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

(a) Max speed centre par hoti hai, $x=0$ par, jahan "root ke andar bacha hua distance" sabse zyada hota hai: $v_{\max }=\omega A=5\times 0.08=0.4\text{ m/s}(\text{at }x=0)$ (b) Max acceleration extremes par hoti hai, $x=\pm A$ par, jahan restoring pull sabse strong hota hai: $|a{|}_{\max }={\omega }^{2}A=25\times 0.08=2{\text{ m/s}}^2(\text{at }x=\pm A)$ Yeh swap notice karo: speed wahan sabse badi hoti hai jahan acceleration zero hoti hai, aur vice versa.

Position form seedha use karo — koi time nahi chahiye: $v=\omega \sqrt{A^2-x^2}=5\sqrt{0.08^2-0.048^2}=5\sqrt{0.0064-0.002304}$ $=5\sqrt{0.004096}=5\times 0.064=0.32\text{ m/s}$

$a=-{\omega }^{2}x=-25\times 0.048=-1.2{\text{ m/s}}^2$ Minus ka matlab hai yeh $x$ ke opposite direction mein point karti hai. Kyunki $x$ positive hai (centre ke right mein), $a$ left taraf, centre ki aur wapas point karti hai — yeh ek restoring acceleration hai.

Velocity formula se shuru karo aur $x$ ke liye solve karo: $v=\omega \sqrt{A^2-x^2}\Rightarrow \frac{v}{\omega }=\sqrt{A^2-x^2}$ Dono sides square karo (yeh root hata deta hai): $\frac{v^2}{{\omega }^2}=A^2-x^2\Rightarrow x^2=A^2-\frac{v^2}{{\omega }^2}$ $x^2=0.10^2-\frac{0.48^2}{8^2}=0.01-\frac{0.2304}{64}=0.01-0.0036=0.0064$ $x=\pm \sqrt{0.0064}=\pm 0.08\text{ m}$ Dono isliye kyunki particle centre ke left aur right dono taraf $0.48$ m/s speed reach karta hai.

$\frac{v(\frac{1}{2}A)}{v(0)}=\frac{\omega \sqrt{A^2-(\frac{1}{2}A{)}^2}}{\omega \sqrt{A^2-0}}=\frac{\sqrt{A^2-\frac{1}{4}{A}^2}}{\sqrt{A^2}}=\frac{\sqrt{\frac{3}{4}{A}^2}}{A}=\frac{\sqrt{3}}{2}\approx 0.866$ Toh aadha bahar aane par bhi 87% top speed milti hai — speed centre ke paas slowly girती hai, phir edge ke paas tezi se gir jaati hai.

(a) Formula $x^2$ use karta hai, isliye $x$ ka sign speed ko affect nahi karta: $v=10\sqrt{0.06^2-0.036^2}=10\sqrt{0.0036-0.001296}=10\sqrt{0.002304}=10\times 0.048=0.48\text{ m/s}$ (b) Acceleration sign rakhti hai: $a=-{\omega }^{2}x=-100\times (-0.036)=+3.6{\text{ m/s}}^2$ Positive isliye kyunki $x$ negative hai — phir se centre ki taraf wapas point kar raha hai.

$v^2={\omega }^2(A^2-x^2)$ se shuru karo aur standard ellipse form mein rearrange karo (divide karo taaki right side $1$ ho): $v^2+{\omega }^2{x}^2={\omega }^2{A}^2\Rightarrow \frac{x^2}{A^2}+\frac{v^2}{(\omega A{)}^2}=1$ Yeh $\frac{x^2}{a^2}+\frac{v^2}{b^2}=1$ hai, ek ellipse jisme

• horizontal semi-axis $a=A=0.10$ m (max displacement),
• vertical semi-axis $b=\omega A=8\times 0.10=0.80$ m/s (max speed).

Jab tum ellipse ke around move karte ho tum motion ki har state sweep karte ho — outbound (top half, $v>0$) aur inbound (bottom half, $v<0$).

$v^2={\omega }^2(A^2-x^2)$ dono points par use karo: $v_1^2={\omega }^2(A^2-x_1^2),v_2^2={\omega }^2(A^2-x_2^2)$ $A^2$ hatane ke liye subtract karo (isliye do points kaafi hain): $v_1^2-v_2^2={\omega }^2(x_2^2-x_1^2)$ $0.09-0.0256={\omega }^2(0.0036-0.0016)\Rightarrow 0.0644={\omega }^2(0.002)$ ${\omega }^2=32.2\Rightarrow \omega \approx 5.675\text{ rad/s}$ Ab $A$ nikalne ke liye pehli equation mein back-substitute karo: $A^2=x_1^2+\frac{v_1^2}{{\omega }^2}=0.0016+\frac{0.09}{32.2}=0.0016+0.002795=0.004395$ $A\approx 0.0663\text{ m}$

Dono magnitudes equal set karo: $\omega \sqrt{A^2-x^2}={\omega }^2|x|$ $\omega$ se divide karo aur square karo: $A^2-x^2={\omega }^2{x}^2\Rightarrow A^2=x^2(1+{\omega }^2)\Rightarrow x=\frac{A}{\sqrt{1+{\omega }^2}}$ $x=\frac{0.10}{\sqrt{1+16}}=\frac{0.10}{\sqrt{17}}=\frac{0.10}{4.1231}\approx 0.02425\text{ m}$ (Note: yeh physically special point nahi hai — yeh chosen units par depend karta hai — lekin yeh dono formulas ko equate karne ki practice karata hai.)

Effective spring constant $k=m{\omega }^2=0.20\times 64=12.8$ N/m. Dekho Energy in SHM — kinetic and potential. (a) Total energy: $E=\frac{1}{2}k{A}^2=\frac{1}{2}(12.8)(0.10^2)=0.064\text{ J}$ (b) $x=0.06$ par speed: $v=8\sqrt{0.10^2-0.06^2}=8\sqrt{0.0064}=8(0.08)=0.64$ m/s. $KE=\frac{1}{2}m{v}^2=\frac{1}{2}(0.20)(0.64^2)=\frac{1}{2}(0.20)(0.4096)=0.04096\text{ J}$ (c) Potential energy: $PE=\frac{1}{2}k{x}^2=\frac{1}{2}(12.8)(0.06^2)=\frac{1}{2}(12.8)(0.0036)=0.02304\text{ J}$ Check: $KE+PE=0.04096+0.02304=0.064$ J $=E$. ✔

Max speed tab hoti hai jab $\cos (\omega t)=\pm 1$, yaani $\omega t=0$. Max displacement tab hota hai jab $\sin (\omega t)=\pm 1$, yaani $\omega t=\frac{\pi }{2}$. Phase mein gap $\frac{\pi }{2}$ hai — ek quarter turn. Kyunki ek full period $2\pi$ phase ka hota hai: $\frac{\Delta t}{T}=\frac{\pi /2}{2\pi }=\frac{1}{4}$ Toh velocity displacement se quarter period ($90^{\circ }$) aage hoti hai, bilkul parent table ki tarah. Dekho Phase and phase difference.

Average speed $=$ (distance travelled) $÷$ (time taken). Centre se extreme tak distance $A=0.10$ m hai. Quarter cycle ka time hai $t=\frac{T}{4}=\frac{1}{4}\cdot \frac{2\pi }{\omega }=\frac{\pi }{2\omega }=\frac{\pi }{16}\approx 0.19635\text{ s}$ (using Angular frequency ω and time period T). Isliye $\bar{v}=\frac{A}{T/4}=\frac{0.10}{0.19635}\approx 0.5093\text{ m/s}$ $v_{\max }=\omega A=0.80$ m/s se compare karo: average peak ka lagbhag $64\%$ hai — kyunki particle slow extreme ke paas crawl karte hue zyada time spend karta hai.

Acceleration turant $\omega$ pin kar deti hai, kyunki known $x$ par $|a|={\omega }^2|x|$ hota hai: $16={\omega }^2(0.04)\Rightarrow {\omega }^2=400\Rightarrow \omega =20\text{ rad/s}$ Ab speed formula use karo: $v^2={\omega }^2(A^2-x^2)\Rightarrow 0.6^2=400(A^2-0.04^2)$ $0.36=400A^2-400(0.0016)=400A^2-0.64$ $400A^2=1.00\Rightarrow A^2=0.0025\Rightarrow A=0.05\text{ m}$

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)$ aur $PE=\frac{1}{2}m{\omega }^2{x}^2$. Unhe equal set karo: $A^2-x^2=x^2\Rightarrow A^2=2x^2\Rightarrow x=\frac{A}{\sqrt{2}}$ $A=0.10$ m ke liye: $x=0.10/\sqrt{2}\approx 0.07071$ m. Is point par $PE=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2\cdot \frac{A^2}{2}=\frac{1}{4}m{\omega }^2{A}^2=\frac{1}{2}E$. Toh dono exactly total ka half rakhte hain. ✔

L3-Q1 se general ellipse $\frac{x^2}{A^2}+\frac{v^2}{(\omega A{)}^2}=1$ se compare karo.

• $A^2=0.09\Rightarrow A=0.30$ m.
• $(\omega A{)}^2=9\Rightarrow \omega A=3\Rightarrow \omega =3/0.30=10$ rad/s.
• $v_{\max }=\omega A=3$ m/s (vertical semi-axis).
• $x=-0.15$ m par acceleration: $a=-{\omega }^{2}x=-100(-0.15)=+15$ m/s² (centre ki taraf wapas point karta hai).

Kya tum bina notes ke yeh kar sakte ho: (1) bata sako ki $v$ aur $a$ max/min kahan hain? (2) $x$ ke liye $v=\omega \sqrt{A^2-x^2}$ ko invert karo? (3) $(x,v)$ ellipse ko pehchano aur uspe se $A$, $\omega$ padho? (4) sirf acceleration se $\omega$ nikalo? (5) jahan $KE=PE$ ho woh locate karo? Agar koi shaky lag raha hai, woh level dobara karo.