Exercises — Hooke's law — spring force F = −kx

WHAT karte hain: numbers ko $F=-kx$ mein daalte hain. WHY: question directly $k$ aur $x$ deta hai — yeh raw definition hai. $F=-(150)(0.04)=-6\text{N}$ Minus humein direction batata hai (spring $x=0$ ki taraf pull karta hai, matlab ek rightward stretch ke left mein). Question sirf magnitude maang raha tha, toh sign drop karte hain: $|F|=k|x|=\boxed{6\text{N}}$.

WHAT: negative $x$ substitute karo. Kyunki question sign aur direction maang raha hai, hum pura vector form $F=-kx$ rakhte hain. $F=-kx=-(500)(-0.03)=+15\text{N}$ WHY positive hai: humne spring ko left mein compress kiya ($x<0$), toh yeh right ki taraf push karta hai ($F>0$). Do minus signs (ek formula mein, ek $x$ mein) cancel ho jaate hain. Yahi minus sign ka poora point hai: jab bhi $x$ flip hota hai, $F$ bhi flip ho jaata hai. Answer: $\boxed{+15\text{N, right ki taraf point karta hai (equilibrium ki taraf back)}}$.

WHAT: $U=\frac{1}{2}k{x}^2$ use karo. WHY squared: energy $x^2$ pe depend karti hai, toh $x$ ka sign matter nahi karta — same size ki stretch aur squeeze same energy store karti hai. $U=\frac{1}{2}(500)(0.03{)}^2=\frac{1}{2}(500)(0.0009)=0.225\text{J}$ Answer: $\boxed{0.225\text{J}}$. Energy yahan kabhi negative nahi hoti.

WHAT: rest pe, spring ka upward pull gravity ke downward pull ko balance karta hai. WHY: "at rest" ka matlab net force zero hai, toh restoring force magnitude $kx$ weight $mg$ ke barabar hai. Hum yahan magnitudes ke saath kaam karte hain kyunki dono forces size mein equal aur direction mein opposite hain — signs balance mein cancel ho jaate hain. $kx=mg\Rightarrow k=\frac{mg}{x}=\frac{(0.8)(9.8)}{0.16}=49\text{N/m}$ Answer: $\boxed{49\text{N/m}}$. Conservation of mechanical energy baad mein dekho agar hum isko giraate hain.

WHAT: $U=\frac{1}{2}k{x}^2$ ko $x$ ke liye invert karo. WHY: hume energy aur stiffness pata hai, displacement chahiye — toh $x$ ke liye solve karte hain. $x^2=\frac{2U}{k}=\frac{2(2)}{250}=0.016\Rightarrow x=\sqrt{0.016}=0.1265\text{m}$ Answer: $\boxed{x\approx 0.126\text{m}}$ (lagbhag 12.6 cm). Hum positive root liya kyunki "kitna stretch karte ho" ek length hai.

WHAT: pehle data point se $k$ find karo, phir doosre ke liye use karo. WHY / sign note: dono stretches positive lengths hain aur question sirf force ka size maang raha hai, toh hum poore mein magnitude form $|F|=kx$ use karte hain (minus sign $F=-kx$ mein sirf yeh batata hai ki pull equilibrium ki taraf hai — direction, jo yahan nahi maanga gaya). $k=\frac{|F|}{x}=\frac{18}{0.09}=200\text{N/m}$ $|F|=kx=(200)(0.15)=30\text{N}$ Answer: $\boxed{30\text{N}}$. (Notice karo $|F|\propto x$: $x$ $\frac{0.15}{0.09}=1.67\times$ badha, aur $|F|$ bhi: $\frac{30}{18}=1.67$.)

WHAT: add ki gayi energy dono $U$ values ka difference hai (figure mein shaded strip). WHY subtract: stored energy $0$ se $x$ tak line ke neeche ka area hai. $x_1$ se $x_2$ jaane par sirf unke beech ki strip add hoti hai. $U_2=\frac{1}{2}(400)(0.10{)}^2=2.0\text{J},U_1=\frac{1}{2}(400)(0.05{)}^2=0.5\text{J}$ $\Delta U=U_2-U_1=2.0-0.5=1.5\text{J}$ Pehle $0.05\text{m}$ ne sirf $0.5\text{J}$ store kiya, lekin doosre (equal-length) stretch ne $1.5\text{J}$ store kiya — teen guna zyada. Answer: $\boxed{1.5\text{J}}$. Seekh: baad ke stretches bahut zyada costly hote hain, kyunki jo force tum lad rahe ho woh already badi hai. Work done by a variable force dekho.

WHAT: har spring ke liye $x=\sqrt{2U/k}$ solve karo. WHY: equal energy lekin alag stiffness — softer spring ko same energy bank karne ke liye zyada aage jaana padega. $x_A=\sqrt{\frac{2(1.5)}{300}}=\sqrt{0.01}=0.10\text{m}$ $x_B=\sqrt{\frac{2(1.5)}{100}}=\sqrt{0.03}=0.1732\text{m}$ Softer spring B zyada stretch hoti hai: $\boxed{x_A=0.10\text{m},x_B\approx 0.173\text{m}}$. Kyunki $x\propto 1/\sqrt{k}$ hai, $k$ ko 3 factor se cut karne par stretch $\sqrt{3}\approx 1.73\times$ zyada ho jaati hai.

WHAT: spring ki saari stored energy block ki kinetic energy ban jaati hai. WHY: frictionless ⇒ mechanical energy conserved hai. Spring ka $U$ $\frac{1}{2}m{v}^2$ mein khaali ho jaata hai. Conservation of mechanical energy aur Elastic potential energy dekho. $\underset{\text{spring PE}}{\underbrace{\frac{1}{2}k{x}^2}}=\underset{\text{block KE}}{\underbrace{\frac{1}{2}m{v}^2}}$ $\frac{1}{2}(800)(0.05{)}^2=\frac{1}{2}(0.2)v^2$ $1.0=0.1v^2\Rightarrow v^2=10\Rightarrow v=\sqrt{10}=3.162\text{m/s}$ Answer: $\boxed{v\approx 3.16\text{m/s}}$.

WHAT: half speed pe block ke paas apni max kinetic energy ka $\frac{1}{4}$ hota hai (kyunki KE $\propto v^2$), toh $\frac{3}{4}$ energy spring mein abhi bhi hai. WHY: total energy $E=\frac{1}{2}k{x}_0^2=1.0\text{J}$ fixed hai. Agar KE $=\frac{1}{4}E$ hai, toh spring PE $=\frac{3}{4}E$. $\frac{1}{2}k{x}^2=\frac{3}{4}E=\frac{3}{4}(1.0)=0.75\text{J}$ $x^2=\frac{2(0.75)}{800}=0.001875\Rightarrow x=0.0433\text{m}$ Answer: $\boxed{x\approx 0.0433\text{m}}$. Note karo yeh halfway ($0.025$ m) nahi hai — kyunki energy $x^2$ pe depend karti hai, spring apni energy sabse tezi se end ke paas deta hai.

Kyunki $F=-kx$ exactly restoring-force law hai, mass Simple Harmonic Motion perform karta hai.

(a) WHAT/WHY: $\omega =\sqrt{k/m}$ $ma=-kx$ set karne se aata hai (Newton's law with spring force). Yeh "swinging ki rate" hai, rad/s mein. $\omega =\sqrt{\frac{200}{0.5}}=\sqrt{400}=20\text{rad/s}$

(b) WHAT/WHY: max speed $x=0$ pe hoti hai, jahan saari energy kinetic hai. $\frac{1}{2}k{A}^2=\frac{1}{2}m{v}_{\max }^2$ set karo amplitude $A=0.10\text{m}$ ke saath. $v_{\max }=A\sqrt{\frac{k}{m}}=A\omega =(0.10)(20)=2\text{m/s}$

(c) WHAT/WHY: energy conservation use karo — total energy $x=0.06$ pe spring PE aur KE mein split hoti hai. $\frac{1}{2}k{A}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$ $\frac{1}{2}(200)(0.10{)}^2=\frac{1}{2}(200)(0.06{)}^2+\frac{1}{2}(0.5)v^2$ $1.0=0.36+0.25v^2\Rightarrow v^2=\frac{0.64}{0.25}=2.56\Rightarrow v=1.6\text{m/s}$ Answers: $\boxed{\omega =20\text{rad/s},v_{\max }=2\text{m/s},v(0.06)=1.6\text{m/s}}$.

Yahan $x\equiv r-r_0$ exactly parent note ka "natural length se displacement" hai, sirf ek coil ki jagah bond ke liye. Yahan Conservative forces and potential energy ka tool $F=-\frac{dU}{dr}$ real kaam karta hai.

(a) WHAT/WHY — $F=-dU/dr$ apply karo aur $r_0$ pe evaluate karo. Energy differentiate karo. Brevity ke liye $s=r_0/r$ likhte hue, chain rule use karo $\frac{d}{dr}(r_0/r{)}^n=-\frac{n}{r}(r_0/r{)}^n$: $\frac{dU}{dr}=U_0\left[-\frac{12}{r}(r_0/r{)}^{12}+\frac{12}{r}(r_0/r{)}^6\right]=\frac{12U_0}{r}\left[(r_0/r{)}^6-(r_0/r{)}^{12}\right]$ $F=-\frac{dU}{dr}=\frac{12U_0}{r}\left[(r_0/r{)}^{12}-(r_0/r{)}^6\right]$ $r=r_0$ pe $r_0/r=1$ hai, toh bracket $1-1=0$ hai. Force $r_0$ pe vanish hoti hai — confirm karta hai ki yeh equilibrium hai (well ka bottom), exactly jaise spring ke liye $x=0$ hota hai. ✓

(b) WHAT/WHY — stiffness curvature $U^{\prime \prime }(r_0)$ hai. Exactly jaise parent note ke Taylor argument mein, kisi bhi energy well ka bottom $U\approx \frac{1}{2}k{x}^2$ jaisa dikhta hai, jiski curvature $\frac{d^{2}U}{d{x}^2}=k$ hoti hai. Kyunki $x=r-r_0$ constant se shift karta hai, $\frac{d^{2}U}{d{x}^2}=\frac{d^{2}U}{d{r}^2}$ hai, toh $k=U^{\prime \prime }(r_0)$. $dU/dr$ ko ek baar aur differentiate karo (phir se $\frac{d}{dr}(r_0/r{)}^n=-\frac{n}{r}(r_0/r{)}^n$ term by term use karo): $\frac{d^{2}U}{d{r}^2}=U_0\left[\frac{12\cdot 13}{r^2}(r_0/r{)}^{12}-\frac{6\cdot 7}{r^2}(r_0/r{)}^6\right]=\frac{U_0}{r^2}\left[156(r_0/r{)}^{12}-42(r_0/r{)}^6\right]$ $r=r_0$ pe ratios $1$ hain, toh: $k=U^{\prime \prime }(r_0)=\frac{U_0}{r_0^2}[156-42]=\frac{114U_0}{r_0^2}?$ Ruko — do exponents $12$ aur $6$ hain, lekin second-derivative ke saamne coefficient force ko differentiate karne se aani chahiye, raw energy powers se nahi. $F$ se cleanly redo karo: $F(r)=\frac{12U_0}{r}\left[(r_0/r{)}^{12}-(r_0/r{)}^6\right]$ differentiate karo aur $r_0$ pe force ka slope evaluate karo, kyunki $k=-\frac{dF}{dr}{|}_{r_0}$ hai. $r=r_0$ pe leading $1/r$ factor first order mein constant hai aur surviving derivative bracket pe act karta hai, deta hai $\frac{dF}{dr}{|}_{r_0}=\frac{12U_0}{r_0}\cdot \frac{1}{r_0}[-12+6]=-\frac{72U_0}{r_0^2}$. Isliye $k=-\frac{dF}{dr}{|}_{r_0}=\frac{72U_0}{r_0^2}.$ Factor 72 sirf $12\times 6$ hai — Lennard-Jones energy mein do exponents ka product. ✓

(c) WHAT/WHY — numbers plug karo. $k=\frac{72(1.6\times 10^{-19})}{(3.0\times 10^{-10}{)}^2}=\frac{1.152\times 10^{-17}}{9.0\times 10^{-20}}=128\text{N/m}$ Answer: $\boxed{k\approx 128\text{N/m}}$. Kamal ki baat hai, ek single molecular bond ki stiffness ek chhote lab spring se comparable hai — kyunki us scale pe forces tiny displacements ke relative enormous hoti hain. Yahi precisely wajah hai ki equilibrium ke paas Interatomic forces Hooke's law reproduce karta hai, aur isliye solids ring, vibrate, aur macroscopic springs ki tarah elastic energy store karte hain.