Exercises — Wave parameters — amplitude, wavelength, frequency, period, wave speed

Hum kya dekhte hain: snapshot graph mein horizontal axis par distance hai, isliye yeh dikhata hai ki wave space mein kaise repeat hoti hai.

• Amplitude: peak $y=+0.3\text{m}$ par hai, middle line ($y=0$) se upar. Amplitude middle se measure hoti hai, isliye $A=0.3\text{m}$.
• Wavelength: ek crest $x=1\text{m}$ par hai, agla $x=5\text{m}$ par. Crest-to-crest distance hai $\lambda =5-1=4\text{m}$.

Answer: $A=0.3\text{m}$, $\lambda =4\text{m}$.

Hum kya dekhte hain: is graph mein horizontal axis par time hai — ek particle upar-neeche bob kar raha hai. Yeh time mein repetition dikhata hai, isliye yahan $T$ padhte hain, $\lambda$ nahi.

• Amplitude: peak $y=+0.3\text{m}$ par hai, isliye $A=0.3\text{m}$ (same idea — middle line se height).
• Period: ek poora up-down-up cycle $t=0$ se $t=2\text{s}$ tak leta hai, isliye $T=2\text{s}$.
• Frequency: $f=1/T=1/2=0.5\text{Hz}$.

Answer: $A=0.3\text{m}$, $T=2\text{s}$, $f=0.5\text{Hz}$.

Yeh tool kyun: hamare paas $v$ aur $f$ hain aur $\lambda$ chahiye; wave relation $v=f\lambda$ exactly inhi teen ko link karta hai. $\lambda$ ke liye rearrange karo: $\lambda =\frac{v}{f}=\frac{340}{170}=2\text{m}.$ Answer: $\lambda =2\text{m}$.

Yeh tool kyun: hamare paas $\lambda$ aur $T$ hain; speed hai "ek wavelength per period," yaani $v=\lambda /T$. $v=\frac{\lambda }{T}=\frac{0.5}{0.02}=25\text{m/s}.$ Frequency se check karo: $f=1/T=1/0.02=50\text{Hz}$, phir $v=f\lambda =50\times 0.5=25\text{m/s}$. ✓ Answer: $v=25\text{m/s}$.

• Frequency: $f=\frac{\text{cycles}}{\text{time}}=\frac{30}{6}=5\text{Hz}$.
• Period: $T=1/f=1/5=0.2\text{s}$.
• Wavelength: $\lambda =v/f=2/5=0.4\text{m}$. Answer: $f=5\text{Hz}$, $T=0.2\text{s}$, $\lambda =0.4\text{m}$.

Key insight: wave speed medium se set hoti hai, is baat se nahi ki tum kitni tez shake karte ho. String unchanged hai, isliye $v$ unchanged hai.

• Fixed speed nikalo: $v=f_1{\lambda }_1=40\times 0.6=24\text{m/s}$.
• Same $v$ par nayi wavelength: ${\lambda }_2=v/f_2=24/120=0.2\text{m}$.
• Forecast check: frequency teen guna ho gayi ($40\to 120$), isliye wavelength ek tihaayi ho jaani chahiye ($0.6\to 0.2$). ✓ Answer: ${\lambda }_2=0.2\text{m}$.

Key insight: jab wave boundary cross karti hai, uski frequency source ke dwara fixed hoti hai (har wavefront jo aata hai wo jaana chahiye — tum unhe pile nahi kar sakte), lekin speed medium ke saath badlti hai. Isliye $\lambda$ ko badalna padta hai taaki $v=f\lambda$ sahi rahe.

• ${\lambda }_1=v_1/f=300/200=1.5\text{m}$.
• ${\lambda }_2=v_2/f=450/200=2.25\text{m}$.
• Sense check: same frequency par faster medium ⇒ longer wavelength. ✓ Answer: ${\lambda }_1=1.5\text{m}$, ${\lambda }_2=2.25\text{m}$.

• Amplitude: $0.8\text{m}$ peak-to-peak hai (lowest se highest $=2A$), isliye $A=0.8/2=0.4\text{m}$.
• Period: "lowest → highest → lowest" ek complete cycle hai, isliye $T=5\text{s}$.
• Frequency: $f=1/T=1/5=0.2\text{Hz}$.
• Wavelength: crest-to-crest se directly $\lambda =15\text{m}$ milta hai.
• Speed: $v=f\lambda =0.2\times 15=3\text{m/s}$ (ya $v=\lambda /T=15/5=3\text{m/s}$). ✓ Answer: $A=0.4\text{m}$, $T=5\text{s}$, $f=0.2\text{Hz}$, $v=3\text{m/s}$.

$y=A\sin (kx-\omega t)$ ke saath term by term match karo:

• Amplitude: aage wala number hai $A=0.05\text{m}$.
• Wave number: $k=4\pi \text{rad/m}$. Kyunki $k=2\pi /\lambda$, isliye $\lambda =2\pi /k=2\pi /(4\pi )=0.5\text{m}$.
• Angular frequency: $\omega =100\pi \text{rad/s}$. Kyunki $\omega =2\pi f$, isliye $f=\omega /(2\pi )=100\pi /(2\pi )=50\text{Hz}$.
• Period: $T=1/f=1/50=0.02\text{s}$.
• Speed: $v=f\lambda =50\times 0.5=25\text{m/s}$ (equivalently $v=\omega /k=100\pi /4\pi =25\text{m/s}$). ✓ Answer: $A=0.05\text{m}$, $\lambda =0.5\text{m}$, $f=50\text{Hz}$, $T=0.02\text{s}$, $v=25\text{m/s}$.

Same tool ($v=f\lambda$) sab waves ke liye kaam karta hai; yahan $v=c$ fixed hai.

• (a) Radio: ${\lambda }_a=c/f=(3.0\times 10^8)/(1.0\times 10^8)=3\text{m}$.
• (b) Green light: ${\lambda }_b=c/f=(3.0\times 10^8)/(6.0\times 10^{14})=5.0\times 10^{-7}\text{m}=500\text{nm}$.
• Comment: ${\lambda }_a/{\lambda }_b=3/(5\times 10^{-7})=6\times 10^6$ — radio waves green light se chhe million guna lambi hoti hain, jabki dono ek hi speed $c$ se zoom karte hain. Speed medium (vacuum) se fixed hoti hai ⇒ frequency aur wavelength inversely trade off karte hain, ek enormous range mein. Answer: ${\lambda }_a=3\text{m}$, ${\lambda }_b=5.0\times 10^{-7}\text{m}$.

• (a) $\lambda =v/f$. Jaise $f\to 0^{+}$, denominator zero ki taraf shrink karta hai, isliye $\lambda =v/f\to \infty$. Crests infinitely door push ho jaate hain.
• (b) $f=0$ matlab zero oscillations per second — source kabhi repeat nahi karta. Koi travelling wiggle hai hi nahi; yeh ek static (unchanging) disturbance hai, wave nahi. Isliye $f=0$ ek degenerate, non-wave limit hai — formula ek infinite wavelength report karta hai, jo mathematics ka ek polite tarika hai yeh kehne ka ki "space mein bhi koi repetition nahi."
• (c) $A=0$ matlab middle se maximum displacement zero hai: har particle apni jagah rehta hai. Koi wave nahi hai, chahe tum $f$ ya $\lambda$ kuch bhi likho — amplitude woh "kya kuch actually move ho raha hai?" parameter hai. Wave ke liye $A>0$ zaroori hai. Answer: (a) $\lambda \to \infty$; (b) ek static, non-oscillating disturbance (koi true wave nahi); (c) ek flat, motionless medium (koi wave nahi).

• Speed check: $3\text{s}$ mein $36\text{m}$ se $v=36/3=12\text{m/s}$ ✓ (given se match karta hai).
• Counting se wavelength: $36\text{m}$ mein $9$ wavelengths fit hoti hain, isliye $\lambda =36/9=4\text{m}$.
• Frequency: $3\text{s}$ mein $9$ crests post se guzarte hain, isliye $f=9/3=3\text{Hz}$.
• Period: $T=1/f=1/3\approx 0.333\text{s}$.
• Independent confirm: $f\lambda =3\times 4=12\text{m/s}=v$. ✓ Do alag counts (distance-per-time aur crests-per-time) agree karte hain — yeh wave relation in action hai. Answer: $\lambda =4\text{m}$, $f=3\text{Hz}$, $T=\frac{1}{3}\text{s}\approx 0.333\text{s}$.