Exercises — Spring-mass system — horizontal, vertical

Neeche har symbol parent note mein build kiya gaya tha. Quick meanings:

• $m$ = mass (kg), $k$ = spring stiffness / "stiffness" newtons per metre mein (N/m).
• $x$ = equilibrium se displacement (woh rest point jahan net force zero hai), metres mein.
• $\omega =\sqrt{k/m}$ = angular frequency (rad/s) — cosine cycle ke kitne radians per second guzarte hain.
• $T=2\pi \sqrt{m/k}$ = period (s) = ek poora aage-peechhe jaane ka time.
• $f=1/T$ = frequency (Hz) = oscillations per second (Hz).
• $A$ = amplitude (m) = mass equilibrium se kitna door tak jaata hai.
• $v_{max}=A\omega$ = top speed, centre par milti hai.
• $y_0=mg/k$ = hanging spring ka static stretch; $g=9.8{\text{m/s}}^2$ poore mein.

KYA karte hain: seedha definitions mein plug karo — yeh pure recognition hai. KYUN: yeh teeno quantities ek hi number ke teen alag chehra hain. $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{50}{2.0}}=\sqrt{25}=5.0\text{rad/s}$ $T=\frac{2\pi }{\omega }=\frac{2\pi }{5.0}=1.257\text{s}$ $f=\frac{1}{T}=\frac{1}{1.257}=0.796\text{Hz}$ Sanity check: $2\pi \sqrt{m/k}=2\pi \sqrt{2/50}=2\pi (0.2)=1.257$ s. ✓

KYA karte hain: metres mein convert karo, phir static-stretch shortcut use karo. YEH tool KYUN: hume na $m$ diya gaya hai na $k$ — lekin parent note ne dikhaya tha ki $T=2\pi \sqrt{y_0/g}$, jisme sirf stretch chahiye, kyunki $k$ aur $m$ dono $y_0=mg/k$ ke andar chhup jaate hain. $y_0=2.5\text{cm}=0.025\text{m}$ $T=2\pi \sqrt{\frac{y_0}{g}}=2\pi \sqrt{\frac{0.025}{9.8}}=2\pi \sqrt{2.551\times 10^{-3}}=2\pi (0.05051)=0.317\text{s}$

KYA: pehle se $\omega =5.0$ rad/s use karo. (a) Max speed — centre par milti hai jahan sari energy kinetic hoti hai (Energy in SHM): $v_{max}=A\omega =0.15\times 5.0=0.75\text{m/s}$ (b) Max acceleration — turning points par milti hai jahan displacement sabse bada hota hai. $\ddot{x}=-{\omega }^{2}x$ se, acceleration ka size $x=\pm A$ par peak karta hai: $a_{max}={\omega }^{2}A=(5.0{)}^2\times 0.15=25\times 0.15=3.75{\text{m/s}}^2$ (c) Total energy — full stretch par sab spring PE hai: $E=\frac{1}{2}k{A}^2=\frac{1}{2}(50)(0.15{)}^2=\frac{1}{2}(50)(0.0225)=0.5625\text{J}$ Cross-check: $\frac{1}{2}m{v}_{max}^2=\frac{1}{2}(2.0)(0.75{)}^2=0.5625$ J. ✓ Do raaste, ek hi energy.

KYA: pehle period nikalo, phir $k$ solve karne ke liye $T=2\pi \sqrt{m/k}$ invert karo. KYUN invert: humein $T$ aur $m$ pata hai, $k$ chahiye; algebra same formula rearrange karta hai. $T=\frac{9.0}{12}=0.75\text{s}$ $T=2\pi \sqrt{m/k}$ dono sides square karo: $T^2=4{\pi }^2\frac{m}{k}\Rightarrow k=\frac{4{\pi }^{2}m}{T^2}=\frac{4{\pi }^2(0.30)}{(0.75{)}^2}=\frac{11.84}{0.5625}=21.1\text{N/m}$

KYA (a): rest par spring force gravity ko balance karta hai, toh $k{y}_0=mg$. $k=\frac{mg}{y_0}=\frac{1.2\times 9.8}{0.08}=\frac{11.76}{0.08}=147\text{N/m}$ (b) Period — vertical hai, lekin derivation gravity cancel kar deta hai, toh horizontal formula use karo: $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{1.2}{147}}=2\pi \sqrt{8.163\times 10^{-3}}=2\pi (0.09035)=0.568\text{s}$ (c) Max speed $\omega =\sqrt{k/m}=\sqrt{147/1.2}=\sqrt{122.5}=11.07$ rad/s ke saath: $v_{max}=A\omega =0.05\times 11.07=0.553\text{m/s}$ (d) Total (net) force — equilibrium se measure karein toh net restoring force $F_{net}=-kx$ hai. Turning points $x=\pm A$ par size mein sabse badi hoti hai, aur yeh equilibrium ki taraf point karti hai, toh dono ends par opposite signs leta hai. Neeche ko positive maanein: $F_{net}=-kx\Rightarrow F_{net}\text{ranges from }-kA\text{ to }+kA$ $F_{net,\text{at bottom }(x=+A)}=-kA=-147\times 0.05=-7.35\text{N (upar ki taraf point karta hai, equilibrium ki taraf)}$ $F_{net,\text{at top }(x=-A)}=+kA=+147\times 0.05=+7.35\text{N (neeche ki taraf point karta hai, equilibrium ki taraf)}$ Toh net force $-7.35$ N se $+7.35$ N tak range karta hai (har turning point par magnitude $7.35$ N, equilibrium par zero). Agar tumhe sirf spring force chahiye (jo gravity se ladhti hai), toh extension $y_0-A$ se $y_0+A$ tak range karta hai: $F_{spring,min}=k(y_0-A)=147(0.03)=4.41\text{N (top par)}$ $F_{spring,max}=k(y_0+A)=147(0.13)=19.11\text{N (bottom par)}$

KYA: same spring matlab same $k$, toh $T=2\pi \sqrt{m/k}\propto \sqrt{m}$ — constant $2\pi /\sqrt{k}$ ratio mein cancel ho jaata hai. KYUN ratio: humein kabhi $k$ ki zaroorat nahi; dono periods divide karne se sirf masses bachte hain. $\frac{T_2}{T_1}=\frac{2\pi \sqrt{m_2/k}}{2\pi \sqrt{m_1/k}}=\sqrt{\frac{m_2}{m_1}}=\sqrt{\frac{0.90}{0.40}}=\sqrt{2.25}=1.5=\frac{3}{2}$ Phir: $T_2=1.5\times T_1=1.5\times 0.50=0.75\text{s}$

KYA/KYUN: block contact tab khota hai jab surface usse neeche nahi kheench sakti — yaani top par surface ka khud ka downward acceleration gravity ke barabar ho jaata hai. Contact tab khoota hai jab peak downward acceleration $g$ tak pahunch jaata hai. SHM mein sabse badi acceleration $a_{max}={\omega }^{2}A$ hoti hai aur yeh equilibrium ki taraf point karti hai; top par yeh neeche point karti hai. Ise $g$ ke barabar set karo: ${\omega }^{2}A=g\Rightarrow \omega =\sqrt{\frac{g}{A}}=\sqrt{\frac{9.8}{0.04}}=\sqrt{245}=15.65\text{rad/s}$ $f=\frac{\omega }{2\pi }=\frac{15.65}{2\pi }=2.49\text{Hz}$ Khoobsurat baat: mass $M$ cancel ho jaata hai — yeh kabhi appear hi nahi karta. Koi bhi block same frequency par uthega, kyunki required aur available forces dono $M$ ke saath scale karti hain.

KYA: total energy apni full-stretch value $\frac{1}{2}k{A}^2$ ke barabar hai. $v$ solve karo. $\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2$ $2/m$ se multiply karo aur ${\omega }^2=k/m$ use karo: $v^2=\frac{k}{m}(A^2-x^2)={\omega }^2(A^2-x^2)$ $\boxed{v(x)=\pm \omega \sqrt{A^2-x^2}}$ Ends check karo: $x=0$ par, $v=\omega \sqrt{A^2}=A\omega =v_{max}$ ✓ (centre par sabse tez). $x=\pm A$ par, $v=\omega \sqrt{A^2-A^2}=0$ ✓ (turning points par momentarily ruka hua). $\pm$ yeh reflect karta hai ki mass har interior point se ek baar dono direction mein guzarta hai.

Sirf stretch wale vertical problem ke liye kaun sa formula? ::: $T=2\pi \sqrt{y_0/g}$ Speed maximum kahan hai, acceleration maximum kahan hai? ::: Speed max centre par ($x=0$); acceleration max ends par ($x=\pm A$). Kis $x$ par KE = PE? ::: $x=\pm A/\sqrt{2}$ Parallel vs series effective stiffness? ::: Parallel $k_1+k_2$ (stiffer); series $1/k_s=1/k_1+1/k_2$ (softer). Phase constant ka sign kaise fix karte hain? ::: $v(0)$ ka sign use karo; position akele ambiguous hai. Position ka function speed? ::: $v=\omega \sqrt{A^2-x^2}$