cycles fit in one second. Step 2 — get $\omega$. Each cycle sweeps $2\pi$ radians, and there are $4$ cycles each second: $\omega =2\pi f=2\pi (4)=8\pi \approx 25.13\text{rad/s}.$

Step 1 — get $T$. Period is the time to sweep one whole turn ($2\pi$ rad) at rate $\omega$: $T=\frac{2\pi }{\omega }=\frac{2\pi }{15}\approx 0.4189\text{s}.$ Step 2 — get $f$. Divide the radian-rate by radians-per-cycle: $f=\frac{\omega }{2\pi }=\frac{15}{2\pi }\approx 2.387\text{Hz}.$ Check: $f=1/T=1/0.4189\approx 2.387\text{Hz}$ ✔ — the two routes agree.

Match to $A\cos (\omega t+\varphi )$:

• Amplitude $A=2\text{m}$ (the number in front).
• $\omega =10\pi \text{rad/s}$ (coefficient of $t$).
• $f=\frac{\omega }{2\pi }=\frac{10\pi }{2\pi }=5\text{Hz}$.
• $T=\frac{1}{f}=\frac{1}{5}=0.2\text{s}$. The phase constant $\varphi =\pi /3$ shifts where the motion starts but does not affect $\omega$, $f$, or $T$ — see Phase and Phase Difference.

• $\omega =2\pi f$ is linear (directly proportional) in $f$. Triple $f$ → $\omega$ is tripled.
• $T=1/f$ is inversely proportional to $f$. Triple $f$ → $T$ becomes one third of its old value. One quantity that shares $f$'s "cycles" character scales with it; the one measuring "time per cycle" scales against it.

${\omega }_A=4\pi$, ${\omega }_B=6\pi$. Since $T=2\pi /\omega$, a larger $\omega$ gives a smaller $T$. So $A$ (smaller $\omega$) has the larger period. $\frac{T_A}{T_B}=\frac{2\pi /{\omega }_A}{2\pi /{\omega }_B}=\frac{{\omega }_B}{{\omega }_A}=\frac{6\pi }{4\pi }=\frac{3}{2}=\mathrm{1.5.}$ $A$'s period is $1.5\times$ B's.

• $T=1/f\to \infty$: one "cycle" takes forever — the motion never completes a repeat.
• $\omega =2\pi f\to 0$: the phase angle stops advancing. At exactly $f=0$ there is no oscillation — the object just sits still (or drifts steadily). $T=1/0$ is undefined, which is the maths telling you "there is no finite cycle time". So $f=0$ is a degenerate limit, not a real oscillation. This mirrors Uniform Circular Motion with zero angular speed: the point on the circle never moves.

Step 1 — $\omega$ from the physics. $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{0.2}}=\sqrt{250}\approx 15.81\text{rad/s}.$ Why this tool? For a spring, the restoring force sets how sharply the phase advances; $\sqrt{k/m}$ is that rad/s rate — it hands you $\omega$ directly, not $f$. Step 2 — convert to measurable rhythm. $f=\frac{\omega }{2\pi }=\frac{15.81}{2\pi }\approx 2.516\text{Hz},T=\frac{2\pi }{\omega }=\frac{2\pi }{15.81}\approx 0.3974\text{s}.$

Step 1 — $T$. $T=2\pi \sqrt{\frac{L}{g}}=2\pi \sqrt{\frac{1.0}{9.8}}=2\pi (0.3194)\approx 2.007\text{s}.$ Step 2 — $f$ and $\omega$. $f=\frac{1}{T}=\frac{1}{2.007}\approx 0.4983\text{Hz},\omega =\frac{2\pi }{T}=\frac{2\pi }{2.007}\approx 3.130\text{rad/s}.$ Sanity check via the direct formula $\omega =\sqrt{g/L}=\sqrt{9.8}\approx 3.130\text{rad/s}$ ✔.

Step 1 — speed. The relation $v=f\lambda$ says: each second the source emits $f$ full waves, each of length $\lambda$, so the front advances $f\lambda$ metres per second. $v=f\lambda =250\times 1.36=340\text{m/s}.$ Step 2 — period. Pure time-rhythm, ignore space: $T=\frac{1}{f}=\frac{1}{250}=0.004\text{s}=4\text{ms}.$ The frequency $f$ is the bridge: it turns the time-rhythm ($T$) into the spatial pattern ($\lambda$) via the speed $v$.

Step 1 — $\omega$ from speed & radius. For circular motion the linear speed is $s=\omega R$ (a full turn covers circumference $2\pi R$ in time $T$). Solve for the angular rate: $\omega =\frac{s}{R}=\frac{1.2}{0.30}=4.0\text{rad/s}.$ Step 2 — $T$ and $f$. $T=\frac{2\pi }{\omega }=\frac{2\pi }{4.0}\approx 1.571\text{s},f=\frac{\omega }{2\pi }=\frac{4.0}{2\pi }\approx 0.6366\text{Hz}.$ Step 3 — amplitude of the shadow. The shadow swings between $+R$ and $-R$, so its amplitude equals the radius: $A=R=0.30\text{m}.$ This is exactly the parent note's picture: oscillation is the projection of a lap, so the circle's radius becomes the SHM amplitude and the lap-rate becomes $\omega$.

Step 1 — $f$ from the count. Cycles per second: $f=\frac{30}{12}=2.5\text{Hz}.$ Step 2 — $\omega$. $\omega =2\pi f=2\pi (2.5)=5\pi \approx 15.71\text{rad/s}.$ Step 3 — $A$ and $\varphi$. Amplitude is the maximum displacement, $A=8\text{cm}$. It starts at maximum at $t=0$: $\cos (\varphi )=1\Rightarrow \varphi =0$. $\boxed{x(t)=8\cos (5\pi t)\text{cm}}$ Check: $T=1/f=0.4\text{s}$; 30 cycles $\times 0.4\text{s}=12\text{s}$ ✔.

The error: $f$ must be in cycles per second (Hz) before using $\omega =2\pi f$. The student left $f$ in cycles per minute. Step 1 — convert. $f=\frac{90\text{cycles}}{60\text{s}}=1.5\text{Hz}.$ Step 2 — $\omega$ and $T$. $\omega =2\pi f=2\pi (1.5)=3\pi \approx 9.425\text{rad/s},T=\frac{1}{f}=\frac{1}{1.5}\approx 0.6667\text{s}.$ The student's $180\pi$ was 60× too big — exactly the minute-to-second factor.

Which levels felt automatic, which needed the reveal? ::: L1–L2 should be reflex; if L3–L5 needed the solution, re-read the unit-anchoring fixes in each [!mistake] callout. The one factor behind every conversion error? ::: The $2\pi$ (radians per cycle) — omit or misplace it and the answer is off by $\approx 6.28$ or by $60$ (minute slip).