Kisi bhi trap ke samajh aane se pehle, humein ek axis pe teen points aur ek sign convention pin karni hogi. Figure dekho: ek hanging spring ko hatched ceiling se draw kiya gaya hai; red block wo mass m hai, jo rest mein baitha hai. Do horizontal dashed lines reference heights mark karti hain — upar wali dashed line wo hai jahan spring ka free end koi mass nahi hone pe hoga (natural length), aur neeche wali dashed line wo hai jahan mass actually rest karta hai (equilibrium). Left pe kala double-headed arrow dono ke beech ka gap measure karta hai, jise y0 label kiya gaya hai. Right pe downward kala arrow positive x direction mark karta hai aur dikhata hai ki x=0 ko equilibrium pe rakha gaya hai, natural length pe nahi.
Recall
T=2πm/k actually kahan se aata hai (poori derivation, taaki page stand alone kare)
Equilibrium se measured restoring force F=−kx aur Newton's law F=mx¨ se shuru karo:
mx¨=−kx⇒x¨=−mkx.
Ye kehta hai: acceleration position ka ek fixed negative multiple hai. Us multiple ko ω2 kaho, yaani ω2=k/m define karo, to x¨=−ω2x.
Ab ek wobble x(t)=Acos(ωt+ϕ)guess karo (cosine hi wo single shape hai jiska second derivative khud ka negative copy hota hai). Twice differentiate karo:
x˙=−Aωsin(ωt+ϕ),x¨=−Aω2cos(ωt+ϕ)=−ω2x.✓
Ye kisi bhi amplitude A aur phase ϕ ke liye fit hota hai, confirm karta hai ω=k/m.
Cosine tab repeat hoti hai jab uska angle 2π se aage badhta hai: ωT=2π, isliye
T=ω2π=2πkm.
Ye wahi formula hai jo is page ka har trap probe karta hai.
Recall Gravity term kyun cancel hoti hai (yahan bhi re-derive kiya gaya hai)
Raw position y ko natural length (upar wali dashed line) se neeche measure karo. Do forces kaam karte hain: gravity +mg (neeche) aur spring −ky (upar, stretch ke proportional). Newton:
my¨=mg−ky.
Ab from-equilibrium coordinate x=y−y0 mein switch karo jahan y0=mg/k hai. Kyunki y0 ek constant hai, y¨=x¨, aur:
mx¨=mg−k(x+y0)==0(mg−ky0)−kx=−kx.
Constant mg shift mein poori tarah absorb ho gaya — dynamics mein koi g-dependent cheez nahi bachi. Upar wala figure exactly ye shift dikhata hai: gravity ne origin ko upar wali dashed line se neeche wali tak relocate kar diya, aur us naye origin ke baare mein physics horizontal case jaisi hi hai.
Stiffness k ko double karne se period half ho jaati hai.
Galat — T∝1/k, isliye Tnew/Told=k/2k=1/2≈0.71. T ko half karne ke liye k ko quadruple karna padega.
Ek vertical spring-mass system wohi spring horizontally se zyada slowly oscillate karta hai.
Galat — dono T=2πm/k dete hain. Gravity constant hai, isliye wo sirf rest point relocate karta hai; naye equilibrium ke baare mein restoring term phir bhi −kx hai (upar wala re-derivation box dekho).
Amplitude A badhane se ideal spring-mass system ki period badh jaati hai.
Galat — derivation box se, solution x(t)=Acos(ωt+ϕ) mein ω=k/m hai andar koi A nahi, isliye T=2πm/k amplitude-independent hai ("isochronous"). Physically, x→cx scale karne se x¨=−(k/m)x same factor c se scale hoti hai, to poori wobble sirf taller hoti hai with identical timing.
Moon par ek horizontal spring-mass oscillator ka period Earth pe jaisa hi hoga.
Sahi — horizontal period T=2πm/k mein koi g nahi hai, isliye gravity strength irrelevant hai.
Moon par ek vertical spring rest mein kam stretch hoti hai lekin Earth jaisa hi period rakhti hai.
Sahi — static extension y0=mg/k kamzor g ke saath shrink hoti hai, lekin dynamics phir bhi mx¨=−kx mein reduce hoti hai, isliye T unchanged rehta hai.
Equilibrium point par net force zero hai, isliye mass momentarily wahan ruk jaati hai.
Galat — net force equilibrium par zero hoti hai, lekin yahi woh jagah hai jahan speed maximum hoti hai. Zero force ka matlab zero acceleration hai, zero velocity nahi.
Extreme positions (turning points, x=±A) par mass ki speed zero hai lekin acceleration non-zero hai.
Sahi — speed zero hoti hai jab wo reverse karti hai, jabki displacement sabse badi hoti hai isliye ∣F∣=k∣x∣ aur acceleration apne peak par hoti hai.
Formula T=2πy0/g dikhata hai ki vertical period gravity par depend karti hai.
Galat — g appear karta hai lekin sirf isliye kyunki y0=mg/k mein m/k chupi hoti hai; y0=mg/k substitute karne par 2π(mg/k)/g=2πm/k milta hai aur g cancel ho jaata hai. Ye ek measurement shortcut hai, g-dependence nahi.
Parallel mein do identical springs ek spring akele se longer period deti hain.
Galat — parallel effective stiffness double karti hai (keff=2k; neeche note dekho), aur stiffer matlab faster, isliye period 2 se short hoti hai: T=2πm/2k.
Neeche ke teen "wrong equations" sab figure ke top pe same axis pe rehte hain — fix hamesha iske baare mein hoti hai ki x kis reference point se measure kiya ja raha hai.
"Hanging spring ke liye main x ko natural (unstretched) length se measure karta hun, tab mx¨=−kx."
Galat reference point — natural length se equation mx¨=mg−kx padhti hai, jisme abhi bhi ek leftover mg hai. Constant cancel hone ke liye equilibrium se measure karna zaroori hai (wo point jo natural length se y0=mg/k neeche hai).
"Gravity ek downward force add karta hai, isliye vertical equation mx¨=−kx+mg hai aur gravity clearly motion ko affect karti hai."
+mg real hai sirf jab x natural length se measure kiya jata hai; origin ko equilibrium tak neeche shift karne se wo ek constant mein absorb ho jaata hai jo cancel ho jaata hai (ky0=mg), oscillation term purely −kx reh jaata hai.
"Heavier mass zyada move karta hai, isliye wohi spring par heavier block zyada fast oscillate karta hai."
Ulta — T∝m, isliye zyada mass matlab spring ko resist karne wali zyada inertia aur slower (longer-period) oscillation.
"Kyunki F=−kx, acceleration constant hai jaise ek falling body, isliye hum v2=u2+2as use kar sakte hain (jahan u initial speed, v final speed, s distance hai)."
Nahi — wo constant-acceleration formulas assume karte hain ki a fixed hai, lekin yahan a=−kx/mposition ke saath change karta hai kyunki ye x par depend karta hai. Constant-acceleration kinematics SHM par kabhi apply nahi hoti; energy conservation ya x(t)=Acos(ωt+ϕ) use karo.
"F=−kx mein minus sign ka matlab sirf hai ki spring push karta hai; main ise drop kar sakta hun aur phir bhi T=2πm/k paaoonga."
Minus sign hi poori physics hai — ye force ko restoring banata hai (equilibrium ki taraf wapas point karta hai). Ise drop karo aur mx¨=+kx runaway exponential growth deta hai (x∼ek/mt), oscillation nahi.
"Series mein springs keff=k1+k2 ki tarah add hoti hain, isliye series stiffer hai."
Ye parallel rule hai. Series mein same force ke liye stretches add hoti hain, deta hai keff1=k11+k21, jo ki kisi bhi spring se softer hai, isliye series zyada slowly oscillate karta hai.
Period amplitude par kyun depend nahi karti, tumhari gut feeling ke unlike?
Equation of motion likho x¨=−(k/m)x: acceleration position ka fixed multiple−k/m hai, isliye poori motion x→cx scale karne se x¨→cx¨ identically scale hoti hai — same sinusoid, sirf taller. Formally x(t)=Acos(ωt+ϕ) mein ω=k/m hai andar koi A nahi, isliye period kisi bhi A ke liye fixed hai.
Restoring force displacement ke proportional kyun honi chahiye (sirf oppose karna kaafi nahi) true SHM ke liye?
Sirf strictly linear −kx equation x¨=−ω2x produce karta hai jiska solution ek pure sinusoid hai with ek single, amplitude-independent frequency. Koi bhi doosra restoring law anharmonic motion deta hai jiska period amplitude ke saath drift karta hai.
Stiffer spring oscillation ko faster kyun bana sakta hai jabki "move karna mushkil" hota hai?
Stiffness k square root ke neeche baithti hai T∝1/k ke roop mein: ek stiffer spring mass ko per unit displacement zyada force se wapas kheenchti hai, isliye wo ghar zyada jaldi accelerate karta hai, period short hoti hai.
Gravity equilibrium position ko kyun move karta hai lekin period ki value ko nahi?
Gravity ek constant force hai, isliye wo sirf ek naya balance point set karta hai jahan ky0=mg; oscillation us point se move karne par spring force mein change se drive hoti hai, jo abhi bhi −kx hai, same T=2πm/k deta hai.
Energy conservation Newton's law jaisa hi ω kyun deta hai?
E=21mv2+21kx2 differentiate karo aur dE/dt=0 set karo jo v(mx¨+kx)=0 mein factor hota hai; kyunki v=0 generally, tum mx¨+kx=0 recover karte ho — identical SHM equation, hence same ω=k/m.
Speed exactly centre par maximum aur ends par zero kyun hai?
Saari energy spring PE (21kx2) aur KE (21mv2) ke beech trade hai. Centre par x=0 isliye PE zero hai aur KE poora budget hold karta hai (max speed); extremes par x=±A isliye PE maximal hai aur KE, aur speed, zero drop kar jaati hai.
Agar mass (ideally) zero ho to period ka kya hoga?
T=2πm/k→0 — ek massless block infinitely fast oscillate karta. Ye unphysical hai aur flag karta hai ki real spring ki apni mass tab dominate karti.
Kya hoga jab k→0 (infinitely floppy spring)?
T=2πm/k→∞ — vanishing restoring force ke saath mass ko wapas kheenchne ke liye almost kuch nahi, isliye "oscillation" forever leti hai. Limit mein ye oscillator rehna band kar deta hai.
Agar ek horizontal spring pe surface par friction ho, kya motion abhi bhi is period ke saath SHM hai?
Nahi — friction ek velocity-dependent force add karta hai, ise damped oscillator mein turn kar deta hai; amplitude decay hoti hai aur period ideal 2πm/k se thoda lengthen hoti hai.
Kya koi oscillation hogi agar mass exactly equilibrium par zero velocity se shuru kare?
Nahi — net force aur velocity dono zero hain, isliye wo simply wahi reh jaata hai (amplitude A=0). Oscillation ke liye ya to initial displacement ya initial push chahiye.
Spring ki period ko pendulum ki period se ek degenerate limit par compare karo: dusre planet par jaane se kaunsa affect hoga?
Sirf pendulum (T=2πL/g) g ke saath badlega; spring ka T=2πm/k mein koi g nahi, isliye ye kahi bhi identically tick karta hai.
Agar ek vertical spring itna stretch ho ki upar jaate waqt slack ho jaaye (spring sirf pull kar sakti hai), kya motion phir bhi pure SHM hai?
Nahi — jab spring slack ho jaati hai to force −kx nahi rehta (cycle ke ek hisse ke liye sirf gravity/free flight hai), isliye linear restoring law break ho jaata hai aur motion sinusoidal nahi rehti.
Recall Ek-line self-test (formal version)
Physical quantities se fill in karo: "gravity sirf ___ shift karta hai, kabhi nahi ___." ::: equilibrium position (y0=mg/k) ; periodT=2πm/k.
