Exercises — ω, T, f relationships

"$0.6\text{s}$ chalti hai" — yeh phrase ek cycle ki duration hai — seconds per cycle. Yahi period $T$ ki definition hai. $T=0.6\text{s}$ Yeh $f$ nahi hai (woh hota "cycles per second", ek count) aur $\omega$ nahi hai (woh hota "rad/s"). Units decide karti hain: plain "seconds" wala answer hamesha $T$ hota hai.

Standard form $x=A\cos (\omega t+\varphi )$ se compare karo. Angular frequency $\omega$ literally woh number hai jo cosine ke andar $t$ ko multiply kar raha hai. $\omega =6\text{rad/s}$ Kuch calculate nahi karna — bas pattern-match karo. ($A=5\text{cm}$, $\varphi =0$.)

Step 1 — $f$ nikalo. Frequency matlab cycles per second, seconds-per-cycle ka reciprocal: $f=\frac{1}{T}=\frac{1}{0.25}=4\text{Hz}.$ Reciprocal kyun? Agar ek cycle quarter second mein hoti hai, toh ek second mein char cycles fit hongi. Step 2 — $\omega$ nikalo. Har cycle $2\pi$ radians sweep karti hai, aur $4$ cycles hain har second: $\omega =2\pi f=2\pi (4)=8\pi \approx 25.13\text{rad/s}.$

Step 1 — $T$ nikalo. Period woh time hai jisme ek poora turn ($2\pi$ rad) rate $\omega$ par sweep ho: $T=\frac{2\pi }{\omega }=\frac{2\pi }{15}\approx 0.4189\text{s}.$ Step 2 — $f$ nikalo. Radian-rate ko radians-per-cycle se divide karo: $f=\frac{\omega }{2\pi }=\frac{15}{2\pi }\approx 2.387\text{Hz}.$ Check: $f=1/T=1/0.4189\approx 2.387\text{Hz}$ ✔ — dono routes agree karte hain.

$A\cos (\omega t+\varphi )$ se match karo:

• Amplitude $A=2\text{m}$ (aage wala number).
• $\omega =10\pi \text{rad/s}$ ($t$ ka coefficient).
• $f=\frac{\omega }{2\pi }=\frac{10\pi }{2\pi }=5\text{Hz}$.
• $T=\frac{1}{f}=\frac{1}{5}=0.2\text{s}$. Phase constant $\varphi =\pi /3$ sirf shift karta hai ki motion kahan se shuru hoti hai, lekin $\omega$, $f$, ya $T$ ko affect nahi karta — dekho Phase and Phase Difference.

• $\omega =2\pi f$ linear hai (directly proportional) $f$ mein. $f$ triple karo → $\omega$ triple ho jaata hai.
• $T=1/f$ inversely proportional hai $f$ se. $f$ triple karo → $T$ apni purani value ka ek-tehai ho jaata hai. Jo quantity $f$ ke "cycles" character ko share karti hai woh uske saath scale karti hai; jo "time per cycle" measure karti hai woh uske ulta scale karti hai.

${\omega }_A=4\pi$, ${\omega }_B=6\pi$. Kyunki $T=2\pi /\omega$ hai, bada $\omega$ chhota $T$ deta hai. Toh $A$ (chhota $\omega$) ka period bada hai. $\frac{T_A}{T_B}=\frac{2\pi /{\omega }_A}{2\pi /{\omega }_B}=\frac{{\omega }_B}{{\omega }_A}=\frac{6\pi }{4\pi }=\frac{3}{2}=\mathrm{1.5.}$ $A$ ka period B ke period ka $1.5\times$ hai.

• $T=1/f\to \infty$: ek "cycle" mein hamesha ke liye time lagta hai — motion kabhi repeat nahi karti.
• $\omega =2\pi f\to 0$: phase angle advance hona band ho jaata hai. Exactly $f=0$ par koi oscillation nahi hoti — object bas baitha rehta hai (ya steadily drift karta hai). $T=1/0$ undefined hai, jo maths ka tarika hai yeh bolne ka ki "koi finite cycle time nahi hai". Toh $f=0$ ek degenerate limit hai, real oscillation nahi. Yeh Uniform Circular Motion se mirror karta hai jab zero angular speed ho: circle pe point kabhi nahi hilta.

Step 1 — $\omega$ physics se. $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{0.2}}=\sqrt{250}\approx 15.81\text{rad/s}.$ Yeh tool kyun? Spring ke liye restoring force decide karta hai ki phase kitni sharply advance karti hai; $\sqrt{k/m}$ wahi rad/s rate hai — yeh tumhe seedha $\omega$ deta hai, $f$ nahi. Step 2 — measurable rhythm mein convert karo. $f=\frac{\omega }{2\pi }=\frac{15.81}{2\pi }\approx 2.516\text{Hz},T=\frac{2\pi }{\omega }=\frac{2\pi }{15.81}\approx 0.3974\text{s}.$

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 — $f$ aur $\omega$. $f=\frac{1}{T}=\frac{1}{2.007}\approx 0.4983\text{Hz},\omega =\frac{2\pi }{T}=\frac{2\pi }{2.007}\approx 3.130\text{rad/s}.$ Sanity check direct formula se $\omega =\sqrt{g/L}=\sqrt{9.8}\approx 3.130\text{rad/s}$ ✔.

Step 1 — speed. $v=f\lambda$ relation kehta hai: har second source $f$ poori waves emit karta hai, har ek $\lambda$ length ki, toh front $f\lambda$ metres per second advance karta hai. $v=f\lambda =250\times 1.36=340\text{m/s}.$ Step 2 — period. Pure time-rhythm, space ignore karo: $T=\frac{1}{f}=\frac{1}{250}=0.004\text{s}=4\text{ms}.$ Frequency $f$ bridge hai: yeh time-rhythm ($T$) ko speed $v$ ke through spatial pattern ($\lambda$) mein convert karta hai.

Step 1 — $\omega$ speed & radius se. Circular motion mein linear speed hai $s=\omega R$ (ek poora turn circumference $2\pi R$ cover karta hai time $T$ mein). Angular rate ke liye solve karo: $\omega =\frac{s}{R}=\frac{1.2}{0.30}=4.0\text{rad/s}.$ Step 2 — $T$ aur $f$. $T=\frac{2\pi }{\omega }=\frac{2\pi }{4.0}\approx 1.571\text{s},f=\frac{\omega }{2\pi }=\frac{4.0}{2\pi }\approx 0.6366\text{Hz}.$ Step 3 — shadow ka amplitude. Shadow $+R$ aur $-R$ ke beech swing karta hai, isliye uska amplitude radius ke barabar hai: $A=R=0.30\text{m}.$ Yahi parent note ki picture hai: oscillation ek lap ka projection hai, isliye circle ka radius SHM amplitude ban jaata hai aur lap-rate $\omega$ ban jaata hai.

Step 1 — $f$ count se. 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 — $A$ aur $\varphi$. Amplitude maximum displacement hai, $A=8\text{cm}$. Yeh $t=0$ pe maximum pe start karta hai: $\cos (\varphi )=1\Rightarrow \varphi =0$. $\boxed{x(t)=8\cos (5\pi t)\text{cm}}$ Check: $T=1/f=0.4\text{s}$; 30 cycles $\times 0.4\text{s}=12\text{s}$ ✔.

Error yeh hai: $\omega =2\pi f$ use karne se pehle $f$ cycles per second (Hz) mein hona chahiye. Student ne $f$ ko cycles per minute mein hi rakha. Step 1 — convert karo. $f=\frac{90\text{cycles}}{60\text{s}}=1.5\text{Hz}.$ Step 2 — $\omega$ aur $T$. $\omega =2\pi f=2\pi (1.5)=3\pi \approx 9.425\text{rad/s},T=\frac{1}{f}=\frac{1}{1.5}\approx 0.6667\text{s}.$ Student ka $180\pi$ 60× zyada bada tha — exactly minute-to-second factor.

Kaun se levels automatic lage, kaun se mein reveal ki zaroorat padi? ::: L1–L2 reflex hona chahiye; agar L3–L5 mein solution chahiye pada, toh har [!mistake] callout mein unit-anchoring fixes dobara padho. Har conversion error ke peeche ek factor? ::: $2\pi$ (radians per cycle) — ise chhodo ya galat jagah rakho aur answer $\approx 6.28$ ya $60$ (minute slip) se off ho jaata hai.